69
Classical Mechanics A Critical Introduction Larry Gladney Solutions Manual for

for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

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Page 1: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Classical Mechanics A Critical IntroductionLarry Gladney

Solutions Manual for

Solutions Manual forldquoClassical Mechanics A Critical Introductionrdquo

Solutions Manual forldquoClassical Mechanics A Critical Introductionrdquo

Larry Gladney

Hindawi Publishing Corporationhttpwwwhindawicom

Solutions Manual for ldquoClassical Mechanics A Critical IntroductionrdquoLarry Gladney

Hindawi Publishing Corporation410 Park Avenue 15th Floor 287 pmb New York NY 10022 USANasr City Free Zone Cairo 11816 Egypt

Copyright copy 2011 2012 Larry Gladney This text is distributed under the Creative CommonsAttribution-NonCommercial license which permits unrestricted use distribution and reproductionin any medium for noncommercial purposes only provided that the original work is properly cited

ISBN 978-977-454-110-0

Contents

Preface vii

Solutions Manual 11 Kinematics 12 Newtonrsquos First andThird Laws 73 Newtonrsquos Second Law 114 Momentum 235 Work and Energy 296 Simple Harmonic Motion 377 Static Equilibrium of Simple Rigid Bodies 408 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 459 Remarks on Gravitation 57

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 2: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Solutions Manual forldquoClassical Mechanics A Critical Introductionrdquo

Solutions Manual forldquoClassical Mechanics A Critical Introductionrdquo

Larry Gladney

Hindawi Publishing Corporationhttpwwwhindawicom

Solutions Manual for ldquoClassical Mechanics A Critical IntroductionrdquoLarry Gladney

Hindawi Publishing Corporation410 Park Avenue 15th Floor 287 pmb New York NY 10022 USANasr City Free Zone Cairo 11816 Egypt

Copyright copy 2011 2012 Larry Gladney This text is distributed under the Creative CommonsAttribution-NonCommercial license which permits unrestricted use distribution and reproductionin any medium for noncommercial purposes only provided that the original work is properly cited

ISBN 978-977-454-110-0

Contents

Preface vii

Solutions Manual 11 Kinematics 12 Newtonrsquos First andThird Laws 73 Newtonrsquos Second Law 114 Momentum 235 Work and Energy 296 Simple Harmonic Motion 377 Static Equilibrium of Simple Rigid Bodies 408 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 459 Remarks on Gravitation 57

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 3: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Solutions Manual forldquoClassical Mechanics A Critical Introductionrdquo

Larry Gladney

Hindawi Publishing Corporationhttpwwwhindawicom

Solutions Manual for ldquoClassical Mechanics A Critical IntroductionrdquoLarry Gladney

Hindawi Publishing Corporation410 Park Avenue 15th Floor 287 pmb New York NY 10022 USANasr City Free Zone Cairo 11816 Egypt

Copyright copy 2011 2012 Larry Gladney This text is distributed under the Creative CommonsAttribution-NonCommercial license which permits unrestricted use distribution and reproductionin any medium for noncommercial purposes only provided that the original work is properly cited

ISBN 978-977-454-110-0

Contents

Preface vii

Solutions Manual 11 Kinematics 12 Newtonrsquos First andThird Laws 73 Newtonrsquos Second Law 114 Momentum 235 Work and Energy 296 Simple Harmonic Motion 377 Static Equilibrium of Simple Rigid Bodies 408 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 459 Remarks on Gravitation 57

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 4: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Solutions Manual for ldquoClassical Mechanics A Critical IntroductionrdquoLarry Gladney

Hindawi Publishing Corporation410 Park Avenue 15th Floor 287 pmb New York NY 10022 USANasr City Free Zone Cairo 11816 Egypt

Copyright copy 2011 2012 Larry Gladney This text is distributed under the Creative CommonsAttribution-NonCommercial license which permits unrestricted use distribution and reproductionin any medium for noncommercial purposes only provided that the original work is properly cited

ISBN 978-977-454-110-0

Contents

Preface vii

Solutions Manual 11 Kinematics 12 Newtonrsquos First andThird Laws 73 Newtonrsquos Second Law 114 Momentum 235 Work and Energy 296 Simple Harmonic Motion 377 Static Equilibrium of Simple Rigid Bodies 408 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 459 Remarks on Gravitation 57

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 5: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Contents

Preface vii

Solutions Manual 11 Kinematics 12 Newtonrsquos First andThird Laws 73 Newtonrsquos Second Law 114 Momentum 235 Work and Energy 296 Simple Harmonic Motion 377 Static Equilibrium of Simple Rigid Bodies 408 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 459 Remarks on Gravitation 57

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 6: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Preface

This instructorrsquos manual has complete solutions to problems for the open-access textbookldquoClassical Mechanics A Critical Introductionrdquo by Professor Michael Cohen Students arestrongly urged to solve the problems before looking at the solutions

The responsibility for errors rests with me and I would appreciate being notified ofany errors found in the solutions manual

Larry Gladney

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 7: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Solutions Manual

1 Kinematics

11 Kinematics Problems Solutions

111 One-Dimensional Motion

(11) (a) We find the area under the velocity versus time graph to get the distance traveled(ie the displacement) From Figure 11 we can easily divide the area under the velocityversus time plot into 3 areas Adding the areas gives the answer

area 1 = 12 (35ms minus 10ms) (10 s minus 0) = 125marea 2 = (10ms minus 0) (20 s minus 0) = 200marea 3 = (10ms minus 0) (40 s minus 20 s) = 200m

(11)

total distance from 119905 = 0 to 119905 = 40 = 525m(b) The acceleration is just the slope of the velocity versus time graph so

119886 = V (119905 = 60) minus V (119905 = 40)60 s minus 40 s = minus05ms2 (12)

(c) We need the total displacement of the car at 119905 = 60 secondsarea 119905 = 40 997888rarr 60 = 12 (60 s minus 40 s) (10ms minus 0) = 100m (13)

The total displacement is therefore 525 + 100 = 625m and the average velocity is

Vavg = 119909 (119905 = 60) minus 119909 (119905 = 0)60 s minus 0 = 104ms (14)

(12) (a)We start by describing the positions of the antelope and cheetah for the initialtime and the time when the cheetah catches the antelope (the ldquofinalrdquo time) When firstbeginning problem-solving in physics it can help to have a table listing the known andunknown information All units are meters ms or ms2 as appropriate Since we wantthese units we need to do some conversions

101 kmh = (101 times 105mh) ( 1 h3600 s) = 281ms88 kmh = (880 times 104mh) ( 1 h3600 s) = 244ms (15)

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 8: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

2 Solutions Manual

Velo

city

(ms

)

5

1

3

10

15

20

30

25

35

10 30 40 5020 60

Time (s)

2

Figure 11 Measuring the displacement of a car from the velocity versus time graph

cheetah antelopeinitial position 1199090119888 = 0 1199090119886 = 500final position 119909119888 = 119909119886 = 119909119888initial velocity V0119888 = 281 V0119886 = 244final velocity V119888 = 281 V119886 = 244acceleration 119886119888 = 0 119886119886 = 119886119888 = 0

(16)

The problem specifies that we get the time at which the positions of the cheetah andantelope are the same Using (111c) we have

119909119888 = 1199091198861199090119888 + V0119888119905 = 1199090119886 + V0119886119905 997904rArr119905 = 1199090119886

V0119888 minus V0119886

= 500m(281 minus 244) ms 119905 = 135 s

(17)

The distance traveled by the cheetah is

119909119888 = V0119888119905 = (281ms) (135 s) = 379m (18)

(b) The situation is the same as for part (a) except that now the starting position ofthe antelope is unknown and the time of capture is 119905 = 20 s Thus

119909119888 = 1199091198861199090119888 + V0119888119905 = V0119888119905 = 1199090119886 + V0119886119905 997904rArr

1199090119886 = (V0119888 minus V0119886) 119905= (281ms minus 244ms) (200 s) = 740m

(19)

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 9: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Kinematics 3

(13) (a) We do not know the height of the bottom of the window above the releasepoint of the throw but this does not matter as the acceleration is constant and known andthe time of travel across a known distance is also known Therefore if V0 is the speed ofthe ball as it passes the bottom of the window 119905 is the time for the ball to cross from thebottom to the top of the window 1199100 is the position of the bottom of the window and 1199101is the position of the top of the window then

1199101 = 1199100 + V0119905 minus 121198921199052 997904rArrV0 = (1199101 minus 1199100) + (12) 1198921199052119905

= (300m) + (12) (980ms2) (0400 s)20400 s V0 = 946ms

(110)

This gives us the position of the top of the ballrsquos flight since the final speed is zero So

V2119891 = 0 = V20 minus 2119892 (119910119891 minus 1199100) 997904rArr119910119891 minus 1199100 = V202119892 = (946ms)22 (980ms2) = 457m (111)

Therefore the maximum height above the top of the window is

119910119891 minus 1199101 = (457 minus 300) m = 157m (112)

(b) The speed of the ball when it first reaches the top of the window is

V1 = V0 minus 119892119905 = 946ms minus (980ms2) (0400 s) = 554ms (113)

The time interval Δ119905 between passing the top of the window on the way up (V1) andon the way down (we call this Vdown) is

Vdown = Vup minus 119892Δ119905 997904rArrΔ119905 = minusV1 minus V1minus119892

= minus2 (554ms2)minus980ms2 Δ119905 = 113 s

(114)

(14) We solve the problem in a reference frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905minus(12)1198921199052The height of the

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 10: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

4 Solutions Manual

ball above the floor is 119910119887minus119910119891 = V0119905minus(12)(119892+119860)1199052This is maximumwhen 119905 = V0(119892+119860)and has the value ℎmax = V202(119892 + 119860)

We see that the maximum height above the floor which the ball reaches is the same asthe maximum height it would reach if it were projected upward with velocity V0 in a non-accelerating box on a planet where the acceleration of gravity is 119892+119860 (rather than 119892) Weshall see (Example 32) that the upward force which the floor exerts on a person of mass119898in an accelerating elevator (this is the force which a scale measures) is119898(119892 + 119860) which iswhat the scale would read if the elevator were not accelerating but were on a planet wherethe acceleration of gravity is (119892 + 119860)

(15) (a)The walkway is less than 200m in length If we call the length of the walkway119909 then for the race to end in a tie Miriam must run the distance 200 minus 119909 m at a rate of6ms so that sum of time on the walkway (determined by her speed of 6 + 2 = 8ms) andoff the walkway equals Alisonrsquos 200m divided by her constant speed of 7ms So

200m700ms = 119909800ms + 200m minus 119909600ms 997904rArr (115)

A common factor for the numbers 6 7 and 8 is 168 (7 sdot 8 sdot 3) so multiply through bythis factor (ie 168ms) and solve for 119909

24 (200m) = 21119909 + 28 (200m minus 119909) 997904rArr7119909 = (28 minus 24) (200m) 997904rArr

119909 = 114m(116)

(b) Now we must look at the times separately for Alison and Miriam

119905Miriam = 200m600ms = 333 s119905Alison = 114m(700 minus 200) ms + (200 minus 114) m700ms = 351 s (117)

Miriam wins the race

112 Two and Three Dimensional Motion

(16) We make use of equations from the chapter(a) We make use of (116)

V (119905) = 119889 119903119889119905 = [(4119905 minus 7) minus 2119905119895] ms 997904rArrV (119905 = 2) = [ minus 4119895] ms (118)

(b)

119886 (119905) = 119889V119889119905 = [4 minus 2119895] ms2 = 119886 (119905 = 5 s) (119)

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 11: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Kinematics 5

(c) The average velocity is given by (11)

Vav = 119903 (119905 = 3) minus 119903 (119905 = 1)3 minus 1= (18 minus 21) minus (9 minus 1) 1198952 ms

Vav = (minus15 minus 4119895) ms(120)

(17) Note that the slope of the hill can be described in terms of

minus tan 1205791 = 119910ℎ119909ℎ (121)

with 1205791 = 100∘ and (119909ℎ 119910ℎ) being coordinates of the hill surface We adopt the usualconvention of up being the +119910 direction so we need to have the angle 1205791 of the hill asdirected below the horizontal The position of the skier as a function of time assumingthe takeoff point is set as coordinate (0 +600m) is

119909119904 = V0 cos 1205792119905119910119904 = 600m + V0 sin 1205792119905 minus 121198921199052

(122)

Eliminating 119905 from the two equations gives

119910119904 = 600m + 119909119904 tan 1205792 minus 1198922 1199092119904(V0 cos 1205792)2 (123)

The definition of ldquolandingrdquo is to have 119909119904 = 119909ℎ simultaneously with 119910119904 = 119910ℎ Letrsquos justdrop the subscripts then and define 119909 as horizontal distance from the takeoff point Then

119910 = minus119909 tan 1205791 = 600m + 119909 tan 1205792 minus 11989211990922(V0 cos 1205792)2 997904rArr0 = 600m + 119909 (tan 150∘ + tan 100∘) minus (980ms2) 11990922(300ms sdot cos 150∘)2= 600m + 119909 (044428) minus (583534 times 10minus3mminus1) 1199092 997904rArr

119909 = minus044428 plusmn radic0197381 + 4 (000583534) (600)2 (minus000583534mminus1)= 878m

(124)

Note that the other root from the square root gives the answer corresponding towherethe trajectory would start on the incline if the ramp were not present

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 12: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

6 Solutions Manual

(18) The period is derived from the speed of a point on the outer rim and this is inturn determined by the centripetal acceleration value If 119903 is the radius to the outer rimand the period is 120591 then

119886 = 1198925 = V2119903 997904rArrV = radic1199031198925

120591 = 2120587119903V

= 2120587radic5119903119892= 2 (3141593)radic 5 (1000m)981ms2

120591 = 142 s

(125)

(19) The radius can be derived from the definition of centripetal acceleration as

| 119886| = V2119903 997904rArr119903 = V2119886= [(300 times 105mh)(1 h3600 sh)]2005 (981ms2)

119903 = 142 times 104m = 142 km

(126)

The answer looks anomously large In reality engineers must bank roads (just as anairplane banks in flight to make a turn) to avoid such a large radius of curvature

(110) We need the radius of the circular path to calculate the acceleration Once wehave this since we know the velocity we can use (118)

119903 = 119871 sin 120579 997904rArr| 119886| = V2119903

= (121ms)2(120m) sin 200∘| 119886| = 357ms2

(127)

The direction of the acceleration is always toward the center of the circular path

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 13: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos First andThird Laws 7

yy

xx

T T N

W1 W2

120579

Figure 21 Free-body diagram for Problem (21)(a)

2 Newtonrsquos First and Third Laws

21 Newtonrsquos First and Third Laws of Motion Solutions

(21) (a) The free-body diagrams for the two blocks for this case are as followsThe first law equations for the weights are

119879 minus1198821 = 0 997904rArr 119879 = 1198821 = 100 newtons119879 minus 1198822 sin 120579 = 0 997904rArr

1198822 = 119879sin 120579 = 1198821

sin 120579 = 100 newtonssin 30∘ = 200 newtons

(21)

(b) Now depending on the weight of 1198822 it may tend to be pulled up the plane orslide down the plane Let1198822min correspond to the lowest weight before1198822 slides up theincline and 1198822max to the maximum weight before 1198822 slides down the incline The free-body diagrams for these two cases are shown in Figures 21 and 22 Note that only for thesetwo special cases is the friction force at its maximum magnitude 119891 = 120583119904119873 For the twocases the first law equation for1198821 remains the same as for part (a) and hence we continueto have 119879 = 1198821 For the minimum1198822 value the second law equations for1198822 are

119910 119873 minus1198822 cos 120579 = 0 997904rArr 119873 = 1198822 cos 120579119909 minus 119879 + 119891 +1198822 sin 120579 = 0 997904rArr1198821 minus 120583119904119873 minus1198822min sin 120579 = 0 997904rArr1198821 = 1198822min (sin 120579 + 120583119904 cos 120579) 997904rArr1198822min = 1198821

sin 120579 + 120583119904 cos 120579= 100 newtonssin 30∘ + 04 cos 30∘ 1198822min = 118 newtons

(22)

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 14: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

8 Solutions Manual

y y

xx

TT

TN

N

W1W2

W2

120579120579

ff

Figure 22 Free-body diagram for Problem (21)(b)

y

x

W1

T1

T2

1205791

1205792

(a)W2

T21205792

F

(b)

T1

1205791

W1 + W2

F

(c)

Figure 23 Free-body diagrams for Problem (22)

For the maximum 1198822 the normal force expression is unchanged but now the netforce along the incline is defined by

1198822max sin 120579 minus 119891 minus 119879 = 0 997888rarr1198822max sin 120579 minus 1205831199041198822 cos 120579 minus1198821 = 0 997888rarr

1198822max = 1198821sin 120579 minus 120583119904 cos 120579

= 100 newtonssin 30∘ minus 04 cos 30∘ 1198822max = 651 newtons

(23)

(22) We can use free-body diagrams (a and b) shown in Figure 23 to look at thehorizontal and vertical forces acting This will lead to 4 equations (one in 119909 and one in 119910for each of the two weights) but 4 unknowns (the tensions 1198791 and 1198792 and the two angles)Sincewe are only interested in the angles it is actually easier to consider free-body diagram(c) This has a system including both weights and hence the gravitational force acting is1198821 +1198822 in magnitude 2 is an internal force and hence does not get shown and the onlyexternal forces besides the weights are 1 and Hence our force equations are

119909 119865 minus 1198791 sin 1205791 = 0 997904rArr 1198791 sin 1205791 = 119865119910 1198791 cos 1205791 minus1198821 minus1198822 = 0 997904rArr 1198791 cos 120579 = 1198821 +1198822 (24)

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 15: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos First andThird Laws 9

Horizontal 120579 = 200∘

y

x

W

FN

fk fair

θ

θW

FN

fk

fair

Hill

Figure 24 Free-body diagram for Problem (23)Divide the first equation by the second to get

1198791 sin 12057911198792 cos 1205791 = tan 1205791 = 1198651198821 +1198822 997904rArr1205791 = tanminus1 1198651198821 +1198822

(25)

To find 1205792 consider free-body diagram (b)

119909 119865 minus 1198792 sin 1205792 = 0 997904rArr 1198792 sin 1205792 = 119865119910 1198792 cos 1205792 minus1198822 = 0 997904rArr 1198792 cos 1205792 = 1198822 (26)

Again divide the first equation by the first to get

1198792 sin 12057921198792 cos 1205792 = tan 1205792 = 1198651198822 997904rArr1205792 = tanminus1 1198651198822

(27)

(23) The free-body diagram is shown in Figure 24 The force equations are belowNote that since the velocity is constant the net force acting on the skier is zero

119909 119882 sin 120579 minus 119891air minus 119891119896 = 0119891air = 0148V2 = 119882 sin 120579 minus 120583119896119882 cos 120579

119910 119865119873 minus119882 cos 120579 = 0 997904rArr119865119873 = 119882 cos 120579 = (620N) cos 200∘ = 583N

(28)

where 119891air is the friction due to air drag corresponding to the terminal velocity Now wecan go back to the 119909 equation to get

0148V2 = (620N) [sin 200∘ minus (0150) (0940)] 997904rArrV = radic 125N0148N(ms)2 = 290ms (29)

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 16: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

10 Solutions Manual

y

x

1205791

12057911205792

1205792

W

F1

F2

Figure 25 Force diagram for Problem (24)

(24) The force diagram is shown in Figure 25 for this situation The 119909 and 119910 forceequations are

119909 1198651 sin 1205791 minus 1198652 sin 1205792 = 0 997904rArr 1198652 = 1198651 sin 1205791sin 1205792 119910 1198651 cos 1205791 + 1198652 cos 1205792 minus119882 = 0 997904rArr1198651 [cos 1205791 + sin 1205791cot1205792] minus 119882 = 0

(210)

Replacing 1198652 in the 119910 equation with the expression from the 119909 equation yields

1198651 = 119882cos 1205791 + sin 1205791cot 1205792 997904rArr

1198652 = 1198651sin 1205792 sin 1205791 = 119882

cos 1205792 + cot 1205791 sin 1205792 (211)

Note that you get a check on the answer by seeing that you get 1198652 by exchangingsubscripts in the formula for 1198651

(25) The force diagram show in Figure 26 lays out all the angles we need The linebetween the center of the 2119863 pipe and the center of either of the smaller pipes makes anangle 120579 with respect to the vertical where

sin 120579 = 11986331198632 = 23 997904rArr cos 120579 = radic1 minus sin2120579 = radic53 (212)

From the force diagram we see in the 119909 direction that equilibrium requires thehorizontal components of 1 and 2 (the normal forces of the bottom pipes on the 2119863pipe) to be equal so

1198651 sin 120579 = 1198652 sin 120579 997904rArr 1198651 = 1198652 (213)

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 17: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 11

D

y

x

120579

3D

2D

D2D2 3D2

FW1 FW2W2

W1 W1Fb1 Fb2

F1F2

Figure 26 Free-body diagrams for Problem (25)Note that this also clear by symmetry By Newtonrsquos Third Law these are also the

magnitudes of the normal forces of the 2119863 pipe on either of the smaller pipes The 119910components must then give

1198651 cos 120579 + 1198652 cos 120579 minus1198822 = 021198651 cos 120579 = 1198822 997904rArr

1198651 = 1198652 = 11988222 cos 120579 = 311988222radic5 (214)

In equilibrium the vertical and horizontal net force on each of the pipes is zero Inparticular the net horizontal force on each of the small pipes can be written as

1198651198821 minus 1198651 sin 120579 = 0 997904rArr1198651198821 = 1198651198822 = 121198822 tan 120579 = 1198822radic5

(215)

3 Newtonrsquos Second Law

31 Newtonrsquos Second Law of Motion Problem Solutions

(31) (a)The free-body diagram for this case is as followsThe pulley is massless and hencethe net force acting on it must be zero so if up is the positive direction (the direction ofthe unit vector shown in Figure 31) then

119865 minus 2119879 = 0 997904rArr 119879 = 12119865 = 50 newtons (31)

The acceleration of each block in an inertial frame is the vector sumof the accelerationof the center of the disk and the acceleration of that block relative to the center of the diskCall the latter 1198861 for 1198981 and 1198862 for 1198982 Since the string is unstretched we must insist that

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 18: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

12 Solutions Manual

m2

m1

k

F F

T

T T

T

m2g

m1g

Figure 31 Free-body diagram for Problem (31)(a)

1198861 = minus 1198862 If the acceleration of the center of the disk is 119886 then writing 1198862 = 1198862 we have1198861 = minus1198862 Now we can write the second law equations for the blocks

119879 minus 1198981119892 = 1198981 (119886 minus 1198862) 119879 minus 1198982119892 = 1198982 (119886 + 1198862) (32)

We can rewrite these equations as

1198791198981 minus 119892 = 119886 minus 11988621198791198982 minus 119892 = 119886 + 1198862

(33)

Summing the two equations we can solve for 119886 and then for 1198862 as follows119879( 11198981 + 11198982) minus 2119892 = 2119886 997904rArr

119886 = 11986504 ( 11198981 + 11198982) minus 119892= (250 newtons) [ 1500 kg + 1200 kg] minus 980ms2

119886 = 770ms2

(34)

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 19: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 13

If we instead subtract the first equation from the second then

21198862 = 11986502 [ 11198982 minus 11198981 ] 997904rArr1198862 = 11986504 [1198981 minus 119898211989811198982 ]

= (250 newtons) [ 300 kg100 kg2 ] 1198862 = 750ms

(35)

(b) The tension is already derived as 500 newtons

(32) We solve the problem in an inertial frame which has the velocity which theelevator had when the ball was released (we assume that the ball does not hit the ceilingof the elevator) If we take 119905 = 0 at the instant when the ball was released and denote thevertical coordinate by 119910 then 119910floor = (12)1198601199052 and 119910ball = V0119905 minus (12)1198921199052 The height ofthe ball above the floor is 119910ball minus 119910final = V0119905 minus (12)(119892 + 119860)1199052 This is maximum when119905 = V0(119892 + 119860) and has the value

ℎmax = V202 (119892 + 119860) (36)

We see that the maximum height above the floor which the ball reaches is the sameas the maximum height it would reach if it were projected upward with velocity V0 in anon-accelerating box on a planet where the acceleration of gravity is 119892 + (rather than 119892where 119892 = minus119892 where points vertically upward from the earth) We have already seen(Example 32) that the upward force which the floor exerts on a person of mass 119898 in anaccelerating elevator (ie the force which a scale measures) is 119898(119892 + 119860) which is whatthe scale would read if the elevator were not accelerating but were on a planet where theacceleration of gravity is ( 119892 + )

Quite generally we can show that if a box has acceleration (and is not rotating)relative to an inertial frame we can treat axes attached to the box as though they were aninertial frame provided that when we list the forces acting on an object in the box we addto the list a fictional force minus119898 This extra ldquoforcerdquo has the same form as the gravitationalforce119898 119892 We call it ldquofictionalrdquo because it is not due to the action of any identifiable pieceof matter

Proof If the axes of the inertial frame are 119895 and axes attached to the box are 101584011989510158401015840 thena particle which has acceleration 1198861015840 relative to the primed axes has acceleration 1198861015840 + relative to the inertial frame The equation of motion of the particle is = 119898( 1198861015840 + )where is the total force acting on the particle We can rewrite this as 1015840 = 119898 1198861015840 where 1015840is the sum of the real force and the fictional force minus119898

(33) (a) If the board is not slipping then the acceleration of the boy in an inertialframe is also 119886 The boy must therefore have force = 119898boy 119886 acting on him and he must

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 20: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

14 Solutions Manual

Initial Final

m

mM M

D D

120579 120579

ΔxM

Figure 32 Initial and final positions for the sliding wedge and block for Problem (34)

exert a force of the same magnitude on the board The minimum acceleration of the boywhich will cause slipping is thus

119891smax = 120583119904 (119898board + 119898boy) 119892 = 119898boy119886min 997904rArr119886min = 120583119904119892[1 + 119898board119898boy

]= (0200) (980ms2) [1 + 300 kg500 kg]

119886min = 314ms2

(37)

(b)The boyrsquos acceleration exceeds 119886min Let be the direction of the boyrsquos accelerationThe board is slipping on the ice so call its acceleration 119886bd = minus119886bd The acceleration of theboy is therefore (400 minus 119886bd) (in units of meterss2) and the force exerted by the board onthe boy must be

119865boy119909 = 119898boy (400 minus 119886bd) (38)

By Newtonrsquos Third Law the net horizontal force on the board is thus

bd = minus119898bd119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2 minus 119886bd) 997904rArrminus (119898bd + 119898boy) 119886bd = 120583119896 (119898boy + 119898bd) 119892 minus 119898boy (400ms2) 997904rArr119886bd = minus (0100) (800 kg) (980ms2) minus (500 kg) (400ms2)800 kg

= 152ms2(39)

The acceleration of the boy relative to the ice is

119886boy = (400 minus 152)ms2 = 248ms2 (310)

(34) The situation initially and finally are shown in Figure 32 The free-bodydiagrams for thewedge and the block are shown in Figure 33We choose the inertial frame

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 21: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 15

120579

120579

FMm

FMm

FMt

mgMg

Figure 33 Free-body diagrams for the sliding wedge and block for Problem (34)of the earth with the 119909-axis along the horizontal (to the right) and the 119910-axis verticallyupward The 2nd law equations are therefore

119909 119865119872119898 sin 120579 = 119898119886119909 minus119865119872119898 sin 120579 = 119872119860119910 119865119872119898 cos 120579 minus 119898119892 = 119898119886119910 (311)

where 119886119909 and 119886119910 are the 119909 and 119910 components of the acceleration of the block and 119860 isthe horizontal acceleration of the wedge We know that the table and gravity maintainno vertical acceleration for the wedge With 119865119872119898 119886119909 119886119910 and 119860 all being unknown weneed one more equation to solve for everythingThe remaining equation is the constraintthat the block maintain contact with the wedge during the whole journey (otherwise thehorizontal force on the wedge would cease) If the wedge were stationary then the blockmoving down the wedge a distance 119904 would yield Δ119909 = 119904 cos 120579 and Δ119910 = 119904 sin 120579 In themoving frame of the wedge free to slide (we use primes for the coordinates in this frame)the ratio of the Δ1199091015840 and Δ1199101015840 distances must always be Δ1199101015840Δ1199091015840 = minus tan 120579 for the block tomaintain contact with the surface of the wedge Hence in the inertial frame of the earthwe have

Δ119909 = Δ1199091015840 + Δ119909119872 = minus Δ119910tan 120579 + Δ119909119872 997904rArr

119886119909 = minus 119886119910tan 120579 + 119860 (312)

where we just took the second derivative with respect to time to get the accelerations Nowwe have from our force equations

119886119909 = 119865119872119898 sin 120579119898 119860 = minus119865119872119898 sin 120579119872

119886119910 = 119865119872119898 cos 120579119898 minus 119892(313)

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 22: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

16 Solutions Manual

120572

120572

120579120579

120579

y

x

m

mg

T

Figure 34 Free-body diagram for Problem (35)(a)

Substituting into our equation coupling 119886119909 and 119886119910 yields119886119909 = minus 119886119910

tan 120579 + 119860119865119872119898 sin 120579119898 = minus119865119872119898 cos 120579 minus 119898119892119898 tan 120579 minus 119865119872119898 sin 120579119872 997904rArr

119865119872119898 = 119898119872 cos 120579119872 + 119898sin2120579119892 997904rArr119860 = minus119898 sin 120579 cos 120579119872 + 119898sin2120579119892119886119909 = 119872 sin 120579 cos 120579119872 + 119898sin2120579 119892

119886119910 = minus(119872 + 119898) sin2120579119872 + 119898sin2120579 119892

(314)

Call the distance traveled by the wedge in the time Δ119905 needed for the block to reachthe bottom of the wedge 119909119872119891 We determine them from the vertical distance minus119863 sin 120579traveled by the block

minus119863 sin 120579 = 12119886119910Δ1199052 997904rArr Δ119905 = radicminus2119863 sin 120579119886119910 997904rArr

Δ119905 = radic 2119863(119872 + 119898sin2120579)(119898 +119872)119892 sin 120579 119909119872119891 = 12119860Δ1199052 = 1198602 sdot 2119863 sin 1205791003816100381610038161003816100381611988611991010038161003816100381610038161003816 = minus 119898119872 +119898119863 cos 120579

(315)

(35) (a) The free-body diagram is shown in Figure 34 The 119909 and 119910 components ofthe forces are parallel and perpendicular to the incline respectively The string keeps theacceleration along 119909 of the pendulum bob the same as the acceleration of the rectangularbox down the incline that is 119892 sin 120579 since the entire system of box plus 119898 can be

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 23: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 17

considered as subject only to gravity and the normal force of the incline For a systemconsisting only of119898 we have along 119909

119879 sin120572 + 119898119892 sin 120579 = 119898119892 sin 120579 (316)

The only solution for non-zero tension is 120572 = 0 Once a steady state is achieved thetension points along the normal to the ceiling so that there is zero acceleration relative tothe ceiling and119898119892 sin 120579 provides the acceleration parallel to the incline so as to maintainthe same acceleration for119898 as for the system of the box plus119898

(b) With friction acting the system of box plus 119898 will fall with an acceleration lessthan 119892 sin 120579 If the system has total mass119872 then

119910 119865normal minus119872119892 cos 120579 = 0 997904rArr 119865normal = 119872119892 cos 120579119909 119872119892 sin 120579 minus 120583119865normal = 119872119886119909

119872119892 sin 120579 minus 120583119872119892 cos 120579 = 119872119886119909 997904rArr119886119909 = 119892 (sin 120579 minus 120583 cos 120579)

(317)

Themass119898 as a system is now subjected to only the forces considered in part (a) butthe acceleration along 119909 has to be the new value just derived We see from this that thetension must pull up along 119909 to keep the 119898119892 sin 120579 component from accelerating 119898 downalong the incline faster than the rectangular box Hence Figure 34 shows the string in thewrong place It has to be downhill of the normal to the ceiling Letrsquos confirm by solving for120572

119910 119879 cos120572 minus 119898119892 cos 120579 = 0 997904rArr 119879 = 119898119892 cos 120579cos120572119909 119879 sin120572 + 119898119892 sin 120579 = 119898119886119909 = 119898119892 (sin 120579 minus 120583 cos 120579)

119898119892 tan120572 cos 120579 + 119898119892 sin 120579 = 119898119892 sin 120579 minus 120583119898119892 cos 120579 997904rArrtan120572 cos 120579 = minus120583 cos 120579 997904rArr120572 = tanminus1 (minus120583) = minustanminus1120583

(318)

The minus sign reflects that the string must hang downhill of the normal (whichshould be obvious if 120583 is very large and consequently the acceleration of the box is verysmall)

(36) Forces in the vertical direction (ie perpendicular to the turntable)must balanceand hence we see immediately that119865119873 = 119898119892where 119865119873 is the normal force of the turntableon the coin and 119898119892 is the magnitude of the gravitational force acting on the coin Inthe horizontal (plane of the turntable surface) direction the only force acting is staticfriction (static since the coin maintains position with respect to the turntable) Hencestatic friction must satisfy the centripetal condition to keep the coin traveling in a circle at

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 24: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

18 Solutions Manual

Side view

Top view

r

Coin

r

Coin

120596120596

mg

FN

fs

fs

Figure 35 Free-body diagram for Problem (36)

a constant speed of V = 120596119903 The maximum radius the coin can be located from the centerof rotation is therefore determined by the maximum static frictional force hence

119891smax = 120583119904119865119873 = 120583119904119898119892 = 119898V2119903 997904rArr120583119904119892 = 1205962119903 997904rArr

120583119904 = 1205962119903119892 = (33 sdot 212058760 s)2 (015m)98ms2 = 018(319)

(37) If we assume circular orbits are possible with any 119899th power of the radius forcelaw then we can run the derivation of Keplerrsquos Third Law in reverse Let the period of thecircular orbit be 119879 Then if the proportionality constant in KeplerrsquosThird Law is 119896 we find

1198792 = 119896119903119899 997904rArr 412058721199032V2

= 119896119903119899 (320)

Since the constant 119896 can be defined as we want we absorb the factor of 41205872 and seethat

119896V2 = 1199032minus119899 997904rArr119896V2119903 = 1199031minus119899

119865grav = 11989610158401199031minus119899(321)

So the force law is proportional to the 1 minus 119899th power of the radius We see that if1 minus 119899 = minus2 then 119899 = 3 as we expect(38) Figure 36 shows just the curve We have idealized the sled to just a point at

position 120579 from the start of the curve We can also characterize the position in terms ofthe arc-length 119904 from the start of the curved path We are ignoring gravitational force but

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 25: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 19

120579

R

m2Rfk

s

Figure 36 Diagram for Problem (38)must consider the kinetic friction (119891119896) and the normal force of the track on the sled thatsatisfies the condition

119865119873 = 119898V2119877 (322)

to keep the sled following the curved trackThe tangential equation of motion is therefore

119898119886 = 119898119889V119889119905 = minus119891119904 = minus120583119898V2119877 (323)

We want the velocity as a function of angular displacement not time so we make useof the chain rule of differentiation

119889V119889119905 = 119889V119889119904 sdot 119889119904119889119905 = 119889V119889119904 sdot V (324)

where V is the tangential velocity of the sledThe differential equation is easy to solve if weset the total distance of the circular track as just 119877120579 that is we now let 120579 be the value ofthe inclination of the ramp leading into the curved track

119889V119889119905 = V119889V119889119904 = minus120583V2119877 997904rArr

intV119891

V0

119889VV

= int1198771205790

minus120583119877119889119904ln V|V119891V0 = minus120583119877 (119877120579) 997904rArrln

V119891V0

= minus120583120579 997904rArrV119891 = V0119890minus120583120579

(325)

(39) The hint steers us to consider the inertial frame to be that where the conveyorbelt is at rest that is a frame moving with respect to the floor at velocity In the floorrsquosframe the puckwould execute a curved path as it slows its velocity relative to the conveyorbelt This makes calculating the effect of friction tricky In the rest frame of the conveyorbelt the puck moves along a straight line since there are no forces acting perpendicular to

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 26: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

20 Solutions Manual

Dd

Rest frame of conveyor belt

x

y120579120579

0

Figure 37 Diagram for Problem (39)

the path of the puck in this frame (except for gravity which points downward and hencehas no direct effect on the puckrsquos motion relative to the conveyor belt) The puck goesdiagonally backward in this frame and since the frictional force is bounded the velocityof the puck (relative to the floor) is still V0 just after the puck is on the belt

The kinetic friction force is 120583119898119892with119898 being themass of the puck so the accelerationalong the path has magnitude 120583119892 Looking at the 119910 component of the puckrsquos motion theacceleration due to friction is

119886119910 = minus120583119892 cos 120579 = 120583119892 V0radicV20 + 1198812 (326)

The minimum value of V0 has the puck just reaching the other edge of the belt as its119910 velocity goes to zero so

V2119910 = V20 + 21198861199101199100 = V20 minus 2120583119892V0119863radicV20 + 1198812 997904rArr

V40 = (2120583119892119863)2V20V20 + 1198812

(327)

We could also get to this equation by considering motion along the diagonal In thatcase we have

0 = V20 + 1198812 minus 2120583119892 119863cos 120579 997904rArr

= radicV20 + 1198812 minus 2120583119892119863V0

997904rArrV40 = (2120583119892119863)2V20

V20 + 1198812 (328)

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 27: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Newtonrsquos Second Law 21

x

y

M

120579

12057990∘ minus 120579

Nm

Nm

NM

aMaMmg

Mg

Figure 38 Diagram for Problem (310)

Change variables to let 119906 = V20 and note that

1199062 = (2120583g119863)2119906119906 + 1198812 997904rArr119906 (119906 + 1198812) = (2120583119892119863)2 997904rArr

119906 = minus1198812 + radic1198814 + 16(120583119892119863)22= minus(600ms)2 + radic(1296m4s4) + 16[(02)(98)(3)]2m4s22

119906 = 350m2s2 997904rArrV0 = 187ms

(329)

(310) The free-body diagrams for the wedge and the block on top of the wedge areshown in Figure 38 We know by looking at the forces acting that there is no horizontalmotion of the block in the inertial frame of the fixed incline since there are no forces witha horizontal component acting on the block Hence the block maintains contact with thewedge only if their vertical accelerations match while the wedge slides also horizontallyrelative to the incline or the block Let the acceleration of the wedge along the incline havemagnitude 119886119872 The direction of 119886119872 is along the incline at angle 120579 below the horizontalLet 119872 be the normal force of the incline on the wedge and 119898 be the normal force ofthe wedge on the block The equations of motion are

119898 119873119898 minus 119898119892 = minus119898119886119872 sin 120579119872 119910119873119872 cos 120579 minus 119873119898 minus119872119892 = minus119872119886119872 sin 120579

119872 119909119873119872 sin 120579 = 119872119886119872 cos 120579(330)

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 28: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

22 Solutions Manual

The simultaneous solution for these three equations in the three unknowns119873119872119873119898and 119886119872 is

119873119872 = (119898 +119872)119872119892 cos 120579119872 + 119898sin2120579 119873119898 = 119898119872119892cos2120579119872 + 119898sin2120579

119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (331)

The acceleration 119886119872 parallel to the incline is the solution requestedWe can use less algebra to get the solution if we note that the acceleration of the block

along the vertical direction must be

119886119910 = 119886119872 sin 120579 (332)

and its acceleration along the direction of the incline is

119886119910 cos (90∘ minus 120579) = 119886119872sin2120579 (333)

Then looking only at the components of the force on the wedge and block which areparallel to the incline we have

119898119892 sin 120579 minus 119873119898 sin 120579 = 119898119886119872sin2120579119872119892 sin 120579 + 119873119898 sin 120579 = 119872119886119872 (334)

Adding the two equations gives the wedge acceleration immediately

(119898 +119872)119892 sin 120579 = 119886119872 (119872 + 119898sin2120579) 997904rArr119886119872 = (119898 +119872)119892 sin 120579119872 + 119898sin2120579 (335)

(311) For119872 to be in equilibriumwe should have the tension in the string equal to theweight119872119892The conditions forminimum andmaximum radius of the toy car are thereforeset when the maximum static frictional force points inward and outward respectivelywith respect to the center of the circular path (see Figure 39) We see this because thecentripetal condition requires according to Figure 39

+ 119891119904 = minus119898V2119903 997904rArr minus (119879 plusmn 119891119904) = minus119898V2119903 (336)

where minus is the direction toward the center of the circular path and therefore the directionof the tension The direction of the tension is fixed along minus so the direction (ie thesign) needed for 119891119904 is determined by the frictional force magnitude such that the vector

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 29: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Momentum 23

k

i

Mg

T

T

fsmax fsmax

Figure 39 Free-body diagram for Problem (311)equation above is true In scalar form the two conditions for the friction are determinedby

119879 + 120583119898119892 = 119898V2119903min

119879 minus 120583119898119892 = 119898V2119903max997904rArr

(337)

Since 119879 = 119872119892 dividing the first equation by the second yields

119903max119903min= (119872 + 120583119898) 119892(119872 minus 120583119898) 119892 = 119872 + 120583119898119872 minus 120583119898 (338)

4 Momentum

41 Momentum Problem Solutions

(41) Note that the definition of center of mass location with respect to the origin is

= sum119894119898119894 119903119894sum119894119898119894 (41)

For each particle 119899 in the system

10038161003816100381610038161003816 minus 119903119899100381610038161003816100381610038162 = (sum119894119898119894 119903119894119872 minus 119903119899) sdot (sum119895119898119895 119903119894119872 minus 119903119899)= 11198722 [sum

119894

119898119894 ( 119903119894 minus 119903119899)] sdot [[sum119895 119898119895 ( 119903119895 minus 119903119899)]] (42)

Now ( 119903119894 minus 119903119899) sdot ( 119903119895 minus 119903119899) is unchanged under all possible motions of the rigid body ofwhich119898119894119898119895 and119898119899 are part Therefore the distance of the CM from all particles (index119899) in the body is unchanged under all motions of the body Hence CM is a point of thebody

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 30: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

24 Solutions Manual

(42)The total force on the floor at any given amount of rope drop 119910 that is wheneverthe top of the rope has fallen a distance 119910 below its initial release point and therefore thereis a total length of rope 119910 already on the floor is

119865 (119910) = 119889119901119889119905 + 119898119892 (43)

where 119889119901119889119905 represents the force necessary to bring an infinitesimal length of rope withmass 119889119898 to a halt if it has acquired speed V in its fall 119898 is the mass of rope already onthe table and is therefore equal to 120582119910 with 120582 = 119872119871 The instantaneous velocity for anyinfinitesimal length of rope 119889119898 can be called Vwhere V2 = 2119892119910 since we consider 119889119898 to bein free-fall Therefore 119889119901 = 119889119898V and 119889119898 = 120582 sdot 119889119910 with 119889119910 being the infinitesimal lengthcorresponding to mass 119889119898 Thus

119865 (119910) = V119889119898119889119905 + 120582119910119892

= V sdot (120582119889119910119889119905 ) + 120582119910119892= V2120582 + 120582119910119892

119865 (119910) = 2119892119910 sdot 120582 + 120582119910119892 = 3120582119892119910(44)

To answer part (b) we just note that the maximum force occurs when 119910 is maximumthat is 119910 = 119871 so the maximum force is 3119872119892 and occurs when the last bit of rope hits thetable This makes sense because this last 119889119898 hits with the maximum speed (since it fallsfrom the greatest height 119871) and the (nearly) maximum weight of rope is already on thetable

(43) (a) Linear momentum conservation demands

1 = minus2 (45)

where 1 and 2 are the momenta of fragments 1 and 2 respectively The fragments mustfly off in directions that are collinear So the horizontal velocities can be determined fromthe distances traveled and the times given

V1119909 = 120m100 s = 120msV2119909 = 240m400 s = 600ms (46)

Given that the sum of the 119909momenta for the fragments must add to zero we have

1199011119909 = 1198981V1119909 = 1199012119909 = 1198982V2119909 997904rArr (E1)1198982 = 1198981 120600 = 21198981 (E2)

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 31: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Momentum 25

When the skyrocket reaches its highest point its velocity is zero so the verticalmomenta of the fragments must also be equal in magnitude and opposite signs

1198981V1119910 = minus1198982V2119910 997904rArr V1119910 = minus2V2119910 (47)

If we use the explosion site as the origin of our coordinate system then the verticaldrop of the fragments is described by

minusℎ = V1119910 (10 s) minus 98ms22 (100 s2)minusℎ = (10V1119910 minus 490) m

minusℎ = V2119910 (4 s) minus 98ms22 (16 s2)= (minus2V1119910 minus 784)m 997904rArr

10V1119910 minus 490 = minus2V1119910 minus 784 997904rArrV1119910 = 343ms 997904rArr V2119910 = minus172ms 997904rArr

ℎ = (4 s) (1715ms) + 784m= 147m

(48)

(b) Given that fragment 1 goes up we get the maximum height from that fragmentrsquostravel With the ground as the origin we have

0 = V21119910 minus 2 (98ms2) (119910max minus 147m) 997904rArr119910max = 147m + (343ms)2196ms2 = 207m (49)

(44) We must have momentum conservation along the directions parallel andperpendicular to the original shell direction (ie east) In addition mass is conservedThe three equations that result from these conditions are as follows

mass 1198981 + 1198982 = 119872 = 300 kg119909 119872V0 = 1198981V1 cos 1205791 + 1198982V2 cos 1205792119910 0 = 1198981V1 sin 1205791 minus 1198982V2 sin 1205792

(410)

From the 119910 equation and mass conservation equation we get

V2 = 11989811198982 V1 sin 1205791sin 1205792 (411)

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 32: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

26 Solutions Manual

N

Ex

y

1205791

1205792

1205791 1205792 both gt 0

Figure 41 Diagram for Problem (44)

Plugging into the 119909 equation gives

119872V0 = 1198981V1 cos 1205791 + 1198981V1 sin 1205791cot 1205792 997904rArr1198981 = 119872V0

V1 [cos 1205791 + sin 1205791cot 1205792]= (300 kg) (350ms)(900ms) [cos 200∘ + (sin 200∘) (cot 400∘)]

1198981 = 0866 kg 997904rArr 1198982 = 300 minus 0866 = 2134 kg(412)

Therefore

V2 = 11989811198982 V1 sin 1205791sin 1205792= 0866 kg2134 kg (900ms) sin 200∘

sin 400∘ V2 = 194ms

(413)

Note Many students miss this problem because they have two equations (119909 and 119910momenta) in 3 unknowns and did not realize the constraint on the sum of the massesof the fragments constitutes the third equation

(45)The thrust is a force withmagnitude equal to the time rate ofmomentum changeof the exhaust gasThe time rate of change of themass of the rocket is the time rate of massof fuel expended as exhaust gas so

1003816100381610038161003816100381610038161003816100381611988911987211988911990510038161003816100381610038161003816100381610038161003816 = fuel burned

burn time= 2300 times 106 kg minus 1310 times 105 kg150 s = 1446 times 104 kgs (414)

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 33: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Momentum 27

The thrust magnitude 119879 is therefore119879 = 119906 10038161003816100381610038161003816100381610038161003816119889119872119889119905

10038161003816100381610038161003816100381610038161003816 997904rArr119906 = 119879|119889119872119889119905|= 3402 times 107N1446 times 104 kgs 119906 = 235 times 103ms

(415)

We should note that the ideal rocket equation (429) is not so accurate here since wehave to consider the effect of gravity The modification comes in (427) where instead ofhaving momentum be conserved we must look at the change in momentum due to anexternal gravitational force that is

(119872V minus119872119892119889119905) = (119872 + 119889119872) (V + 119889V) minus 119889119872 (V minus 119906) 997904rArr119872V minus119872119892119889119905 = 119872V +119872119889V + V119889119872 + 119889119872119889V minus V119889119872 + 119906119889119872 997904rArr

119872119889V = minus119906119889119872 minus119872119892119889119905(416)

where we drop the product of differentials and we note that 119889119872 is negative Dividingthrough by119872 as in Chapter 4 and eliminating 119889119905 by noting that

119889119905 = 119889119872119889119872119889119905 (417)

yields

119889V = minus119906119889119872119872 minus 119892 119889119872119889119872119889119905 (418)

Integrating then gives (remember that 119889119872 lt 0)intV

V0

119889V1015840 = minus119906int1198721198720

11988911987210158401198721015840 minus 119892119889119872119889119905 int119872

1198720

1198891198721015840 997904rArrV minus V0 = minus119906 ln 1198721198720 minus 119892119872 minus1198720119889119872119889119905

(419)

The ideal rocket equation modified by gravity is

V = V0 minus 119906 ln( 1198721198720) minus 119892119872 minus1198720119889119872119889119905 (420)

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 34: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

28 Solutions Manual

There are two assumptionsmade hereThe first is that 67 kilometers is a small enoughaltitude relative to the earthrsquos radius that we can continue to use 119892 = 980ms2 as thegravitational acceleration We have also assumed that at time 119905 = 0

1198720119889V119889119905 = minus119906119889119872119889119905 minus1198720119892 gt 0 997904rArr minus119906119889119872119889119905 gt 1198720119892 (421)

otherwise the rocket does not initially lift off but just sits on the launch pad burning fueluntil the total mass is sufficiently small enough for the thrust to lift the rocket

The relevant numbers for this problem require us to get themass at the end of the firststage firing These numbers are

1198720 = 280 times 106 kg119872 = launch mass minusmass of 1st stage = (28 times 106 minus 23 times 106) kg

119906 = exhaust velocity = 235 times 103ms119905 = time for first stage = 150 s

(422)

Therefore the velocity of the rocket at 67 kilometers altitude would be

V = 0 minus (235 times 103ms) ln(500 times 105 kg280 times 106 kg) minus (980ms2) (150 s)= 2578 times 103ms

(423)

(46) We know the velocity of the cannonball but in a non-inertial frame As thecannonball accelerates along the barrel of the cannon the fact that the cannonball isaccelerating means that a force is being exerted on it by the cannon and hence the cannonmust have a force exerted on it by the cannonball according to Newtonrsquos third law Asthe cannonball exits the cannon it is moving at angle 120572 with respect to the horizontal inthe inertial frame (fixed with respect to the ground) The speed of the flatcar is calculablesince the momentum of the system along the horizontal (119909) is conserved as there are noexternal forces acting along 119909 Call the speed of the flatcar along the 119909-axis 119906 just as thecannonball leaves the barrelThe appropriate vector diagram for finding the velocity of thecannonball is Figure 42 where 1015840 is the final velocity of the cannonball with respect to theground Momentum conservation along the yields (assuming to the right is the positive 119909direction)

1198981198811015840 cos120572 = 119872119906119898119881 cos 120579 minus 119898119906 = 119872119906 997904rArr

119906 = 119898119881 cos 120579119898 +119872 (424)

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 35: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Work and Energy 29

120579 120572

V

u

V998400

Figure 42 Velocity vectors for Problem (46)The angle 120572 is derived from the vector diagram by noting that 1198811015840119910 = 119881119910 that is the

vertical components are equalThehorizontal component of 1015840 wehave already calculatedabove Thus

1198811015840119910 = 1198811015840 sin120572 = 119881 sin 1205791198811015840119909 = 1198811015840 cos120572 = 119872119906119898 997904rArr

11988110158401199101198811015840119909 = tan120572 = 119898119872119906 sdot 119881 sin 120579 997904rArr= 119898119881 sin 120579119872 sdot 119898 +119872119898119881 cos 120579 tan120572 = 119898 +119872119872 tan 120579

(425)

5 Work and Energy

51 Work and Conservation of Energy Problem Solutions

(51) Assume that the racket and ball meet in a one-dimensional elastic collision asconserving kinetic energy in the collision leaves more energy for the ball and hencecorresponds to the highest velocity it can have In such a case we can use the velocityexchange formula that is Vball minus Vracket = minus(119906ball minus119906racket) where 119906 indicates initial velocityand V indicates final velocity just after the collision Using this we see that the relativevelocity of racket and ball becomes

Vball minus Vracket = minus (minus119906ball minus 119906racket) (51)

where theminus1199060 occurs since the ball ismoving opposite to the direction of the racket (whichwe assume is positive 119909 direction) However the racket is held by the player and its velocitydoes not change (or at least not by an appreciable amount) So Vracket = 1199061 Therefore

Vball minus 119906racket = 119906ball + 119906racket 997904rArr Vballmax = 119906ball + 2119906racket (52)

Even if youdid not know the velocity exchange formula you can visualize the collisionin an inertial frame moving with the same constant velocity as the racket In this framethe racket looks like a stationary wall and the ball approaches the wall with speed 119906ball +

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 36: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

30 Solutions Manual

119906racket and rebounds with equal speed in the opposite direction Relative to the groundthe velocity of the rebounding ball is therefore 119906ball + 2119906racket

(52) Let 119906 be the velocity of the block relative to the wedge We use this velocity sinceits direction is always known relative to the wedge For the system consisting of the blockand wedge but not the table on which the wedge rests we have (assuming the 119909-axis isalong the horizontal with +119909 being to the right)

sum119875119909 = 0 = minus119872119881 + 119898 (119906 cos 120579 minus 119881) 997904rArr119906 cos 120579 = 119872 + 119898119898 119881 997904rArr119906 = 119872 +119898119898 119881

cos 120579 997904rArr119906 sin 120579 = 119872 + 119898119898 119881 tan 120579

(53)

where 119881 is the speed of the wedge By use of energy conservation and the substitution of119906 cos 120579 from the momentum equation we have

KE0 + 1198800 = KEfinal + 119880final

119898119892ℎ = 121198721198812 + 12119898 [(119906 cos 120579 minus 119881)2 + (119906 sin 120579)2] 997904rArr2119898119892ℎ = 1198721198812 + 119898[(119872119898119881 + 119881 minus 119881)2 + (119872 + 119898119898 119881 tan 120579)2]

= 1198812 [119872 + 1198722119898 + (119872 + 119898)2119898 tan2120579] 997904rArr21198982119892ℎ = 1198812 [119872 (119872 + 119898) + (119872 + 119898)2tan2120579] 997904rArr21198982119892ℎ119872 + 119898 = 1198812 [119872 + (119872 + 119898) tan2120579]

= 1198812 [119872cos2120579 +119872sin2120579 + 119898 sin2120579cos2120579 ] 997904rArr

119881 = radic 21198982119892ℎcos2120579(119872 + 119898) (119872 + 119898 sin2120579)

(54)

(53) (a) The jumperrsquos maximum velocity occurs at the point where the bungee cordexerts a force which just counteracts gravity (at which point the jumperrsquos acceleration is

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 37: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Work and Energy 31

zero) After this point the acceleration is upward and the jumper decelerates until comingto rest momentarily and then being accelerated upward

119865jumper = 119898119892 minus 119896119909 = 0 997904rArr119909 = 119898119892119896= (800 kg) (980ms2)200 Nm

119909 = 392m(55)

So the maximum velocity occurs at 539 meters below the bridge(b) The maximum velocity occurs when the bungee cord is stretched by 392 meters

so setting the coordinate system origin at 500 meters below the bridge

KE0 + PE0 = KE119891 + PE119891

0 + 119898119892 (500) = 12119898V2max + 119898119892 (minus392m) + 12119896(392m)2 997904rArrVmax = radic2 (980ms2) (5392m) minus (200Nm) (392m)2800 kg

= 324ms(56)

(c) The maximum acceleration occurs for the maximum net force acting on thejumper Until the 500 meters is reached only gravity acts and the jumperrsquos accelerationis 980ms2 Once the bungee cord is stretched by more than 392 meters the net force isupwardThe question is then whether the net force of 119896119909minus119898119892 gt |119898119892| for any point beforethe velocity goes to zero since at the stop point the cord is at its maximum length for thisjumper and exerting the maximum upward force Hence we want to find the maximumstretch of the rope 119909max Let119898119892119896 = 120575 as a shorthand

119898119892 (500 + 119909) = 121198961199092max 997904rArr100120575 + 2120575119909max = 1199092max 997904rArr

1199092max minus 2120575119909max minus 100120575 = 0 997904rArr119909max = 120575[1 + radic1 + 100120575 ]

(57)

The last step is just the usual quadratic equation solution simplified The value of 120575 is120575 = (800 kg) (980ms2)200Nm = 392m (58)

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 38: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

32 Solutions Manual

So

119909max = (392m) [1 + radic1 + 100m392m] = 241m (59)

This indeed leads to an acceleration upward which is greater than 980ms2 inmagnitude

119896119909max minus 119898119892 = (200Nm) (241m) minus (800 kg) (980ms2)= 4036N gt 119898 10038161003816100381610038161198921003816100381610038161003816 (510)

119896119909 is monotonically increasing so the maximum displacement always corresponds to themaximum upward force and hence the largest acceleration

(d) The value of the maximum acceleration is derived from

119865 = 119896119909max minus 119898119892= 119898119892[1 + radic1 + 100120575 minus 1] 997904rArr

119886max = (98ms2) [radic1 + 100m392m] = 504ms2(511)

Being subjected to 5119892rsquos is an exhilirating experience(e)The maximum distance from the bridge is 500m plus 119909max so 741m You need a

tall bridge(54)TheHookersquos Law constant of a rope whose unstretched length is 10 (in arbitrary

length units) is119870This means that if equal and opposite forces of magnitude 119865 are appliedto both ends the rope will stretch to length 1 + 119865119870 that is 119865 = 119870 sdot 119909 (change in length)Now suppose we have a piece of the same kind of rope with unstretched length 119871 where 119871is an integer number of unit lengthsWe can put marks on the rope conceptually dividingit into 119871 (note that 119871 is an integer with no units) equal length segments If equal andopposite forces of magnitude 119865 are applied to the two ends every segment will be inmechanical equilibrium and will have equal and opposite forces 119865 pulling on its two endsThus the increase in length of each segment is 119865119870 and the increase in length of the ropeis 119871119865119870 The Hookersquos Law constant 119896 of the rope is defined by 119865 = 119896 sdot 119909 (with 119909 being thechange in length) Thus 119896 = 119865(119871119865119870) = 119870119871

If you are a quibbler andwant to see how to extend the argument to non-integer valuesof 119871 divide the unstretched rope conceptually into very many very short pieces each oflengthΔWith negligible error you can assume that 1Δ and 119871Δ are integersThe numberof segments in a piece of rope of unit length is 1Δ If the Hookersquos Law constant for eachlittle segment is 120572 then the preceding argument shows that the Hookersquos Law constant fora piece of rope of (unstretched) unit length is 120572(1Δ) = 120572Δ Thus 119870 = 120572Δ The numberof segments in a rope of length 119871 is 119871Δ and thus the Hookersquos Law constant for the rope is119896 = 120572(119871Δ) = 120572Δ119871 = 119870119871minus QED

A bungee jumper ofmass119872 has a number of different ropes all cut from the same bigspool but of different lengths 119871 She chooses a rope ties one end to a rail on the bridge and

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 39: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Work and Energy 33

the other end to her harness and jumps We want to show that the maximum subsequenttension is the same for all of the ropes that is that 119879max does not depend on 119871

Clearly the maximum tension occurs when the rope is maximally stretched that iswhen the jumper is at her low point and has zero velocity Let the length of the rope at thispoint be 119871 + 119909 We take the jumperrsquos gravitational potential energy as zero at the bridgeand thus it equals minus119872119892(119871 + 119909) at the low point Because the ropersquos mass is much less thanthe jumperrsquos we neglect its gravitational PE The PE of the unstretched rope is zero andthe PE of the stretched rope is (12)1198961199092 The jumperrsquos KE is zero on the bridge and zeroat the low point

Thus 0 = minus119872119892(119871+119909)+(12)1198961199092We could solve this quadratic equation for its positiveroot and then calculate the maximum tension 119879max = 119896119909 = (119870119871)119909 If the statement ofthe problem is correct then 119879max does not depend on 119871 that is 119909119871 does not depend on119871 We can see this even without solving the quadratic equation Let 119910 = 119909119871 Then 119909 = 119910119871and our equation is

0 = minus119872119892 (119871 + 119910119871) + 12 (119870119871) (119910119871)2 997904rArr= minus119872119892 (1 + 119910) + 121198701199102

(512)

Note that 119871 has disappeared so its clear that 119910 and 119879max do not depend on 119871 This factis known to many jumpers and rock climbers

(55) We seek to prove that if linear momentum is conserved in one frame that is fora system of 119899 particles

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 (513)

where the 119894 and 119891 subscripts refer to the velocities before and after the collision then itis true in all other inertial frames We can write down the initial and final momenta inany other inertial frame moving with velocity with respect to the first inertial frame asfollows

119899sum119895=1

119898119895V119895119894 minus 119899sum119895=1

119898119895 = 119899sum119895=1

119898119895V119895119891 minus 119899sum119895=1

119898119895119899sum119895=1

119898119895 (V119895119894 minus ) = 119899sum119895=1

119898119895 (V119895119891 minus )119899sum119895=1

119898119895V 1015840119895119894 = 119899sum119895=1

119898119895V 1015840119895119891(514)

Hence momentum is also conserved in the new inertial frame We now turn to theargument for kinetic energy associated with motion Kinetic energy conservation is

119899sum119895=1

12119898119895V2119895119894 =119899sum119895=1

12119898119895V2119895119891 (515)

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 40: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

34 Solutions Manual

For a different inertial frame 1198601015840 as stated above119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895(V119895119894 minus )2

= 119899sum119895=1

12119898119895 [V2119895119894 minus 2 sdot V119895119894 + 1198812] 997904rArr= 119899sum119895=1

12119898119895 [V2119895119891 minus 2119881 sdot V119895119894 + 1198812] (516)

where the last line is from our previous expression for kinetic energy conservation in theoriginal inertial frame Momentum conservation in the original inertial frame means

119899sum119895=1

119898119895V119895119894 = 119899sum119895=1

119898119895V119895119891 997904rArr sdot 119899sum119895=1

119898119895V119895119894 = sdot 119899sum119895=1

119898119895V119895119891 997904rArr2 sdot 119899sum119895=1

12119898119895V119895119894 = 2 sdot 119899sum119895=1

12119898119895V119895119891(517)

Therefore we can rewrite our previous initial kinetic energy for the new inertial frameas

119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895 [V2119895119891 minus 2 sdot V119895119891 + 1198812]= 119899sum119895=1

12119898119895(V119895119891 minus )2119899sum119895=1

12119898119895V1015840 2119895119894 =119899sum119895=1

12119898119895V1015840 2119895119891(518)

so kinetic energy is conserved in this frame(56)The collision of two particles produces two outgoing momentum vectorsThose

two vectors define a plane hence we have a two-dimensional problem Momentumconservation for identical mass particles with one at rest requires that

119898V1119894 = 119898V1119891 + 119898V2119891 997904rArrV21119894 = V21119891 + V22119891 + 2V1119891 sdot V2119891 (519)

but kinetic energy conservation requires that

12119898V21119894 = 12119898V21119891 + 12119898V22119891 997904rArr V21119894 = V21119891 + V22119891 (520)

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 41: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Work and Energy 35

1

F Fx2x2

(a)

FF

FFF

F F F

L

111

1

(b)

Figure 51 Forces acting on sections of a rope under tension

To make both momentum and kinetic energy conservation equations true requires

2V1119891 sdot V2119891 = 0 (521)

which in turn requires the final momentum vectors to be orthogonal that is particlesmoving perpendicular to one another

(57) The collision exerts an impulsive force on the platform which due to inertiabrings the block to the same velocity as the platform before the spring can act to exertany more force than it exerts at equilibrium with just the platform alone We measure themaximum compression of the spring from the equilibrium position of the platform andspring before the collision The speed of the block just before the collision comes frommechanical energy conservation so if119898 = 0500 kg and 1199100 = 0600m then

12119898V2 = 1198981198921199100 997904rArrV = radic21198921199100 = radic2 (980ms2) (0600m)= 343ms

(522)

We use linear momentum conservation to find the speed of the platform and blockimmediately after the collision Let 119872 be the mass of the platform V the initial speed ofthe block just before the collision and 119881 the speed of the block and platform just after thecollision Then

119898V = (119898 +119872)119881 997904rArr119881 = 119898V119898 +119872 = (0500 kg) (343ms)150 kg

= 114ms(523)

After the collision we can use energy conservation Let 119910 be the distance the springcompresses from the equilibrium position (which is to say the equilibrium position at

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 42: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

36 Solutions Manual

x

0

01 D 05 D 09 Dx

uD minus x

R2R1

Figure 52 Gravitational potential for Problem (58)which the platform on top of the spring is at rest before the collision with the block) Thespring is already compressed by a distance119872119892119896 before the block hits the platform so

119870119891 + 119880119891 = 1198700 + 119880012119896(119872119892119896 + 119910)2 = 12 (119898 +119872)1198812 + (119898 +119872)119892119910 + 12(119872119892119896 )2 997904rArr

0 = 1198961199102 minus 2119892119898119910 minus (119898 +119872)1198812 997904rArr119910 = 119898119892 + radic11989821198922 + 1198961198812 (119898 +119872)119896= (490N) + radic(490N)2 + (120Nm) (114ms)2 (150 kg)120Nm

119910 = 0175m = 175 cm

(524)

(58) (a)The qualitative figure is below where we represent 119909 as the distance from oneof the planets along the line between the planets We are free to place the origin of thecoordinate system anywhere we like An easy choice is on the line between the planetsFor the figure below it is obvious that100381610038161003816100381610038161 minus 11990310038161003816100381610038161003816 = 119863 minus 119909 100381610038161003816100381610038162 minus 11990310038161003816100381610038161003816 = 119909 (525)

where 119909 is a positive distance The potential energy function is

119906 (119909) = minus 119866119872[1119909 + 1119863 minus 119909]= minus 119866119872[ 119863119909 (119863 minus 119909)]

119906 (119909) = minus 119866119872119863(119909119863) (1 minus 119909119863) (526)

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 43: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Simple Harmonic Motion 37

We plot 119906(119909) in arbitrary units of energy(b) This problem is somewhat unphysical as the stations cannot stably be at rest with

respect to the planets at their positions Nevertheless we take the positions and note thatthe initial total mechanical energy of the projectile must exceed the maximum potentialenergy at1198632 to make it to the other stationWe know the position of themaximum since

119889119889119909 [ 1(119909119863) (1 minus 119909119863)] = 0 997904rArr 119909 = 1198632 (527)

and you can easily verify that this extremum is a maximum Along the line between theplanets if the initial kinetic energy is such that the kinetic energy is non-zero at1198632 thenthe projectile ldquofallsrdquo to the other station

119906Alpha = minus119866119898119872[ 11198634 + 131198634] = minus161198661198981198723119863 997904rArr1198700 + 119906Alpha = 0 + 119906 (119909 = 1198632)

12119898V2 minus 161198661198721198983119863 = minus4119866119872119898119863 997904rArrV0 = radic81198661198723119863

(528)

6 Simple Harmonic Motion

61 Simple Harmonic Motion Problem Solutions

(61) We showed that the equation of motion is the same as that for a frame where thegravitational acceleration is modified to 1198921015840 where 1198921015840 = 119892 minus 119886 The period of the pendulummust therefore be

period = 2120587radic ℓ1198921015840 = 2120587radic ℓradic1198922 + 1198862 (61)

(62) The motion of the particle is identical to that of the bob of a simple pendulumThe tension in this case is replaced by the normal force of the bowl surface on theparticle We could use the forces acting on the particle to derive the equation of motionLetrsquos instead take the approach of Section 64 and look at the mechanical energy FromFigure 61 we see that the height of the particle above the lowest point of the sphericalsurface is

119910 = 119877 minus 119877 cos 120579 = 119877 (1 minus cos 120579) (62)

There is no friction so all the forces are conservative and we can write

KE + PE = 119864 997904rArr 12119898V2 + 12119898119892119910 = constant (63)

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 44: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

38 Solutions Manual

120579

R

y

Figure 61 Solution 62

where 119898 is the mass of the particle We note that the angular speed of the particle at anyinstant is 120596 = V119877 We take the derivative of the energy equation with respect to time tofind

119889119889119905119864 = 0 = 12119898(2V119889V119889119905 ) + 119898119892119877(0 minus sin 120579 sdot 119889120579119889119905 )= 119898119877120596 sdot 119889119889119905 (119877120596) + 119898119892119877 sin 120579 sdot 120596

0 = 1198981198772120596120572 + 119898120596119892119877 sin 120579 997904rArr120572 = minus119892119877 sin 120579

(64)

where 120572 of course is the angular acceleration For small displacements from the lowestpoint we need 120579 sim 0 and hence sin 120579 sim 120579 Therefore

120572 ≃ minus119892119877120579 (65)

This has the same form as (63) if we replace 1205962 by (119892119877) hence we see immediatelythat the equation of motion predicts simple harmonic oscillation with a period of

119879 = 2120587120596 = 2120587radic119892119877 = 2120587radic119877119892 (66)

(63)The position of the particle can be determined as 119909 as shown in Figure 610Thisposition is

119909 = 119903 sin 120579 (67)

The force on the particle at this position has magnitude

119865 = 119866119898119872(119903)1199032 = 119898119892119903119877 (68)

where 119872(119903) = 11987211990331198773 since the mass density is uniform (and hence equals119872[(43)1205871198773]) and 119892 = 1198661198721198772 The vector component of directed perpendicular to

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 45: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Simple Harmonic Motion 39

the tunnel is offset by the normal forces holding the particle in the tunnelThe componentparallel to the tunnel surface is

119865119909 = minus119898119892119903119877 sdot sin 120579 (69)

where the minus sign indicates that the direction of the force always points towards thecenter of the tunnel At this point 120579 = 0 and the parallel-directed force is zero hencethis point represents an equilibrium position Hence the equation describing the particlersquosmotion is

119898119886119909 = 119865119909 = minus119898119892119903119877 sin 120579 = minus119898119892119909119877 997904rArr119886119909 = 11988921199091198891199052 = minus119892119877119909 (610)

This has the same form as (63) so we determine this motion to be simple harmonicoscillation along the tunnel with angular frequency

120596 = radic119892119877 (611)

The time to cross from one end of the tunnel to the other is half a period of oscillation

12119879 = 12 sdot 2120587 sdot radic119877119892= (314159)radic 637 times 106m981ms2

119905 = 2530 s = 422 minutes

(612)

Note that the result is independent of the mass of the particle or length of the tunnelAnother interesting note is that the period is equal to that of a satellite in a circular orbitskimming the rooftops on earth

(64) For the top block the normal force due to the bottom block is119898119892Themaximumhorizontal force that can be exerted on the top block is given by the maximum staticfrictional force of 120583119898119892 Therefore the maximum acceleration possible for the top blockbefore slipping occurs is 120583119892 To get the maximum acceleration for the amplitude givenwe refer to (613) In this case we are free to set the reference point 1199090 = 0 The angularfrequency is given by

1205962 = 119896119898 +119872 (613)

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 46: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

40 Solutions Manual

From the equation ofmotion (613) we see that themaximumaccelerationmagnitudeoccurs at the amplitude positions (the maximum magnitude for 119909) so

119886max = minus1205962 sdot 119909max = minus1205962 sdot 119860120583min119892 = 119896119860119898 +119872 997904rArr

120583min = 0136(614)

7 Static Equilibrium of Simple Rigid Bodies

71 Static Equilibrium Problem Solutions

(71) If we take the ladder as a whole as the system then the only external forces are thevertical forces from the floor (which cannot exert horizontal forces as there is no friction)and the weights Hence

1198651119892 + 1198652119892 minus1198821 minus1198822 = 0 997904rArr 1198651119892 + 1198652119892 = 1198821 +1198822 (71)

In the horizontal direction for rods 1 and 2 we find

1 119879 minus 1198651119909 = 0 997904rArr 119879 = 11986511199092 1198652119909 minus 119879 = 0 997904rArr 119879 = 1198652119909 = 1198651119909 (72)

We could also get this from Newtonrsquos Third Law applied to the hinge Evaluating thetorque about the CM of rod 1 we get assuming the rods are of length ℓ

1 1198651119909 ℓ2 sin 120579 minus 1198651119892 ℓ2 cos 120579 = 02 1198652119909 ℓ2 sin 120579 minus 1198652119892 ℓ2 cos 120579 = 0 (73)

Adding these two equations gives

12 (1198651119909 + 1198652119909) sin 120579 = 12 (1198651119892 + 1198652119892) cos 120579 997904rArr2119879 sin 120579 = (1198821 +1198822) cos 120579 997904rArr

119879 = 1198821 +11988222 cot 120579(74)

(72) The situation shown in Figure 72 depicts the block at an angle where themaximum static frictional force is just sufficient to prevent sliding and the angle wherethe block is just at unstable equilibrium for tippingThe problem asks us to determine thecondition whereby these angles are the same that is the value of 120583 = 120583119888 that determinesthe angle for these two conditions (tipping and sliding) be the sameThe tipping conditionis straightforward as the equilibrium cannot be maintained if the box has its center of

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 47: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Static Equilibrium of Simple Rigid Bodies 41

120579 120579

F1x F2x

T T

W1 W2

F1g F2g

y

x

Figure 71 Force diagrams on rods of the ladder in Problem (71)

Critical angle for sliding

h = 010mw = 006m

120579

mg

fsmax

FN

(a)Critical angle for tipping

120579c

mg

FN

(b)

Figure 72 A block on an inclined board at the critical angle for sliding or tipping in Problem (72)gravity unsupported by the bottom of the box So the unstable equilibrium occurs for anangle at which the right bottom corner is direcly underneath the center of gravity hence

tan 120579119888 = 1199082ℎ2 = 119908ℎ (75)

with 119908 being the width and ℎ the height of the box and 120579119888 is the critical angle atwhich tipping occurs The critical angle for sliding is determined by the maximum staticfriction force and the component of the gravitational force pointing down the inclineThemaximum static friction is determined from the normal force 119865119873

119865119873 = 119898119892 cos 120579 997904rArr 119891119904max = 120583119865119873 = 120583119898119892 cos 120579critical angle 120583119898119892 cos 120579119888 = 119898119892 sin 120579119888 997904rArr

tan 120579119888 = 120583(76)

For the tipping and sliding critical angles to be the same we need

120583119888 = tan 120579119888 = 119908ℎ = 0600 (77)

So the general condition is that if the static friction coefficient is greater than 120579119888 fortipping then increasing 120579 causes tipping otherwise the box begins to slide before tippingcan occur That is to say if 120583 gt 119908ℎ it tips and if 120583 lt 119908ℎ it slides

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 48: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

42 Solutions Manual

B

B

C

FABy

F

F

F

F

W

W998400

W998400

W998400

120579AB120579BC

120579BC

y

x

90∘ minus 120579BC

Figure 73 Two rods hung from a ceiling in Problem (73)

FAx

FAy

FBy

FBx

FCy

FCx

120579

M1g

M3g

M2g

y

x

90∘ minus 120579

L22

L22

L12 L12

Figure 74 Force diagram for Problem (74)(73) For rod BC we have the torque about an axis through point B as

120591B = 0 = 1198651198711015840 cos 120579BC minus1198821015840 11987110158402 sin 120579BC 997904rArrtan 120579BC = 21198651198821015840 997904rArr120579BC = tanminus1 21198651198821015840

(78)

For rod AB we have the torque about the hinge at the ceiling as

0 = 119865119871 cos 120579AB minus1198821015840119871 sin 120579AB minus1198821198712 sin 120579AB 997904rArr120579AB = tanminus1 2119865119882 + 21198821015840

(79)

(74) The force diagram is shown in Figure 74 By choosing point C for the axis ofrotation we can quickly determine the vertical force acting on beam 1 at point A as the

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 49: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Static Equilibrium of Simple Rigid Bodies 43

hinge atC and theweight of1198721 have no lever armatCThe tension in the string attached atthe right end of beam 1must be equal to theweight1198723119892 since thatmass is in equilibrium

119865A11991011987112 minus1198723119892 = 0 997904rArr 119865A119910 = 1198723119892 (710)

Now we use point A as the axis and look at the torque equation for beam 1 with 119865C119909and 119865C119910 being the force exerted by the hinge at point C on beam 1

0 = 119865C119910 11987112 minus119872111989211987112 minus11987231198921198711 997904rArr119865C119910 = (21198723 +1198721) 119892 (711)

We could also have gotten this from the vertical force equilibrium equation for beam1 119865C119910 is not a force component we wanted but all other axis points have contributionsfrom two unknown force components Looking at the 119910 force equation for beam 2 andnoting that by Newtonrsquos Second Law C119910 points opposite to the direction of C119910 on beam1

minus119865C119910 minus1198722119892 + 119865B119910 = 0 997904rArr 119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (712)

Now calculating the torque about point C for beam 2 gives us

0 = 119865B1199091198712 sin 120579 +119872211989211987122 cos 120579 minus 119865B1199101198712 cos 120579 997904rArr119865B119909 = (119865B119910 minus 121198722119892) cot 120579 = (21198723 +1198721 + 121198722)119892cot 120579 (713)

Nowwe see that the only horizontal force components acting on beam 2 are 119865B119909 and119865C119909 so these must be equal in magnitude (remember that C119909 points to the left on beam2) This is also true for A119909 and C119909 so these must also be equal in magnitude hence

119865A119909 = 119865C119909 = 119865B119909 = (21198723 +1198721 + 121198722)119892cot 120579 (714)

If we follow the rule stated of minimizing the algebra then the shorter path to theanswers is to treat 1 and 2 and1198723 as one rigid body The torque about point A is then

0 = 119872111989211987112 +11987231198921198711 +119872211989211987114 minus 119865B11990911987112 tan 120579 997904rArr119865B119909 = (1198721 + 21198723 + 121198722)119892cot 120579 (715)

For this system there are no external forces acting at C so the only horizontal forcesare at A and B hence

119865A119909 = 119865B119909 (716)

Now we can consider beam 1 on its own where

119865C119909 = 119865A119909 (717)

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 50: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

44 Solutions Manual

BC

20∘

120572fs

Mg

Mg

FN

A

Figure 75 Free-body diagram for Problem (75)

As before calculating the torque about C for beam 1 yields 119865A119910 = 1198723119892 while thetorque about A for that beam gives

119865C119910 = (21198723 +1198721) 119892 (718)

and the force equations for beam 1 give

119865B119910 = 119865C119910 +1198722119892 = (21198723 +1198721 +1198722) 119892 (719)

(75) In the text it was proven that the net result of summing the gravitional torque oneach piece of the hoop is that the force acts as if it were acting at the center of mass thatis the geometric center of the hoopWe can note that that the net force on the hoop in thedirection parallel to the incline must be

119872119892 sin 20∘ +119872119892 sin 20∘ minus 119891119904 = 0 997904rArr 119891119904 = 2119872119892 sin 20∘ (720)

since the net force on the hoopmust be zero If we look at the torque about an axis throughthe center of the hoop we see that the weight of the hoop has no lever arm and the normalforce points along the radial vector to point A where the normal force acts Hence thetorque due to the point mass and the torque due to the static friction force must cancelabout this axis We call the radius of the hoop 119877 so

119872119892119877 sin (120572 + 20∘) minus 119891119904119877 = 0 997904rArr119872119892119877 sin (20∘ + 120572) = 2119872119892119877 sin 20∘ 997904rArr

(20∘ + 120572) = sinminus1 [2 sin 20∘] 997904rArr120572 = 232∘

(721)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 51: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 45

R

+

Mg

T

Figure 81 Forces acting on a falling cylinder wrapped with a string in Problem (81)8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies

81 Rotational Motion Problem Solutions

(81) (a) The forces acting are shown in Figure 81 The torque about an axis through thecenter of mass is

120591 = 119879119877 = 119868120572 = 121198721198772 119886CM119877 997904rArr119879 = 12119872119886CM

(81)

where119879 is the tension in the string andwe have used the rollingwithout slipping conditionthat the rotational acceleration equals the linear acceleration of the center of mass dividedby the radius of the cylinder Nowwe can look at the force equation for the center of mass

119872119892 minus 119879 = 119872119886CM119872119892 minus 12119872119886CM = 119872119886CM 997904rArr

119886CM = 23119892(82)

(b)The force diagram is unchanged but now the acceleration of the center of mass ofthe cylinder is zero with respect to a stationary observerTherefore the tensionmust equalthe weight that is 119879 = 119872119892 The torque about an axis through the center of mass is then

119879119877 = 119868120572119872119892119877 = 121198721198772120572 997904rArr

120572 = 2119892119877 (83)

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 52: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

46 Solutions Manual

(M + m)g

fr ff

Figure 82 Force and torque diagram for Problem (82)The rolling without slipping condition is assumed to still apply that is to say the

cylinder does not slip on the string so the hand needs to accelerate upward at value

119886hand = 119877120572 = 2119892 (84)

(82) The torque applied by the bicycle chain is considered known We do not needto know the radius at which it is applied Since the angular velocity of both wheels isincreasing and the only torque on the front wheel (around its center) is due to the staticfrictional force that force must be directed backward Since the net force on the bicycleand rider must be forward the static frictional force exerted by the ground on the rearwheel must be forward Thus for the rear wheel

120591 minus 119891119903119877 = 119868120572 = 1198981198772 119886119877 (85)

while for the front wheel we have

119891119891119877 = 119868120572 = 1198981198772 119886119877 (86)

The static frictional forces are the external forces from the ground acting on thesystem of bicycle plus rider so

119891119903 minus 119891119891 = (119872 + 2119898) 119886[ 120591119877 minus 119898119886] minus 119898119886 = (119872 + 2119898) 119886 997904rArr

119886 = 120591(119872 + 4119898)119877(87)

(83) (a) The net torque about the end of the rod attached to the string is zero so

120591net = 0 = 119898119892ℓ2 cos 120579 minus 119865119873ℓ cos 120579 997904rArr119865119873 = 12119898119892 cos 120579 (88)

where we note that the force exerted by the floor is only along the vertical as the smoothfloor cannot exert any horizontal force component on the rod

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 53: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 47

+

Mg

T

FN

120579

985747

90∘ minus 120579

Figure 83 Forces acting on a rod suspended from the ceiling and in contact with the floor in Problem(83)

(b) When the string is cut there are still no horizontal components to either thecontact force from the floor or gravity so the 119909-component of the center of mass positionstays constantThis means the contact point with the floor is subject to acceleration to theleft We cannot use this point for calculations of torque For an axis through the center ofmass though we can calculate the torque

120591CM = minus119865119873 ℓ2 cos 120579 = minus119868CM120572 = minus 112119898ℓ2120572 (89)

The vertical acceleration of the center of mass is the result of the normal force andgravity so

119865119873 minus 119898119892 = minus119898119886CM = minus119898 1003816100381610038161003816100381610038161003816100381610038161198892119910 (119905)1198891199052

100381610038161003816100381610038161003816100381610038161003816 (810)

assuming 119910 is the vertical (up) direction and 119886CM is the magnitude of the acceleration ofthe CM To relate the angular acceleration to the vertical acceleration we note that thevertical position of the floor as an origin determines that

119910 (119905) = ℓ2 sin 120579 (119905) 997904rArr119889119910 (119905)119889119905 = ℓ2 cos 120579 (119905) 119889120579 (119905)119889119905 997904rArr

1198892119910 (119905)1198891199052 = ℓ2 [minus sin 120579 (119905) (119889120579(119905)119889119905 )2 + cos 120579 (119905) 1198892120579 (119905)1198891199052 ](811)

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 54: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

48 Solutions Manual

For the time 119905 = 0 we have the initial linear and angular velocity of the CM as zeroso 119889120579(119905)119889119905|119905=0 = 0 Hence the vertical acceleration of the CM of the rod at 119905 = 0 is

1198892119910(119905)1198891199052100381610038161003816100381610038161003816100381610038161003816119905=0 = minus119886CM = minus120572ℓ2 cos 120579 (812)

We can use the force and torque equations to solve for 119886CM that is the verticalacceleration

119865119873 = 119898ℓ1205726 cos 120579= 119898ℓ6 cos 120579 sdot 2119886CMℓ cos 120579

119898 (119892 minus 119886CM) = 1198983cos2120579119886CM 997904rArr119892 = 119886CM [ 13cos2120579 + 1] 997904rArr

119886CM = 3119892cos21205791 + 3cos2120579 997904rArr 119886CM = minus 3119892cos21205791 + 3cos2120579119910

(813)

and the normal force is

119873 = 119898119886CM3cos2120579119910 = 1198981198921 + 3cos2120579119910 (814)

(84)We know that the static frictional force acts up along the incline on the cylinderThe normal force of the incline on the cylinder is 119898119892 cos 120579 with 119898 being the mass of thecylinder The maximum static frictional force for any given angle 120579 is therefore

119891119904max = 120583119898119892 cos 120579 (815)

which decreases as 120579 increases The maximum torque that could be exerted about anaxis through the center of the cylinder also decreases This torque must provide theangular acceleration that matches the condition for rolling without slipping of the linearacceleration of the center of mass of the cylinder Hence if 119877 is the radius of the cylinder

120591max = 119891119904max119877 = 119868120572120583119898119892119877 cos 120579 = 121198981198772 119886CMmax119877 997904rArr

119886CMmax = 2120583119892 cos 120579(816)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 55: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 49

But the acceleration of the center of mass must also satisfy Newtonrsquos Second LawAlong the direction of down the incline

119886CMmax = 119898119892 sin 120579 minus 119891119904max1198982120583119892 cos 120579max = 119892 sin 120579max minus 120583119892 cos 120579max 997904rArr120579max = tanminus1 (3120583)

(817)

Contrast this result with the fact that the steepest incline on which a block can restwithout slipping has 120579max = tanminus1120583

(85) The hubcaprsquos CM is originally moving along with the wheel with linear velocityV0 The rotational velocity is originally therefore

1205960 = V0119877 (818)

The impact with the road subjects the hubcap to an impulse which changes the linearvelocity of the center of mass of the hubcap from V0 to V119891 and the angular velocity fromV0119877 to 120596119891 = V119891119903 Note that if the direction of the impulse slows (or quickens) the linearvelocity it must speed up (or reduce) the angular velocity given the direction of the torqueabout the center of mass

Δ119901 = 119898(V119891 minus V0) 997904rArrΔL = minusΔ119901 sdot 119903

121198981199032 (120596119891 minus 1205960) = minus119898119903 (V119891 minus V0) 997904rArr1199032 (V119891119903 minus V0119877 ) = V0 minus V119891 997904rArr

3V119891 = V0 [2 + 119903119877] 997904rArrV119891 = V0 [23 + 1199033119877] 997904rArr

120596119891 = V119891119903 = V0119877 [23 119877119903 + 13]

(819)

Note that V119891 lt V0 while 120596119891 gt 1205960 We should expect that the linear velocity woulddecrease since the lowest point of the hubcap has forward velocity V0 minus 1205960119903 = V0(1 minus119903119877) gt 0 before it hits the road and therefore the frictional force is directed backwardThis direction of frictional force also contributes torque about the CMwhich increases therotational velocity

(86) (a) The force diagram above shows that if V0 lt V119888 then the condition thatmaintains the sphere on the small radius 119903 of the curved edge of the cube (in red) is

119865119873 minus 119898119892 cos 120579 = minus 119898V20119877 + 119903 (820)

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 56: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

50 Solutions Manual

r

R

FN

120579

120579

y

x

mg

Figure 84 Forces acting on a solid sphere rolling over the edge of a cube in Problem (86)This is the centripetal condition required for a circular path Assuming that 119903 ≪ 119877

we can ignore 119903 and just assume the radius of the curved path is just 119877 the radius of thesphere The definition of ldquomaintain contactrdquo is that the normal force not be zero so if theinitial velocity of the center ofmass of the sphere is too high119865119873would have to be less thanzero (non-physical) to satisfy the equation Hence the sphere leaves the cube without anydeviation in the direction of V0 Therefore we want to evaluate V119888 for 120579 = 0

119898119892 cos 120579 minus 119898V2119888119877 = 0 997904rArr V119888 = radic119877119892 cos 120579 = radic119877119892 (821)

(b) Now we allow for 120579 gt 0 The linear velocity of the center of mass and the angularvelocity about the center of mass increase as the spherersquos center drops as it rolls over theedge The initial kinetic energy can be calculated as

1198700 = 12119898V20 + 121198681205962CM= 12119898V20 + 12 sdot 251198981198772 sdot V201198772

1198700 = 710119898V20

(822)

It is more convenient to consider the motion as pure rotation about the stationarypoint (assumes no slipping) of contact of the sphere with the cube edge Then usingthe parallel-axis theorem and the rolling without slipping condition that angular speedaround the center of mass is the same as angular speed of the center of mass around theaxis through the point of rolling contact we have

1198700 = 12119868edge1205962edge= 12 (251198981198772 + 1198981198772) V201198772

1198700 = 710119898V20

(823)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 57: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 51

The speed increases as the sphere rolls over the edge as the center of mass goes downConsidering the cubersquos top surface as the zero point of the vertical position we calculatethis increase in speed as

minusΔ119880 = 119898119892119877 minus 119898119892119877 cos 120579 = Δ119870 = 710119898V2119891 minus 710119898V20 997904rArr119892119877 (1 minus cos 120579) = 710V2119891 minus 710V20

(824)

From our analysis in part a of this problem we know that there is a maximum speedcorresponding to maintaining contact that is when the normal force goes to zero

119865119873 minus 119898119892 cos 120579 = minus119898V2119877 997904rArr0 minus 119898119892 cos 120579119888 = minus119898V2119891119877 997904rArr

V2119891 = 119892119877 cos 120579(825)

Plugging this into our previous energy equation we see

119892119877 (1 minus cos 120579119888) = 710119892119877 cos 120579119888 minus 710V20 997904rArr1710119892119877 cos 120579119888 = 119892119877 + 710V20 997904rArr120579119888 = cosminus1 [ 717 V20119892119877 + 1017]

(826)

(c) As before the normal force goes to zero at the critical angle 120579119888 so if V0 = 0 theformula from part (b) gives us

120579119888 = cosminus1 1017 = 540∘ (827)

(d) Consider Figure 85 where we show the sphere just leaving the edge of the cubeWe know the critical angle 120579119888 for which this happens if V0 = 0 From the figure we see thatif in the time it takes for the spherersquos center of mass to travel horizontally by a distanceΔ119909 the sphere has traveled down vertically less than a distance Δ119910 = 119877 cos 120579 then the leftedge of the sphere misses the last point of contact of sphere and cube and therefore thereis no bump The free-fall formulae for the center of mass are

Δ119909 = VCM cos 120579119888 (Δ119905)Δ119910 = VCM sin 120579119888 (Δ119905) + 12119892(Δ119905)2

(828)

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 58: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

52 Solutions Manual

r

R

y

x

mg

VCM

Δx

Δy

120579c

Figure 85 Figure of a sphere just leaving the edge of a cube

The first equation can be solved for Δ119905 and this can be substituted into the Δ119910equation

Δ119905 = Δ119909VCM cos 120579119888 997904rArr

Δ119910 = Δ119909 tan 120579119888 + 1198922 (Δ119909)2V2CMcos2120579119888

= Δ119909 tan 120579119888 + 7(Δ119909)220119877cos2120579119888 (1 minus cos 120579119888) (829)

where in the last equationweused the result for VCM frompart (c) Sincewe have calculatedcos 120579119888 = 1017 and we find from Figure 85 that we want Δ119909 = 119877(1 minus sin 120579119888) to calculatethe distance fallen Δ119910 then

Δ119910 = Δ119909 tan 120579119888 + 7 (Δ119909) 119877 (1 minus sin 120579119888)20119877cos2120579119888 (1 minus cos 120579119888)= Δ119909[tan 120579119888 + 7 (1 minus sin 120579119888)20cos2120579 (1 minus cos 120579119888)]= Δ119909 [13748 + 7 (019131)20 (03402) (041175)]

Δ119910 = 18447Δ119909= 18447119877 (1 minus sin 120579119888)

Δ119910 = 0353119877

(830)

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 59: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 53

This is the distance the leftmost edge point of the sphere falls (since the sphere is asolid body) in the time it takes for this same point to just pass the edge of the cube Tobump the edge the sphere would have to fall vertically a distance

Δ119910 = 119877 cos 120579 = 0588119877 (831)

in this time The sphere does not fall this far before the leftmost point clears the edge sono bump occurs

Alternatively we could just as well look at the position of the center of the sphere as afunction of time after the sphere loses contact with the cube edge The velocity of the CMat this position we calculated in part b so call this 0

11988120 = 119892119877 cos 120579119888 (832)

where again we know cos 120579119888 = 1017 The initial 119909 and 119910 velocities of the CM just afterthe sphere loses contact are

1198810119909 = 1198810 cos 120579119888 = radic119892119877(cos 120579119888)321198810119910 = minus 1198810 sin 120579119888 = minusradic119892119877 (sin 120579119888) (cos 120579119888)12 (833)

The position of the CM as a function of time is

119909 (119905) = 1199090 + 1198810119909119905 = 119877 sin 120579119888 + radic119892119877(cos 120579119888)32119905119910 (119905) = 1199100 + 1198810119910119905 minus 121198921199052

= 119877 cos 120579119888 minus radic119892119877 cos 120579119888 (sin 120579119888) 119905 minus 121198921199052(834)

Therefore the square of the distance of the CM from the last point of contact as afunction of time is

119877(119905)2 = 1199092 (119905) + 1199102 (119905)= 1198772sin2120579119888 + 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 + 119892119877cos31205791198881199052 + 1198772cos2120579119888

minus 2radic1198921198773 sin 120579119888(cos 120579119888)32119905 minus 119892119877 cos 1205791198881199052+ 119892119877 cos 120579119888sin21205791198881199052 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

119877(119905)2 = 1198772 + radic1198923119877 cos 120579119888 sin 1205791198881199053 + 119892211990544

(835)

since 119877(119905)2 gt 1198772 for all 119905 gt 0 we have shown that the sphere always misses bumping theedge

(87) (a) The velocity of this point is due to a combination of rotational andtranslational motion The translational motion is due to the motion of the center of mass

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 60: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

54 Solutions Manual

(CM) in response to the momentum transfer of the impulse and the fact that the rod is arigid body Thus the translational velocity of the point 1199011015840 is

119872VCM = minus119875 997904rArr Vtrans = VCM = minus 119875119872 (836)

where 119872 is the mass of the rod and we assume the impulse is in the minus119909 direction Therotation of 1199011015840 about the center of mass comes from the angular momentum imparted bythe impulse so we calculate 120596CM the angular velocity about the center of mass as follows

L = 119868CM120596CM = 119875 sdot 119904119872119871212 120596CM = 119875 sdot 119904 997904rArr

120596CM = 121198751199041198721198712(837)

The velocity of1199011015840 due to rotation is thus in the +119909 direction (since it is on the oppositeside of the CM from the impulse) and hence

Vrot = 120596CM sdot 119910 = 121198751199041198721198712 119910 (838)

Thus the net velocity of 1199011015840 isV1199011015840 = Vtrans + Vrot = 121198751199041198721198712 119910 minus 119875119872 (839)

(b) The magic value of 119904makes V1199011015840 = 0 soV1199011015840 = 0 997904rArr

121198751199041198721198712 119910 = 119875119872 997904rArr119904 = 119871212119910

(840)

If 119910 = 0400119871 then

119904 = 119871212 (0400119871) = 0208119871 (841)

Note that 119901 and 1199011015840 are interchangeable (119904 sdot 119910 = 119871212) If the impulse is delivered atdistance 04119871 from the CM then a point 0208119871 from the CM on the opposite side willnot recoil

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 61: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Rotational Motion Angular Momentum and Dynamics of Rigid Bodies 55

L2

M+ m

120579max

(L2) cos 120579max

m

x

y

L2

L

M

Figure 86 Problem (88)(c) Calculate the angularmomentumabout the axle after the impulseUse the parallel-

axis theorem to calculate the rotational inertia 119868axle about the axleLaxle = 119875 (119889 + 1198891015840)

119868axle120596axle = 119875 (119889 + 1198891015840)(1198721198862 +1198721198892) 120596axle = 119875 (119889 + 1198891015840) 997904rArr

120596axle = 119875 (119889 + 1198891015840)119872 (1198862 + 1198892)

(842)

The velocity of the CM is derived from the fact that it executes purely rotationalmotion about the fixed axle

VCM = 120596axle119889 = 119875119889 (119889 + 1198891015840)119872 (1198862 + 1198892) (843)

The linear momentum of the CM comes from the combination of the impulses andfrom the axle We know these are in opposite directions as otherwise the bat would fly offin the direction of

119875 minus 119875axle = 119872VCM 997904rArr119875axle = 119875[1 minus 1198892 + 119889 sdot 11988910158401198862 + 1198892 ]

= 119875[1198862 minus 119889 sdot 11988910158401198862 + 1198892 ](844)

For no impulse at the axle we need

119875axle = 0 997904rArr 1198862 = 119889 sdot 1198891015840 (845)

(88) (a) The collision is inelastic so mechanical energy is not conserved The hingeexerts external forces on the rod so the linear momentum of the rod and clay is notconserved in the collision either Angular momentum of the rod and clay about an axisthrough the hinge and perpendicular to the 119909 minus 119910 plane is conserved since the hinge can

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 62: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

56 Solutions Manual

apply no torque about that axis The angular velocity just after the collision is thereforecalculable as follows

Linitial = Lfinal = 119868total120596119898V sdot 1198712 = [119868rod about end + 119898 sdot (1198712)

2]120596119898V1198712 = [131198721198712 + 141198981198712]120596 997904rArr

120596 = 6119898V119871 [4119872 + 3119898]

(846)

After the collision rotational mechanical energy is conserved as again the hingeprovides no external torque for an axis through the hinge and gravity acting on the centerof the mass of the system of rod and clay is conservative So we can find the maximumangle of swing from equilibrium as follows

Δ119870 = minusΔ119880grav

0 minus 12119868total1205962 = minus (119898 +119872) 1198921198712 (1 minus cos 120579max) 997904rArr31198982V22 (4119872 + 3119898) = 12 (119898 +119872)119892119871 (1 minus cos 120579max) 997904rArr120579max = cosminus1 [1 minus 31198982V2119892119871 (119898 +119872) (3119898 + 4119872)]

(847)

Note what happens if the term in the square brackets is less than minus1 In that case therod would rotate full circle forever (if the ceiling werenrsquot there)

(b) Gravity plays no role in the collision since it is a finite force (and can thereforeproduce no impulse) and even if it could produce an impulse that impulse would beorthogonal to the momentum change in this problem We take our system to consist ofthe rod and clay The linear momentum of the system just before the collision is

initial = 119898V (848)

and just after the collision

after = (119872 + 119898) CM

= (119872 + 119898) 1198712120596after = 119872 + 1198984119872 + 3119898 (3119898V)

(849)

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 63: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Remarks on Gravitation 57

The only external impulse on the system is exerted by the hinge Call this impulseΔhinge Then

Δhinge = after minus initial

= 119898V [3 (119872 + 119898)3119898 + 4119872 minus 1] = minus 119898119872V3119898 + 4119872

(850)

The impulse given by the rod to the hinge is

Δhinge = 119898119872V3119898 + 4119872 (851)

We should note that the total impulse given to the hinge by the rod and ceiling is zero

9 Remarks on Gravitation

91 Remarks on Gravity Problem Solutions

(91) The kinetic energy of a circular orbit is half the gravitational potential energy of theorbit So

119870 + 119880 = 12119880 997904rArr119870 = minus12119880 = 1198661198721198641198982 (119877119864 + 300 times 105m) 997904rArr

V = radic 119866119872119864119877119864 + 300 times 105= radic(667 times 10minus11m3 (kg sdot s2)) (597 times 1024 kg)667 times 106m

V = 773 times 103ms

(91)

The period of such an orbit is easy to get from Keplerrsquos Third Law

period = [412058721199033119866119872119864 ]12

= [[4(314159)2(667 times 106)3(667 times 10minus11m3(kg sdot s2)) (597 times 1024 kg)]]

12

period = 5420 s = 151 hours

(92)

(92) (a)The orbit diagram for this case is as followsTheHohmann orbit goes furtheraway from the sun than earth sowe need to increase the total energy at the perihelion of the

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 64: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

58 Solutions Manual

Figure 91 Orbit diagram for Problem (92)(a)transfer orbit That means increasing the kinetic energy and therefore we fire the rocketsso as to speed up in the direction of earth in its orbit

(b) To determine the speed necessary for the launch from low earth orbit we can useenergy conservation Since the earthrsquos orbit is nearly circular we can approximate (thespacecraft mass is119898 the earthrsquos speed in its orbit is V0 and the radius of earthrsquos orbit fromthe sun is 119877earth)

119870 = 1211988012119898V20 = 119866119872sun1198982119877earth997904rArr

V0 = radic119866119872sun119877earth

= radic(667 times 10minus11N sdotm2kg2) (199 times 1030 kg)150 times 1011mV0 = 297 times 104ms

(93)

Of course the ship is in orbit so the actual speed is higher than this but the correctionis about 25 Letrsquos use the V0 above

For the elliptical orbit (the transfer orbit) the mechanical energy is still conserved asis the angular momentum Now we use V119901 as the speed of the probe relative to the sun atthe perihelion point (the launch from earth)

119864perihelion = 119864aphelion 997904rArr12119898V2peri minus 119866119872sun119877earth

= 12119898V2aph minus 119866119872sun119898119877Mars

Lperi = Laph

119898Vperi119877earth = 119898Vaph119877Mars 997904rArrVaph = Vperi

119877earth119877Mars

(94)

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 65: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

Remarks on Gravitation 59

θ

Position of Mars at rendezvous

Position of Mars at launch

Angular displacement of Mars during probe travel

Figure 92 Positions of Mars for Hohmann transfer orbit

Combining the angularmomentum result with the energy conservation result we seethat

12V2peri minus 119866119872sun119877earth= 12V2peri119877

2earth1198772Mars

minus 119866119872sun119877Mars997904rArr

V2peri [1198772Mars minus 1198772Mars21198772Mars] = 119866119872sun [119877Mars minus 119877earth119877earth119877Mars

] 997904rArrVperi = radic119866119872sun

2119877Mars119877earth (119877Mars + 119877earth)= radic 2 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg) (228 times 1011m)(150 times 1011m) (228 + 150) times 1011m

Vperi = 327 times 104ms

(95)

So the launch speed must be

Vlaunch = (327 minus 297) times 104ms = 293 times 103ms ≃ 3 kms (96)

To get the time needed we can just take advantage of our derivation of Keplerrsquos ThirdLaw for elliptical orbits

119879 = 12radic 41205872G119872sun

[119903earth + 119903Mars2 ]2

= radic (3141593)2(378 times 1011m)38 (667 times 10minus11N sdotm2kg2) (199 times 1030 kg)119879 = 224 times 107 s ≃ 85 months

(97)

(c) Since Mars needs to be at the apogee point when the spacecraft arrives there weneed Mars to be at angle 120579 as shown in the figure below when the probe is launched

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation
Page 66: for Classical Mechanicsdownloads.hindawi.com/books/9789774541100.pdf · 2SolutionsManual Velocity (m/s) 5 1 3 10 15 20 30 25 35 10 30 40 5020 60 Time (s) 2 Figure1.1. Measuringthedisplacementofacarfromthevelocityversustimegraph

60 Solutions Manual

The angle we want to calculate is the difference between the launch position of Marsand its position at the apogee of the transfer orbit

120579 = 120587 minus (224 times 107 s) (2120587Martian year)(687 daysMartian year) (24 sdot 3600 sday) 120579 = 0770 radians = 441∘

(98)

  • 1pdf
    • Preface
    • Solutions Manual
      • 1 Kinematics
      • 2 Newtons First and Third Laws
      • 3 Newtons Second Law
      • 4 Momentum
      • 5 Work and Energy
      • 6 Simple Harmonic Motion
      • 7 Static Equilibrium of Simple Rigid Bodies
      • 8 Rotational Motion Angular Momentum and Dynamics of Rigid Bodies
      • 9 Remarks on Gravitation