ClassB Solutions to Miscellaneous Problems

7/30/2019 ClassB Solutions to Miscellaneous Problems

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APPENDIX B

Solutions to Miscellaneous Problems

1.

By proportions,x

m

l

h

x

n

k

h== and and therefore .1=+

l

h

k

h

Therefore by Pythagoras:

.111

2222=

+

xbxah

Everything but x is known in this equation, which can therefore be solved forx. There

are several ways of solving it; heres a suggestion. If we put in the numbers, theequation becomes

.01

100

1

64

13

22=

+

xx

PutX = 100 x2, and the equation becomes

.011

36

13 =

+

XX

a = 8 m

b = 10 m

h =3 m

k

l

x

m n


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This can be written ,01)(3)( =+= BAXf whereA andB are obvious functions of

X. Differentiation with respect toXgives )()(' 3323 BAXf += and Newton-Raphson

iteration (X = Xf/f') soon givesX, from which it is then found thatx = 6.326 182 m.

2.

In the corotating frame the bob is in equilibrium under the action of three forces its

weight, the tension in the string and the centrifugal force. (If you dont like rotatingreference frames and centrifugal force, it will be easy for you to do it properly.)

Resolve the forces perpendicular to the string: ,sincos..sin 20 = mgml and theproblem is finished.

3. (a) Raising or lowering the board doesnt apply any torques to the system, so the

angular momentumL is conserved. That is,

= .sin22mlL is constant. (1)

We also have that ..cos = lg (2)

i. Eliminate from these equations. This gives:

,tansin2

233

gm

Ll = (3)

which is constant.

l0

T

ml0sin.2

mg


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ii. Eliminate lfrom equations (1) and (2). This gives:

,cot2

23

L

mg= (4)

which is constant.

iii. Eliminate from equations (1) and (2). This gives:

.223 gm

Ll =

(5)

(Check the dimensions of all the equations.) Then we can getL/m from equation (1) andhence

( ) ,sin 222023 gll =

which is constant.

(b) i. .m675023.0tansintansin 333033 == ll

Although we are asked to plot vertically versus l horizontally, it is easier, whenworking out numerical values, to calculate las a function of. That is,

.tansin

142287.03

=l

(The number in the numerator is the cube root of 0.023675.)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

Length of string, metres

Semiverticalangle

,degrees


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Forl= 40 cm = 0.4 m, the semivertical angle is given by

.923369.0tansin3 =

The solution to this is .'3145o=

(See section 1.4 of Celestial Mechanics if you need to know how to solve the equation

f(x) = 0.)

(b) ii. .cotcot 2323 =

With the given data, this is .tan385.199 23 =

0 10 20 30 40 50 60 700

2

4

6

8

10

12

Angularspeed,rad

s-1

Semivertical angle, degrees

(b) iii. ( ) ( ) .cossinsin22

0

422

0

2

0

322

0

23

== lllll

That is, ( ) ,223 gal = where, with the given initial data,

a = 0.48168 m2

s1

and g2

= 96.04 m2

s4

.


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Although we are asked to plot vertically versus l horizontally, it is easier, whenworking out numerical values, to calculate las a function of. That is,

.4

22

+

=

agl

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10

12

14

16

18

20

Angularspeed,ra

ds-1

Length of string, metres

To solve the above equation for might be slightly easier with the substitution of u for1/:

.0242 =+ lauug

With l= 0.6 m, this gives u = 0.226121 rad1

s, and hence = 4.422 rad s1. As in part(b) i, it is necessary to know how to solve the equation f(x) = 0. See section 1.4 ofCelestial Mechanics if you need to know how.


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4. There are no horizontal forces, because the table is smooth. Therefore the centre of

mass of the rod falls vertically.

From energy considerations

( ) .constant2231

212

21 =++ mgymlym && (1)

But = cosly and therefore ..sin = && ly

.cos6)1sin3( 22 Cgl =++ & (2)

Initially .6,0 gC === &

.)1sin3(

)cos1(62

2

+

=l

g& (3)

Also, since ,cosandsin 22222222 == &&&& lxly

we obtain1sin3

)cos1(sin62

22

+

=gl

y& (4)

mg

Ny

x

l


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and .1sin3

)cos1(cos62

22

+

=gl

x& (5)

Of course y&& and increase monotonically with ;but x& starts and finishes at zero, andmust go through a maximum. With c = cos , equation (5) can be written

,34

)1(62

2

c

cglcx

=& (6)

and by differentiating 2x& with respect to c, we see that 2x& is greatest at an angle givenby

,08123 3 =+ cc (7)

the solution of which is .'5037o=

If the length of the rod is 1 m (l= 0.5 m) and x& = 1 m s1

, equation (6) becomes

,044.294.26 2 =+ cc (8)

and the two solutions are .'5280and'1517 oo=

The reader who has done all the problems so far will be aware of the importance of being

able instantly to solve the equationf(x) = 0. If you have not already done so, you should

write a computer or calculator program that enables you to do this instantly and at amoments notice. See section 1.4 of Celestial Mechanics if you need to know how.

If you want to find the normal reaction Nof the table on the lower end of the rod, you

could maybe start with the vertical equation of motion .mgNym =&& Differentiateequation (4): =yy &&&2 whatever, and the use equation (4) again fory& . This looks likerather heavy and uninteresting algebra to me, so I shant pursue it. There may be a better

way...


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5. In the figure below I have marked in red the forces on the rod, namely its weight mgand the horizontal and vertical components Xand Yof the reaction of the hinge on the

rod. I have also marked, in green, the transverse and radial components of the

acceleration of the centre of mass. The transverse component is &&l and the radial

component is the centripetal acceleration .2

&l

From consideration of the moment of the force mgabout the lower end of the rod, it is

evident that the angular acceleration is

,4

sin3

l

g =&& (1)

and by writing dd /as &&&& and integrating (with initial conditions 0== & ), or fromenergy considerations, we obtain the angular speed:

.2

)cos1(32

l

g =& (2)

The horizontal and vertical equations of motion are:

)sincos( 2 = &&&mlX (3)

and ).cossin( 2 += &&&mlYmg (4)

(As ever, check the dimensions - and count the dots!)

Y

mgl&&l

2&l

X


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After substitution for 2and &&& we find

)2cos3(sin43 = mgX (5)

and .)cos31( 2

4

1 = mgY (6)

The results follow immediately.

6. Call the length of the rod 2l. Initially the height above the table of its centre of mass

is lcos 40, and its gravitational potential energy is mg lcos 40. When it hits the tableat angular speed , its kinetic energy is ( ) .22

3222

34

212

21 == mlmlI Therefore,

.sdeg9.271srad746.42

40cos3 11O

===l

g

To find the time taken, you can use equation 9.2.10:

( ).

2

0

0

90

40

=VE

dIt (7)

Here, === cos)(,40cos, o23

4 mglVmglEmlI and therefore

.

cos40cos3

2

o

90

40

= dg

lt (8)

The magnitude of the quantity before the integral sign is 0.184428 s. To find the value ofthe integral requires either that you be an expert in elliptic integrals or (more likely and

more useful) that you know how to integrate numerically (see Celestial Mechanics 1.2.)

I make the value of the integral 2.187314, so that the time taken is 0.4034 seconds.When integrating, note that the value of the integrand is infinite at the lower limit. How

to deal with this difficulty is dealt with in Celestial Mechanics 1.2. It cannot be glossedover.


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7. Here is the diagram. The forces are the weight mgof the rod, and the force of thetable on the rod. However, I have resolved the latter into two components the normal

reactionNof the table on the rod, and the frictional force F, which may be either to the

left or the right, depending on whether rod is tending to slip towards the right or the left.

The magnitude ofFis less than Nas long as the rod is not jus about the slip. When therod is just about to slip,F= N, being the coefficient of limiting static friction.

Just as in Problem 5, the equations of motion, as long as the rod does not slip, are

)2cos3(sin43

= mgF (1)

and .)cos31( 241 = mgN (2)

.)cos31(

)2cos3(sin32

=N

F(3)

The figure below showsF/Nas a function of. One sees that, as the rod falls over,F/Nincreases, and, as soon as it attains a value of, the rod will slip. We see, however, thatF/N reaches a maximum value, and by calculus we can determine that it reaches a

maximum value of 3706.0128/1015 = when '.0635)(cos o1191 == If 1tan the particle will slide down the plane unless helped byan extra force. I have drawn the three forces acting on the particle. Its weight mg. The

reactionR of the plane on the particle; if the particle is in limiting static equilibrium, this

reaction will make an angle (the angle of friction) with the plane such that tan = .It therefore makes an angle with the vertical. Finally, the additional force Pneeded; we do not initially know the direction of this force.

When three (or more) coplanar forces are in equilibrium and are drawn head-to-tail, they

form a closed triangle (polygon). I draw the triangle of forces below.

R

P

mg


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It will be clear from the triangle thatPis least when the angle between P and Ris 90o:

The least value ofPis therefore .)sincoscos(sin mg But =tan and

therefore22 1

1cosand

1sin

+=

+

= .

.1

)cos(sin

2min

=

mg

P

R

P

mg

P

R

mg


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11.

As the cylinder rolls down the plane, the wedge, because its base is smooth, will slide

towards the left. Since there are no external horizontal forces on the system, the centre ofmass of the system will not move horizontally (or, rather, it wont accelerate

horizontally.)

As usual, we draw a large diagram, using a ruler , and we mark in the forces in red andthe accelerations in green, after which well apply F = ma to the cylinder, or to thewedge, or to the system as a whole, in two directions. It should be easy and

straightforward.

I have drawn the linear acceleration s&& of the cylinder down the slope, and its angular

acceleration && . I have drawn the linear acceleration x&& of the wedge, which is alsoshared with the cylinder. I have drawn the gravitational force mgon the cylinder. There

is one more force on the cylinder, namely the reaction of the wedge on the cylinder. But

Im not sure in which direction to draw it. Is it normal to the plane? That would mean

there is no frictional force between the cylinder and the plane. Is that correct(remembering that both the cylinder and the wedge are accelerating)? Of course I could

calculate the moment of the force mgabout the point of contact of the cylinder with the

plane, and then I wouldnt need to concern myself with any forces at that point of contact.

But then that point of contact is not fixed. Oh, dear, Im getting rather muddled andunsure of myself.

This problem, in fact, is ideally suited to a lagrangian rather than a newtonian treatment,

and that is what we shall do. Lagrange proudly asserted that it was not necessary to draw

any diagrams in mechanics, because it could all be done analytically. We are not quiteso talented as Lagrange, however, so we still need a large diagram drawn with a ruler.

x

s

m, a2mkI =

M

mg

s&&

x&&

x&&

as /&&&& =


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But, instead of marking in the forces and accelerations in red and green, we mark in the

velocities in blue.

No frictional or other nonconservative forces do any work, so we can use Lagranges

equations of motion for a conservative holonomic system; .

=

q

V

q

T

q

T

dt

d

&

The speed of the wedge is x& and the speed of the centre of mass of the cylinder is

,cos222 + xsxs &&&& and the angular speed of the cylinder is ./as&

The kinetic energy of the system is

,)()cos2( 221

2

2

2122

21 xM

a

smkxsxsmT &

&&&&& +

++=

or ,)(cos1 2212

2

2

21 xMmxsms

a

kmT &&&& ++

+=

and the potential energy is

.sinconstant = mgsV

Application of Lagranges equation to the coordinatex gives us

x

sm, a 2mkI =

M

mg

s&

x&

x&

as /&& =


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xMmsm &&&& )(cos +=

and application of Lagranges equation to the coordinates gives us

.sincos1 2

2

+=

+ gxsa

k&&&&

Elimination ofs&& from these two equations gives us

.

cos1)(

cossin

2

2

2

++

=

ma

kMm

mgx&&

You can also easily find an expression fors&& is you wish.

12.x

y

V0


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There is no acceleration normal to the plane, and therefore .cos = mgN The frictionalforceFacts along the tangent to the path and is equal to N, ormgcos , where is the

coefficient of moving friction. We are told to ignore the difference between thecoefficients of moving and limiting static friction. Since the particle was originally at

rest in limiting static friction, we must have = tan . ThereforeF = mgsin . Thetangential equation of motion is

+= Fsm && whatever the component ofmg is in the tangential direction in the slopingplane.

The component ofmg down the plane would be (look at the left hand drawing) mgsin

, and so its tangential component (look at the right hand drawing) is .sinsin mg Sowe have, for the tangential equation of motion,

,sinsinsin += mgmgsm &&

or ).sin1(sin = gs&&

N

mg

Fs


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We are seeking a relation between Vand , so, in the now familiar fashion, we write

sds

VdV &&for , so the tangential equation of motion is

).sin1(sin = gds

dV

V (1)

We also need the equation of motion normal to the trajectory. The component ofmg in

that direction is cossinmg , and so the normal equation of motion is

.cossin2

=

mgmV

Here is the radius of curvature of the path, which is the reciprocal of the curvature./ dds The normal equation of motion is therefore

.cossin2 =

gds

dV (2)

Divide equation (1) by equation (2) to eliminate s and thus get a desired differential

equation between Vand :

.cos

)sin1(1

=d

dV

V(3)

This is easily integrated; a convenient (not the only) way is to multiply top and bottom by1 + sin . In any case we soon arrive at

,constant)sin1ln(ln ++=V (4)

and with the initial condition V = V0 when = 0, this becomes

.sin1

0

+=

VV (5)

In the limit, as 90, .021 VV The particle is then moving at constant velocity

and is in equilibrium under the forces acting upon it just when it was initially at rest.

13. M1 = mass of complete sphere of radius a.

M2 = mass of missing inner sphere of radiusxa.

M = mass of given hollow sphere.


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We have ,/and 31212 xMMMMM == and therefore

.1

and1 3

3

231 x

MxM

x

MM

=

=

Also ).( 2212

5222

2522

152 xMMaaxMaMI ==

Hence .1

13

52

52

x

xMaI

=

Ifx = 0,I = ,252Ma as expected. Ifx 1, you may have to use de lHpitals rule to

show thatI ,232Ma as expected.

14. M1 = mass of mantle.

M2 = mass of coreM = mass of entire planet.

We have ,)1(

and3

3

2

121

x

xs

M

MMMM

=+= and therefore

.)1(

)1(and

)1( 33

3

133

3

2xsx

xsMM

xsx

xMM

+

=+

=

Also ,1

13

52

15222

252

mantlecorex

xaMaxMIII

+=+=

where I have made use of the result from the previous problem. On substitution of theexpressions forM1 andM2, we quickly obtain

.)1(

)1(3

52

52

xss

xssMaI

++

= (1)

A hollow planet would correspond to 1/s = 0. Divide top and bottom by s and it isimmediately seen that the expression for a hollow planet would be identical to the

expression obtained for the previous problem.


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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Momentofinertia

Core size

s = 0.2

0.4

0.6

0.8

Note that bothx = 0 andx = 1 correspond to a uniform sphere, so that in either case,

;252MaI = for all other cases, the moment of inertia is less than .2

52Ma

The core size for minimum moment of inertia is easily found by differentiation of the

above expression forI, and the required expression follows after some algebra. Fors =0.6, the equation becomes ,04159 52 = xx of which the only positive real root is

,382736.0=x which corresponds to a moment of inertia of 0.90376 % .252Ma Note

that. fors = 0.6, the moment of inertia, expressed in units of ,252Ma varies very little as

the core size goes from 0 to 1, so that measurement of the moment of inertia places very

little restriction on the possible core size.

The inverse of equation (1) is

,0)1()1()1( 35 =+ sIxsIxs (2)

whereIis expressed in units of .252Ma ForI = 0.911, there are two positive real roots

(look at the graph); they are x = 0.64753 and 0.81523. ForI = 0.929, the roots are

0.55589 and 0.87863. Thus the core size could be anything between 0.55589 and

0.64753 or between 0.81523 and 0.87863 a rather large range of uncertainty. Even ifIwere known exactly (which does not happen in science), there would be two solutions for

x.


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15. This is just a matter of geometry. If, when you make a small angular displacement,you raise the centre of mass of the brick the equilibrium is stable. For, while the brick is

in its vertical position, it is evidently at a potential minimum, and you have to do work to

raise the centre of mass. If, on the other hand, your action in making a small angular

displacement results in a lowering of the centre of mass, the equilibrium is unstable.

When the brick is in its vertical position, the height h0 of its centre of mass above thebase of the semicylinder is just

.0 lRh +=

When it is displaced from the vertical by an angle , the point of contact between brickand semicylinder is displaced by a distance R, and, by inspection of the drawing, thenew height h is

.cossincos ++= lRRh

.)cos1)((sin0 += lRRhh

If you Maclaurin expand this as far as 2, you arrive at

.)( 221

0 lRhh

This is positive, and therefore the equilibrium is stable, ifl < R , or 2l < 2R , i.e. if the

length of the brick is less than the diameter of the semicylinder.


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R

l

R

lcos

R cos

Rsin


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31

16. As in the previous question, it is just a matter of geometry. If rolling the Thing

results in raising its centre of mass, the equilibrium is stable. Initially, the height of thecentre of mass is h0 = b + l.

After rolling, the dashed line, which joins the centres and is of length a + b, makes an

angle with the vertical. The short line joining the centre of mass of the Thing to thecentre of curvature of its bottom is of length la and it makes an angle + with thevertical. The height of the centre of mass is therefore now

.)cos()(cos)( +++= albah

The centre of mass has therefore rise through a height

.)cos()(cos)(0 lbalbahh +++=

Also, the two angles are related by a = b, so that

.])}/(1cos[{)(cos)(0 lbabalbahh +++=

Maclaurin expand the cosines to 2 and you should get

.])/1)(([ 2221

0 abalbahh +++=

l

b

a


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For stability this must be positive, and hence .111

bal+>

Ifa = b, this becomes .21 al<

For a hollow semicylinder, .363.0)/21( aal == Stable

For a hollow hemisphere, .5.0 al= Borderline stable

For a solid semicylinder, .576.0)]3(/41[ aal == Unstable

For a solid hemisphere, .625.085 aal == Unstable

17. We need to find the height h of the centre of mass above the level of the pegs as a

function of . See drawing on next page.

Angles: BAC = 45o

= 45o+

Distances: AB = 2ka

AC = 2ka cos (45o )

EF = 2ka cos (45o ) cos (45o+ ) = akcos 2DC = 2a

DF = cos2a h = DF EF = )2coscos2( ka

h0 = height of centre of mass above pegs when = 0o

= )2( ka

k

k

h

hy

==

2

2coscos2

0

dy/d will show that maxima and minima ofy ( and of the potential energy), and henceequilibria, occur for = 0o and for ,)8/(1cos k= which is possible only if .8/1>k A second differentiation, or a graph ofh versus , will show which extrema are maximaand which are minima. In particular the second derivative at = 0o is zero for

.354.08/1 ==k I draw below graphs of y : or ofh/a : for several k, and also agraph of the unstable equilibrium versus k for the range .500.0354.0 k Ifk=0.45, = ! 38.2.


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A B

C

D

E

F

h


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34

-40 -30 -20 -10 0 10 20 30 400.7

0.8

0.9

1

1.1

1.2

1.3

y

degrees

0.00

0.20

0.354

0.40

0.45

0.50

0.55

0.60

k


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-40 -30 -20 -10 0 10 20 30 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

h/a

degrees

k

0.6

0.8

1.0

1.2

2

0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.50

5

10

15

20

25

30

35

40

45

thetadegrees,unstableequilibrium

k


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The right-hand drawing is a position of equilibrium, but it is unstable. What happens if it

tips counterclockwise? What happens if it tips clockwise?

k = 0.2

unstable

k = 0.45

stable

= 38.2unstable


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18.

NP

a OC

a

R

mg

k= 0

stable


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There are three forces acting on the hemisphere: Its weight mg. The reaction Nof the

wall, which is perpendicular to the wall since the wall is smooth. The reactionR of the

floor, which acts at an angle to the floor, where = tan . Three forces in equilibrium

must act through a point; therefore all three forces act through the point P. It is thus clearthat

.3

8

OC

OPsin

83

=

==

a

a

If '.4841, o41 == If .90, o

83 == If ,

83> the hemisphere can rest in any

position, the equilibrium not being limitingstatic equilibrium.

19.

a

P

mg

C


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39

When the rod makes an angle to the horizontal, the distance from its centre of mass C

to the point of contact P is a. The moment of inertia about P is ( ) .22

31 + amml Thetorque about P is mga cos . In an earlier version of this solution, I had written that theequation of motion is therefore.

.cos)( 22231 =+ mgaalm &&

However, I am indebted to Eyal Gruss of Israel for pointing out to me that the torque is

equal to the rate of change of ,&I and bothIand & change with time. Thus the equationof motion is properly

.cos)( 22231 =++ mgaalmI &&&&

Since ( ) ,2231 += ammlI we have .2 2 = && maI Thus the equation of motion is

.cos)(2 2223122 =++ mgaalmma &&&

To first order in , 1cos and .23122 la


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40

with our first order simple harmonic approximation that the displacement is always

negligible compared with .3a

l

20.

I have drawn only one force, namely the weight of the hemisphere at C. I am going to

consider the rotational equation of motion about A, so I shant be concerned with the

force at A. The moment of the force mgabout A is .mgx This is equal to the rate of

change of angular momentum about A. That is to say, it is equal to the rate of change of

the product &I whereIis the moment of inertia of the hemisphere about A, and & is theangular speed of C around A. I am indebted to Eyal Gruss of Israel for pointing out to

me that bothIand & change with time (an important point that I had missed in an earlier

posting of the solution to this problem), and therefore the equation of motion is

).( += &&&& IImgx (1)

The minus sign arises because the clockwise torque results in a decrease in .

O

A

a

b

x

a8

3

C

mg


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41

I am going to try to express all the variables, && and,,, xII in terms of the single angle

and its time derivative & , in order to obtain an equation of motion as an equation in and the time.

The moments of inertia about O, C and A, respectively:

( ) ( ) ,,, 22832

522

832

522

52 mbammaammama +

where I have made good use of the parallel axes theorem. The distance b is given by the

cosine rule to triangle OCA, and hence we find, after a little algebra, for the moment of

inertia about A:

.)cos(43

572 = maI (2)

From this we also find that

.sin243 = && maI (3)

We also need

.8

sin3 =

ax (4)

The relation between and is found by application of the sine rule to triangle OCA:

.)sin(sin

83 aa

+=

(5)

If we now make use of the small angle approximations ,1cosandsin which are good to first order in (i.e., we are assuming that 1


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42

to which we can now add .and53

53 == &&&&&& (10)

On substitution of these into equation (1), we obtain for the equation of motion:

.26

25

13

15 2

= a

g&&&

(11)

If we can determine that the term 2

13

15 & is negligible compared with the term ,

26

25

a

g

then equation (11) is the equation of simple harmonic motion with angular frequency

a

g

26

25= and period ,

04.12

25

262

g

a

g

a= which was required.

If 2

13

15 & is to be negligible compared with ,

26

25

a

gwe require that .

6

52

a

g


43/78

43

22.

11

1 )cos(

NF

mgN

=

=

22

2 )cos(

NF

mgN

=

+=

Left-hand particle: .)]sin()cos([ += mgT

Right-hand particle: .)]cos()[sin( ++= mgT

),sin()sin()]cos()[cos( +=++

and, by the sum and difference trigonometrical formulae, we obtain

,sincos2coscos2 =

from which .tan =

23.

T

T + T NN

y +

N1

N2

T

F1

T + F2

mg

mg

y


44/78

44

Consider a portion of the rope between and . There are four forces on this portion.The tension T at . The tension T + T at + (T is negative). The normal

reaction Nof the cylinder on the rope. The frictional force Nof the cylinder on therope. Note that the rope is about to slip downwards, so the friction force is upwards asshown.

We have )sin()2(21 += TTN

and .)cos()cos()(21

21 =++ TNTT

To first order, these become

= TN

and .NT =

= TT

and hence by integration .= eMgF

24

2ay

2ax

C

H

X'-axis

Y'-axis

2a(x +y 1)


45/78

45

. Area of square = 24a

Area of rectangle = )1(4 2 xa

Area of triangle = )1(2 2 +yxa

Area of trapezoid = )1(2

2

yxa +

The weight of the cube is 8a3sg, and it acts downward through C, the centre of

mass. The hydrostatic upthrust is 4a3(1 x +y)gand it acts upward through the

centre of buoyancy H. Here is the density of the fluid, and s is the density ofthe wood. We evidently must find theX'-coordinate of C and of H. Lets first ofall find theX- and Y- coordinates (see the next figure).

TheX- and Y- coordinates of C are trivial and quite easy respectively:

)21(CC xaYaX ==

You are going to have to work quite hard at it to find theX- and Y- coordinates ofH, the centre of buoyancy, which is the centroid of the trapezoid. After some

algebra you should find

)1(3

)22242(2

)1(3

)21(2 22

yx

ayxyxyxY

yx

ayxX HH +

++=

++

=

X-axis

Y-axis

C

H


46/78

46

To find theX'- coordinates of C and of H, we use the usual formulas for rotation

of axes, being sure to get it the right way round:

,cossin

sincos

'

'

=

Y

X

Y

X

together with .1tan += yx

Take moments about the axle (origin):

).1(4'8 33 yxgasgXa C +=

After a little more algebra, you should eventually arrive at

.

)2232(3

3236273

2

23322

xyxyx

xyyxyxyxs

++

++++=

25.

Let the radii of the cylinder and sphere be a and b respectively, and the mass of the

sphere beM. The angles and are related by a = b. I have drawn the three forceson the sphere, namely its weight, the normal reaction of the cylinder on the sphere, and

a

b

Mg

FN


47/78

47

the frictional force on the sphere. The transverse acceleration of the centre of the sphere

is + &&)( ba and the centripetal acceleration is .)( 2+ &ba The equations of motion are:

+= &&)(sin baMFMg (1)

and .)(cos 2+= &baMNMg (2)

The angular acceleration of the sphere about its centre is ,)/1( +=+ &&&&&& ba and itsrotational inertia is 2Mb

2/5. The torque that is causing this angular acceleration is Fb,

and therefore the rotational equation of motion is

.)/1(252 += &&baMbFb (3)

Elimination ofFbetween equations (1) and (3) yields

.sin)(7

5 +

=ba

g&& (4)

Write dd /as &&&& in the usual way and integrate with initial conditions ,0== & orfrom energy considerations:

.)cos1()(7

102 +

=ba

g& (5)

Substitute for 2and &&& into equation (2) to obtain

.)10cos17( = MgN (6)

This is zero, and the sphere leaves the cylinder, when cos = 10/17, = 53o 58' .


48/78

48

26.

Surface density = g cm2

Original sandwich:

Mass = 54 g

x-coordinate of centre of mass = 3 cm

y-coordinate of centre of mass = 4 cm

Bite:

Mass = 221 3 =14.137 166 94 g

Distance of centre of mass from hypotenuse =

=

43

3

4= 1.273 239 545 cm

x-coordinate of centre of mass =

=

5

165.4sin

45.4 = 3.481 408 364 cm

y-coordinate of centre of mass =

=

5

126cos

46 = 5.236 056 273 cm

9 cm

12 cm


49/78

49

Remainder:

Mass = (54 14.137 166 94) = 39.862 833 06 g

x-coordinate of centre of mass =

y-coordinate of centre of mass = y

Moments:

39.862 833 06 + 14.137 166 94 3.481 408 364 = 54 3. x = 2.829 270 780 cm

39.862 833 06y + 14.137 166 94 5.236 056 273 = 53 4. y = 3.561 638 436 cm

This point is very close to the edge of the bite. The centre of the bite is at (4.5, 6), and its

radius is 3. Its equation is therefore

.025.47129or,9)6()5.4( 2222 =++=+ yxyxyx

The line x = 2.829 270 780 cuts the circle where .013336791.29122 =+ yy Thelower of the two points of intersection is aty = 3.508 280 941 cm. The centre of mass isslightly higher than this and is therefore just inside the bite.

27.

F

TT


50/78

50

Consider a portion of the band within the angle . Its mass is .2m

When the band is

spinning at angular speed and its radius is r, the centrifugal force on that portion is

.2

2

= mrF (I leave it to the philosophers and the schoolteachers to debate as to

whether there really is such thing as centrifugal force I want to get this problemdone, and Im referring to a co-rotating frame.) The y-component of this force is

.2

cos2

mr

Also, the tension in the band when its radius is ris T = 2k(r a).

Consider the equilibrium of half of the band. The y-component of the centrifugal force

on it is .cos2

222

2

=

+

mrd

mrThe opposing force is 2T = 4k(r a). Equating

these gives .)(4 22

mr

ark =

28. Let the distance AB be land the distance AC be c. Let the mass of the rod be m.

Consider an elemental portion x of the rod at P at a distance x from A. Its weight is

.l

xmWhen the rod is about to move, it will experience a frictional force

,

l

xmgf

= which will be in the direction shown if P is to the left of C, and in the

opposite direction if P is to the right of C. When the rod is just about to move (but has notyet done so) it is still in equilibrium. Consider the moment about A of the frictional

forces on the rod. The clockwise moment of the frictional forces on AC must equal the

counterclockwise moment of the frictional forces on CB. Thus

= l

c

c

dxxl

mgdxx

l

mg.

0

A P BC

xx

f

F


51/78

51

.2/lc =

The net force on the rod is

,0 +

l

c

c

dxl

mgdx

l

mgF

and this is zero, and therefore

( ) .12)2( mgl

lcmgF =

=

29. The cone slips when tan > .

It tips when C (the centre of mass) is to the left of M.

The distance OC is h/4. (See Chapter 1, section 1.7). Therefore it tips when ./4tan ha>

Thus it slips if ha /4

O

C

M


52/78

52

30.

When the block is just about to tip, the reaction of the table on the block acts at A and it

is directed towards the point K, because, when three coplanar forces are in equilibrium

they must act through a single point. The angle is given by tan = a/x. However, by

the usual laws of friction, the block will slip as soon as tan = . Thus the block willslip if < a/x, and it will tip if > a/x. Expressed otherwise, it will slip ifx < a/ andit will tip ifx > a/. The greatest possible value ofx is 2a; therefore the block willinevitably slip if < .

31.

When or if the cylinder is just about to tip, it is about to lose contact with the left handpeg. The only forces on the cylinder are the torque, the weight, and the reactionR of the

right hand peg on the cylinder, which must be vertical and equal to mg. But the greatest

possible angle that the reactionR can make with the surface of the cylinder is the angle of

friction given by tan = . From geometry, we see that sin = k, or

2a

x

K

A

2ka

mg

R


53/78

53

.1/tan 2kk = Thus the cylinder will slip before it tips if 21/ kk

If the cylinder tips (which it will do if 21/ kk > ), the clockwise torque at thatmoment will equal the counterclockwise torque of the couple (R and mg), which is mgka.

Thus the torque when the cylinder tips is

TIP: .mgak= (1)

When or if the cylinder is just about to slip, the forces are as shown above, in which Ihave resolved the reactions of the pegs on the cylinder into a normal reaction (towards

the axis of the cylinder) and a frictional force, which, when slipping is about to occur, is

equal to times the normal reaction. The equilibrium conditions are

,0sin)(cos)( 2121 =++ NNNN

0cos)(sin)( 2121 =++ mgNNNN

and .)( 21 =+ aNN

We can find N1 + N2 by eliminatingN1 N2 from the first two equations, and then,

writing 21 k for cos , we find that, when slipping is about to occur,

2ka

mgN2N1

N1

N2


54/78

54

SLIP .1

1

1 22 kmga

+

= (2)

I have drawn below the functions

kmga

= (tip) and22 1

1

1 kmga

+= (slip)

fork= 0.1, 0.3, 0.5, 2/1 and 0.9. The horizontal lines are the tip functions, and the

curves are the slip functions. As long as 21/ kk the cylinder will tip.

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

0.1

0.3

0.5

1/2

0.9

k

t

orque/(mga)


55/78

55

32.

Well leave to the philosophers the question as to whether centrifugal force really

exists, and well work in a co-rotating reference frame, so that the car, when referred tothat frame, is in static equilibrium under the six forces shown. Clearly,N1 andN2 = mg

andF1 +F2 = mv2/R.

The car slips whenF1 +F2 = (N1 +N2); that is, when .gR=v

The car tips when ;/2 mgdRhm =v that is, when ./ hdgR=v

That is, it will slip or tip according as to whether < d/h or > d/h.

For example suppose d= 60 cm, h = 80 cm,g= 9.8 m s2

, R = 30 m, = 0.8.

In that case, d/h = 0.75, so it will tip at v = 14.8 m s1

= 53.5 km hr1

.

But if it rains, reducing to 0.7, it will slip at v = 14.3 m s1 = 51.6 km hr1 .

h

2d

mg

N1

F2

N2

F1

mv2/R


56/78

56

33.

I have drawn in green the radial and transverse components of the acceleration of the

centre of mass 2&a and &&a respectively. I have drawn in red the weight of the rod andthe normal and frictional components of the force of the table on the rod at A, Nand F

respectively.

The following are the equations of motion:

Normal: .cos Nmgma =&&

(1)

Lengthwise: .sin2 Fmgma +=& (2)

Rotation: .cos2 = gak && (3)

Here kis the radius of gyration about A, given by

.22312 alk += (4)

From equations (1), (3) and (4), we obtain

.3

.cos22

2

+=

al

lmgN (5)

The space integral (see Chapter 6, section 6.2) of equation (3), with initial condition

0=& when = 0, results in

A

mg

&&a

2&a

N

F


57/78

57

.sin2

2

2 =k

ga& (6)

This can also be obtained by equation the loss of potential energy, ,sin mga to the gain

in kinetic energy, .2221 &mk

Combining this with equations (2) and (4) leads to

.3

9.sin

22

22

+

+=

al

almgF (7)

At the instant of slipping,F = N, and hence, from equations (5) and (7) we find

.)/(91

tan 2la+ =

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

a/l


58/78

58

34. I derive 22 xl

ggx +=v by two different methods one from energy

considerations, the other from angular momentum considerations. First, energy.

If the table top is taken to be the zero level for potential energy, the initial potentialenergy was ...

81

41

21 mgllgm =

When the length of the dangling portion is ,21 xl+ the potential energy is

.2

)(2

)(..2

21

812

21

21

212

1

l

mgxmgxmglxl

l

mgxlgm

l

xl=+=+

+

The loss of potential energy is therefore .2

2

21

l

mgxmgx +

This is equal to the gain in kinetic energy ,

2

2

1v

m and therefore

.22 xl

ggx +=v

Another method:

Consider a point A. Anywhere will do, but I have chosen it to be a distance lbelow the

level of the table and lto the left of the table edge. The moment of momentum (= angular

momentum) of the chain about this point is ,xmlml &=v and its rate of change is

xl21

xl+21

v

v

A

mgl

xl

+21


59/78

59

therefore .xml&& The torque about A is .)(212

1

mgxlmgll

xl+=

+These are equal, and

so .)(21 xlgxl +=&& Write

dx

dx

vv=&& in the usual way, and integrate (with v= 0 whenx

= 0) and the result 22 xl

ggx +=v follows.

To find the relation betweenx and twe can use the energy equation 9.2.9 for conservative

systems

( ).

2 0 =x

x xVE

dxmt

Herex0 = 0 and we have already seen that .22)(2

mgxxl

mgxVE += Upon integrating

this expression, we obtain, after a little algebra and calculus,

.ln21

2

21

+++=

l

lxxlx

g

lt

The converse of this is the required expression

.4

)1(/

2/

t

t

lg

lg

e

el

x

=

Differentiation of this with respect to time produces the third required expression:

.4

)1(/

/4

t

t

lg

lg

e

egl =v

You may verify from these last two equations, if you wish, that .22 xl

ggx +=v

35. (a)

D

d1

d2

d3

1

2

4

3


60/78

60

The maximum overhang of book1 is d1 = w.

The centre mass of1+ 2 is at 3w/2 from the left hand side (LHS) of2, so d2 = w/2.

The distance of the centre of mass of 1+2+3 is at 5w/2 from the LHS of 3, so d3 = w/3.

Thus .38.1)1(31

21

321 wwdddD&=++=++=

(b) In a similar manner we find that, given n + 1 books, the maximum overhang is

.)1( 131

21 wD

n++++= KK

I dont know if there is a simple expression for the sum to n terms of this harmonic series.

Please let me know if you know of one or can find one. Therefore I used a computer tosolve

101 131

21 =++++

nKK

by brute force. I got n = 12367, so you would need 12368 books.

(c) The harmonic series is divergent and has no finite limit, so there is no finite

limit to the possible overhang.

You might wish to speculate on any practical limitations on constructing such a pile

of books. For example, we have been assuming a uniform gravitational field but thiswill no longer be valid once the overhang becomes comparable to the radius of Earth.

This will, however, need quite a large number of books.

36. In the solution that follows, a prime )'( will be used to denote differentiation with

respect to x, and .'dx

dyyp == I shall also make use of an auxiliary variable

.sinh 1p= The initial conditions arey = 0, x = a, p = 0, = 0. The final conditionsarex = 0, p = , = , y to be determined.


61/78

61

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x

y

vt

(x,y)

At time t, they-coordinate of the Man is vt. If (x , y) are the coordinates of the Dog at

that time, the slope of the path taken by the Dog is

,x

ytp

=v

(1)

so that .pxyt =v (2)

The speed of the Dog is

.1 2

dt

dxp

dt

dsA +==v (3)

[This comes from .)/(1 2 dxdxdyds += The minus sign is necessary because

)/( dtdx is negative, andAv, the speed (not velocity!) of the Dog is necessarily positive.]

Now ,'/1)/( tdtdx = so equation (3) can be written


62/78

62

.1' 2ptA +=v (4)

If we can eliminate tbetween equations (2) and (4), we will obtain a relation between the

slopep andx, and hence potentially a relation betweeny andx.

Differentiate equation (2) with respect tox (recalling that py =' ):

.'' xpt =v (5)

It is now simple to eliminate 't from equations (4) and (5):

.1' 2pxAp += (6)

On separating the variables and integrating, we obtain

.1 2 x

dx

p

dpA =

+(7)

With initial conditionsp = 0 when x = a, this gives us

,)/ln(sinh 1 axpA = (8)

or ,= Aaex (9)

where .sinh 1p= (10)

Equation (9), with (10), gives us the relation betweenx and the slope,p. Note thatp and

hence are negative, so that equation says thatx < a.

Our next task will be to find a relation betweeny andp (or betweeny and ).

From equation (10) we have

,sinh dxdy = (11)

and from equation (9) we have

.= daAedx A (12)

From these we obtain the differential relation betweeny and :

,sinh = daAedy A (13)


63/78

63

or ( ) .)1()1(21 = + deeaAdy AA (14)

Integrate this, with initial condition = 0 wheny = 0, to obtain

.1

2

11 2

)1()1(

21

++=

+

AA

e

A

eaAy

AA

(15)

Equation (9) and (15) are parametric equations to the path of the Dog, though it is easy to

eliminate and writey explicitly as a function ofx:

( ) ( ).

1

2

112

/11/11

21

+

+=

+

AAAaAy

A

axA

ax

(16)

The figure was drawn fora = 1,A = 2, for which equation (16) reduces to

].2)3([ 2/131 += xxy (17)

The distance walked by the Man is found by putting = in equation 15. Thus

,12

=A

aAy (18)

and the time taken is

.

)1(2

=A

aAt

v

(19)

37. Let lbe the length of the string.

a.

Kinetic energy of the upper mass = .)( 22122

21 rmmr &+

Kinetic energy of the lower mass = .221 rm&

Potential energy of the lower mass = ).( rlmg =

Total energy of the system = ).()( 22221 rlmgrmmr + & (1)

Initial total energy of the system = ).()( 202

21 almgma (2)

Energy is conserved and therefore, by equating (1) and (2), we obtain


64/78

64

.)( 22212

0

2

212 += raragr& (3)

Angular momentum is also conserved, and therefore

.022 = ar (4)

On elimination ofrbetween equations (3) and (4) we obtain, after some algebra,

.12

1 0

0

20

2

+=g

a

ga

r&(5)

b. If ,/and 02

0 == ga it is trivial to show that

./121

23

2

=ga

r&(6)

c. Algebra and calculus show that /121

23 is negative for all positive

except for = 1, when it reaches a maximum value of zero.

d. If ,/and2 02


./12

2

=ga

r&

(7)

Algebra and calculus show that /12 reaches a maximum value for

,961629.02/1/ 3/20 === at which time .118110.0)/(2 =gar& That is, when

.841331.0 gar =& Equation (4) (conservation of angular momentum) shows that

.921259.12/ 3 aaar ===

Solution of 0/12 = shows that the speed is zero when = 1 (the initialcondition) and when 966381.0/ 0 == (the equilibrium value). Equation (4)

(conservation of angular momentum) shows that .034618.1/ aar ==

e. If ,/and 0212



65/78

65

./141

45

2

=ga

r&(8)

Algebra and calculus show that /141

45 reaches a maximum value for

,401587.12/3/2

0 === at which time .449059.0)/(2

=gar& That is, when.822243.0 gar =& Equation (4) (conservation of angular momentum) shows that

.701793.02// 3 aaar ===

Solution of 0/141

45 = shows that the speed is zero when = 1 (the initial

condition) and when 447438.2/ 0 == (the equilibrium value). Equation (4)

(conservation of angular momentum) shows that .388640.0/ aar ==

How much further can we go with this question? By elimination ofrbetween equations(3) and (4) we obtained a relation between r& and . By elimination of betweenequations (3) and (4) we can get a relation between r& and r. It will be of the form

,/2

rBgrAr =& (9)

where .and 4042

0

2

21 =+= aBagaA If you can integrate this, you then get a

relation between rand t. I havent given much though as to whether you can get integrate

equation (9) analytically (if anyone manages it, please let me know), but at least anumerical integration will certainly be possible.

In another variation of the question, you can start with an equilibrium situation in which

,20 ga = and then add an extra mass m (orM, if you want to make it more general) andthen follow the motion from there. I leave that to you.

38. Lets look at the rod from above when it is twisted in the horizontal plane through a

small angle .


66/78

66

Each of the points where the threads are attached to the rod is displaced horizontally

through a distance .21 D (Since is small and ,LD


67/78

67

The equation of motion is therefore

=L

mgDI

4

2

&&

and consequently the periodPof small oscillations is

.44

22 mg

IL

DmgD

ILP

==

If the rod is uniform and of length 2l, its moment of inertia is ,231 ml and in that case the

period of small oscillations is

.3

4

g

L

D

lP

=

There is no need to remind the reader to check the dimensions of these equations.

39. When the yo-yo has fallen through a distancex, it has lost potential energyMgh, and

it has gained translational kinetic energy 221

vm and gained rotational kinetic energy

,221 I where ./av= Therefore ,)/( 221221 aIMMgx vv += from which

...22

22 x

IMa

gMa

+=v

Thus, from the usual equations for constant linear acceleration, the acceleration is

.2

2

gIMa

Ma

+

The net downward force is ,PMg wherePis the tension in the string. This is equal toM times the acceleration, from which we obtain

.2

MgIMa

IP

+=


68/78

68

40 (a)

I have drawn four forces on the yo-yo. Its weightMg. The tensionPin the string. The

normal reactionNof the table on the yo-yo. And the frictional forceFof the table on the

yo-yo. As long as the yo-yo is in contact with the table and there is no vertical

acceleration, we must have P g+ = .

Let us suppose that there is no slipping between the yo-yo and the table, so that the yo-yo

rolls to the left. We note that there is a net forceF to the left, and a net counterclockwise

torque about C equal to .FbPa Thus the yo-yo accelerates to the left at a rate F/Mit

and experiences a counterclockwise angular acceleration ./)( IaFbPa If there is noslipping, these must be related by ./

I

FbPabMF

= Thus, if there is no slipping,

.2

PMbI

MabF

+= (1)

The linear acceleration to the left must beF/M, or

.2MbI

abP

+(2)

__________________________

Mg

P

N

C Q

AF

Inner radius aOuter radius b


69/78

69

Alternative derivation:

There is a net counterclockwise torque about A equal toPa. The moment of inertia with

respect to A is .2MbI + Therefore there is an angular acceleration about A equal to

.2MbI

Pa

+ Therefore the linear acceleration of C to the left is ,2MbI

abP

+ and the

frictional forceFisMtimes this, or .2

PMbI

MabF

+=

End of Alternative Derivation.__________________________

However, if the yo-yo is just about to slip, F g P= = ( ). Upon substitution of this into equation (1), we see that the yo-yo will just slip if

.)(

)(2

2

MabMbI

MbIMgP

+++

= (3)

That is, the yo-yo will roll to the left without slipping if

.))(( 2MbIPMg

MabP

+> (4)

Its linear acceleration is then given by equation (2), namely

.2MbI

abP

+(2)

On the other hand, the yo-yo will rotate counterclockwise with no rolling if

.))(( 2MbIPMg

MabP

+


70/78

70

40(b)

I have drawn the four forces on the yo-yo. Its weightMg. The normal reaction of thetable on the yo-yo, which is also of magnitude Mg. The tensionPin the string. And the

frictional forceFof the table on the yo-yo.

At this point it may not be immediately obvious whetherFacts to the left or the right.For example, let us suppose that the coefficient of friction is zero. The forcePwill result

in a translation of the yo-yo to the right together with a clockwise rotation of the yo-yo.

So, in which direction does the point A on the circumference of the yo-yo move to theleft or the right? It is hard to say, but one might suppose, qualitatively, that, if the

moment of inertia is large, the induced rotation will be sluggish, so that A moves to the

right. Whereas ifIis small, the induced rotation will be rapid, and A will move to the leftin spite of the translational motion of the centre of mass to the right. From this we might

conclude that, if ,0 Fwill act to the right ifIis small, andFwill act to the left ifIislarge. The following analysis shows that this qualitative expectation is correct.

(The reader might find some of the Problems in Section 8.2 of Chapter 8 to be helpful at

this point, particularly Problem 2.5.)

For the time being, I have drawnFas if acting towards the left.

Let us suppose there is no slipping and that the yo-yo rolls.

The sum of the clockwise moments of the forces about A is ),( baP + and the moment of

inertia about A is .2MbI + The yo-yo therefore undergoes an initial clockwise angular

P

F

Mg

Mg

A


C


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acceleration about A equal to ,)(2MbI

baP

++

and, therefore (if there is no slipping), an initial

linear acceleration of C to the right equal to

.

)(2MbI

baPb

+

+(1)

The above linear acceleration must equal ,/)( MFP from which we obtain

.2

PMbI

MabIF

+

= (2)

This shows that the frictional force F acts to the left, as shown, if ;MabI > but if,MabI < the frictional forceFacts to the right. This is in agreement with our qualitative

expectations, namely thatFwill act to the left ifIis large, and to the right ifIis small.

Let is consider three cases in turn: .and, MabIMabIMabI =

(i) .MabI > In this case,Facts to the left, as drawn. Provided ,MgF < there willbe no slipping at A, and the yo-yo will roll to the right without slipping. On recalling

equation (2), we see that the yo-yo will roll to the right without slipping, with a linearacceleration given by equation (1) if

.2

+

>

Mg

P

MbI

MabI(3)

The linear acceleration to the right is given by equation (1), namely

.)(

2MbI

baPb

++

(1)

However, if ,2

+


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,MgF = (10)

and it acts to the right.

The net force to the right is then ,MgP + so the linear acceleration to the right is

.M

MgP +(11)

The net clockwise moment of the forces about the centre of mass C is .MgbPa Theyo-yo therefore undergoes a clockwise angular acceleration about C of

.I

MgbPa (12)

The linear acceleration to the left of the point A on the circumference of the yo-yo is

,MgP

I

MgbPab

+

and, because of condition (8), some algebra will show

that this is necessarily positive, as expected.

.MabI = In this case, Fis zero. Whatever the coefficient of friction, even zero, theyo-yo will undergo a linear acceleration MP/ to the right (Verify that this is consistent

with equation (1)), and a clockwise angular acceleration about C equal to ./IPa The

linear acceleration to the right of the point A on the circumference of the yo-yo is

,I

Pab

P

which is zero. The initial linear velocity of the point A is therefore zero.

40 (c)

P

F

Mg

Mg

A



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I have drawn the four forces on the yo-yo. Its weightMg. The normal reaction of the

table on the yo-yo, which is also of magnitude Mg. The tensionPin the string. And thefrictional forceFof the table on the yo-yo. On this occasion (unlike in Problem 40 (b))

there is no question about the direction ofF, which is towards the left.

Let us suppose there is no slipping.

The sum of the clockwise moments of the forces about A is ),( abP and the moment of

inertia about A is .2MbI + The yo-yo therefore undergoes an initial clockwise angular

acceleration about A equal to ,)(2MbI

abP

+

and therefore (if there is no slipping) an initial

linear acceleration to the right equal to

.)(

2MbI

abPb

+

(1)

Additional string therefore becomes wrapped around the axle. (Yes, it really does! Itried it!)

The above linear acceleration must equal ,/)( MFP from which we obtain

.2

PMbI

MabIF

++

= (2)

Provided ,MgF < there will be no slipping at A, and the yo-yo will roll to the rightwithout slipping. Thus the yo-yo will roll to the right without slipping, with a linear

acceleration given by equation (1) if

.2

++

>Mg

P

MbI

MabI(3)

However, if ,2

++


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.M

MgP (6)

Because of condition (4), this is necessarily positive.

The net counterclockwise moment of the forces about the centre of mass C is

.MgbPa The yo-yo therefore undergoes a counterclockwise angular accelerationabout C of

.I

MgbPa (7)

By virtue of condition (4), both of these expressions ((6) and (7)) are necessarily positive.

40(d)

I have drawn the four forces on the yo-yo. Its weightMg. The normal reactionNof thetable on the yo-yo. The tensionPin the string. And the frictional forceFof the table on

the yo-yo. From geometry we can find that the perpendicular distance from A to the lineof the string is .cos+ ba

Let us suppose there is no slipping. (And we shall also suppose that ,sin MgP


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The sum of the clockwise moments of the forces about A is ),cos( + baP and the

moment of inertia about A is .2MbI + The yo-yo therefore undergoes an initial

clockwise angular acceleration about A equal to ,)cos(

2MbI

baP

++

and, therefore, if there

is no slipping, it undergoes an initial linear acceleration to the right equal to

.)cos(

2MbI

baPb

++

(1)

The above linear acceleration must equal ,/)cos( MFP from which we obtain

.cos

2P

MbI

MabIF

+

= (2)

The frictional force F therefore acts to the left, as drawn, if IMab /cos > , and to the

right if IMab /cos PMgF If slipping does not occur, the yo-yorolls to the right. If slipping occurs, C accelerates to the right, and the yo-yo undergoes aclockwise angular acceleration. This may result inFbeing directed to either left or right,

as in Problem 40(b). Apart from the addition of a cos is many of the equations, thesolutions to Problems 40(b) and 40(d) should be broadly similar. Problem 40(e),

however, has an additional point of interest.

40(e)

P

F

Mg

N

A

Inner radius a

Outer radius bString makes angle

with horizontal.

C


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I have drawn the four forces on the yo-yo. Its weightMg. The normal reactionNof the

table on the yo-yo. The tensionPin the string. And the frictional forceFof the table onthe yo-yo. From geometry we can find that the perpendicular distance from A to the line

of the string is .cos ab

The clockwise moment ofPabout A is ).cos( abP The moment of inertia about A is.2MbI + If there is no slipping, the yo-yo turns about A with a clockwise angular

acceleration

.)cos(

2MbI

abP

+

(1)

The linear acceleration of C to the right is therefore

.)cos(

2MbI

abPb

+

(2)

This must be equal to ,cos

M

FP from which we find that

,cos

2P

MbI

MabIF

++

= (3)

acting to the left.

However, we can see, either from the drawing, or from equation (1), that, if the string is a

little steeper, so that ,/cos ba


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78

The perpendicular distance from A to the line of the string is now .cos ba

The counterclockwise moment ofPabout A is ).cos( baP The moment of inertia

about A is .2MbI + If there is no slipping, the yo-yo turns about A with acounterclockwise angular acceleration

.)cos(

2MbI

baP

+

(4)

The linear acceleration of C to the left is therefore

.)cos(

2MbI

baPb

+

(5)

This must be equal to ,

cos

M

PF from which we find that

,cos

2P

MbI

MabIF

++

= (6)

acting to the left.

In either case, slipping occurs if ).sin( > PMgF

If ,/cos ba= the four forces act through a point (A), and, in the absence of slipping,the four forces are in static equilibrium. Equation (3) or (6) becomes ,cos= PF whichis otherwise obvious. The yo-yo will roll neither clockwise nor counterclockwise. When

I tried this with an actual physical model, I found that, when ),/(cos1

ba it rolled equally easily to the left. But when ),/(cos 1 ba= I could pullon the string quite hard and nothing moved. Only when I pulled with a force greater than

+

=

+

22sincos aba

bMgMg, did I manage to drag the yo-yo along the table

to the right without rotation

Documents

ClassB Solutions to Miscellaneous Problems