Example Problem #1.38

Example Problem #1.38• For components of C, we have:

Cx = (3.1 km)(cos45o) = 2.2 km

Cy = (3.1 km)(sin45o) = 2.2 km

3.1

km

o

Cx

Cy

• Components of total displacement R are thus:

Rx = Ax + Bx + Cx = 0 + 4.0 km + 2.2 km = 6.2 km

Ry = Ay + By + Cy = 2.6 km + 0 km + 2.2 km = 4.8 km• Magnitude of R is:

|R| = (Rx2 + Ry

2)1/2 = [(6.2 km)2 + (4.8 km)2]1/2 = 7.8 km

• Direction of R found from trig. functions:

sin = 4.8 km/7.8 km = sin–1(4.8/7.8) = 38o

Rx

Ry

R

Example Problem #2.33(a) max. speed attained when acceleration = 0 constant speed achieved (aqua region)

v = v0 + at look at times when t = 0 (v0 = 0) and t = t1 (v = max. and const.)

v = at1 a = 20.0 m/s2 t1 = 15 min = 900 s

v = (20.0 m/s2)(900 s) = 18,000 m/s = 18.0 km/s(b) 1st leg of journey (represented by red line in graphs):

x – x0 = ½(v0 + v)t x0 = 0, v0 = 0

x = x1 = vt/2 v = 18.0 km/s t = 900 s x1 = 8100 km

3rd leg of journey (yellow line):

x – x0 = ½(v0 + v)t x3 – x2 = ½(v2 + v3)t v2 = 18.0 km/s, v3 = 0, t = 900 s x3 – x2 = 8100 km leg 1 + leg 3 = 16,200 km

Example Problem #2.33(b continued) leg 2 distance = 384,000 km – 16,200 km = 367,800 km

fraction of total distance = leg 2 dist. / total dist. = 367,800 km / 384,000 km = 0.958

(c) Find time ship was traveling at constant speed (leg 2, aqua region):

x – x0 = ½(v0 + v)t x2 – x1 = ½(v1 + v2)t

x1 = 8100 km, x2 = 384,000 km – 8100 km = 375,900 km, v1 = 18.0 km/s, v2 = 18.0 km/s

t = 2(x2 – x1) / (v1 + v2) = 20,433 s = tleg2

ttotal = tleg1 + tleg2 + tleg3 = 900 s + 20,433 s + 900 s = 22,233 s

= 370.56 min = 6.18 hr

Free Fallin’How long would it take for Tom Petty to go “Free-Fallin’” from the top of the Sears Tower?

+y

443 m

y = y0 + v0t – (1/2)gt2

– 443 m = 0 + 0 – (1/2) gt2

886 m / g = t2 t2 = 90.4 s2

t = 9.5 s t = 9.5 s (neg. value has no physical meaning)

How fast will he be moving just before he hits the ground?

v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s

(negative sign means downward direction)

Example Problem #3.12v0x = v0.

v0y = 0.

x = x0 + v0xt.

y = y0 + v0yt – 0.5gt2.

x

y

Let x0 = 0 and y0 = 0

What must v0 be so that x = 1.75 m when y = –9.00 m?

Time it takes to travel 9.00 m vertically: y = – 0.5gt2 = – 0.5(9.8 m/s2)t2 = –9.00 m t = 1.36 s

Speed to travel 1.75 m horizontally: x = v0t v0 = x / t = 1.75 m / 1.36 s = 1.29 m/s

Example Problem #3.34

atan = 0.5 m/s2

arad

(a) arad = v2 / R = (3 m/s)2 / (14.0 m) = 0.643 m/s2

atan = 0.5 m/s2

a = [(arad)2 + (atan)2]1/2 = [(0.643 m/s2)2 + (0.5 m/s2)2]1/2 = 0.814 m/s2

Direction of a determined by: tan = arad / atan = tan–1 (0.643/0.5) = 52.10 (up from horizontal)

(b)

v

a52.10

Example Problem #4.39

(a) Since the two crates are connected by the light rope, they move together with the same acceleration of 2.50 m/s2.

(b) m = 4.00 kg

x

y

T

mg Nm

N2: Fx = T = max = (4.00 kg)(2.50 m/s2) = 10.0 N

(c) M = 6.00 kg

x

y

T

Mg NM

FNet force points in the +x – direction (same direction as acceleration), making force F larger in magnitude

Example Problem #4.39 (continued)

(d) M = 6.00 kg

x

y

T

Mg NM

F

N2: Fx = F – T = Max

F = Max + T = (6.00 kg)(2.50 m/s2) + 10.0 N = 25.0 N

Example Problem #5.7

TBy

TBx

400

Free–body diagram of wrecking ball:

x

y

TA

TB

mgmg = (4090 kg)(9.8 m/s2) = 40000 N

(a) Fy = 0 TBy – mg = 0 TBcos400 – mg = 0

TB = mg / cos400 = 5.23 104 N

(b) Fx = 0 TBx – TA = 0 TBsin400 – TA = 0

TA = TBsin400 = 3.36 104 N

Example Problem #5.10

110

v = constant Piano moving at const. velocity a = 0 (piano in equilibrium)

Free–body diagram:

110xy F

N

W

N

W

Wx

Wy110

In general:

Fx = 0 F – Wx = 0 F = Wx

F = mg sin110 = (180 kg)(9.8 m/s2) sin110 = 336.6 N

xy

VI.B. Dynamics ProblemsVI.B. Dynamics Problems1. Dynamics problems involve bodies which have

a nonzero acceleration

2. From Newton’s 2nd Law:

3. In component form:

4. Summary of Problem–Solving Strategy:a) Similar to strategy given for statics problems

(bodies in equilibrium)

b) Exception: In Step #3, set as appropriate

amF

xx maF yy maF

xx maF yy maF

Problem #5.16

Free–body diagram:

W

N

Wx

Wy

x

y

Apply Newton’s 2nd Law in x – direction:

Fx = max Wx = max Wsin= max

mg sin= max sin = ax / g

From 1 – D motion with constant acceleration:

v2 = v02 + 2a(x – x0) a = (v2 – v0

2) / 2(x – x0)

Analysis of Swinging Pail of WaterTop:

vt

+y

vb

Bottom:

Free–body diagrams of water:

mg

Fp

(top)

mg

Fp

(bottom)

Force exerted on water by pail at top:

Fy = may = m(–vt2 / r) – Fp – mg = m(–vt

2 / r) Fp = m (vt

2 / r) – mg

Analysis of Swinging Pail of WaterMinimum value of vt for water to remain in pail:

Minimum force pail can exert is zero, so set Fp = 0 and solve for minimum speed vt,min:

mg = m(–vt,min2 / r) vt,min

2 = rg vt,min = (rg)1/2

Force exerted on water by pail at bottom:

Fy = may = m(vb2 / r) Fp – mg = m(vb

2 / r) Fp

= m(vb2 / r) + mg

Remember that centripetal force is not an external force Remember that centripetal force is not an external force acting on a body – it is just the name of the net force acting acting on a body – it is just the name of the net force acting on a body undergoing circular motion (so there is no arrow on a body undergoing circular motion (so there is no arrow for centripetal force on a free–body diagram)for centripetal force on a free–body diagram)

Example Problem #5.97Free–body diagram of car:

x

y

R

W

N

f

(a) Fx = max = m(v2/R) f = m(v2/R) R = mv2/f f = sN = sW = smg R = mv2/smg = v2/sg = (35.8 m/s)2/(0.76)(9.8 m/s2) = 171.7 m (about 563 ft.)(b) From above, vmax

2 = sgR vmax = (sgR)1/2 = [(0.20)(9.8 m/s2)(171.7 m)]1/2 = 18.34 m/s = 41.0 mph

(c) vmax = (sgR)1/2 = [(0.37)(9.8 m/s2)(171.7 m)]1/2 = 24.95 m/s = 55.8 mph

The posted speed limit is evidently designed for wet road conditions.

Example Problem #6.23(a) Free–body diagram of 12–pack :

x

y

Work–Energy Theorem: Wtot = K2 – K1 = ½ m(v22 – v1

2) = ½ mv2

2 (v1 = 0 starts from rest)

Wtot = WN + Wmg + WF = WF

WF = Fs cos00 = Fs = (36.0 N)(1.20 m) = 43.2 J = ½ mv22

v2 = [2(43.2 J) / m]1/2 = [2(43.2 J) / 4.30 kg]1/2 = 4.48 m/s

s F = 36 N

N

mg

fk

(b) Free–body diagram of 12–pack :

x

y

F = 36 NN

mgs

Example Problem #6.23 (continued)

Wtot = Wf + WF

Wf = fs cos1800 = –fs = – (kN)s = –kmgs = –(0.30)(4.30 kg)(9.8 m/s2)(1.20 m) = –15.17 J

Wtot = –15.17 J + 43.2 J = 28.03 J

v2 = [2(28.03 J) / 4.30 kg]1/2 = 3.61 m/s

Example Problem

fmg

N

(a) Wtot = K2 – K1 = ½ m(v22 – v1

2) = –½ mv02 (v2 = 0)

Wtot = Wf + WN + Wmg = Wf

Wf = fs cos1800 = –fs = – (kN)s = –kmgs

–kmgs = –½ mv02 s = v0

2 / 2kg

(b) For v0 = 80.0 km/h = 22.2 m/s: k = v0

2 / 2gs = (22.2 m/s)2 / 2(9.8 m/s2)(91.2 m) = 0.28

For v0 = 60.0 km/h = 16.7 m/s: s = (16.7 m/s)2 / 2(0.28)(9.8 m/s2) = 50.8 m

v0 v = 0

s

Example Problem #6.36

x = 0

x = – 0.025 m

(Spring compressed)x = 0

(Spring relaxed)

v2

(a) W = 1/2 kx2 = 1/2 (200 N/m)(0.025 m)2 = 0.062 J

(b) Wtot = K2 – K1 = 1/2 m(v22 – v1

2) = 1/2 mv22 (block

initially at rest when spring is compressed)

Wtot = WN + Wmg + WS = WS = 0.062 J

0.062 J = 1/2 mv22 v2 = [2(0.062 J) / m]1/2

= [2(0.062 J) / (4.00 kg)]1/2 = 0.177 m/s

1. Consider the fall of Tom Petty from the Sears Tower again

a) Once Tom Petty begins to fall, gravity does work on him, accelerating him toward the ground

b) Work done by gravity on Tom Petty:

c) More generally:

0 (y2)

h (y1)

s

y

a

Free – body diagram (neglect air resistance):

W = Mg

IX.B. Gravitational Potential EnergyIX.B. Gravitational Potential Energy

mghmgssFsFsFW gggg cos

(IX.B.1)

21 yymgWg (IX.B.2)

1. Product of weight mg and vertical height y is defined as the gravitational potential energy U = mgy

a) Has units of (kg)(m/s2)(m) = J

2. Work done by gravity can thus be interpreted as a change in the gravitational potential energy:

UUUUUyymgWg 122121

Path independent – all that matters is change in vertical position

IX.B. Gravitational Potential EnergyIX.B. Gravitational Potential Energy

(II.B.3)

Design of a Loop–the–Loop Roller Coaster Suppose we wish to design the following Loop–the–Loop roller coaster:

What is the minimum value of H such that the roller coaster What is the minimum value of H such that the roller coaster cars make it safely around the loop? (Assuming cars fall cars make it safely around the loop? (Assuming cars fall under influence of gravity only.)under influence of gravity only.)

12 H – 2R

y2 = 2Ry1 = R

Conservation of mechanical energy: 1/2mv1

2 + mgy1 = 1/2mv22 + mgy2

Assume that roller coaster starts from rest at top of hill. Then we have: mgH = 1/2mv2

2 + mg(2R)

v22 = 2mg(H – 2R) / m = 2g(H – 2R)

H

y

0

Design of a Loop–the–Loop Roller CoasterFor car to make it safely over the loop: acar = arad g (remember water in bucket)

v22 / R g

2g(H – 2R) / R g

H R/2 + 2R, or

H 5R/2.

Example Problem

45020 m

300

20 mGeorge0 +y

Conservation of mechanical energy: 1/2mv1

2 + mgy1 = 1/2mv2

2 + mgy2

y1 = –(20 m)cos450 = –14.14 m y2 = –(20 m)cos300 = –17.32 m v1 = 0 (George starts from rest)

y1 y2

Example Problem #7.12 (continued)So:

1/2 mv22 = mgy1 – mgy2

v22 = 2g(y1 – y2)

v2 = [2g(y1 – y2)]1/2

v2 = [2(9.8 m/s2)(–14.14 m – (–17.32 m))]1/2

v2 = 7.89 m/s

Example Problem #7.18

pt. 1y1 = 0 Ug

= 0

pt. 2

y2 = 22.0 m(a) Conservation of mechanical energy:

K1 + U1,g + U1,el = K2 +U2,g + U2,el

K1 = 0 (pebble initially at rest) U1,g = 0 (by choice) K2 = 0 (pebble at rest at max. height) U2,el = 0 (slingshot in relaxed position) U1,el = U2,g = mgy2 = (0.01 kg)(9.8 m/s2)(22.0 m) = 2.16 J

(b) U1,el = 2.16 J = U2,g = mgy

y = 2.16 J / mg = 2.16 J / (0.025 kg)(9.8 m/s2) = 8.82 m

(c) No air resistance, no deformation of the rubber band

Example Problem #7.43x = 0

x1 = –0.20 m

Point 1 (Spring compressed) Point 2 (Block stopped)

With friction doing work on the block, we have:

½ mv12 + ½ kx1

2 + Wf = ½ mv22 + ½ kx2

2 0 0 0

Wf = – ½ kx12 = – ½ (100 N/m)(–0.20 m)2 = –2 J

Also, Wf = – f s = – f (1.00 m) = –2 J f = 2 N = kN = kmg

k = 2 N / mg = 2 N / (0.50 kg)(9.8 m/s2) = 0.41

s = 1.0 m

Example Problem #7.73

Pt. A

300

From the Work – Energy Theorem:

KA + UA,g + UA,el + Wf = KB + UB,g + UB,el

(Spring compressed)

0 00

Pt. B

300

v = 7 m/s

L = 6 m

KB = ½ mvB2 = ½ (1.50 kg)(7.00 m/s)2 = 36.75 J

UB,g = mgH = mgLsin(300) = (1.50 kg)(9.8 m/s2)(6.00 m)sin(300) = 44.1 J

H

Wf = – f s = – kNL = – kmgcos(300)L = – (0.50)(1.50 kg)(9.8 m/s2)cos(300)(6.00 m) = – 38.19 J

N

mg

f

UA,el = 36.75 J + 44.1 J – (–38.19 J) = 119.0 J

UA,el = KB + UB,g – Wf

1. Inelastic collisionsa) Classic example: car crash where cars stick together after

collision

b) vf?

Conservation of momentum: afterbefore PP

iibefore vmvmP 1111 0

fafter vmmP

21

X. E. CollisionsX. E. Collisions

fi vmmvm

2111

if vmm

mv 1

21

1

(X.E.1,2)

(X.E.3)

(X.E.4)

a) Limiting behavior: m1 => 0:

b) Limiting behavior: v1 => 0:

c) Limiting behavior: m1 = m2:

d) In inelastic collisions, the KE of system after collision

< KE of system before collision (Why?)

• K.E. before collision = K1 = 1/2 m1v1i2

• K.E. after collision = K2 = 1/2 (m1 + m2)vf2

= 1/2 (m1 + m2)[m1 / (m1 + m2)]2v1i2

• So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1

X. E. CollisionsX. E. Collisions

vf => 0

vf => 0

vf = 1/2 v1i

2. Elastic collisionsa) Forces between colliding bodies are conservativeb) Kinetic energy is conserved (may be temporarily converted

to elastic potential energy)c) Momentum is conserved

X. E. CollisionsX. E. Collisions

1. Consider a head-on collision where one object is at rest at t=0:

2. Conservation of kinetic energy gives:

1/2 mAvA12 = 1/2mAvA2

2 + 1/2mBvB22 (X.F.1)

3. Conservation of momentum gives:

mAvA1 = mAvA2 + mBvB2 (X.F.2)

4. Combining F.1 and F.2 yields:

mB, vB1 = 0

mA, vA1mA, vA2 mB, vB2

12 ABA

BAA v

mm

mmv

12

2A

BA

AB v

mm

mv

X. F. Elastic CollisionsX. F. Elastic Collisions

(X.F.3,4)

5. Limiting Case: mA = mB

vA2 = 0 and vB2 = vA1

6. Limiting Case: mA << mB

vA2 – vA1

vB2 << vA1

Similar to result of ping-pong ball (mA) striking stationary bowling ball (mB)

7. Limiting Case: mA >> mB

vA2 vA1

vB2 2vA1

Similar to result of bowling ball (mA) striking stationary ping-pong ball (mB)

X. F. Elastic CollisionsX. F. Elastic Collisions

8. “Grazing” collision:

In this case, we need to remember to take both In this case, we need to remember to take both magnitude and direction into accountmagnitude and direction into account

a) Use the component form of conservation of momentum:

Px,initial = Px,final

Py,initial = Py,final

Note: True for elastic or inelastic collisions

x

mB, vB1 = 0

mA, vA1

mA, vA2

mB, vB2

X. F. Elastic CollisionsX. F. Elastic Collisions

(X.F.5,6)

Example ProblemConservation of Momentum:

ice

BEFORE

ice

AFTER

“Odd – Job”x

y36.90

vhat

vOJafterbefore PP

x–direction: 0 = phat,x + pOJ,x

y–direction: 0 = phat,y + pOJ,y

We are interested in the horizontal recoil velocity of the bad guy, so from the x – component equation we have:

pOJ,x = – phat,x mOJvOJ = – mhatvhat,x = – mhatvhat cos36.90

vOJ = – mhatvhat cos36.90 / mOJ = – (4.50 kg)(22.0 m/s)cos36.90 / 120 kg = – 0.66 m/s

Example Problem #8.20

BEFORE

vA = 0

A B

vB = 0

A B

AFTER

vB’vA’

(a) Conservation of momentum: Ptotal,before = Ptotal,after mAvA + mBvB = mAvA’ + mBvB’ 0 = mAvA’ + mBvB’ mAvA’ = –mBvB’ vA’ = –mBvB’ / mA = –[(3.00 kg)(1.20 m/s)] / 1.00 kg,

= –3.6 m/s (b) Conservation of mechanical energy:

Kbefore + Uel,before = Kafter + Uel,after 1/2mAvA

2 + 1/2mBvB2 + Uel,before

= 1/2mAvA’2 + 1/2mBvB’2 + Uel,after Uel,before = 1/2(1 kg)(–3.6 m/s)2 + 1/2(3 kg)(1.2m/s)2 = 8.6 J

0 00

Example Problem #8.50(a) xcm = (m1x1 + m2x2) / (m1 + m2)

= (1800 kg)(40.0 m) / 3000 kg = 24.0 m ahead of m1

(or 16.0 m behind m2)x = 0 (b) P = m1v1 + m2v2 = (1200 kg)(12.0 m/s)

+ (1800 kg)(20.0 m/s) = 5.04 104 kgm/s

(c) vcm = (m1v1 + m2v2) / (m1 + m2) = [(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s)/3000 kg = 16.8 m/s

Consider a solid disk rolling without slipping down an inclined plane. What is its velocity at the bottom of the plane if it starts from rest (~no friction).

Ei = mgh.

Ef = 1/2mv2 + (1/2)I2.

But:

I = 1/2mR2, and

v/R

So:

mgh = 1/2mv2 + (1/2)(1/2mR2v/R)2 = (3/4)mv2, or

v = SQRT(4/3gh).

Example ProblemExample Problem

h

Example ProblemA uniform solid ball of mass m and radius R rolls without slipping down a plane inclined at an angle q. Using dynamics (Newton’s 2nd Law), find the final speed.

vcm

E1 = mgh = mgsin

E2 = 1/2mv2 + 1/2I2 = 1/2mv2 + 1/2(2/5mR2)(v/R)2

= 1/2mv2( 1+ 2/5) = 7/10mv2 = mgsin;

v = 1.2(gsin)1/2

h