respostas

437

Chapter 7 Conservation of Energy Conceptual Problems *1 • Determine the Concept Because the peg is frictionless, mechanical energy is conserved as this system evolves from one state to another. The system moves and so we know that ∆K > 0. Because ∆K + ∆U = constant, ∆U < 0. correct. is )( a

2 • Determine the Concept Choose the zero of gravitational potential energy to be at ground level. The two stones have the same initial energy because they are thrown from the same height with the same initial speeds. Therefore, they will have the same total energy at all times during their fall. When they strike the ground, their gravitational potential energies will be zero and their kinetic energies will be equal. Thus, their speeds at impact will be equal. The stone that is thrown at an angle of 30° above the horizontal has a longer flight time due to its initial upward velocity and so they do not strike the ground at the same time. correct. is )( c

3 • (a) False. Forces that are external to a system can do work on the system to change its energy. (b) False. In order for some object to do work, it must exert a force over some distance. The chemical energy stored in the muscles of your legs allows your muscles to do the work that launches you into the air. 4 • Determine the Concept Your kinetic energy increases at the expense of chemical energy. *5 • Determine the Concept As she starts pedaling, chemical energy inside her body is converted into kinetic energy as the bike picks up speed. As she rides it up the hill, chemical energy is converted into gravitational potential and thermal energy. While freewheeling down the hill, potential energy is converted to kinetic energy, and while braking to a stop, kinetic energy is converted into thermal energy (a more random form of kinetic energy) by the frictional forces acting on the bike. *6 • Determine the Concept If we define the system to include the falling body and the earth, then no work is done by an external agent and ∆K + ∆Ug + ∆Etherm= 0. Solving for the change in the gravitational potential energy we find ∆Ug = −(∆K + friction energy).

Chapter 7

438

correct. is )( b

7 •• Picture the Problem Because the constant friction force is responsible for a constant acceleration, we can apply the constant-acceleration equations to the analysis of these statements. We can also apply the work-energy theorem with friction to obtain expressions for the kinetic energy of the car and the rate at which it is changing. Choose the system to include the earth and car and assume that the car is moving on a horizontal surface so that ∆U = 0. (a) A constant frictional force causes a constant acceleration. The stopping distance of the car is related to its speed before the brakes were applied through a constant-acceleration equation.

0. where220

2 =∆+= vsavv

0. where2

20 <

−=∆∴ a

avs

Thus, ∆s ∝ 20v and statement (a) is false.

(b) Apply the work-energy theorem with friction to obtain:

smgWK ∆−=−=∆ kf µ

Express the rate at which K is dissipated: t

smgtK

∆∆

−=∆∆

kµ

Thus, v

tK

∝∆∆

and therefore not constant.

Statement (b) is false.

(c) In part (b) we saw that: K ∝ ∆s

Because ∆s ∝ ∆t: K ∝ ∆t and statement (c) is false.

Because none of the above are correct: correct. is )( d

8 • Picture the Problem We’ll let the zero of potential energy be at the bottom of each ramp and the mass of the block be m. We can use conservation of energy to predict the speed of the block at the foot of each ramp. We’ll consider the distance the block travels on each ramp, as well as its speed at the foot of the ramp, in deciding its descent times. Use conservation of energy to find the speed of the blocks at the bottom of each ramp:

0=∆+∆ UK or

0topbottopbot =−+− UUKK

Conservation of Energy

439

Because Ktop = Ubot = 0:

0topbot =−UK

Substitute to obtain:

02bot2

1 =− mgHmv

Solve for vbot: gHv 2bot = independently of the shape of the ramp.

Because the block sliding down the circular arc travels a greater distance (an arc length is greater than the length of the chord it defines) but arrives at the bottom of the ramp with the same speed that it had at the bottom of the inclined plane, it will require more time to arrive at the bottom of the arc. correct. is )(b

9 •• Determine the Concept No. From the work-kinetic energy theorem, no total work is being done on the rock, as its kinetic energy is constant. However, the rod must exert a tangential force on the rock to keep the speed constant. The effect of this force is to cancel the component of the force of gravity that is tangential to the trajectory of the rock. Estimation and Approximation *10 •• Picture the Problem We’ll use the data for the "typical male" described above and assume that he spends 8 hours per day sleeping, 2 hours walking, 8 hours sitting, 1 hour in aerobic exercise, and 5 hours doing moderate physical activity. We can approximate his energy utilization using activityactivityactivity tAPE ∆= , where A is the surface area of his body, Pactivity is the rate of energy consumption in a given activity, and ∆tactivity is the time spent in the given activity. His total energy consumption will be the sum of the five terms corresponding to his daily activities. (a) Express the energy consumption of the hypothetical male: act. aerobicact. mod.

sittingwalkingsleeping

EE

EEEE

++

++=

Evaluate Esleeping:

( )( )( )( )J1030.2

s/h3600h8W/m40m26

22

sleepingsleepingsleeping

×=

=

∆= tAPE

Evaluate Ewalking:

( )( )( )( )J1030.2

s/h3600h2W/m160m26

22

walkingwalkingwalking

×=

=

∆= tAPE

Evaluate Esitting:

( )( )( )( )J1046.3

s/h3600h8W/m60m26

22

sittingsittingsitting

×=

=

∆= tAPE

Chapter 7

440

Evaluate Emod. act.:

( )( )( )( )J1030.6

s/h3600h5W/m175m26

22act. mod.act. mod.act. mod.

×=

=

∆= tAPE

Evaluate Eaerobic act.:

( )( )( )( )J1016.2

s/h3600h1W/m300m26

22act. aerobicact. aerobicact. aerobic

×=

=

∆= tAPE


J105.16

J1016.2J1030.6J1046.3J1030.2J1030.2

6

66

666

×=

×+×+

×+×+×=E

Express the average metabolic rate represented by this energy consumption:

( )( ) W191s/h3600h24J1016.5 6

av =×

=∆

=t

EP

or about twice that of a 100 W light bulb.

(b) Express his average energy consumption in terms of kcal/day: kcal/day3940

J/kcal4190J/day1016.5 6

=×

=E

(c) kcal/lb22.5lb175kcal3940

= is higher than the estimate given in the statement of the

problem. However, by adjusting the day's activities, the metabolic rate can vary by more than a factor of 2. 11 • Picture the Problem The rate at which you expend energy, i.e., do work, is defined as power and is the ratio of the work done to the time required to do the work. Relate the rate at which you can expend energy to the work done in running up the four flights of stairs and solve for your running time:

PWt

tWP ∆

=∆⇒∆

∆=

Express the work done in climbing the stairs:

mghW =∆

Substitute for ∆W to obtain: P

mght =∆


441

Assuming that your weight is 600 N, evaluate ∆t:

( )( ) s6.33W250

m3.54N600=

×=∆t

12 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation .2

0 mcE =

(a) Relate the rest mass consumed to the energy produced and solve for and evaluate m:

202

0 cEmmcE =⇒= (1)

( ) kg1011.1m/s10998.2

J1 1728

−×=×

=m

(b) Express the energy required as a function of the power of the light bulb and evaluate E:

( )( )

J1047.9

hs3600

dh24

yd365.24

y10W10033

10×=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛×

== PtE

Substitute in equation (1) to obtain:

( ) g05.1m/s10998.2

J1047.928

10

µ=×

×=m

*13 • Picture the Problem There are about 3×108 people in the United States. On the assumption that the average family has 4 people in it and that they own two cars, we have a total of 1.5×108 automobiles on the road (excluding those used for industry). We’ll assume that each car uses about 15 gal of fuel per week. Calculate, based on the assumptions identified above, the total annual consumption of energy derived from gasoline:

( ) J/y103.04galJ102.6weeks52

weekautogal15auto101.5 1988 ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛×⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅×

y

Express this rate of energy use as a fraction of the total annual energy use by the US:

%6J/y105

J/y103.0420

19

≈×

×

Remarks: This is an average power expenditure of roughly 9x1011 watt, and a total cost (assuming $1.15 per gallon) of about 140 billion dollars per year. 14 • Picture the Problem The energy consumption of the U.S. works out to an average power consumption of about 1.6×1013 watt. The solar constant is roughly 103 W/m2 (reaching

Chapter 7

442

the ground), or about 120 W/m2 of useful power with a 12% conversion efficiency. Letting P represent the daily rate of energy consumption, we can relate the power available at the surface of the earth to the required area of the solar panels using IAP = . Relate the required area to the electrical energy to be generated by the solar panels:

IAP = where I is the solar intensity that reaches the surface of the Earth.

Solve for and evaluate A: ( )

211

2

13

m1067.2W/m120

W101.62

×=

×==

IPA

where the factor of 2 comes from the fact that the sun is only up for roughly half the day.

Find the side of a square with this area: km516m1067.2 211 =×=s

Remarks: A more realistic estimate that would include the variation of sunlight over the day and account for latitude and weather variations might very well increase the area required by an order of magnitude. 15 • Picture the Problem We can relate the energy available from the water in terms of its mass, the vertical distance it has fallen, and the efficiency of the process. Differentiation of this expression with respect to time will yield the rate at which water must pass through its turbines to generate Hoover Dam’s annual energy output. Assuming a total efficiencyη, use the expression for the gravitational potential energy near the earth’s surface to express the energy available from the water when it has fallen a distance h:

mghE η=

Differentiate this expression with respect to time to obtain:

[ ]dtdVgh

dtdmghmgh

dtdP ηρηη ===

Solve for dV/dt: gh

PdtdV

ηρ= (1)

Using its definition, relate the dam’s annual power output to the energy produced: t

EP∆∆

=

Substitute numerical values to obtain: ( )( ) W1057.4

h/d24d365.24hkW104 8

9

×=⋅×

=P


443

Substitute in equation (1) and evaluate dV/dt: ( )( )( )

L/s1010.1

m211m/s9.81kg/L12.0W1057.4

6

2

8

×=

×=

dtdV

The Conservation of Mechanical Energy 16 • Picture the Problem The work done in compressing the spring is stored in the spring as potential energy. When the block is released, the energy stored in the spring is transformed into the kinetic energy of the block. Equating these energies will give us a relationship between the compressions of the spring and the speeds of the blocks.

Let the numeral 1 refer to the first case and the numeral 2 to the second case. Relate the compression of the spring in the second case to its potential energy, which equals its initial kinetic energy when released:

( )( )211

2112

1

2222

1222

1

18

34

vm

vm

vmkx

=

=

=

Relate the compression of the spring in the first case to its potential energy, which equals its initial kinetic energy when released:

2112

1212

1 vmkx =

or 21

211 kxvm =

Substitute to obtain: 21

222

1 18kxkx =

Solve for x2: 12 6xx =

17 • Picture the Problem Choose the zero of gravitational potential energy to be at the foot of the hill. Then the kinetic energy of the woman on her bicycle at the foot of the hill is equal to her gravitational potential energy when she has reached her highest point on the hill.

Equate the kinetic energy of the rider at the foot of the incline and her gravitational potential energy when she has reached her highest point on the hill and solve for h:

gvhmghmv2

22

21 =⇒=

Relate her displacement along the d = h/sinθ

Chapter 7

444

incline d to h and the angle of the incline: Substitute for h to obtain:

gvd2

sin2

=θ

Solve for d:

θsin2

2

gvd =

Substitute numerical values and evaluate d:

( )( ) m4.97

sin3m/s9.812m/s10

2

2

=°

=d

and correct. is )( c

*18 • Picture the Problem The diagram shows the pendulum bob in its initial position. Let the zero of gravitational potential energy be at the low point of the pendulum’s swing, the equilibrium position. We can find the speed of the bob at it passes through the equilibrium position by equating its initial potential energy to its kinetic energy as it passes through its lowest point.

Equate the initial gravitational potential energy and the kinetic energy of the bob as it passes through its lowest point and solve for v:

hgv

mvhmg

∆=

=∆

2

and

221

Express ∆h in terms of the length L of the pendulum: 4

Lh =∆

Substitute and simplify:

2gLv =

19 • Picture the Problem Choose the zero of gravitational potential energy to be at the foot of the ramp. Let the system consist of the block, the earth, and the ramp. Then there are


445

no external forces acting on the system to change its energy and the kinetic energy of the block at the foot of the ramp is equal to its gravitational potential energy when it has reached its highest point.

Relate the gravitational potential energy of the block when it has reached h, its highest point on the ramp, to its kinetic energy at the foot of the ramp:

221 mvmgh =

Solve for h: g

vh2

2

=

Relate the displacement d of the block along the ramp to h and the angle the ramp makes with the horizontal:

d = h/sinθ

Substitute for h: g

vd2

sin2

=θ

Solve for d:

θsin2

2

gvd =


( )( ) m89.3

sin40m/s9.812m/s7

2

2

=°

=d

20 • Picture the Problem Let the system consist of the earth, the block, and the spring. With this choice there are no external forces doing work to change the energy of the system. Let Ug = 0 at the elevation of the spring. Then the initial gravitational potential energy of the 3-kg object is transformed into kinetic energy as it slides down the ramp and then, as it compresses the spring, into potential energy stored in the spring. (a) Apply conservation of energy to relate the distance the spring is compressed to the initial potential energy of the block:

0ext =∆+∆= UKW

and, because ∆K = 0, 02

21 =+− kxmgh

Solve for x:

kmghx 2

=

Chapter 7

446

Substitute numerical values and evaluate x:

( )( )( )

m858.0

N/m400m5m/s9.81kg32 2

=

=x

(b) The energy stored in the compressed spring will accelerate the block, launching it back up the incline:

m. 5 ofheight a torising path, its retrace block will The

21 • Picture the Problem With Ug chosen to be zero at the uncompressed level of the spring, the ball’s initial gravitational potential energy is negative. The difference between the initial potential energy of the spring and the gravitational potential energy of the ball is first converted into the kinetic energy of the ball and then into gravitational potential energy as the ball rises and slows … eventually coming momentarily to rest.

Apply the conservation of energy to the system as it evolves from its initial to its final state:

mghkxmgx =+− 221

Solve for h: x

mgkxh −=2

2

Substitute numerical values and evaluate h:

( )( )( )( )

m05.5

m05.0m/s9.81kg0.0152m0.05N/m600

2

2

=

−=h

22 • Picture the Problem Let the system include the earth and the container. Then the work done by the crane is done by an external force and this work changes the energy of the system. Because the initial and final speeds of the container are zero, the initial and final kinetic energies are zero and the work done by the crane equals the change in the gravitational potential energy of the container. Choose Ug = 0 to be at the level of the deck of the freighter.

Apply conservation of energy to the system:

UKEW ∆+∆=∆= sysext

Because ∆K = 0:

hmgUW ∆=∆=ext


447

Evaluate the work done by the crane: ( )( )( )

kJ314

m8m/s9.81kg4000 2

ext

−=

−=

∆= hmgW

23 • Picture the Problem Let the system consist of the earth and the child. Then Wext = 0. Choose Ug,i = 0 at the child’s lowest point as shown in the diagram to the right. Then the child’s initial energy is entirely kinetic and its energy when it is at its highest point is entirely gravitational potential. We can determine h from energy conservation and then use trigonometry to determine θ.

Using the diagram, relate θ to h and L:

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −

= −−

Lh

LhL 1coscos 11θ

Apply conservation of energy to the system to obtain:

02i2

1 =− mghmv

Solve for h: g

vh2

2i=


⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

gLv

21cos

2i1θ

Substitute numerical values and evaluate θ :

( )( )( )

°=

⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

6.25

m6m/s9.812m/s3.41cos 2

21θ

*24 •• Picture the Problem Let the system include the two objects and the earth. Then Wext = 0. Choose Ug = 0 at the elevation at which the two objects meet. With this choice, the initial potential energy of the 3-kg object is positive and that of the 2-kg object is negative. Their sum, however, is positive. Given our choice for Ug = 0, this initial potential energy is transformed entirely into kinetic energy.

Apply conservation of energy: 0gext =∆+∆= UKW

or, because Wext = 0,

Chapter 7

448

∆K = −∆Ug

Substitute for ∆K and solve for vf; noting that m represents the sum of the masses of the objects as they are both moving in the final state:

g2i2

12f2

1 Umvmv ∆−=−

or, because vi = 0,

mU

v gf

2∆−=

Express and evaluate ∆Ug:

( )( )( )

J91.4m/s9.81

m0.5kg2kg302

ig,fg,g

−=×

−−=

−=∆ UUU

Substitute and evaluate vf: ( ) m/s1.40

kg5J4.912

f =−−

=v

25 •• Picture the Problem The free-body diagram shows the forces acting on the block when it is about to move. Fsp is the force exerted by the spring and, because the block is on the verge of sliding, fs = fs,max. We can use Newton’s 2nd law, under equilibrium conditions, to express the elongation of the spring as a function of m, k and θ and then substitute in the expression for the potential energy stored in a stretched or compressed spring.

Express the potential energy of the spring when the block is about to move:

221 kxU =

Apply ,m∑ = aF rrunder equilibrium

conditions, to the block: ∑

∑

=−=

=−−=

0cosand

0sin

n

maxs,sp

θ

θ

mgFF

mgfFF

y

x

Using fs,max = µsFn and Fsp = kx, eliminate fs,max and Fsp from the x equation and solve for x:

( )k

mgx θµθ cossin s+=


449

Substitute for x in the expression for U:

( )

( )[ ]k

mg

kmgkU

2cossin

cossin

2s

2s

21

θµθ

θµθ

+=

⎥⎦⎤

⎢⎣⎡ +

=

26 •• Picture the Problem The mechanical energy of the system, consisting of the block, the spring, and the earth, is initially entirely gravitational potential energy. Let Ug = 0 where the spring is compressed 15 cm. Then the mechanical energy when the compression of the spring is 15 cm will be partially kinetic and partially stored in the spring. We can use conservation of energy to relate the initial potential energy of the system to the energy stored in the spring and the kinetic energy of block when it has compressed the spring 15 cm. Apply conservation of energy to the system:

0=∆+∆ KU or

0ifis,fs,ig,fg, =−+−+− KKUUUU

Because Ug,f = Us,I = Ki = 0:

0ffs,ig, =++− KUU


( ) 02212

21 =+++− mvkxxhmg

Solve for v:

( )m

kxxhgv2

2 −+=

Substitute numerical values and evaluate v:

( )( ) ( )( ) m/s00.8kg2.4

m0.15N/m3955m0.15m5m/s9.8122

2 =−+=v

Chapter 7

450

*27 •• Picture the Problem The diagram represents the ball traveling in a circular path with constant energy. Ug has been chosen to be zero at the lowest point on the circle and the superimposed free-body diagrams show the forces acting on the ball at the top and bottom of the circular path. We’ll apply Newton’s 2nd law to the ball at the top and bottom of its path to obtain a relationship between TT and TB and the conservation of mechanical energy to relate the speeds of the ball at these two locations. Apply ∑ = radialradial maF to the ball

at the bottom of the circle and solve for TB:

RvmmgT

2B

B =−

and

RvmmgT

2B

B += (1)

Apply ∑ = radialradial maF to the ball

at the top of the circle and solve for TT:

RvmmgT

2T

T =+

and

RvmmgT

2T

T +−= (2)

Subtract equation (2) from equation (1) to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−

+=−

Rvmmg

RvmmgTT

2T

2B

TB

mgRvm

Rvm 2

2T

2B +−= (3)

Using conservation of energy, relate the mechanical energy of the ball at the bottom of its path to its mechanical energy at the top of the

circle and solve for Rvm

Rvm

2T

2B − :

( )Rmgmvmv 22T2

12B2

1 +=

mgRvm

Rvm 4

2T

2B =−

Substitute in equation (3) to obtain: mgTT 6TB =−


451

28 •• Picture the Problem Let Ug = 0 at the lowest point in the girl’s swing. Then we can equate her initial potential energy to her kinetic energy as she passes through the low point on her swing to relate her speed v to R. The FBD show the forces acting on the girl at the low point of her swing. Applying Newton’s 2nd law to her will allow us to establish the relationship between the tension T and her speed.

Apply ∑ = radialradial maF to the girl

at her lowest point and solve for T:

RvmmgT

RvmmgT

2

2

and

+=

=−

Equate the girl’s initial potential energy to her final kinetic energy

and solve for Rv2

:

gRvmvRmg =⇒=

22

21

2

Substitute for v2/R2 and simplify to obtain:

mgmgmgT 2=+=

29 •• Picture the Problem The free-body diagram shows the forces acting on the car when it is upside down at the top of the loop. Choose Ug = 0 at the bottom of the loop. We can express Fn in terms of v and R by apply Newton’s 2nd law to the car and then obtain a second expression in these same variables by applying the conservation of mechanical energy. The simultaneous solution of these equations will yield an expression for Fn in terms of known quantities.

Apply ∑ = radialradial maF to the car

at the top of the circle and solve for Fn:

RvmmgF

2

n =+

and

mgRvmF −=

2

n (1)

Chapter 7

452

Using conservation of energy, relate the energy of the car at the beginning of its motion to its energy when it is at the top of the loop:

( )RmgmvmgH 2221 +=

Solve for Rvm

2

: ⎟⎠⎞

⎜⎝⎛ −= 22

2

RHmg

Rvm (2)

Substitute equation (2) in equation (1) to obtain:

⎟⎠⎞

⎜⎝⎛ −=

−⎟⎠⎞

⎜⎝⎛ −=

52

22n

RHmg

mgRHmgF

Substitute numerical values and evaluate Fn:

( ) ( ) ( ) N1067.15m7.5m232m/s9.81kg1500 42

n ×=⎥⎦

⎤⎢⎣

⎡−=F ⇒ correct. is )( c

30 • Picture the Problem Let the system include the roller coaster, the track, and the earth and denote the starting position with the numeral 0 and the top of the second hill with the numeral 1. We can use the work-energy theorem to relate the energies of the coaster at its initial and final positions.

(a) Use conservation of energy to relate the work done by external forces to the change in the energy of the system:

UKEW ∆+∆=∆= sysext

Because the track is frictionless, Wext = 0:

0=∆+∆ UK and

00101 =−+− UUKK


001202

1212

1 =−+− mghmghmvmv

Solve for v0: ( )01210 2 hhgvv −+=

If the coaster just makes it to the top of the second hill, v1 = 0 and:

( )010 2 hhgv −=


453

Substitute numerical values and evaluate v0:

( )( )m/s9.40

m5m9.5m/s9.812 20

=

−=v

(b) hills. two theof heights in the

difference on theonly depends speed required that theNote No.

31 •• Picture the Problem Let the radius of the loop be R and the mass of one of the riders be m. At the top of the loop, the centripetal force on her is her weight (the force of gravity). The two forces acting on her at the bottom of the loop are the normal force exerted by the seat of the car, pushing up, and the force of gravity, pulling down. We can apply Newton’s 2nd law to her at both the top and bottom of the loop to relate the speeds at those locations to m and R and, at b, to F, and then use conservation of energy to relate vt and vb. Apply radialradial maF =∑ to the rider at the bottom of the circular arc:

RvmmgF

2b=−

Solve for F to obtain: R

vmmgF2b+= (1)

Apply radialradial maF =∑ to the rider at the top of the circular arc:

Rvmmg

2t=

Solve for 2t v : gRv =2

t

Use conservation of energy to relate the energies of the rider at the top and bottom of the arc:

0tbtb =−+− UUKK or, because Ub = 0,

0ttb =−− UKK


022t2

12b2

1 =−− mgRmvmv

Solve for 2bv : gRvb 52 =

Substitute in equation (1) to obtain: mg

RgRmmgF 65

=+=

i.e., the rider will feel six times heavier than her normal weight.

Chapter 7

454

*32 •• Picture the Problem Let the system consist of the stone and the earth and ignore the influence of air resistance. Then Wext = 0. Choose Ug = 0 as shown in the figure. Apply the law of the conservation of mechanical energy to describe the energy transformations as the stone rises to the highest point of its trajectory.

Apply conservation of energy:

0ext =∆+∆= UKW

and 00101 =−+− UUKK

Because U0 = 0:

0101 =+− UKK


02212

21 =+− mgHmvmvx

In the absence of air resistance, the horizontal component of vr is constant and equal to vx = vcosθ. Hence:

( ) 0cos 2212

21 =+− mgHmvvm θ

Solve for v:

θ2cos12

−=

gHv


( )( ) m/s2.2753cos1

m24m/s9.8122

2

=°−

=v

33 •• Picture the Problem Let the system consist of the ball and the earth. Then Wext = 0. The figure shows the ball being thrown from the roof of a building. Choose Ug = 0 at ground level. We can use the conservation of mechanical energy to determine the maximum height of the ball and its speed at impact with the ground. We can use the definition of the work done by gravity to calculate how much work was done by gravity as the ball rose to its maximum height. (a) Apply conservation of energy: 0ext =∆+∆= UKW


455

or 01212 =−+− UUKK

Substitute for the energies to obtain:

012212

1222

1 =−+− mghmghmvmv

Note that, at point 2, the ball is moving horizontally and:

θcos12 vv =

Substitute for v2 and h2:

( )0

cos

1

212

1212

1

=−

+−

mghmgHmvvm θ

Solve for H: ( )1cos

22

2

1 −−= θg

vhH

Substitute numerical values and evaluate H:

( )( )( )

m0.31

140cosm/s9.812

m/s30m21 22

2

=

−°−=H

(b) Using its definition, express the work done by gravity:

( )( ) ( )ii

g i

hHmgmghmgH

UUUW hH

−−=−−=

−−=∆−=

Substitute numerical values and evaluate Wg:

( )( )( )J7.31

m12m31m/s9.81kg0.17 2g

−=

−−=W

(c) Relate the initial mechanical energy of the ball to its just-before-impact energy:

2f2

1i

2i2

1 mvmghmv =+

Solve for vf: i2if 2ghvv +=

Substitute numerical values and evaluate vf

( ) ( )( )m/s7.33

m12m/s9.812m/s30 22f

=

+=v

Chapter 7

456

34 •• Picture the Problem The figure shows the pendulum bob in its release position and in the two positions in which it is in motion with the given speeds. Choose Ug = 0 at the low point of the swing. We can apply the conservation of mechanical energy to relate the two angles of interest to the speeds of the bob at the intermediate and low points of its trajectory. (a) Apply conservation of energy: 0ext =∆+∆= UKW

or

.zeroequalandwhere0

if

ifif

KUUUKK =−+−

0if =−∴ UK

Express Ui: ( )0i cos1 θ−== mgLmghU

Substitute for Kf and Ui: ( ) 0cos1 0

2f2

1 =−− θmgLmv

Solve for θ0:

⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

gLv

21cos

21

0θ

Substitute numerical values and evaluate θ0:

( )( )( )

°=

⎥⎦

⎤⎢⎣

⎡−= −

0.60

m0.8m/s9.812m/s2.81cos 2

21

0θ

(b) Letting primed quantities describe the indicated location, use the law of the conservation of mechanical energy to relate the speed of the bob at this point to θ :

.0where0

i

ifif

==−+−

KU'UK'K

0iff =−+∴ U'U'K

Express 'U f : ( )θcos1f −== mgLmgh'U '

Substitute for iff and, U'U'K : ( ) ( )

( ) 0cos1cos1

0

2f2

1

=−−−+

θθ

mgLmgL'vm


457

Solve for θ : ( )⎥⎦

⎤⎢⎣

⎡+= −

0

2f1 cos

2'cos θθ

gLv


( )( )( )

°=

⎥⎦

⎤⎢⎣

⎡°+= −

3.51

60cosm0.8m/s9.812

m/s4.1cos 2

21θ

*35 •• Picture the Problem Choose Ug = 0 at the bridge, and let the system be the earth, the jumper and the bungee cord. Then Wext = 0. Use the conservation of mechanical energy to relate to relate her initial and final gravitational potential energies to the energy stored in the stretched bungee, Us cord. In part (b), we’ll use a similar strategy but include a kinetic energy term because we are interested in finding her maximum speed. (a) Express her final height h above the water in terms of L, d and the distance x the bungee cord has stretched:

h = L – d − x (1)

Use the conservation of mechanical energy to relate her gravitational potential energy as she just touches the water to the energy stored in the stretched bungee cord:

0ext =∆+∆= UKW

Because ∆K = 0 and ∆U = ∆Ug + ∆Us, ,02

21 =+− kxmgL

where x is the maximum distance the bungee cord has stretched.

Solve for k: 2

2xmgLk =

Find the maximum distance the bungee cord stretches:

x = 310 m – 50 m = 260 m.

Evaluate k: ( )( )( )( )

N/m40.5m260

m310m/s9.81kg6022

2

=

=k

Chapter 7

458

Express the relationship between the forces acting on her when she has finally come to rest and solve for x:

0net =−= mgkxF

and

kmgx =

Evaluate x: ( )( ) m109

N/m5.40m/s9.81kg60 2

==x

Substitute in equation (1) and evaluate h:

m151m109m50m310 =−−=h

(b) Using conservation of energy, express her total energy E:

0isg ==++= EUUKE

Because v is a maximum when K is a maximum, solve for K::

( ) 221

sg

kxxdmg

UUK

−+=

−−= (1)

Use the condition for an extreme value to obtain:

valuesextremefor 0=−= kxmgdxdK

Solve for and evaluate x: ( )( ) m109N/m5.40

m/s9.81kg60 2

===k

mgx

From equation (1) we have: ( ) 2

212

21 kxxdmgmv −+=

Solve for v to obtain:

( )m

kxxdgv2

2 −+=

Substitute numerical values and evaluate v for x = 109 m:

( )( ) ( )( ) m/s3.45kg60

m109N/m5.4m109m50m/s9.8122

2 =−+=v

Because ,02

2

<−= kdx

Kd x = 109 m corresponds to Kmax and so v is a maximum.


459

36 •• Picture the Problem Let the system be the earth and pendulum bob. Then Wext = 0. Choose Ug = 0 at the low point of the bob’s swing and apply the law of the conservation of mechanical energy to its motion. When the bob reaches the 30° position its energy will be partially kinetic and partially potential. When it reaches its maximum height, its energy will be entirely potential. Applying Newton’s 2nd law will allow us to express the tension in the string as a function of the bob’s speed and its angular position.

(a) Apply conservation of energy to relate the energies of the bob at points 1 and 2: 0

or0

1212

ext

=−+−

=∆+∆=

UUKK

UKW

Because U1 = 0:

02212

1222

1 =+− Umvmv

Express U2: ( )θcos12 −= mgLU

Substitute for U2 to obtain:

( ) 0cos1212

1222

1 =−+− θmgLmvmv

Solve for v2: ( )θcos12212 −−= gLvv


( ) ( )( )( ) m/s52.3cos301m3m/s9.812m/s4.5 222 =°−−=v

(b) From (a) we have:

( )θcos12 −= mgLU

Substitute numerical values and evaluate U2:

( )( )( )( )J89.7

cos301m3m/s9.81kg2 22

=

°−=U

(c) Apply ∑ = radialradial maF to the bob to

obtain:

LvmmgT

22cos =− θ

Solve for T: ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

LvgmT

22cosθ

Chapter 7

460

Substitute numerical values and evaluate T:

( ) ( ) ( ) N3.25m3m/s3.52cos30m/s9.81kg2

22 =⎥

⎦

⎤⎢⎣

⎡+°=T

(d) When the bob reaches its greatest height:

( )

0and

cos1

max1

maxmax

=+

−==

UK

mgLUU θ

Substitute for K1 and Umax ( ) 0cos1 max

212

1 =−+− θmgLmv

Solve for θmax:

⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

gLv

21cos

211

maxθ

Substitute numerical values and evaluate θmax:

( )( )( )

°=

⎥⎦

⎤⎢⎣

⎡−= −

0.49

m3m/s9.812m/s4.51cos 2

21

maxθ

37 •• Picture the Problem Let the system consist of the earth and pendulum bob. Then Wext = 0. Choose Ug = 0 at the bottom of the circle and let points 1, 2 and 3 represent the bob’s initial point, lowest point and highest point, respectively. The bob will gain speed and kinetic energy until it reaches point 2 and slow down until it reaches point 3; so it has its maximum kinetic energy when it is at point 2. We can use Newton’s 2nd law at points 2 and 3 in conjunction with the law of the conservation of mechanical energy to find the maximum kinetic energy of the bob and the tension in the string when the bob has its maximum kinetic energy.

(a) Apply ∑ = radialradial maF to the

bob at the top of the circle and solve for 2

3v :

Lvmmg

23=

and gLv =2

3


461

Use conservation of energy to express the relationship between K2, K3 and U3 and solve for K2:

0where0 22323 ==−+− UUUKK

Therefore,

( )Lmgmv

UKKK

2232

1

33max2

+=

+==

Substitute for 2

3v and simplify to

obtain:

( ) mgLmgLgLmK 25

21

max 2 =+=

(b) Apply ∑ = radialradial maF to the

bob at the bottom of the circle and solve for T2:

LvmmgTF

22

2net =−=

and

LvmmgT

22

2 += (1)

Use conservation of energy to relate the energies of the bob at points 2 and 3 and solve for K2:

0where0 22323 ==−+− UUUKK

( )Lmgmv

UKK

2232

1

332

+=

+=

Substitute for 2

3v and K2 and solve

for 22v :

( ) ( )LmggLmmv 2212

221 +=

and gLv 52

2 =

Substitute in equation (1) to obtain: mgT 62 =

38 •• Picture the Problem Let the system consist of the earth and child. Then Wext = 0. In the figure, the child’s initial position is designated with the numeral 1; the point at which the child releases the rope and begins to fall with a 2, and its point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. While one could use the law of the conservation of energy between points 1 and 2 and then between points 2 and 3, it is more direct to consider the energy transformations between points 1 and 3. Given our choice of the zero of gravitational potential energy, the initial potential energy at point 1 is transformed into kinetic energy at point 3.

Chapter 7

462

Apply conservation of energy to the energy transformations between points 1 and 3:

0ext =∆+∆= UKW

zero.areandwhere0

13

1313

KUUUKK =−+−

Substitute for K3 and U1; ( )[ ] 0cos1232

1 =−+− θLhmgmv

Solve for v3: ( )[ ]θcos123 −+= Lhgv


( ) ( )( )[ ] m/s91.8cos231m10.6m3.2m/s9.812 23 =°−+=v

*39 •• Picture the Problem Let the system consist of you and the earth. Then there are no external forces to do work on the system and Wext = 0. In the figure, your initial position is designated with the numeral 1, the point at which you release the rope and begin to fall with a 2, and your point of impact with the water is identified with a 3. Choose Ug = 0 at the water level. We can apply Newton’s 2nd law to the forces acting on you at point 2 and apply conservation of energy between points 1 and 2 to determine the maximum angle at which you can begin your swing and then between points 1 and 3 to determine the speed with which you will hit the water.

(a) Use conservation of energy to relate your speed at point 2 to your potential energy there and at point 1:

0ext =∆+∆= UKW

or 01212 =−+− UUKK

Because K1 = 0: ( )[ ] 0cos1

222

1

=+−−+

mghmgLmghmv

θ

Solve this equation for θ :

⎥⎦

⎤⎢⎣

⎡−= −

gLv

21cos

221θ (1)

Apply ∑ = radialradial maF to

LvmmgT

22=−


463

yourself at point 2 and solve for T: and

LvmmgT

22+=

Because you’ve estimated that the rope might break if the tension in it exceeds your weight by 80 N, it must be that: ( )

mLv

Lvm

N80or

N80

22

22

=

=

Let’s assume your weight is 650 N. Then your mass is 66.3 kg and:

( )( ) 2222 /sm55.5

66.3kgm4.6N80

==v

Substitute numerical values in equation (1) to obtain: ( )( )

°=

⎥⎦

⎤⎢⎣

⎡−= −

2.20

m4.6m/s9.812/sm5.551cos 2

221θ

(b) Apply conservation of energy to the energy transformations between points 1 and 3:

0ext =∆+∆= UKW

zeroareandwhere0

1

31313

KUUUKK =−+−

Substitute for K3 and U1 to obtain: ( )[ ] 0cos1232

1 =−+− θLhmgmv

Solve for v3:

( )[ ]θcos123 −+= Lhgv


( ) ( )( )[ ] m/s39.6cos20.21m4.6m8.1m/s9.812 23 =°−+=v

Chapter 7

464

40 •• Picture the Problem Choose Ug = 0 at point 2, the lowest point of the bob’s trajectory and let the system consist of the bob and the earth. Given this choice, there are no external forces doing work on the system. Because θ << 1, we can use the trigonometric series for the sine and cosine functions to approximate these functions. The bob’s initial energy is partially gravitational potential and partially potential energy stored in the stretched spring. As the bob swings down to point 2 this energy is transformed into kinetic energy. By equating these energies, we can derive an expression for the speed of the bob at point 2.

Apply conservation of energy to the system as the pendulum bob swings from point 1 to point 2:

( )θcos12212

221 −+= mgLkxmv

Note, from the figure, that x ≈ Lsinθ when θ << 1:

( ) ( )θθ cos1sin 2212

221 −+= mgLLkmv

Also, when θ << 1:

2211cosandsin θθθθ −≈≈

Substitute, simplify and solve for v2:

Lg

mkLv += θ2


465

41 ••• Picture the Problem Choose Ug = 0 at point 2, the lowest point of the bob’s trajectory and let the system consist of the earth, ceiling, spring, and pendulum bob. Given this choice, there are no external forces doing work to change the energy of the system. The bob’s initial energy is partially gravitational potential and partially potential energy stored in the stretched spring. As the bob swings down to point 2 this energy is transformed into kinetic energy. By equating these energies, we can derive an expression for the speed of the bob at point 2.

Apply conservation of energy to the system as the pendulum bob swings from point 1 to point 2:

( )θcos12212

221 −+= mgLkxmv (1)

Apply the Pythagorean theorem to the lower triangle in the diagram to obtain: ( ) ( )[ ] [ ] ( )θθθθθθ cos3coscos3sincossin 4

132249222

23222

21 −=+−+=+=+ LLLLx

Take the square root of both sides of the equation to obtain:

( )θcos3413

21 −=+ LLx

Solve for x: ( )[ ]21

413 cos3 −−= θLx

Substitute for x in equation (1):

( )[ ] ( )θθ cos1cos32

21

4132

212

221 −+−−= mgLkLmv

Solve for 2

2v to obtain:

( ) [ ]( ) ( ) ⎥⎦

⎤⎢⎣⎡ −−+−=

−−+−=

2

21

4132

2

21

41322

2

cos3cos12

cos3cos12

θθ

θθ

mk

LgL

LmkgLv

Chapter 7

466

Finally, solve for v2:

( ) ( ) 2

21

413

2 cos3cos12 −−+−= θθmk

LgLv

The Conservation of Energy 42 • Picture the Problem The energy of the eruption is initially in the form of the kinetic energy of the material it thrusts into the air. This energy is then transformed into gravitational potential energy as the material rises. (a) Express the energy of the eruption in terms of the height ∆h to which the debris rises:

hmgE ∆=

Relate the density of the material to its mass and volume:

Vm

=ρ

Substitute for m to obtain: hVgE ∆= ρ

Substitute numerical values and evaluate E:

( )( )( )( ) J1014.3m500m/s9.81km4kg/m1600 16233 ×==E

(b) Convert 3.13×1016 J to megatons of TNT:

TNTMton48.7J104.2

TNTMton1J1014.3J1014.3 151616 =

×××=×

43 •• Picture the Problem The work done by the student equals the change in his/her gravitational potential energy and is done as a result of the transformation of metabolic energy in the climber’s muscles. (a) The increase in gravitational potential energy is: ( )( )( )

kJ2.94

m120m/s9.81kg80 2

=

=

∆=∆ hmgU


467

(b) body. the

in storedenergy chemical from comes work thisdo torequiredenergy The

(c) Relate the chemical energy expended by the student to the change in his/her potential energy and solve for E:

UE ∆=2.0 and

( ) kJ471kJ94.255 ==∆= UE

Kinetic Friction 44 • Picture the Problem Let the car and the earth be the system. As the car skids to a stop on a horizontal road, its kinetic energy is transformed into internal (i.e., thermal) energy. Knowing that energy is transformed into heat by friction, we can use the definition of the coefficient of kinetic friction to calculate its value. (a) The energy dissipated by friction is given by:

thermEsf ∆=∆

Apply the work-energy theorem for problems with kinetic friction:

sfEEEW ∆+∆=∆+∆= mechthermmechext or, because imech KKE −=∆=∆ and

Wext = 0, sfmv ∆+−= 2

i210

Solve for f∆s to obtain:

2i2

1 mvsf =∆

Substitute numerical values and evaluate f∆s:

( )( ) kJ625m/s25kg2000 221 ==∆sf

(b) Relate the kinetic friction force to the coefficient of kinetic friction and the weight of the car and solve for the coefficient of kinetic friction:

mgfmgf k

kkk =⇒= µµ

Express the relationship between the energy dissipated by friction and the kinetic friction force and solve fk:

sEfsfE∆

∆=⇒∆=∆ therm

kktherm

Substitute to obtain: smg

E∆

∆= therm

kµ

Chapter 7

468

Substitute numerical values and evaluate µk: ( )( )( )

531.0

m60m/s9.81kg2000kJ625

2k

=

=µ

45 • Picture the Problem Let the system be the sled and the earth. Then the 40-N force is external to the system. The free-body diagram shows the forces acting on the sled as it is pulled along a horizontal road. The work done by the applied force can be found using the definition of work. To find the energy dissipated by friction, we’ll use Newton’s 2nd law to determine fk and then use it in the definition of work. The change in the kinetic energy of the sled is equal to the net work done on it. Finally, knowing the kinetic energy of the sled after it has traveled 3 m will allow us to solve for its speed at that location.

(a) Use the definition of work to calculate the work done by the applied force:

( )( ) J10430cosm3N40

cosext

=°=

=⋅≡ θFsW sF rr

(b) Express the energy dissipated by friction as the sled is dragged along the surface:

xFxfE ∆=∆=∆ nktherm µ

Apply ∑ = yy maF to the sled and

solve for Fn:

0sinn =−+ mgFF θ

and θsinn FmgF −=

Substitute to obtain: ( )θµ sinktherm FmgxE −∆=∆

Substitute numerical values and evaluate ∆Etherm:

( )( ) ( )( )[( ) ]

J2.70

sin30N40m/s9.81kg8m34.0 2

therm

=

°−=∆E

(c) Apply the work-energy theorem sfEEEW ∆+∆=∆+∆= mechthermmechext


469

for problems with kinetic friction:

or, because UKE ∆+∆=∆ mech and

∆U = 0, thermext EKW ∆+∆=

Solve for and evaluate ∆K to obtain:

J33.8

J70.2J041thermext

=

−=∆−=∆ EWK

(d) Because Ki = 0: 2

f21

f mvKK =∆=

Solve for vf:

mKv ∆

=2

f

Substitute numerical values and evaluate vf:

( ) m/s2.91kg8

J33.82f ==v

*46 • Picture the Problem Choose Ug = 0 at the foot of the ramp and let the system consist of the block, ramp, and the earth. Then the kinetic energy of the block at the foot of the ramp is equal to its initial kinetic energy less the energy dissipated by friction. The block’s kinetic energy at the foot of the incline is partially converted to gravitational potential energy and partially dissipated by friction as the block slides up the incline. The free-body diagram shows the forces acting on the block as it slides up the incline. Applying Newton’s 2nd law to the block will allow us to determine fk and express the energy dissipated by friction.

(a) Apply conservation of energy to the system while the block is moving horizontally:

sfUKEEW

∆+∆+∆=∆+∆= thermmechext

or, because ∆U = Wext = 0, sfKKsfK ∆+−=∆+∆= if 0

Solve for Kf: sfKK ∆−= if

Chapter 7

470

Substitute for Kf, Ki, and f∆s to obtain:

xmgmvmv ∆−= k2i2

12f2

1 µ

Solving for vf yields: xgvv ∆−= k2if 2µ


( ) ( )( )( )m/s6.10

m2m/s9.810.32m/s7 22f

=

−=v

(b) Apply conservation of energy to the system while the block is on the incline:

sfUKEEW

∆+∆+∆=∆+∆= thermmechext

or, because Kf = Wext = 0, sfUK ∆+∆+−= i0

Apply ∑ = yy maF to the block

when it is on the incline:

θθ cos0cos nn mgFmgF =⇒=−

Express f∆s: θµµ cosknkk mgLLFLfsf ===∆

The final potential energy of the block is:

θsinf mgLU =

Substitute for Uf, Ui, and f∆s to obtain:

θµθ cossin0 ki mgLmgLK ++−=

Solving for L yields: ( )θµθ cossin k

2i2

1

+=

gvL

Substitute numerical values and evaluate L:

( )( ) ( )( )

m17.2

cos400.3sin40m/s9.81m/s10.6

2

221

=

°+°=L

47 • Picture the Problem Let the system include the block, the ramp and horizontal surface, and the earth. Given this choice, there are no external forces acting that will change the energy of the system. Because the curved ramp is frictionless, mechanical energy is conserved as the block slides down it. We can calculate its speed at the bottom of the ramp by using the law of the conservation of energy. The potential energy of the block at the top of the ramp or, equivalently, its kinetic energy at the bottom of the ramp is


471

converted into thermal energy during its slide along the horizontal surface. (a) Choosing Ug = 0 at point 2 and letting the numeral 1 designate the initial position of the block and the numeral 2 its position at the foot of the ramp, use conservation of energy to relate the block’s potential energy at the top of the ramp to its kinetic energy at the bottom:

thermmechext EEW ∆+∆=

or, because Wext = Ki = Uf = ∆Etherm = 0, 00 2

221 =∆−= hmgmv

Solve for v2 to obtain: hgv ∆= 22


( )( ) m/s67.7m3m/s9.812 22 ==v

(b) The energy dissipated by friction is responsible for changing the thermal energy of the system:

0thermf =∆+∆+∆=∆+∆+ UKEUKW

Because ∆K = 0 for the slide: ( ) 112f UUUUW =−−=∆−=

Substitute numerical values and evaluate Wf:

( )( )( )J9.58

m3m/s9.81kg2 2f

=

=∆= hmgW

(c) The energy dissipated by friction is given by:

xmgsfE ∆=∆=∆ ktherm µ

Solve for µk: xmg

E∆

∆= therm

kµ

Substitute numerical values and evaluate µk: ( )( )( ) 333.0

m9m/s9.81kg2J58.9

2k ==µ

48 •• Picture the Problem Let the system consist of the earth, the girl, and the slide. Given this choice, there are no external forces doing work to change the energy of the system. By the time she reaches the bottom of the slide, her potential energy at the top of the slide has been converted into kinetic and thermal energy. Choose Ug = 0 at the bottom of the slide and denote the top and bottom of the slide as shown in

Chapter 7

472

the figure. We’ll use the work-energy theorem with friction to relate these quantities and the forces acting on her during her slide to determine the friction force that transforms some of her initial potential energy into thermal energy.

(a) Express the work-energy theorem:

0thermext =∆+∆+∆= EUKW

Because U2 = K1 = Wext = 0:

222

121therm

therm12

or00

mvhmgKUE

EUK

−∆=−=∆

=∆+−=


( )( )( ) ( )( ) J611m/s1.3kg20m3.2m/s9.81kg20 2212

therm =−=∆E

(b) Relate the energy dissipated by friction to the kinetic friction force and the distance over which this force acts and solve for µk:

sFsfE ∆=∆=∆ nktherm µ

and

sFE

∆∆

=n

thermkµ

Apply ∑ = yy maF to the girl and

solve for Fn:

0cosn =− θmgF ⇒ θcosn mgF =

Referring to the figure, relate ∆h to ∆s and θ: θsin

hs ∆=∆

Substitute for ∆s and Fn to obtain:

hmgE

hmg

E∆

∆=

∆∆

=θ

θθ

µ tan

cossin

thermthermk

Substitute numerical values and evaluate µk:


473

( )( )( )( ) 354.0

m3.2m/s9.81kg20tan20J611

2k =°

=µ

49 •• Picture the Problem Let the system consist of the two blocks, the shelf, and the earth. Given this choice, there are no external forces doing work to change the energy of the system. Due to the friction between the 4-kg block and the surface on which it slides, not all of the energy transformed during the fall of the 2-kg block is realized in the form of kinetic energy. We can find the energy dissipated by friction and then use the work-energy theorem with kinetic friction to find the speed of either block when they have moved the given distance. (a) The energy dissipated by friction when the 2-kg block falls a distance y is given by:

gymsfE 1ktherm µ=∆=∆


( )( )( )( )y

yE

N7.13

m/s9.81kg435.0 2therm

=

=∆

(b) From the work-energy theorem with kinetic friction we have:

thermmechext EEW ∆+∆=

or, because Wext = 0, ( )yEE N7.13thermmech −=∆−=∆

(c) Express the total mechanical energy of the system:

( ) therm22

2121 Egymvmm ∆−=−+

Solve for v to obtain: ( )21

therm22mm

Egymv+

∆−= (1)


( )( )( ) ( )( )[ ] m/s98.1kg2kg4

m2N73.13m2m/s9.81kg22 2

=+

−=v

*50 •• Picture the Problem Let the system consist of the particle, the table, and the earth. Then Wext = 0 and the energy dissipated by friction during one revolution is the change in the thermal energy of the system. (a) Apply the work-energy theorem thermext EUKW ∆+∆+∆=

Chapter 7

474

with kinetic friction to obtain: or, because ∆U = Wext = 0, therm0 EK ∆+∆=

Substitute for ∆Kf and simplify to obtain:

( )( ) ( )[ ]208

3

202

1202

121

2i2

12f2

1therm

mv

vmvm

mvmvE

=

−−=

−−=∆

(b) Relate the energy dissipated by friction to the distance traveled and the coefficient of kinetic friction:

( )rmgsmgsfE πµµ 2kktherm =∆=∆=∆

Substitute for ∆E and solve for µk to obtain: gr

vmgrmv

mgrE

πππµ

163

22

20

208

3therm

k ==∆

=

(c) .remaining thelose torevolution

1/3another requireonly it will ,revolution onein lost it Because

i41

i43

KK

51 •• Picture the Problem The box will slow down and stop due to the dissipation of thermal energy. Let the system be the earth, the box, and the inclined plane and apply the work-energy theorem with friction. With this choice of the system, there are no external forces doing work to change the energy of the system. The free-body diagram shows the forces acting on the box when it is moving up the incline.

Apply the work-energy theorem with friction to the system: therm

thermmechext

EUKEEW

∆+∆+∆=∆+∆=

Substitute for ∆K, ∆U, and ∆Etherm to obtain:

LFhmgmvmv nk202

1212

10 µ+∆+−= (1)

Referring to the FBD, relate the normal force to the weight of the box and the angle of the incline:

θcosn mgF =

Relate ∆h to the distance L along the θsinLh =∆


475

incline: Substitute in equation (1) to obtain: 0sin

cos 202

1212

1k

=+

−+

θθµ

mgLmvmvmgL

(2)

Solving equation (2) for L yields: ( )θθµ sincos2 k

20

+=

gvL


( )( ) ( )[ ]

m875.0

sin37cos370.3m/s9.812m/s3.8

2

2

=

°+°=L

Let vf represent the box’s speed as it passes its starting point on the way down the incline. For the block’s descent, equation (2) becomes:

0sincos 2

1212

f21

k

=−

−+

θθµ

mgLmvmvmgL

Set v1 = 0 (the block starts from rest at the top of the incline) and solve for vf :

( )θµθ cossin2 kf −= gLv


( )( ) ( )[ ]] m/s2.49cos370.3sin37m 0.875m/s9.812 2f =°−°=v

52 •••

Picture the Problem Let the system consist of the earth, the block, the incline, and the spring. With this choice of the system, there are no external forces doing work to change the energy of the system. The free-body diagram shows the forces acting on the block just before it begins to move. We can apply Newton’s 2nd law to the block to obtain an expression for the extension of the spring at this instant. We’ll apply the work-energy theorem with friction to the second part of the problem.

(a) Apply ∑ = aF rr

m to the block ∑ =−−= 0sinmaxs,spring θmgfFFx

Chapter 7

476

when it is on the verge of sliding: and

∑ =−= 0cosn θmgFFy

Eliminate Fn, fs,max, and Fspring between the two equations to obtain:

0sincoss =−− θθµ mgmgkd

Solve for and evaluate d: ( )θµθ cossin s+=k

mgd

(b) Begin with the work-energy theorem with friction and no work being done by an external force:

therm

thermmechext

EUUKEEW

sg ∆+∆+∆+∆=∆+∆=

Because the block is at rest in both its initial and final states, ∆K = 0 and:

0therm =∆+∆+∆ EUU sg (1)

Let Ug = 0 at the initial position of the block. Then: θsin

0initialg,finalg,g

mgdmghUUU

=

−=−=∆

Express the change in the energy stored in the spring as it relaxes to its unstretched length:

221

221

initials,finals,s 0

kd

kdUUU

−=

−=−=∆

The energy dissipated by friction is:

θµµ

cosk

nkktherm

mgddFdfsfE

−=−=−=∆=∆


0cossin k2

21 =−− θµθ mgdkdmgd

Finally, solve for µk: ( )s2

1k tan µθµ −=

Mass and Energy 53 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation .2

0 mcE =


477

(a) Relate the rest mass consumed to the energy produced and solve for and evaluate m:

( )( )J1000.9

m/s103kg10113

283

20

×=

××=

=−

mcE

(b) Express kW⋅h in joules: ( )( )( )

J1060.3s/h3600h1J/s101hkW1

6

3

×=

×=⋅

Convert 9×1013 J to kW⋅h: ( )

hkW1050.2

J103.60hkW1J109J109

7

61313

⋅×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⋅×=×

Determine the price of the electrical energy:

( )6

7

105.2$

hkW$0.10hkW102.50Price

×=

⎟⎠⎞

⎜⎝⎛

⋅⋅×=

(c) Relate the energy consumed to its rate of consumption and the time and solve for the latter:

PtE = and

y28,500s109

W100J109

11

13

=×=

×==

PEt

54 • Picture the Problem We can use the equation expressing the equivalence of energy and matter, E = mc2, to find the mass equivalent of the energy from the explosion. Solve E = mc2 for m:

2cEm =

Substitute numerical values and evaluate m: ( )

kg1056.5

m/s102.998J105

5

28

12

−×=

×

×=m

55 • Picture the Problem The intrinsic rest energy in matter is related to the mass of matter through Einstein’s equation .2

0 mcE =

Relate the rest mass of a muon to its rest energy: 20 c

Em =

Chapter 7

478

Express 1 MeV in joules: 1 MeV = 1.6×10−13 J

Substitute numerical values and evaluate m0:

( )( )( )

kg1088.1

m/s103J/MeV101.6MeV105.7

28

28

13

0

−

−

×=

×

×=m

*56 • Picture the Problem We can differentiate the mass-energy equation to obtain an expression for the rate at which the black hole gains energy. Using the mass-energy relationship, express the energy radiated by the black hole:

201.0 mcE =

Differentiate this expression to obtain an expression for the rate at which the black hole is radiating energy:

[ ]dtdmcmc

dtd

dtdE 22 01.001.0 ==

Solve for dm/dt: 201.0 c

dtdEdtdm

=

Substitute numerical values and evaluate dm/dt: ( )( )

kg/s1045.4

m/s10998.201.0watt104

16

28

31

×=

×

×=

dtdm

57 • Picture the Problem The number of reactions per second is given by the ratio of the power generated to the energy released per reaction. The number of reactions that must take place to produce a given amount of energy is the ratio of the energy per second (power) to the energy released per second. In Example 7-15 it is shown that the energy per reaction is 17.59 MeV. Convert this energy to joules:

( )( )

J1028.1J/eV101.6

MeV17.59MeV59.17

13

19

−

−

×=

××

=

The number of reactions per second is:

sreactions/103.56

J/reaction1028.1J/s1000

14

13

×=

× −


479

58 • Picture the Problem The energy required for this reaction is the difference between the rest energy of 4He and the sum of the rest energies of 3He and a neutron. Express the reaction:

nHeHe 34 +→

The rest energy of a neutron (Table 7-1) is:

939.573 MeV

The rest energy of 4He (Example 7-15) is:

3727.409 MeV

The rest energy of 3He is:

2808.432 MeV

Substitute numerical values to find the difference in the rest energy of 4He and the sum of the rest energies of 3He and n:

( )[ ] MeV574.20MeV573.93941.2808409.3727 =+−=E

59 • Picture the Problem The energy required for this reaction is the difference between the rest energy of a neutron and the sum of the rest energies of a proton and an electron. The rest energy of a proton (Table 7-1) is:

938.280 MeV

The rest energy of an electron (Table 7-1) is:

0.511 MeV

The rest energy of a neutron (Table 7-1) is:

939.573 MeV

Substitute numerical values to find the difference in the rest energy of a neutron and the sum of the rest energies of a positron and an electron:

( )[ ]MeV.7820

MeV511.0280.938573.939

=

+−=E

60 •• Picture the Problem The reaction is E+→+ HeHH 422 . The energy released in this reaction is the difference between twice the rest energy of 2H and the rest energy of 4He.

Chapter 7

480

The number of reactions that must take place to produce a given amount of energy is the ratio of the energy per second (power) to the energy released per reaction. (a) The rest energy of 4He (Example 7-14) is:

3727.409 MeV

The rest energy of a deuteron, 2H, (Table 7-1) is:

1875.628 MeV

The energy released in the reaction is:

( )[ ]J103.816MeV847.23

MeV409.3727628.1875212−×==

−=E

(b) The number of reactions per second is:

sreactions/1062.2

J/reaction10816.3J/s1000

14

12

×=

× −

61 •• Picture the Problem The annual consumption of matter by the fission plant is the ratio of its annual energy output to the square of the speed of light. The annual consumption of coal in a coal-burning power plant is the ratio of its annual energy output to energy per unit mass of the coal. (a) Express m in terms of E:

2cEm =

Assuming an efficiency of 33 percent, find the energy produced annually:

( )( )( )( )

( )( )J1084.2

d365.24h/d24s/h3600J/s1033

y1J/s10333

17

9

9

×=

××=

×=∆= tPE


( ) kg16.3m/s103

J1084.228

17

=×

×=m

(b) Assuming an efficiency of 38 percent, express the mass of coal required in terms of the annual energy production and the energy released per kilogram:

( ) ( )kg1004.8

J/kg103.138.0J109.47

/38.09

7

16annual

coal

×=

××

==mE

Em


481

General Problems *62 •• Picture the Problem Let the system consist of the block, the earth, and the incline. Then the tension in the string is an external force that will do work to change the energy of the system. Because the incline is frictionless; the work done by the tension in the string as it displaces the block on the incline is equal to the sum of the changes in the kinetic and gravitational potential energies.

Relate the work done by the tension force to the changes in the kinetic and gravitational potential energies of the block:

KUWW ∆+∆== extforcetension

Referring to the figure, express the change in the potential energy of the block as it moves from position 1 to position 2:

θsinmgLhmgU =∆=∆

Because the block starts from rest:

221

2 mvKK ==∆


221

forcetension sin mvmgLW += θ

and correct. is (c)

63 •• Picture the Problem Let the system include the earth, the block, and the inclined plane. Then there are no external forces to do work on the system and Wext = 0. Apply the work-energy theorem with friction to find an expression for the energy dissipated by friction.

Express the work-energy theorem with friction:


Chapter 7

482

Because the velocity of the block is constant, ∆K = 0 and:

hmgUE ∆−=∆−=∆ therm

In time ∆t the block slides a distance tv∆ . From the figure:

θsintvh ∆=∆

Substitute to obtain: θsintherm tmgvE ∆−=∆

and correct. is )( b

64 • Picture the Problem Let the system include the earth and the box. Then the applied force is external to the system and does work on the system in compressing the spring. This work is stored in the spring as potential energy. Express the work-energy theorem: thermsgext EUUKW ∆+∆+∆+∆=

Because :0thermg =∆=∆=∆ EUK

sext UW ∆=

Substitute for Wext and ∆Us: 221 kxFx =

Solve for x:

kFx 2

=

Substitute numerical values and evaluate x:

( ) cm06.2N/m6800N702

==x

*65 • Picture the Problem The solar constant is the average energy per unit area and per unit time reaching the upper atmosphere. This physical quantity can be thought of as the power per unit area and is known as intensity. Letting Isurface represent the intensity of the solar radiation at the surface of the earth, express Isurface as a function of power and the area on which this energy is incident:

AtE

API ∆∆

==/

surface

Solve for ∆E: tAIE ∆=∆ surface


483

Substitute numerical values and evaluate ∆E:

( )( )( )( )MJ6.57

s/h3600h8m2kW/m1 22

=

=∆E

66 •• Picture the Problem The luminosity of the sun (or of any other object) is the product of the power it radiates per unit area and its surface area. If we let L represent the sun’s luminosity, I the power it radiates per unit area (also known as the solar constant or the intensity of its radiation), and A its surface area, then L = IA. We can estimate the solar lifetime by dividing the number of hydrogen nuclei in the sun by the rate at which they are being transformed into energy. (a) Express the total energy the sun radiates every second in terms of the solar constant:

IAL =

Letting R represent its radius, express the surface area of the sun:

24 RA π=


IRL 24π=


( ) ( )watt1082.3

kW/m1.35m101.5426

2211

×=

×= πL

Note that this result is in good agreement with the value given in the text of 3.9×1026 watt.

(b) Express the solar lifetime in terms of the mass of the sun and the rate at which its mass is being converted to energy:

tnmM

tnNt

∆∆=

∆∆= nuclei H

solar

where M is the mass of the sun, m the mass of a hydrogen nucleus, and n is the number of nuclei used up.

Substitute numerical values to obtain:

tn

tnt

∆∆×

=

∆∆×

×

=−

nucleiH1019.1

nucleuskg/H101.67kg1099.1

57

27

30

solar

For each reaction, 4 hydrogen nuclei are "used up"; so:

( )

138

12

26

s1057.3J104.27J/s103.824

−

−

×=

××

=∆∆

tn

Chapter 7

484

Because we’ve assumed that the sun will continue burning until roughly 10% of its hydrogen fuel is used up, the total solar lifetime should be: y101.06s1033.3

s1057.3nucleiH1019.11.0

1017

138

57

solar

×=×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×= −t

67 • Picture the Problem Let the system include the earth and the Spirit of America. Then there are no external forces to do work on the car and Wext = 0. We can use the work-energy theorem to relate the coefficient of kinetic friction to the given information. A constant-acceleration equation will yield the car’s velocity when 60 s have elapsed. (a) Apply the work-energy theorem with friction to relate the coefficient of kinetic friction µk to the initial and final kinetic energies of the car:

0k202

1221 =∆+− smgmvmv µ

or, because v = 0, 0k

202

1 =∆+− smgmv µ

Solve for µk:

sgv∆

=2

2

kµ


( )( )[ ]( )( ) 208.0

km9.5m/s9.812sh/36001km/h708

2

2

k ==µ

(b) Express the kinetic energy of the car:

221 mvK = (1)

Using a constant-acceleration equation, relate the speed of the car to its acceleration, initial speed, and the elapsed time:

tavv ∆+= 0

Express the braking force acting on the car:

mamgfF =−=−= kknet µ

Solve for a:

ga kµ−=

Substitute for a to obtain: tgvv ∆−= k0 µ

Substitute in equation (1) to obtain: ( )2

k021 tgvmK ∆−= µ

Substitute numerical values and evaluate K:


485

( )[ ( )( )( )] MJ45.3s60m/s9.810.208m/h10708kg1250 22321 =−×=K

68 •• Picture the Problem The free-body diagram shows the forces acting on the skiers as they are towed up the slope at constant speed. Because the power required to move them is ,vF rr

⋅ we need to find F as a function of mtot, θ, and µk. We can apply Newton’s 2nd law to obtain such a function. Express the power required as a function of force on the skiers and their speed:

FvP = (1)

Apply ∑ = aF rrm to the skiers:

∑ =−−= 0sintotk θgmfFFx

and

∑ =−= 0costotn θgmFFy

Eliminate fk = µkFn and Fn between the two equations and solve for F:

θµθ cossin totktot gmgmF +=

Substitute in equation (1) to obtain: ( )( )θµθ

θµθcossin

cossin

ktot

totktot

+=+=

gvmvgmgmP

Substitute numerical values and evaluate P:

( )( )( ) ( )[ ] kW6.4615cos06.015sinm/s2.5m/s9.81kg7580 2 =°+°=P

Chapter 7

486

69 •• Picture the Problem The free-body diagram for the box is superimposed on the pictorial representation shown to the right. The work done by friction slows and momentarily stops the box as it slides up the incline. The box’s speed when it returns to bottom of the incline will be less than its speed when it started up the incline due to the energy dissipated by friction while it was in motion. Let the system include the box, the earth, and the incline. Then Wext = 0. We can use the work-energy theorem with friction to solve the several parts of this problem.

(a) earth. by the exerted box) theof weight (the force nalgravitatio the

and force,friction kinetic a plane, inclined by the exerted force normal thearebox on the acting forces that theseecan weFBD theFrom

(b) Apply the work-energy theorem with friction to relate the distance ∆x the box slides up the incline to its initial kinetic energy, its final potential energy, and the work done against friction:

0cosk212

1 =∆+∆+− θµ xmghmgmv

Referring to the figure, relate ∆h to ∆x to obtain:

θsinxh ∆=∆

Substitute for ∆h to obtain:

0cossin

k

212

1

=∆+

∆+−

θµθ

xmgxmgmv

Solve for ∆x:

( )θµθ cossin2 k

21

+=∆

gvx

Substitute numerical values and evaluate ∆x:

( )( ) ( )[ ]

m0.451

cos600.3sin60m/s9.812m/s3

2

2

=

°+°=∆x


487

(c) Express and evaluate the energy dissipated by friction:

( )( )( )( ) J1.33cos60m0.451m/s9.81kg20.3

cos2

kktherm

=°=

∆=∆=∆ θµ xmgxfE

(d) Use the work-energy theorem with friction to obtain:


or 0therm2121 =∆+−+− EUUKK

Because K2 = U1 = 0 we have: 0therm21 =∆+− EUK

or

0cossin

k

212

1

=∆+

∆−

θµθ

xmgxmgmv

Solve for v1: ( )θµθ cossin2 k1 −∆= xgv


( )( ) ( )[ ] m/s2.52cos600.3sin60m0.451m/s9.812 21 =°−°=v

*70 • Picture the Problem The power provided by a motor that is delivering sufficient energy to exert a force F on a load which it is moving at a speed v is Fv. The power provided by the motor is given by:

P = Fv

Because the elevator is ascending with constant speed, the tension in the support cable(s) is:

( )gmmF loadelev +=

Substitute for F to obtain: ( )gvmmP loadelev +=


( )( )( )kW45.1

m/s2.3m/s9.81kg2000 2

=

=P

Chapter 7

488

71 •• Picture the Problem The power a motor must provide to exert a force F on a load that it is moving at a speed v is Fv. The counterweight does negative work and the power of the motor is reduced from that required with no counterbalance. The power provided by the motor is given by:

P = Fv

Because the elevator is counterbalanced and ascending with constant speed, the tension in the support cable(s) is:

( )gmmmF cwloadelev −+=

Substitute and evaluate P: ( )gvmmmP cwloadelev −+=


( )( )( )kW3.11

m/s2.3m/s9.81kg005 2

=

=P

Without a load: ( )gmmF cwelev −=

and ( )( )( )( )

kW6.77

m/s2.3m/s9.81kg003 2cwelev

−=

−=

−= gvmmP

72 •• Picture the Problem We can use the work-energy theorem with friction to describe the energy transformation within the dart-spring-air-earth system. With this choice of the system, there are no external forces to do work on the system; i.e., Wext = 0. Choose Ug = 0 at the elevation of the dart on the compressed spring. The energy initially stored in the spring is transformed into gravitational potential energy and thermal energy. During the dart’s descent, its gravitational potential energy is transformed into kinetic energy and thermal energy. Apply conservation of energy during the dart’s ascent:


or, because ∆K = 0, 0thermis,fs,ig,fg, =∆+−+− EUUUU

Because 0fs,ig, == UU :

0thermis,fg, =∆+− EUU


489

Substitute for Ug,i and Us,f and solve for ∆Etherm:

mghkxUUE −=−=∆ 221

fg,is,therm


( )( )( )( )( )

J0.602

m24m/s9.81kg0.007

m0.03N/m50002

221

therm

=

−

=∆E

Apply conservation of energy during the dart’s descent:


or, because Ki = Ug,f = 0, 0thermig,f =∆+− EUK

Substitute for Kf and Ug,i to obtain: 0therm

2f2

1 =∆+− Emghmv

Solve for vf: ( )

mEmghv therm

f2 ∆−

=


( )( )( )[ ] m/s3.17kg007.0

J602.0m24m/s81.9kg007.02 2

f =−

=v

*73 •• Picture the Problem Let the system consist of the earth, rock and air. Given this choice, there are no external forces to do work on the system and Wext = 0. Choose Ug = 0 to be where the rock begins its upward motion. The initial kinetic energy of the rock is partially transformed into potential energy and partially dissipated by air resistance as the rock ascends. During its descent, its potential energy is partially transformed into kinetic energy and partially dissipated by air resistance. (a) Using the definition of kinetic energy, calculate the initial kinetic energy of the rock:

( )( )kJ1.60

m/s40kg2 2212

i21

i

=

== mvK

(b) Apply the work-energy theorem with friction to relate the energies of the system as the rock ascends:

0therm =∆+∆+∆ EUK

Because Kf = 0: 0thermi =∆+∆+− EUK

and UKE ∆−=∆ itherm

Chapter 7

490


( )( )( )J619

m50m/s9.81kg2J6001 2therm

=

−=∆E

(c) Apply the work-energy theorem with friction to relate the energies of the system as the rock descends:

07.0 therm =∆+∆+∆ EUK

Because Ki = Uf = 0: 07.0 thermif =∆+− EUK

Substitute for the energies to obtain: 07.0 therm

2f2

1 =∆+− Emghmv

Solve for vf:

mEghv therm

f4.12 ∆

−=


( )( ) ( )

m/s23.4

kg2J6191.4m50m/s9.812 2

f

=

−=v

74 •• Picture the Problem Let the distance the block slides before striking the spring be L. The pictorial representation shows the block at the top of the incline (1), just as it strikes the spring (2), and the block against the fully compressed spring (3). Let the block, spring, and the earth comprise the system. Then Wext = 0. Let Ug = 0 where the spring is at maximum compression. We can apply the work-energy theorem to relate the energies of the system as it evolves from state 1 to state 3.

Express the work-energy theorem: 0sg =∆+∆+∆ UUK

or 0s,1s,3g,1g,3 =−+−+∆ UUUUK


491

Because ∆K = Ug,3 = Us,1 = 0: 0s,3g,1 =+− UU

Substitute for each of these energy terms to obtain:

0221

1 =+− kxmgh

Substitute for h3 and h1:

( ) 0sin 221 =++− kxxLmg θ

Rewrite this equation explicitly as a quadratic equation:

0sin2sin22 =−−k

mgLxk

mgx θθ

Solve this quadratic equation to obtain:

θθθ sin2sinsin 22

kmgL

kmg

kmgx +⎟

⎠⎞

⎜⎝⎛+=

Note that the negative sign between the two terms leads to a non-physical solution. *75 • Picture the Problem We can find the work done by the girder on the slab by calculating the change in the potential energy of the slab. (a) Relate the work the girder does on the slab to the change in potential energy of the slab:

hmgUW ∆=∆=

Substitute numerical values and evaluate W:

( )( )( )J147

m0.001m/s9.81kg101.5 24

=

×=W

(b)

expansion. sgirder' thecauses which ,separation averagelarger a toleading energy, kinetic averagegreater a with rategirder vib in the

atoms therises,girder theof re temperatu theAs girder. n thewarmer thaare which gs,surroundin its fromgirder the toed transferrisenergy The

76 •• Picture the Problem The average power delivered by the car’s engine is the rate at which it changes the car’s energy. Because the car is slowing down as it climbs the hill, its potential energy increases and its kinetic energy decreases. Express the average power delivered by the car’s engine: t

EP∆∆

=av

Chapter 7

492

Express the increase in the car’s mechanical energy:

( )hgvvm

hmgmvmv

UUKKUKE

∆+−=

∆+−=

−+−=∆+∆=∆

22bot

2top2

1

2bot2

12top2

1

bottopbottop


( ) ( ) ( ) ( )( )[ ] MJ41.1m120m/s9.812m/s24m/s10kg1500 22221 =+−=∆E

Assuming that the acceleration of the car is constant, find its average speed during this climb:

m/s172

bottopav =

+=

vvv

Using the vav, find the time it takes the car to climb the hill:

s118m/s17

m2000

av

==∆

=∆v

st

Substitute to determine Pav: kW11.9s118

MJ1.41av ==P

*77 •• Picture the Problem Given the potential energy function as a function of y, we can find the net force acting on a given system from dydUF /−= . The maximum extension of

the spring; i.e., the lowest position of the mass on its end, can be found by applying the work-energy theorem. The equilibrium position of the system can be found by applying the work-energy theorem with friction … as can the amount of thermal energy produced as the system oscillates to its equilibrium position. (a) The graph of U as a function of y is shown to the right. Because k and m are not specified, k has been set equal to 2 and mg to 1. The spring is unstretched when y = y0 = 0. Note that the minimum value of U (a position of stable equilibrium) occurs near y = 5 m.


493

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

y (m)

U (

J)

(b) Evaluate the negative of the derivative of U with respect to y:

( )

mgky

mgykydyd

dydUF

+−=

−−=−= 221

(c) Apply conservation of energy to the movement of the mass from y = 0 to y = ymax:


Because ∆K = 0 (the object starts from rest and is momentarily at rest at y = ymax) and ∆Etherm = 0 (no friction), it follows that:

∆U = U(ymax) – U(0) = 0

Because U(0) = 0: U(ymax) = 0 ⇒ 0max2max2

1 =− mgyky

Solve for ymax:

kmgy 2

max =

(d) Express the condition of F at equilibrium and solve for yeq:

00 eqeq =+−⇒= mgkyF

and

kmgy =eq

(e) Apply the conservation of energy to the movement of the mass from y = 0 to y = yeq and solve for ∆Etherm:


or, because ∆K = 0. fitherm UUUE −=∆−=∆

Chapter 7

494

Because ( ) :00i == UU ( )eq

2eq2

1ftherm mgykyUE −−=−=∆

Substitute for yeq and simplify to obtain: k

gmE2

22

therm =∆

78 •• Picture the Problem The energy stored in the compressed spring is initially transformed into the kinetic energy of the signal flare and then into gravitational potential energy and thermal energy as the flare climbs to its maximum height. Let the system contain the earth, the air, and the flare so that Wext = 0. We can use the work-energy theorem with friction in the analysis of the energy transformations during the motion of the flare. (a) The work done on the spring in compressing it is equal to the kinetic energy of the flare at launch. Therefore:

202

1flarei,s mvKW ==

(b) Ignoring changes in gravitational potential energy (i.e., assume that the compression of the spring is small compared to the maximum elevation of the flare), apply the conservation of energy to the transformation that takes place as the spring decompresses and gives the flare its launch speed:

0s =∆+∆ UK

or 0is,fs,if =−+− UUKK

Because Ki = ∆Ug = Us,f:

0is,f =−UK

Substitute for is,f and UK :

02212

021 =− kdmv

Solve for k to obtain: 2

20

dmvk =

(c) Apply the work-energy theorem with friction to the upward trajectory of the flare:

0thermg =∆+∆+∆ EUK


495

Solve for ∆Etherm:

fifi

gtherm

UUKKUKE

−+−=

∆−∆−=∆

Because Kf = Ui = 0: mghmvE −=∆ 202

1therm

79 •• Picture the Problem Let UD = 0. Choose the system to include the earth, the track, and the car. Then there are no external forces to do work on the system and change its energy and we can use Newton’s 2nd law and the work-energy theorem to describe the system’s energy transformations to point G … and then the work-energy theorem with friction to determine the braking force that brings the car to a stop. The free-body diagram for point C is shown to the right.

The free-body diagram for point D is shown to the right.

The free-body diagram for point F is shown to the right.

(a) Apply the work-energy theorem to the system’s energy transformations between A and B:

0=∆+∆ UK or

0ABAB =−+− UUKK

If we assume that the car arrives at point B with vB = 0, then:

02A2

1 =∆+− hmgmv

where ∆h is the difference in elevation between A and B.

Chapter 7

496

Solve for and evaluate ∆h: ( )( ) m34.7

m/s9.812m/s12

2 2

22A ===∆g

vh

The height above the ground is: m17.3m7.34m10 =+=∆+ hh

(b) If the car just makes it to point B; i.e., if it gets there with vB = 0, then the force exerted by the track on the car will be the normal force:

( )( )kN4.91

m/s9.81kg500 2

ncarontrack

=

=

== mgFF

(c) Apply ∑ = xx maF to the car at

point C (see the FBD) and solve for a:

mamg =θsin

and ( )2

2

m/s4.91

sin30m/s9.81sin

=

°== θga

(d) Apply ∑ = yy maF to the car at

point D (see the FBD) and solve for Fn:

RvmmgF

2D

n =−

and

R

2D

nvmmgF +=

Apply the work-energy theorem to the system’s energy transformations between B and D:

0=∆+∆ UK or

0BDBD =−+− UUKK

Because KB = UD = 0:

0BD =−UK

Substitute to obtain: ( ) 02D2

1 =∆+− hhmgmv

Solve for 2

Dv :

( )hhgv ∆+= 22D

Substitute to find Fn:

( )

( )⎥⎦⎤

⎢⎣⎡ ∆+

+=

∆++=

+=

Rhhmg

Rhhgmmg

vmmgF

21

2R

2D

n


497

Substitute numerical values and evaluate Fn:

( )( ) ( )

upward.directedkN,13.4

m20m17.321m/s9.81kg500 2

n

=

⎥⎦

⎤⎢⎣

⎡+=F

(e) F has two components at point F; one horizontal (the inward force that the track exerts) and the other vertical (the normal force). Apply

∑ = aF rrm to the car at point F:

∑ =⇒=−= mgFmgFFy nn 0

and

∑ ==RvmFFx

2F

c

Express the resultant of these two forces:

( )

22

4F

222

F

2n

2c

gRvm

mgRvm

FFF

+=

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

+=

Substitute numerical values and evaluate F: ( ) ( )

( )( )

kN46.5

m/s9.81m30

m/s12kg500 222

4

=

+=F

Find the angle the resultant makes with the x axis:

( )( )( )

°=⎥⎦

⎤⎢⎣

⎡=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

−−

9.63m/s12

m30m/s9.81tan

tantan

2

21

2F

1

c

n1

vgR

FFθ

(f) Apply the work-energy theorem with friction to the system’s energy transformations between F and the car’s stopping position:

0thermG =∆+− EK

and 2G2

1Gtherm mvKE ==∆

The work done by friction is also given by:

dFsfE braketherm =∆=∆

where d is the stopping distance.

Equate the two expressions for ∆Etherm and solve for Fbrake: d

mvF2

2F

brake =

Chapter 7

498

Substitute numerical values and evaluate Fbrake:

( )( )( ) kN1.44

m252m/s12kg500 2

brake ==F

*80 • Picture the Problem The rate of conversion of mechanical energy can be determined from .vF rr

⋅=P The pictorial representation shows the elevator moving downward just as it goes into freefall as state 1. In state 2 the elevator is moving faster and is about to strike the relaxed spring. The momentarily at rest elevator on the compressed spring is shown as state 3. Let Ug = 0 where the spring has its maximum compression and the system consist of the earth, the elevator, and the spring. Then Wext = 0 and we can apply the conservation of mechanical energy to the analysis of the falling elevator and compressing spring.

(a) Express the rate of conversion of mechanical energy to thermal energy as a function of the speed of the elevator and braking force acting on it:

0brakingvFP =

Because the elevator is moving with constant speed, the net force acting on it is zero and:

MgF =braking

Substitute for Fbraking and evaluate P: ( )( )( )

kW29.4

m/s1.5m/s9.81kg2000 20

=

=

= MgvP

(b) Apply the conservation of energy to the falling elevator and compressing spring:

0sg =∆+∆+∆ UUK

or 0s,1s,3g,1g,313 =−+−+− UUUUKK

Because K3 = Ug,3 = Us,1 = 0: ( ) ( ) 02212

021 =∆+∆+−− ykydMgMv


499

Rewrite this equation as a quadratic equation in ∆y, the maximum compression of the spring:

( ) ( ) 022 20

2 =+−∆⎟⎠⎞

⎜⎝⎛−∆ vgd

kMy

kMgy

Solve for ∆y to obtain: ( )202

22

2 vgdkM

kgM

kMgy ++±=∆

Substitute numerical values and evaluate ∆y:

( )( )

( ) ( )( ) ( )( ) ( )[ ]

m19.5

m/s5.1m5m/s81.92N/m105.1kg2000

N/m105.1m/s81.9kg2000

N/m105.1m/s81.9kg2000

22424

222

4

2

=

+×

+×

+

×=∆y

81 • Picture the Problem We can use Newton’s 2nd law to determine the force of friction as a function of the angle of the hill for a given constant speed. The power output of the engine is given by vF

rr⋅= fP .

FBD for (a):

FBD for (b):

(a) Apply ∑ = xx maF to the car: 0sin f =− Fmg θ ⇒ θsinf mgF =

Evaluate Ff for the two speeds: ( )( )

( )( )N981

sin5.74m/s9.81kg1000and

N491

sin2.87m/s9.81kg1000

230

220

=

°=

=

°=

F

F

(b) Express the power an engine must deliver on a level road in order ( )( ) kW9.82m/s20N49120 ==

=

P

vFP f

Chapter 7

500

to overcome friction loss and evaluate this expression for v = 20 m/s and 30 m/s:

and ( )( ) kW29.4m/s30N98130 ==P

(c) Apply ∑ = xx maF to the car: ∑ =−−= 0sin fx FmgFF θ

Relate F to the power output of the engine and the speed of the car:

vPFFvP == ,Since

Substitute for F and solve for θ :

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −= −

mg

FvP

201sinθ


( )( )

°=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −= −

85.8

m/s9.81kg1000

N491m/s20kW40

sin 21θ

(d) Express the equivalence of the work done by the engine in driving the car at the two speeds:

( ) ( )30302020engine sFsFW ∆=∆=

Let ∆V represent the volume of fuel consumed by the engine driving the car on a level road and divide both sides of the work equation by ∆V to obtain:

( ) ( )Vs

FVs

F∆∆

=∆∆ 30

3020

20

Solve for ( )

Vs

∆∆ 30 :

( ) ( )Vs

FF

Vs

∆∆

=∆∆ 20

30

2030

Substitute numerical values and

evaluate ( )

Vs

∆∆ 30 :

( ) ( )

km/L6.36

km/L12.7N981N49130

=

=∆∆

Vs

82 •• Picture the Problem Let the system include the earth, block, spring, and incline. Then Wext = 0. The pictorial representation to the left shows the block sliding down the incline


501

and compressing the spring. Choose Ug = 0 at the elevation at which the spring is fully compressed. We can use the conservation of mechanical energy to determine the maximum compression of the spring. The pictorial representation to the right shows the block sliding up the rough incline after being accelerated by the fully compressed spring. We can use the work-energy theorem with friction to determine how far up the incline the block slides before stopping.

(a) Apply conservation of mechanical energy to the system as it evolves from state 1 to state 3:

0sg =∆+∆+∆ UUK

or

0s,1s,3

g,1g,313

=−+

−+−

UU

UUKK

Because

0s,1g,313 ==== UUKK : 0s,3g,1 =+− UU

or 02

21 =+∆− kxhmg

Relate ∆h to L + x and θ and substitute to obtain:

( ) θsinxLh +=∆ ( ) 0sin2

21 =+−∴ θxLmgkx

Rewrite this equation in the form of an explicit quadratic equation:

( ) 0sinsin221 =−− θθ mgLxmgkx

Substitute for k, m, g, θ and L to obtain:

( ) 0J24.39N81.9mN50 2 =−−⎟

⎠⎞

⎜⎝⎛ xx

Solve for the physically meaningful (i.e., positive) root:

m989.0=x

(b) Proceed as in (a) but include energy dissipated by friction:

0therms,3g,1 =∆++− EUU

The mechanical energy transformed to thermal energy is given by:

( ) ( )( )xLmg

xLFxLFE+=

+=+=∆θµ

µcosk

nkftherm

Chapter 7

502

Substitute for ∆h and ∆Etherm to obtain:

( )( ) 0cos

sin

k

221

=++

++−

xLmgkxxLmg

θµθ

Substitute for k, m, g, θ, µk and L to obtain:

( ) 0J65.25N41.6mN50 2 =−−⎟

⎠⎞

⎜⎝⎛ xx

Solve for the positive root:

m783.0=x

(c) Apply the work-energy theorem with friction to the system as it evolves from state 3 to state 4:

0therms,3s,4

g,3g,434

=∆+−+

−+−

EUUUUKK

Because 0s,4g,314 ==== UUKK :

0therms,3g,4 =∆+− EUU

or 0' therm

221 =∆++∆− Ekxhmg

Substitute for ∆h′ and ∆Etherm to obtain:

( )( ) 0'cos

sin'

k

221

=++

++−

xLmgkxxLmg

θµθ

Solve for L′ with x = 0.783 m: m54.1'=L

83 •• Picture the Problem The work done by the engines maintains the kinetic energy of the cars and overcomes the work done by frictional forces. Let the system include the earth, track, and the cars but not the engines. Then the engines will do external work on the system and we can use this work to find the power output of the train’s engines. (a) Use the definition of kinetic energy:

( )

MJ17.4

s3600h1

hkm15kg102

26

21

221

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅×=

= mvK

(b) The change in potential energy of the train is: ( )( )( )

J101.39

m707m/s9.81kg10210

26

×=

×=

∆=∆ hmgU

(c) Express the energy dissipated by kinetic friction:

sfE ∆=∆ therm


503

Express the frictional force:

mgf 008.0=

Substitute for f and evaluate ∆Etherm:

( )( )( ) J109.73km62m/s9.81kg1020.008008.0 926therm ×=×=∆=∆ smgE

(d) Express the power output of the train’s engines in terms of the work done by them:

tWP

∆∆

=

Use the work-energy theorem with friction to find the work done by the train’s engines:

thermext EUKW ∆+∆+∆=

or, because ∆K = 0, thermext EUW ∆+∆=

Find the time during which the engines do this work:

vst ∆

=∆

Substitute in the expression for P to obtain:

( )s

vEUP∆∆+∆

= therm


( ) MW1.59km62

s3600h1

hkm15

J109.73J101.39 910 =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

×+×=P

*84 •• Picture the Problem While on a horizontal surface, the work done by an automobile engine changes the kinetic energy of the car and does work against friction. These energy transformations are described by the work-energy theorem with friction. Let the system include the earth, the roadway, and the car but not the car’s engine. (a) The required energy equals the change in the kinetic energy of the car: ( )

kJ116

s3600h1

hkm50kg1200

2

21

221

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

=∆ mvK

(b) The required energy equals the sfE ∆=∆ therm

Chapter 7

504

work done against friction: Substitute numerical values and evaluate ∆Etherm:

( )( ) kJ90.0m300N300therm ==∆E

(c) Apply the work-energy theorem with friction to express the required energy:

EKEKWE

75.0' thermext

+∆=∆+∆==

Divide both sides of the equation by E to express the ratio of the two energies:

75.0'+

∆=

EK

EE

Substitute numerical values and evaluate E′/E:

04.20.75kJ90kJ116'

=+=EE

*85 ••• Picture the Problem Assume that the bob is moving with speed v as it passes the top vertical point when looping around the peg. There are two forces acting on the bob: the tension in the string (if any) and the force of gravity, Mg; both point downward when the ball is in the topmost position. The minimum possible speed for the bob to pass the vertical occurs when the tension is 0; from this, gravity must supply the centripetal force required to keep the ball moving in a circle. We can use conservation of energy to relate v to L and R.

Express the condition that the bob swings around the peg in a full circle:

2

MgRvM >

Simplify to obtain: g

Rv

>2

Use conservation of energy to relate the kinetic energy of the bob at the bottom of the loop to its potential energy at the top of its swing:

( )2221 RLMgMv −=

Solve for v2: ( )RLgv 222 −=

Substitute to obtain: ( ) gR

RLg>

− 22


505

Solve for R:

LR52 <

86 •• Picture the Problem If the wood exerts an average force F on the bullet, the work it does has magnitude FD. This must be equal to the change in the kinetic energy of the bullet, or because the final kinetic energy of the bullet is zero, to the negative of the initial kinetic energy. We’ll let m be the mass of the bullet and v its initial speed and apply the work-kinetic energy theorem to relate the penetration depth to v. Apply the work-kinetic energy theorem to relate the penetration depth to the change in the kinetic energy of the bullet:

iftotal KKKW −=∆= or, because Kf = 0,

itotal KW −=

Substitute for Wtotal and Ki to obtain: 221 mvFD −=

Solve for D to obtain:

FmvD2

2

−=

For an identical bullet with twice the speed we have:

( )221 2' vmFD −=

Solve for D′ to obtain: D

FmvD 42

4'2

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

and correct. is )(c

87 •• Picture the Problem For part (a), we’ll let the system include the glider, track, weight, and the earth. The speeds of the glider and the falling weight will be the same while they are in motion. Let their common speed when they have moved a distance Y be v and let the zero of potential energy be at the elevation of the weight when it has fallen the distance Y. We can use conservation of energy to relate the speed of the glider (and the weight) to the distance the weight has fallen. In part (b), we’ll let the direction of motion be the x direction, the tension in the connecting string be T, and apply Newton’s 2nd law to the glider and the weight to find their common acceleration. Because this acceleration is constant, we can use a constant-acceleration equation to find their common speed when they have moved a distance Y. (a) Use conservation of energy to relate the kinetic and potential energies of the system:

0=∆+∆ UK or

0ifif =−+− UUKK

Because the system starts from rest and Uf = 0:

0if =−UK


21 =−+ mgYMvmv

Chapter 7

506

Solve for v:

mMmgYv

+=

2

(b) The free-body diagrams for the glider and the weight are shown to the right:

Apply Newton’s 3rd law to obtain: T== 21 TT

rr

Apply maFx =∑ to the glider:

MaT =

Apply maFx =∑ to the weight:

maTmg =−

Add these equations to eliminate T and obtain:

maMamg +=

Solve for a to obtain:

Mmmga+

=

Using a constant-acceleration equation, relate the speed of the glider to its initial speed and to the distance that the weight has fallen:

aYvv 220

2 += or, because v0 = 0,

aYv 22 =

Substitute for a and solve for v to obtain:

mMmgYv

+=

2, the same result we

obtained in part (a). *88 •• Picture the Problem We’re given dtdWP /= and are asked to evaluate it under the assumed conditions. Express the rate of energy expenditure by the man:

( )( )W270

m/s3kg1033 22

=== mvP

Express the rate of energy expenditure P′ assuming that his

P'P 51=


507

muscles have an efficiency of 20%: Solve for and evaluate P′: ( ) kW1.35W27055 === PP'

89 •• Picture the Problem The pictorial representation shows the bob swinging through an angle θ before the thread is cut and it is launched horizontally. Let its speed at position 1 be v. We can use conservation of energy to relate v to the change in the potential energy of the bob as it swings through the angle θ . We can find its flight time ∆t from a constant-acceleration equation and then express D as the product of v and ∆t.

Relate the distance D traveled horizontally by the bob to its launch speed v and time of flight ∆t:

tvD ∆= (1)

Use conservation of energy to relate its launch speed v to the length of the pendulum L and the angle θ :

00101 =−+− UUKK or, because U1 = K0 = 0,

001 =−UK


( ) 0cos1221 =−− θmgLmv

Solving for v yields:

( )θcos12 −= gLv

In the absence of air resistance, the horizontal and vertical motions of the bob are independent of each other and we can use a constant-acceleration equation to express the time of flight (the time to fall a distance H):

( )221

0 tatvy yy ∆+∆=∆ or, because ∆y = −H, ay = −g, and v0y = 0,

( )221 tgH ∆−=−

Solve for ∆t to obtain: gHt /2=∆

Substitute in equation (1) and simplify to obtain: ( )

( )θ

θ

cos12

2cos12

−=

−=

HL

gHgLD

which shows that, while D depends on θ, it is independent of g.

Chapter 7

508

90 •• Picture the Problem The pictorial representation depicts the block in its initial position against the compressed spring (1), as it separates from the spring with its maximum kinetic energy (2), and when it has come to rest after moving a distance x + d. Let the system consist of the earth, the block, and the surface on which the block slides. With this choice, Wext = 0. We can use the work-energy theorem with friction to determine how far the block will slide before coming to rest.

(a) The work done by the spring on the block is given by:

221

springspring kxUW =∆=

Substitute numerical values and evaluate Wspring:

( )( ) J0.900cm3N/cm20 221

spring ==W

(b) The energy dissipated by friction is given by:

xmgxFsfE ∆=∆=∆=∆ knktherm µµ


( )( )( )( )J0.294

m0.03m/s9.81kg50.2 2therm

=

=∆E

(c) Apply the conservation of energy between points 1 and 2:

0therms,1s,212 =∆+−+− EUUKK

Because K1 = Us,2 = 0:

0therms,12 =∆+− EUK

Substitute to obtain: 0therm2

212

221 =∆+− Ekxmv

Solve for v2:

mEkxv therm

2

22∆−

=


( )( ) ( )

m/s0.492

kg5J0.2942cm3N/cm20 2

2

=

−=v


509

(d) Apply the conservation of energy between points 1 and 3:

0therms,1s,3 =∆+−+∆ EUUK

Because ∆K = Us,3 = 0: 0therms,1 =∆+− EU

or ( ) 0k

221 =++− dxmgkx µ

Solve for d:

xmg

kxd −=k

2

2µ


( )( )( )( )( )

cm6.17

m0.03m/s9.81kg50.22

cm3N/cm202

2

=

−=d

91 •• Picture the Problem The pictorial representation shows the block initially at rest at point 1, falling under the influence of gravity to point 2, partially compressing the spring as it continues to gain kinetic energy at point 3, and finally coming to rest at point 4 with the spring fully compressed. Let the system consist of the earth, the block, and the spring so that Wext = 0. Let Ug = 0 at point 3 for part (a) and at point 4 for part (b). We can use the work-energy theorem to express the kinetic energy of the system as a function of the block’s position and then use this function to maximize K as well as determine the maximum compression of the spring and the location of the block when the system has half its maximum kinetic energy.

(a) Apply conservation of mechanical energy to describe the energy transformations between state 1 and state 3:

0sg =∆+∆+∆ UUK

or 0s,1s,3g,1g,313 =−+−+− UUUUKK

Because K1 = Ug,3 = Us,1 = 0: 0s,3g,13 =+− UUK

Chapter 7

510

and ( ) 2

21

3 kxxhmgKK −+==

Differentiate K with respect to x and set this derivative equal to zero to identify extreme values:

.valuesextremefor0=−= kxmgdxdK

Solve for x: k

mgx =

Evaluate the second derivative of K with respect to x:

.maximizes

02

2

Kk

mgx

kdt

Kd

=⇒

<−=

Evaluate K for x = mg/k:

kgmmgh

kmgk

kmgmgmghK

2

22

2

21

max

+=

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+=

(b) The spring will have its maximum compression at point 4 where K = 0:

( )

022or

0

max2max

2max2

1max

=−−

=−+

kmghx

kmgx

kxxhmg

Solve for x and keep the physically meaningful root:

kmgh

kgm

kmgx 2

2

22

max ++=

(c) Apply conservation of mechanical energy to the system as it evolves from state 1 to the state in which max2

1 KK = :

0sg =∆+∆+∆ UUK

or 0s,1s,3g,1g,31 =−+−+− UUUUKK

Because K1 = Ug,3 = Us,1 = 0: 0s,3g,1 =+− UUK

and ( ) 2

21 kxxhmgK −+=


511

Substitute for K to obtain: ( ) 221

22

21

2kxxhmg

kgmmgh −+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

Express this equation in quadratic form:

02

22

222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−+−

kmgh

kgmx

kmgx

Solve for the positive value of x:

kmgh

kgm

kmgx 42

2

22

++=

92 ••• Picture the Problem The free-body diagram shows the forces acting on the pendulum bob. The application of Newton’s 2nd law leads directly to the required expression for the tangential acceleration. Recall that, provided θ is in radian measure, s = Lθ. Differentiation with respect to time produces the result called for in part (b). The remaining parts of the problem simply require following the directions for each part.

(a) Apply ∑ = xx maF to the bob: tantan sin mamgF =−= θ

Solve for atan: θsin/tan gdtdva −==

(b) Relate the arc distance s to the length of the pendulum L and the angle θ :

θLs =

Differentiate with respect to time: dtLdvdtds // θ==

(c) Multiply θθ

dd

dtdv by and

substitute for dtdθ

from part (b): ⎟⎠⎞

⎜⎝⎛=

==

Lv

ddv

dtd

ddv

dd

dtdv

dtdv

θ

θθθ

θ

Chapter 7

512

(d) Equate the expressions for dv/dt from (a) and (c):

θθ

singLv

ddv

−=⎟⎠⎞

⎜⎝⎛

Separate the variables to obtain: θθ dgLvdv sin−=

(e) Integrate the left side of the equation in part (d) from v = 0 to the final speed v and the right side from θ = θ0 to θ = 0:

∫∫ −=0

0 0

''sin''θ

θθ dgLdvvv

Evaluate the limits of integration to obtain:

( )02

21 cos1 θ−= gLv

Note, from the figure, that h = L(1 − cosθ0). Substitute and solve for v:

ghv 2=

93 ••• Picture the Problem The potential energy of the climber is the sum of his gravitational potential energy and the potential energy stored in the spring-like bungee cord. Let θ be the angle which the position of the rock climber on the cliff face makes with a vertical axis and choose the zero of gravitational potential energy to be at the bottom of the cliff. We can use the definitions of Ug and Uspring to express the climber’s total potential energy. (a) Express the total potential energy of the climber:

( ) gcord bungee UUsU +=

Substitute to obtain: ( )( )

( ) ⎟⎠⎞

⎜⎝⎛+−=

+−=

+−=

HsMgHLsk

MgHLsk

MgyLsksU

cos

cos

)(

221

221

221

θ

A spreadsheet solution is shown below. The constants used in the potential energy function and the formulas used to calculate the potential energy are as follows:

Cell Content/Formula Algebraic Form B3 300 H B4 5 k B5 60 L B6 85 M B7 9.81 g

D11 60 s D12 D11+1 s + 1


513

E11 0.5*$B$4*(D11−$B$5)^2 +$B$6*$B$7*$B$3*(cos(D11/$B$3)) ( ) ⎟

⎠⎞

⎜⎝⎛+−

HsMgHLsk cos2

21

G11 E11−E61 ( ) ( )m110m60 UU −

A B C D E 1 2 3 H = 300 m 4 k = 5 N/m 5 L = 60 m 6 m = 85 kg 7 g = 9.81 m/s^2 8 9

10 s U(s) 11 60 2.45E+05 12 61 2.45E+05 13 62 2.45E+05 14 63 2.45E+05 15 64 2.45E+05

147 196 2.45E+05 148 197 2.45E+05 149 198 2.45E+05 150 199 2.45E+05 151 200 2.46E+05

The following graph was plotted using the data from columns D (s) and E (U(s)).

238

239

240

241

242

243

244

245

246

50 70 90 110 130 150 170 190 210

s (m)

U (k

J)

Chapter 7

514

*94 ••• Picture the Problem The diagram shows the forces each of the springs exerts on the block. The change in the potential energy stored in the springs is due to the elongation of both springs when the block is displaced a distance x from its equilibrium position and we can find ∆U using ( )2

21 Lk ∆ . We can find the magnitude of the force pulling the block

back toward its equilibrium position by finding the sum of the magnitudes of the y components of the forces exerted by the springs. In Part (d) we can use conservation of energy to find the speed of the block as it passes through its equilibrium position.

(a) Express the change in the potential energy stored in the springs when the block is displaced a distance x:

( )[ ] ( )22212 LkLkU ∆=∆=∆

where ∆L is the change in length of a spring.

Referring to the force diagram, express ∆L:

LxLL −+=∆ 22

Substitute to obtain: ( )222 LxLkU −+=∆

(b) Sum the forces acting on the block to express Frestoring:

22

restoring

2

cos2cos2

xLxLk

LkFF

+∆=

∆== θθ

Substitute for ∆L to obtain: ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=

+−+=

22

22

22restoring

12

2

xLLkx

xLxLxLkF

(c) A spreadsheet program to calculate U(x) is shown below. The constants used in the potential energy function and the formulas used to calculate the potential energy are as follows:

Cell Content/Formula Algebraic Form B1 1 L B2 1 k B3 1 M C8 C7+0.01 x D7 $B$2*((C7^2+$B$1^2)^0.5−$B$1)^2 U(x)


515

A B C D 1 L = 0.1 m 2 k = 1 N/m 3 M = 1 kg 4 5 6 x U(x) 7 0 0 8 0.01 2.49E−07 9 0.02 3.92E−06 10 0.03 1.94E−05 11 0.04 5.93E−05 12 0.05 1.39E−04

23 0.16 7.86E−03 24 0.17 9.45E−03 25 0.18 1.12E−02 26 0.19 1.32E−02 27 0.20 1.53E−02

The following graph was plotted using the data from columns C (x) and D (U(x)).

0

2

4

6

8

10

12

14

16

0.00 0.05 0.10 0.15 0.20

x (m)

U (m

J)

(d) Use conservation of energy to relate the kinetic energy of the block as it passes through the equilibrium position to the change in its potential energy as it returns to its equilibrium position:

UK ∆=mequilibriu or

UMv ∆=221

Chapter 7

516

Solve for v to obtain: ( )

( )MkLxL

MLxLk

MUv

2

22

22

222

−+=

−+=

∆=


( ) ( ) ( ) cm/s86.5kg1N/m12m1.0m1.0m1.0 22 =⎟

⎠⎞⎜

⎝⎛ −+=v

517

Chapter 8 Systems of Particles and Conservation of Momentum Conceptual Problems 1 • Determine the Concept A doughnut. The definition of the center of mass of an object does not require that there be any matter at its location. Any hollow sphere (such as a basketball) or an empty container with any geometry are additional examples of three-dimensional objects that have no mass at their center of mass. *2 • Determine the Concept The center of mass is midway between the two balls and is in free-fall along with them (all forces can be thought to be concentrated at the center of mass.) The center of mass will initially rise, then fall. Because the initial velocity of the center of mass is half of the initial velocity of the ball thrown upwards, the mass thrown upwards will rise for twice the time that the center of mass rises. Also, the center of mass will rise until the velocities of the two balls are equal but opposite. correct. is )(b

3 • Determine the Concept The acceleration of the center of mass of a system of particles is described by ,cm

iexti,extnet, aFF

rrrM== ∑ where M is the total mass of the system.

Express the acceleration of the center of mass of the two pucks: 21

1extnet,cm mm

FM

Fa

+==

and correct. is )(b

4 • Determine the Concept The acceleration of the center of mass of a system of particles is described by ,cm

iexti,extnet, aFF

rrrM== ∑ where M is the total mass of the system.

Express the acceleration of the center of mass of the two pucks: 21

1extnet,cm mm

FM

Fa

+==

because the spring force is an internal force.

correct. is )( b

Chapter 8

518

*5 • Determine the Concept No. Consider a 1-kg block with a speed of 1 m/s and a 2- kg block with a speed of 0.707 m/s. The blocks have equal kinetic energies but momenta of magnitude 1 kg·m /s and 1.414 kg·m/s, respectively. 6 • (a) True. The momentum of an object is the product of its mass and velocity. Therefore, if we are considering just the magnitudes of the momenta, the momentum of a heavy object is greater than that of a light object moving at the same speed. (b) True. Consider the collision of two objects of equal mass traveling in opposite directions with the same speed. Assume that they collide inelastically. The mechanical energy of the system is not conserved (it is transformed into other forms of energy), but the momentum of the system is the same after the collision as before the collision, i.e., zero. Therefore, for any inelastic collision, the momentum of a system may be conserved even when mechanical energy is not. (c) True. This is a restatement of the expression for the total momentum of a system of particles. 7 • Determine the Concept To the extent that the system in which the rifle is being fired is an isolated system, i.e., the net external force is zero, momentum is conserved during its firing. Apply conservation of momentum to the firing of the rifle:

0bulletrifle =+ pp rr

or

bulletrifle pp rr−=

*8 • Determine the Concept When she jumps from a boat to a dock, she must, in order for momentum to be conserved, give the boat a recoil momentum, i.e., her forward momentum must be the same as the boat’s backward momentum. The energy she imparts to the boat is .2 boat

2boatboat mpE =

zero.y essentiall is them toimparts sheenergy that thelarge so isearth theplusdock theof mass theanother, dock to one from jumps sheWhen

*9 ••

Determine the Concept Conservation of momentum requires only that the net external force acting on the system be zero. It does not require the presence of a medium such as air.

Systems of Particles and Conservation of Momentum

519

10 • Determine the Concept The kinetic energy of the sliding ball is 2

cm21 mv . The kinetic

energy of the rolling ball is rel2cm2

1 Kmv + , where its kinetic energy relative to its center of mass is relK . Because the bowling balls are identical and have the same velocity, the

rolling ball has more energy. 11 • Determine the Concept Think of someone pushing a box across a floor. Her push on the box is equal but opposite to the push of the box on her, but the action and reaction forces act on different objects. You can only add forces when they act on the same object. 12 • Determine the Concept It’s not possible for both to remain at rest after the collision, as that wouldn't satisfy the requirement that momentum is conserved. It is possible for one to remain at rest: This is what happens for a one-dimensional collision of two identical particles colliding elastically. 13 • Determine the Concept It violates the conservation of momentum! To move forward requires pushing something backwards, which Superman doesn’t appear to be doing when flying around. In a similar manner, if Superman picks up a train and throws it at Lex Luthor, he (Superman) ought to be tossed backwards at a pretty high speed to satisfy the conservation of momentum. *14 •• Determine the Concept There is only one force which can cause the car to move forward−the friction of the road! The car’s engine causes the tires to rotate, but if the road were frictionless (as is closely approximated by icy conditions) the wheels would simply spin without the car moving anywhere. Because of friction, the car’s tire pushes backwards against the road−from Newton’s third law, the frictional force acting on the tire must then push it forward. This may seem odd, as we tend to think of friction as being a retarding force only, but true. 15 •• Determine the Concept The friction of the tire against the road causes the car to slow down. This is rather subtle, as the tire is in contact with the ground without slipping at all times, and so as you push on the brakes harder, the force of static friction of the road against the tires must increase. Also, of course, the brakes heat up, and not the tires. 16 • Determine the Concept Because ∆p = F∆t is constant, a safety net reduces the force acting on the performer by increasing the time ∆t during which the slowing force acts. 17 • Determine the Concept Assume that the ball travels at 80 mi/h ≈ 36 m/s. The ball stops in a distance of about 1 cm. So the distance traveled is about 2 cm at an average speed of

Chapter 8

520

about 18 m/s. The collision time is ms1m/s18

m0.02≈ .

18 • Determine the Concept The average force on the glass is less when falling on a carpet because ∆t is longer. 19 • (a) False. In a perfectly inelastic collision, the colliding bodies stick together but may or may not continue moving, depending on the momentum each brings to the collision. (b) True. In a head-on elastic collision both kinetic energy and momentum are conserved and the relative speeds of approach and recession are equal. (c) True. This is the definition of an elastic collision. *20 •• Determine the Concept All the initial kinetic energy of the isolated system is lost in a perfectly inelastic collision in which the velocity of the center of mass is zero. 21 •• Determine the Concept We can find the loss of kinetic energy in these two collisions by finding the initial and final kinetic energies. We’ll use conservation of momentum to find the final velocities of the two masses in each perfectly elastic collision. (a) Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V:

afterbefore pp =

or 02 =⇒=− VmVmvmv

Express the loss of kinetic energy for the case in which the two objects have oppositely directed velocities of magnitude v/2:

4

220

2

2

21

if

mv

vmKKK

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=−=∆

Letting V represent the velocity of the masses after their perfectly inelastic collision, use conservation of momentum to determine V:

afterbefore pp =

or vVmVmv 2

12 =⇒=


521

Express the loss of kinetic energy for the case in which the one object is initially at rest and the other has an initial velocity v:

( )42

22

221

2

21

if

mvmvvm

KKK

−=−⎟⎠⎞

⎜⎝⎛=

−=∆

cases.both in same theisenergy kinetic of loss The

(b) Express the percentage loss for the case in which the two objects have oppositely directed velocities of magnitude v/2:

%100241

241

before

==∆

mvmv

KK

Express the percentage loss for the case in which the one object is initially at rest and the other has an initial velocity v:

%50221

241

before

==∆

mvmv

KK

/2. magnitude ofs velocitiedirected oppositely have

objects twohein which t case thefor greatest is loss percentage The

v

*22 •• Determine the Concept A will travel farther. Both peas are acted on by the same force, but pea A is acted on by that force for a longer time. By the impulse-momentum theorem, its momentum (and, hence, speed) will be higher than pea B’s speed on leaving the shooter. 23 •• Determine the Concept Refer to the particles as particle 1 and particle 2. Let the direction particle 1 is moving before the collision be the positive x direction. We’ll use both conservation of momentum and conservation of mechanical energy to obtain an expression for the velocity of particle 2 after the collision. Finally, we’ll examine the ratio of the final kinetic energy of particle 2 to that of particle 1 to determine the condition under which there is maximum energy transfer from particle 1 to particle 2. Use conservation of momentum to obtain one relation for the final velocities:

f2,2f1,1i1,1 vmvmvm += (1)

Use conservation of mechanical energy to set the velocity of

( ) i1,i1,i2,f1,f2, vvvvv =−−=− (2)

Chapter 8

522

recession equal to the negative of the velocity of approach: To eliminate v1,f, solve equation (2) for v1,f, and substitute the result in equation (1):

i1,f2,f1, vvv +=

( ) f2,2i1,f2,1i1,1 vmvvmvm +−=

Solve for v2,f:

i1,21

1f2,

2v

mmm

v+

=

Express the ratio R of K2,f to K1,i in terms of m1 and m2:

( )221

21

1

2

2i,112

1

2i,1

2

21

122

1

i1,

f2,

4

2

mmm

mm

vm

vmm

mm

KK

R

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

Differentiate this ratio with respect to m2, set the derivative equal to zero, and obtain the quadratic equation:

0121

22 =+−

mm

Solve this equation for m2 to determine its value for maximum energy transfer:

12 mm =

.when 2 toed transferrisenergy kinetic s1' of all becausecorrect is )(

12 mmb

=

24 • Determine the Concept In the center-of-mass reference frame the two objects approach with equal but opposite momenta and remain at rest after the collision. 25 • Determine the Concept The water is changing direction when it rounds the corner in the nozzle. Therefore, the nozzle must exert a force on the stream of water to change its direction, and, from Newton’s 3rd law, the water exerts an equal but opposite force on the nozzle. 26 • Determine the Concept The collision usually takes place in such a short period of time that the impulse delivered by gravity or friction is negligible.


523

27 • Determine the Concept No. dtdpF rr

=netext, defines the relationship between the net

force acting on a system and the rate at which its momentum changes. The net external force acting on the pendulum bob is the sum of the force of gravity and the tension in the string and these forces do not add to zero. *28 •• Determine the Concept We can apply conservation of momentum and Newton’s laws of motion to the analysis of these questions. (a) Yes, the car should slow down. An easy way of seeing this is to imagine a "packet" of grain being dumped into the car all at once: This is a completely inelastic collision, with the packet having an initial horizontal velocity of 0. After the collision, it is moving with the same horizontal velocity that the car does, so the car must slow down. (b) When the packet of grain lands in the car, it initially has a horizontal velocity of 0, so it must be accelerated to come to the same speed as the car of the train. Therefore, the train must exert a force on it to accelerate it. By Newton’s 3rd law, the grain exerts an equal but opposite force on the car, slowing it down. In general, this is a frictional force which causes the grain to come to the same speed as the car. (c) No it doesn’t speed up. Imagine a packet of grain being "dumped" out of the railroad car. This can be treated as a collision, too. It has the same horizontal speed as the railroad car when it leaks out, so the train car doesn’t have to speed up or slow down to conserve momentum. *29 •• Determine the Concept Think of the stream of air molecules hitting the sail. Imagine that they bounce off the sail elastically−their net change in momentum is then roughly twice the change in momentum that they experienced going through the fan. Another way of looking at it: Initially, the air is at rest, but after passing through the fan and bouncing off the sail, it is moving backward−therefore, the boat must exert a net force on the air pushing it backward, and there must be a force on the boat pushing it forward. Estimation and Approximation 30 •• Picture the Problem We can estimate the time of collision from the average speed of the car and the distance traveled by the center of the car during the collision. We’ll assume a car length of 6 m. We can calculate the average force exerted by the wall on the car from the car’s change in momentum and it’s stopping time. (a) Relate the stopping time to the assumption that the center of the car travels halfway to the wall with constant deceleration:

( )av

car41

av

car21

21

av

stopping

vL

vL

vd

t ===∆

Chapter 8

524

Express and evaluate vav:

m/s5.122

kmm1000

s3600h1

hkm900

2fi

av

=

××+=

+=

vvv

Substitute for vav and evaluate ∆t: ( )

s120.0m/s12.5m64

1

==∆t

(b) Relate the average force exerted by the wall on the car to the car’s change in momentum:

( )kN417

s0.120km

m1000s3600

h1h

km90kg2000

av =⎟⎟⎠

⎞⎜⎜⎝

⎛××

=∆∆

=tpF

31 •• Picture the Problem Let the direction the railcar is moving be the positive x direction and the system include the earth, the pumpers, and the railcar. We’ll also denote the railcar with the letter c and the pumpers with the letter p. We’ll use conservation of momentum to relate the center of mass frame velocities of the car and the pumpers and then transform to the earth frame of reference to find the time of fall of the car.

(a) Relate the time of fall of the railcar to the distance it falls and its velocity as it leaves the bank:

cvyt ∆

=∆

Use conservation of momentum to find the speed of the car relative to the velocity of its center of mass: 0

or

ppcc

fi

=+

=

umum

pp rr

Relate uc to up and solve for uc:

m/s4

m/s4

cp

pc

−=∴

=−

uu

uu

Substitute for up to obtain: ( ) 0m/s4cpcc =−+ umum


525

Solve for and evaluate uc:

( )

m/s85.1

kg754kg3501

m/s4

1

m/s4

p

cc =

+=

+=

mmu

Relate the speed of the car to its speed relative to the center of mass of the system:

m/s74.10km

m1000s3600

h1h

km32sm.851

cmcc

=

××+=

+= vuv

Substitute and evaluate ∆t: s2.33

m/s10.74m25

==∆t

(b) Find the speed with which the pumpers hit the ground: m/s6.74

m/s4m/s10.74pcp

=

−=−= uvv

injured. bemay theyspeed, at this ground theHitting

*32 •• Picture the Problem The diagram depicts the bullet just before its collision with the melon and the motion of the melon-and-bullet-less-jet and the jet just after the collision. We’ll assume that the bullet stays in the watermelon after the collision and use conservation of momentum to relate the mass of the bullet and its initial velocity to the momenta of the melon jet and the melon less the plug after the collision.

Apply conservation of momentum to the collision to obtain:

( ) 332f1321i1 2 Kmvmmmvm ++−=

Solve for v2f:

132

331i12f

2mmmKmvm

v+−

−=

Express the kinetic energy of the jet of melon in terms of the initial kinetic energy of the bullet:

( ) 21i120

121i12

1101

1101

3 vmvmKK ===

Chapter 8

526

Substitute and simplify to obtain: ( )

( )132

3111i

132

21i120

131i1

2f

1.0

2

mmmmmmv

mmmvmmvm

v

+−−

=

+−−

=

Substitute numerical values and evaluate v2f:

( )( )( )

ft/s1.27

m/s386.0kg0.0104kg0.14kg2.50

kg0.14kg0.01040.1kg0.0104ft3.281

m1sft18002f

−=

−=+−

−⎟⎟⎠

⎞⎜⎜⎝

⎛×=v

Note that this result is in reasonably good agreement with experimental results. Finding the Center of Mass 33 • Picture the Problem We can use its definition to find the center of mass of this system. Apply its definition to find xcm:

( )( ) ( )( ) ( )( ) m233.0kg2kg2kg2

m0.5kg2m0.2kg20kg2

321

332211cm =

++++

=++++

=mmm

xmxmxmx

Because the point masses all lie along the x axis:

0cm =y and the center of mass of this

system of particles is at ( )0,m233.0 .

*34 • Picture the Problem Let the left end of the handle be the origin of our coordinate system. We can disassemble the club-ax, find the center of mass of each piece, and then use these coordinates and the masses of the handle and stone to find the center of mass of the club-ax. Express the center of mass of the handle plus stone system:

stonestick

stonecm,stonestickcm,stickcm mm

xmxmx

+

+=

Assume that the stone is drilled and the stick passes through it. Use symmetry considerations to locate the center of mass of the stick:

cm0.45stickcm, =x


527

Use symmetry considerations to locate the center of mass of the stone:

cm0.89stonecm, =x

Substitute numerical values and evaluate xcm:

( )( ) ( )( )

cm5.78

kg8kg2.5cm89kg8cm54kg2.5

cm

=

++

=x

35 • Picture the Problem We can treat each of balls as though they are point objects and apply the definition of the center of mass to find (xcm, ycm). Use the definition of xcm:

( )( ) ( )( ) ( )( )

m00.2kg1kg1kg3

m3kg1m1kg1m2kg3

cm

=++

++=

++++

=CBA

CCBBAA

mmmxmxmxm

x

Use the definition of ycm:

( )( ) ( )( ) ( )( )

m40.1kg1kg1kg3

0kg1m1kg1m2kg3

cm

=++

++=

++++

=CBA

CCBBAA

mmmymymymy

The center of mass of this system of particles is at:

( )m40.1,m00.2

36 • Picture the Problem The figure shows an equilateral triangle with its y-axis vertex above the x axis. The bisectors of the vertex angles are also shown. We can find x coordinate of the center-of-mass by inspection and the y coordinate using trigonometry. From symmetry considerations:

0cm =x

Chapter 8

528

Express the trigonometric relationship between a/2, 30°, and ycm:

230tan cm

ay

=°

Solve for ycm: aay 289.030tan21

cm =°=

The center of mass of an equilateral

triangle oriented as shown above is at ( )a289.0,0 .

*37 •• Picture the Problem Let the subscript 1 refer to the 3-m by 3-m sheet of plywood before the 2-m by 1-m piece has been cut from it. Let the subscript 2 refer to 2-m by 1-m piece that has been removed and let σ be the area density of the sheet. We can find the center-of-mass of these two regions; treating the missing region as though it had negative mass, and then finding the center-of-mass of the U-shaped region by applying its definition. Express the coordinates of the center of mass of the sheet of plywood:

21

2,cm21cm,1cm mm

xmxmx

−−

=

21

2,cm21cm,1cm mm

ymymy

−−

=

Use symmetry to find xcm,1, ycm,1, xcm,2, and ycm,2:

m0.2,m5.1and

m5.1m,5.1

cm,2cm,2

cm,1cm,1

==

==

yx

yx

Determine m1 and m2:

kgAm

Am

σσ

σσ

2and

kg9

22

11

==

==

Substitute numerical values and evaluate xcm:

( )( ) ( )( )

m50.1kg2kg9

kg5.1kg2m5.1kg9cm

=−−

=σσσσx

Substitute numerical values and evaluate ycm:

( )( ) ( )( )

m36.1kg2kg9

m2kg2m5.1kg9cm

=−−

=σσ

σσy

The center of mass of the U-shaped sheet of plywood is at ( )m1.36m,1.50 .


529

38 •• Picture the Problem We can use its definition to find the center of mass of the can plus water. By setting the derivative of this function equal to zero, we can find the value of x that corresponds to the minimum height of the center of mass of the water as it drains out and then use this extreme value to express the minimum height of the center of mass. (a) Using its definition, express the location of the center of mass of the can + water: mM

xmHMx

+

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

= 22cm

Let the cross-sectional area of the cup be A and use the definition of density to relate the mass m of water remaining in the can at any given time to its depth x:

Axm

AHM

==ρ

Solve for m to obtain: M

Hxm =


⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

⎟⎠⎞

⎜⎝⎛+

=

+

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

=

Hx

Hx

H

MHxM

xMHxHM

x

1

1

2

22

2

cm

(b) Differentiate xcm with respect to x and set the derivative equal to zero for extrema:

021

12

1

21

121

2

21

12

1

21

211

21

21

2cm

=

+

+

−

+

+=

+

++

−

+

++

=+

+=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎪⎪

⎩

⎪⎪

⎨

⎧

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛

Hx

HHx

Hx

HHx

Hx

H

Hx

Hx

dxd

Hx

Hx

Hx

dxd

Hx

H

Hx

Hx

dxdH

dxdx

Simplify this expression to obtain:

0122

=−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

Hx

Hx

Chapter 8

530

Solve for x/H to obtain:

( ) HHx 414.012 ≈−= where we’ve kept the positive solution because a negative value for x/H would make no sense.

Use your graphing calculator to convince yourself that the graph of xcm as a function of x is concave upward at Hx 414.0≈ and that, therefore, the minimum value of xcm occurs at .414.0 Hx ≈ Evaluate xcm at ( )12 −= Hx to obtain:

( )

( )

( )

( )12

121

121

2

2

12cm

−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

=−=

H

HH

HH

HxHx

Finding the Center of Mass by Integration *39 •• Picture the Problem A semicircular disk and a surface element of area dA is shown in the diagram. Because the disk is a continuous object, we’ll use

∫= dmM rrrr

cm and symmetry to find its

center of mass.

Express the coordinates of the center of mass of the semicircular disk:

symmetry.by0cm =x

M

dAyy ∫=

σcm

Express y as a function of r and θ : θsinry =

Express dA in terms of r and θ : drdrdA θ=

Express M as a function of r and θ : 2

21

diskhalf RAM σπσ ==


531

Substitute and evaluate ycm:

RRM

drrMM

drdry

R

R

πσ

σθθσ

π

34

32

2sin

3

0

20 0

2

cm

==

== ∫∫ ∫

40 ••• Picture the Problem Because a solid hemisphere is a continuous object, we’ll use

∫= dmM rrrr

cm to find its center of mass. The volume element for a sphere is

dV = r2 sinθ dθ dφ dr, where θ is the polar angle and φ the azimuthal angle. Let the base of the hemisphere be the xy plane and ρ be the mass density. Then:

θcosrz =

Express the z coordinate of the center of mass:

∫∫=

dV

dVrz

ρ

ρcm

Evaluate ∫= dVM ρ :

( ) 3323

34

21

sphere21

RR

VdVM

πρπρ

ρρ

==

== ∫

Evaluate ∫ dVrρ :

[ ]4

sin2

cossin

42/

02

21

40

2/

0

2

0

3

RR

drddrdVrR

πρθπρ

φθθθρ

π

π π

==

=∫ ∫ ∫ ∫

Substitute and simplify to find zcm:

RRRz 8

33

32

441

cm ==πρπρ

41 ••• Picture the Problem Because a thin hemisphere shell is a continuous object, we’ll use

∫= dmM rrrr

cm to find its center of mass. The element of area on the shell is dA = 2πR2

sinθ dθ, where R is the radius of the hemisphere. Let σ be the surface mass density and express the z coordinate of the center of mass: ∫

∫=dA

dAzz

σ

σcm

Chapter 8

532

Evaluate ∫= dAM σ :

( ) 2221

shellspherical21

24 RR

AdAM

πσπσ

σσ

==

== ∫

Evaluate ∫ dAzσ :

σπ

θθσπ

θθθσπσ

π

π

3

2/

0

3

2/

0

3

2sin

cossin2

R

dR

dRdAz

=

=

=

∫

∫ ∫

Substitute and simplify to find zcm:

RR

Rz 21

2

3

cm 2==

πσσπ

42 ••• Picture the Problem The parabolic sheet is shown to the right. Because the area of the sheet is distributed symmetrically with respect to the y axis, xcm = 0. We’ll integrate the element of area dA (= xdy) to obtain the total area of the sheet and yxdy to obtain the numerator of the definition of the center of mass. Express ycm:

∫

∫= b

b

xdy

xydyy

0

0cm

Evaluate ∫b

xydy0

:

25

0

23

0

21

0

52

1

ba

dyya

ydya

yxydybbb

=

== ∫∫∫

Evaluate ∫b

xdy0

:

23

0

21

0

21

0

32

1

ba

dyya

dya

yxdybbb

=

== ∫∫∫


533

Substitute and simplify to determine ycm:

bb

a

bay 5

3

23

25

cm

32

52

==

Note that, by symmetry:

xcm = 0

The center of mass of the parabolic sheet is at:

( )b53,0

Motion of the Center of Mass 43 • Picture the Problem The velocity of the center of mass of a system of particles is related to the total momentum of the system through cm

iii vvP

rrrMm == ∑ .

Use the expression for the total momentum of a system to relate the velocity of the center of mass of the two-particle system to the momenta of the individual particles:

21

2211iii

cm mmmm

M

m

++

==∑ vv

vv

rrr

r

Substitute numerical values and evaluate cmvr :

( )( ) ( )

( ) ( )[ ]( ) ( ) ji

ji

vvvvv

ˆm/s5.1ˆm/s3

ˆm/s3ˆm/s6

kg6kg3

21

212121

cm

−=

−=

+=+

=rr

rrr

*44 • Picture the Problem Choose a coordinate system in which east is the positive x direction and use the relationship cm

iii vvP

rrrMm == ∑ to determine the velocity of the

center of mass of the system. Use the expression for the total momentum of a system to relate the velocity of the center of mass of the two-vehicle system to the momenta of the individual vehicles:

ct

ccttiii

cm mmmm

M

m

++

==∑ vv

vv

rrr

r

Express the velocity of the truck: ( ) iv ˆm/s16t =r

Chapter 8

534

Express the velocity of the car: ( )iv ˆm/s20c −=r


( )( ) ( )( ) ( ) iiiv ˆm/s00.4

kg1500kg3000

ˆm/s20kg1500ˆm/s16kg3000cm =

+−+

=r

45 • Picture the Problem The acceleration of the center of mass of the ball is related to the net external force through Newton’s 2nd law: cmextnet, aF rr

M= .

Use Newton’s 2nd law to express the acceleration of the ball:

Mextnet,

cm

Fa

rr

=

Substitute numerical values and evaluate cmar :

( ) ( )iia ˆm/s4.2kg1kg1kg3

ˆN12 2cm =

++=

r

46 •• Picture the Problem Choose a coordinate system in which upward is the positive y direction. We can use Newton’s 2nd law cmextnet, aF rr

M= to find the acceleration of the

center of mass of this two-body system.

(a) . is reading

thefall, freein is while;) ( reads scale theinitially Yes;Mg

mgmM +

(b) Using Newton’s 2nd law, express the acceleration of the center of mass of the system:

tot

extnet,cm m

Fa

rr

=


ja ˆcm mM

mg+

−=r

(c) Use Newton’s 2nd law to express the net force acting on the scale while the object of mass m is falling:

( ) cmextnet, )( amMgmMF +−+=

Substitute and simplify to obtain: ( )

Mg

mMmgmMgmMF

=

⎟⎠⎞

⎜⎝⎛

++−+= )(extnet,


535

as expected, given our answer to part (a).

*47 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a) and the work-energy theorem in conjunction with Newton’s 2nd law in parts (b) and (c).

(a) Apply ∑ = yy maF to the

spring when it is compressed a distance d:

∑ =−−= 0springonballpn FgmFFy

Solve for Fn:

( )gmmgmgm

kgmkgmkdgm

FgmF

bpbp

bpp

springonballpn

+=+=

⎟⎠⎞

⎜⎝⎛+=+=

+=

(b) Use conservation of mechanical energy, with Ug = 0 at the position at which the spring is fully compressed, to relate the gravitational potential energy of the system to the energy stored in the fully compressed spring:

0sg =∆+∆+∆ UUK

Because ∆K = Ug,f = Us,i = 0, 0fs,ig, =−UU

or 02

21

b =− kdgdm

Solve for d: k

gmd b2

=

Evaluate our force equation in (a)

with k

gmd b2

= :

( )gmmgmgm

kgmkgmkdgm

FgmF

bpbp

bpp

springonballpn

22

2

+=+=

⎟⎠⎞

⎜⎝⎛+=+=

+=

Chapter 8

536

(c) When the ball is in its original position, the spring is relaxed and exerts no force on the ball. Therefore:

gm

F

p

n readingscale

=

=

*48 •• Picture the Problem Assume that the object whose mass is m1 is moving downward and take that direction to be the positive direction. We’ll use Newton’s 2nd law for a system of particles to relate the acceleration of the center of mass to the acceleration of the individual particles. (a) Relate the acceleration of the center of mass to m1, m2, mc and their accelerations:

cc2211cm aaaa rrrr mmmM ++=

Because m1 and m2 have a common acceleration a and ac = 0:

c21

21cm mmm

mmaa

++−

=

From Problem 4-81 we have:

21

21

mmmm

ga+−

=


( )( )( ) g

mmmmmmm

mmmmmg

mmmma

c2121

221

c21

21

21

21cm

+++−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

(b) Use Newton’s 2nd law for a system of particles to obtain:

cmMaMgF −=−

where M = m1 + m2 + mc and F is positive upwards.

Solve for F and substitute for acm from part (a):

( )

gmmmmm

gmm

mmMg

MaMgF

⎥⎦

⎤⎢⎣

⎡+

+=

+−

−=

−=

c21

21

21

221

cm

4

(c) From Problem 4-81: g

mmmmT

21

212+

=


537

Substitute in our result from part (b) to obtain:

gmTgmgT

gmmmmmF

cc

c21

21

22

22

+=⎥⎦

⎤⎢⎣

⎡+=

⎥⎦

⎤⎢⎣

⎡+

+=

49 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a) and the result of Problem 7-96 part (b) to obtain the scale reading when the ball is dropped from a height h above the cup.

(a) Apply ∑ = yy maF to the spring

when it is compressed a distance d:

∑ =−−= 0springonballpn FgmFFy

Solve for Fn:

( )gmmgmgm

kgmkgmkdgm

FgmF

bpbp

bpp

springonballpn

+=+=

⎟⎠⎞

⎜⎝⎛+=+=

+=

(b) From Problem 7-96, part (b):

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

gmkh

kgm

xb

bmax

211

From part (a):

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

+=+=

gmkhgmgm

kxgmFgmF

bbp

maxpspringonballpn

211

The Conservation of Momentum 50 • Picture the Problem Let the system include the woman, the canoe, and the earth. Then the net external force is zero and linear momentum is conserved as she jumps off the

Chapter 8

538

canoe. Let the direction she jumps be the positive x direction. Apply conservation of momentum to the system:

0canoecanoegirlgirlii =+=∑ vvv rrr mmm

Substitute to obtain: ( )( ) ( ) 0kg57ˆm/s5.2kg55 canoe =+ vi r

Solve for canoevr : ( ) iv ˆm/s83.1canoe −=

r

51 •

Picture the Problem If we include the earth in our system, then the net external force is zero and linear momentum is conserved as the spring delivers its energy to the two objects.

Apply conservation of momentum to the system:

0101055ii =+=∑ vvv rrr mmm


( )( ) ( ) 0kg10ˆm/s8kg5 10 =+− vi r

Solve for 10vr : ( ) iv ˆm/s410 =r

*52 • Picture the Problem This is an explosion-like event in which linear momentum is conserved. Thus we can equate the initial and final momenta in the x direction and the initial and final momenta in the y direction. Choose a coordinate system in the positive x direction is to the right and the positive y direction is upward. Equate the momenta in the y direction before and after the explosion:

( ) 022

2

11

12fy,iy,

=−=

−== ∑∑mvvm

mvmvpp

We can conclude that the momentum was

entirely in the x direction before the particle exploded.

Equate the momenta in the x direction before and after the explosion:

3i

fx,ix,

4 mvmvpp

=∴

=∑ ∑

Solve for v3: 34

1i vv = and correct. is )(c


539

53 • Picture the Problem Choose the direction the shell is moving just before the explosion to be the positive x direction and apply conservation of momentum. Use conservation of momentum to relate the masses of the fragments to their velocities:

fi pprr

=

or 'ˆˆ

21

21 vji

rmmvmv +=

Solve for 'vr : jiv ˆˆ2' vv −=r

*54 •• Picture the Problem Let the system include the earth and the platform, gun and block. Then extnet,F

r= 0 and momentum is conserved within the system.

(a) Apply conservation of momentum to the system just before and just after the bullet leaves the gun:

platformbullet

afterbefore

0or

pp

pp

rr

rr

+=

=

Substitute for platformbullet and pp

rrand

solve for platformvr : platformpbb

ˆ0 vi rmvm +=

and

iv ˆb

p

bplatform v

mm

−=r

(b) Apply conservation of momentum to the system just before the bullet leaves the gun and just after it comes to rest in the block:

afterbefore pp rr=

or platform0 pr= ⇒ 0platform =vr

(c) Express the distance ∆s traveled by the platform:

tvs ∆=∆ platform

Express the velocity of the bullet relative to the platform:

bp

bpb

p

b

bp

bbplatformbrel

1 vm

mmv

mm

vmmvvvv

+=⎟

⎟⎠

⎞⎜⎜⎝

⎛+=

+=−=

Relate the time of flight ∆t to L and vrel: relv

Lt =∆

Chapter 8

540

Substitute to find the distance ∆s moved by the platform in time ∆t:

Lmm

m

vm

mmLv

mm

vLv

mmtvs

bp

b

bp

bpb

p

b

relb

p

bplatform

+=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=∆=∆

55 ••

Picture the Problem The pictorial representation shows the wedge and small object, initially at rest, to the left, and, to the right, both in motion as the small object leaves the wedge. Choose the direction the small object is moving when it leaves the wedge be the positive x direction and the zero of potential energy to be at the surface of the table. Let the speed of the small object be v and that of the wedge V. We can use conservation of momentum to express v in terms of V and conservation of energy to express v in terms of h.

Apply conservation of momentum to the small object and the wedge:

Vi

pp

r

rr

mmv

xx

2ˆ0

orf,i,

+=

=

Solve for :V

r iV ˆ

21 v−=

r (1)

and vV 2

1=

Use conservation of energy to determine the speed of the small object when it exits the wedge: 0

or0

ifif =−+−

=∆+∆

UUKK

UK

Because Uf = Ki = 0: ( ) 02 2

212

21 =−+ mghVmmv


541

Substitute for V to obtain: ( )( ) 02 221

212

21 =−+ mghvmmv

Solve for v to obtain: 3

2 ghv =

Substitute in equation (1) to determine V

r: iiV ˆ

3ˆ

322

1 ghgh−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

r

i.e., the wedge moves in the direction

opposite to that of the small object with a

speed of3gh

.

*56 •• Picture the Problem Because no external forces act on either cart, the center of mass of the two-cart system can’t move. We can use the data concerning the masses and separation of the gliders initially to calculate its location and then apply the definition of the center of mass a second time to relate the positions X1 and X2 of the centers of the carts when they first touch. We can also use the separation of the centers of the gliders when they touch to obtain a second equation in X1 and X2 that we can solve simultaneously with the equation obtained from the location of the center of mass. (a) Apply its definition to find the center of mass of the 2-glider system:

( )( ) ( )( )

m10.1kg0.2kg0.1

m1.6kg0.2m0.1kg0.121

2211cm

=++

=

++

=mm

xmxmx

from the left end of the air track.

Use the definition of the center of mass to relate the coordinates of the centers of the two gliders when they first touch to the location of the center of mass:

( ) ( )

232

131

21

21

2211

kg0.2kg0.1kg0.2kg0.1

m10.1

XX

XXmm

XmXm

+=++

=

++

=

Also, when they first touch, their centers are separated by half their combined lengths:

( ) m0.15cm20cm1021

12 =+=− XX

Thus we have:

m10.1667.0333.0 21 =+ XX and

m0.1512 =− XX

Chapter 8

542

Solve these equations simultaneously to obtain:

m00.11 =X and m15.12 =X

(b)

collision. after thezero bemust it so zero, is system theof momentum initial The No.

Kinetic Energy of a System of Particles *57 • Picture the Problem Choose a coordinate system in which the positive x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass. (a) Find the sum of the kinetic energies:

( )( ) ( )( )J43.5

m/s2kg3m/s5kg3 2212

21

2222

12112

1

21

=

+=

+=

+=

vmvm

KKK

(b) Relate the velocity of the center of mass of the system to its total momentum:

2211cm vvv rrr mmM +=

Solve for :cmvr

21

2211cm mm

mm++

=vvvrr

r

Substitute numerical values and evaluate :cmvr

( )( ) ( )( )

( )i

iiv

ˆm/s50.1

kg3kg3

ˆm/s2kg3ˆm/s5kg3cm

=

+−

=r

(c) The velocity of an object relative to the center of mass is given by:

cmrel vvv rrr−=


543


( ) ( )( )

( ) ( )( )i

iiv

i

iiv

ˆm/s50.3

ˆm/s5.1ˆm/s2

ˆm/s50.3

ˆm/s5.1ˆm/s5

rel2,

rel1,

−=

−−=

=

−=

r

r

(d) Express the sum of the kinetic energies relative to the center of mass:

2rel,222

12rel,112

1rel,2rel,1rel vmvmKKK +=+=

Substitute numerical values and evaluate Krel:

( )( )( )( )

J75.63

m/s5.3kg3

m/s3.5kg32

21

221

rel

=

−+

=K

(e) Find Kcm: ( )( )

rel

2212

cmtot21

cm

J36.75J43.5J6.75

m/s1.5kg6

KK

vmK

−=

−==

==

58 •

Picture the Problem Choose a coordinate system in which the positive x direction is to the right. Use the expression for the total momentum of a system to find the velocity of the center of mass and the definition of relative velocity to express the sum of the kinetic energies relative to the center of mass. (a) Express the sum of the kinetic energies:

2222

12112

121 vmvmKKK +=+=


( )( ) ( )( )J0.06

m/s3kg5m/s5kg3 2212

21

=

+=K

(b) Relate the velocity of the center of mass of the system to its total momentum:

2211cm vvv rrr mmM +=

Solve for cmvr :

21

2211cm mm

mm++

=vvvrr

r

Chapter 8

544


( )( ) ( )( )

( )i

iiv

ˆm/s75.3

kg5kg3


=

++

=r

(c) The velocity of an object relative to the center of mass is given by:

cmrel vvv rrr−=

Substitute numerical values and evaluate the relative velocities:

( ) ( )( )i

iiv

ˆm/s25.1

ˆm/s75.3ˆm/s5rel1,

=

−=r

and ( ) ( )

( )iiiv

ˆm/s750.0

ˆm/s75.3ˆm/s3rel2,

−=

−=r

(d) Express the sum of the kinetic energies relative to the center of mass:

2rel,222

12rel,112

1

rel,2rel,1rel

vmvm

KKK

+=

+=

Substitute numerical values and evaluate Krel:

( )( )( )( )

J75.3

m/s75.0kg5

m/s25.1kg32

21

221

rel

=

−+

=K

(e) Find Kcm: ( )( )

rel

2212

cmtot21

cm

J3.65

m/s75.3kg8

KK

vmK

−==

==

Impulse and Average Force 59 • Picture the Problem The impulse imparted to the ball by the kicker equals the change in the ball’s momentum. The impulse is also the product of the average force exerted on the ball by the kicker and the time during which the average force acts. (a) Relate the impulse delivered to the ball to its change in momentum: 0since if

if

==−=∆=

vmvpppI

Substitute numerical values and evaluate I:

( )( ) sN10.8m/s25kg0.43 ⋅==I


545

(b) Express the impulse delivered to the ball as a function of the average force acting on it and solve for and evaluate avF :

tFI ∆= av

and

kN1.34s0.008sN10.8

av =⋅

=∆

=t

IF

60 • Picture the Problem The impulse exerted by the ground on the brick equals the change in momentum of the brick and is also the product of the average force exerted by the ground on the brick and the time during which the average force acts. (a) Express the impulse exerted by the ground on the brick:

bricki,brickf,brick pppI −=∆=

Because pf,brick = 0: vmpI brickbricki, == (1)

Use conservation of energy to determine the speed of the brick at impact: 0

or0

ifif =−+−

=∆+∆

UUKK

UK

Because Uf = Ki = 0:

0or

0

brick2

brick21

if

=−

=−

ghmvm

UK

Solve for v: ghv 2=

Substitute in equation (1) to obtain: ghmI 2brick=


( ) ( )( )sN76.3

m8m/s9.812kg0.3 2

⋅=

=I

(c) Express the impulse delivered to the brick as a function of the average force acting on it and solve for and evaluate avF :

tFI ∆= av

and

kN89.2s0.0013sN76.3

av =⋅

=∆

=t

IF

*61 • Picture the Problem The impulse exerted by the ground on the meteorite equals the change in momentum of the meteorite and is also the product of the average force exerted by the ground on the meteorite and the time during which the average force acts.

Chapter 8

546

Express the impulse exerted by the ground on the meteorite:

ifmeteorite pppI −=∆=

Relate the kinetic energy of the meteorite to its initial momentum and solve for its initial momentum:

ii

2i

i 22

mKpm

pK =⇒=

Express the ratio of the initial and final kinetic energies of the meteorite:

2

2m

22f

2i

2f

2i

f

i ===pp

pm

p

KK

Solve for pf:

2i

fp

p =

Substitute in our expression for I and simplify:

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=−=

12

12

12

12

i

iii

mK

pppI

Because our interest is in its magnitude, evaluate I :

( )( ) sMN81.112

1J10617kg1030.82 63 ⋅=⎟⎠⎞

⎜⎝⎛ −××=I

Express the impulse delivered to the meteorite as a function of the average force acting on it and solve for and evaluate avF :

tFI ∆= av

and

MN602.0s3

sMN81.1av =

⋅=

∆=

tIF

62 •• Picture the Problem The impulse exerted by the bat on the ball equals the change in momentum of the ball and is also the product of the average force exerted by the bat on the ball and the time during which the bat and ball were in contact. (a) Express the impulse exerted by the bat on the ball in terms of the change in momentum of the ball:

( ) iii

pppIˆ2ˆˆ

if

ifball

mvmvmv =−−=

−=∆=rrrr

where v = vf = vi


547

Substitute for m and v and evaluate I:

( )( ) sN00.6m/s20kg15.02 ⋅==I

(b) Express the impulse delivered to the ball as a function of the average force acting on it and solve for and evaluate avF :

tFI ∆= av

and

kN62.4ms3.1

sN00.6av =

⋅=

∆=

tIF

*63 •• Picture the Problem The figure shows the handball just before and immediately after its collision with the wall. Choose a coordinate system in which the positive x direction is to the right. The wall changes the momentum of the ball by exerting a force on it during the ball’s collision with it. The reaction to this force is the force the ball exerts on the wall. Because these action and reaction forces are equal in magnitude, we can find the average force exerted on the ball by finding the change in momentum of the ball.

Using Newton’s 3rd law, relate the average force exerted by the ball on the wall to the average force exerted by the wall on the ball:

ballon avon wall av FFrr

−=

and ballon avon wall av FF = (1)

Relate the average force exerted by the wall on the ball to its change in momentum:

tm

t ∆∆

=∆∆

=vpFrrr

ballon av

Express xvr∆ for the ball: iiv ˆˆ,i,f xxx vv −=∆

r

or, because vi,x = vcosθ and vf,x = −vcosθ, iiiv ˆcos2ˆcosˆcos θθθ vvvx −=−−=∆

r

Substitute in our expression for

ballon avFr

: ivF ˆcos2

ballon av tmv

tm

∆−=

∆∆

=θ

rr

Chapter 8

548

Evaluate the magnitude of ballon avFr

:

( )( )

N230ms2

cos40m/s5kg0.062

cos2ballon av

=

°=

∆=

tmvF θ

Substitute in equation (1) to obtain: N230on wall av =F

64 •• Picture the Problem The pictorial representation shows the ball during the interval of time you are exerting a force on it to accelerate it upward. The average force you exert can be determined from the change in momentum of the ball. The change in the velocity of the ball can be found by applying conservation of mechanical energy to its rise in the air once it has left your hand. (a) Relate the average force exerted by your hand on the ball to the change in momentum of the ball:

tmv

tpp

tpF

∆=

∆−

=∆∆

= 212av

because v1 and, hence, p1 = 0.

Letting Ug = 0 at the initial elevation of your hand, use conservation of mechanical energy to relate the initial kinetic energy of the ball to its potential energy when it is at its highest point:

0since0

or0

if

fi

===+−

=∆+∆

UKUK

UK

Substitute for Kf and Ui and solve for v2:

ghv

mghmv

2

and0

2

222

1

=

=+−

Relate ∆t to the average speed of the ball while you are throwing it upward:

22av

2

2vd

vd

vdt ===∆


549

Substitute for ∆t and v2 in the expression for Fav to obtain: d

mghF =av

Substitute numerical values and evaluate Fav:

( )( )( )

N1.84

m0.7m40m/s9.81kg0.15 2

av

=

=F

(b) Express the ratio of the weight of the ball to the average force acting on it:

( )( ) %2N84.1

m/s9.81kg0.15 2

avav

<==Fmg

Fw

weight.its neglected have toreasonable isit ball, on theexerted force average theof 2% than less is ball theof weight theBecause

65 •• Picture the Problem Choose a coordinate system in which the direction the ball is moving after its collision with the wall is the positive x direction. The impulse delivered to the wall or received by the player equals the change in the momentum of the ball. We can find the average forces from the rate of change in the momentum of the ball. (a) Relate the impulse delivered to the wall to the change in momentum of the handball:

( )( )( )( )[ ]

( ) wall.into directed ˆsN08.1

ˆm/s01kg0.06

ˆm/s8kg0.06if

i

i

i

vvpI

⋅=

−−

=

−=∆=rrrr

mm

(b) Find Fav from the change in the ball’s momentum:

wall.into N,603

s0.003sN08.1

av

=

⋅=

∆∆

=tpF

(c) Find the impulse received by the player from the change in momentum of the ball:

( )( ) wall.fromaway s,N480.0

m/s8kg0.06ball

⋅=

=∆=∆= vmpI

(d) Relate Fav to the change in the ball’s momentum: t

pF

∆∆

= ballav

Express the stopping time in terms of the average speed vav of the ball avv

dt =∆

Chapter 8

550

and its stopping distance d:

Substitute to obtain: dpvF ballav

av∆

=


( )( )

wall.fromaway N,84.3

m0.5sN480.0m/s4

av

=

⋅=F

66 ••• Picture the Problem The average force exerted on the limestone by the droplets of water equals the rate at which momentum is being delivered to the floor. We’re given the number of droplets that arrive per minute and can use conservation of mechanical energy to determine their velocity as they reach the floor. (a) Letting N represent the rate at which droplets fall, relate Fav to the change in the droplet’s momentum:

tvmN

tp

F∆∆

=∆

∆= droplets

av

Find the mass of the droplets: ( )( )kg103

mL0.03kg/L15−×=

== Vm ρ

Letting Ug = 0 at the point of impact of the droplets, use conservation of mechanical energy to relate their speed at impact to their fall distance:

0or

0

ifif =−+−

=∆+∆

UUKK

UK

Because Ki = Uf = 0: 02f2

1 =− mghmv

Solve for and evaluate v = vf: ( )( )

m/s9.90m5m/s9.8122 2

=

== ghv


( )( )N1095.4

m/s9.90kg103

s60min1

mindroplets10

5

5

av

−

−

×=

××

⎟⎟⎠

⎞⎜⎜⎝

⎛×=

∆⎟⎠⎞

⎜⎝⎛

∆= vm

tNF


551

(b) Calculate the ratio of the weight of a droplet to Fav:

( )( ) 6N104.95m/s9.81kg103

5

25avav

≈×

×=

=

−

−

Fmg

Fw

Collisions in One Dimension *67 • Picture the Problem We can apply conservation of momentum to this perfectly inelastic collision to find the after-collision speed of the two cars. The ratio of the transformed kinetic energy to kinetic energy before the collision is the fraction of kinetic energy lost in the collision. (a) Letting V be the velocity of the two cars after their collision, apply conservation of momentum to their perfectly inelastic collision:

( )Vmmmvmv

pp

+=+

=

21

finalinitial

or

Solve for and evaluate V:

m/s0.202

m/s10m/s302

21

=

+=

+=

vvV

(b) Express the ratio of the kinetic energy that is lost to the kinetic energy of the two cars before the collision and simplify:

( )

12

12

1

22

21

2

222

1212

1

221

initial

final

initial

initialfinal

initial

−+

=

−+

=

−=

−=

∆

vvV

mvmvVm

KK

KKK

KK

Substitute numerical values to obtain: ( )

( ) ( )200.0

1m/s10m/s30

m/s20222

2

initial

−=

−+

=∆

KK

metal. ofn deformatio the and sound, heat, into ed transformisenergy kinetic initial theof 20%

Chapter 8

552

68 • Picture the Problem We can apply conservation of momentum to this perfectly inelastic collision to find the after-collision speed of the two players. Letting the subscript 1 refer to the running back and the subscript 2 refer to the linebacker, apply conservation of momentum to their perfectly inelastic collision:

( )Vmmvm

pp

2111

fi

or+=

=

Solve for V: 1

21

1 vmm

mV+

=

Substitute numerical values and evaluate V:

( ) m/s13.3m/s7kg051kg58

kg58=

+=V

69 • Picture the Problem We can apply conservation of momentum to this collision to find the after-collision speed of the 5-kg object. Let the direction the 5-kg object is moving before the collision be the positive direction. We can decide whether the collision was elastic by examining the initial and final kinetic energies of the system. (a) Letting the subscript 5 refer to the 5-kg object and the subscript 2 refer to the 10-kg object, apply conservation of momentum to obtain:

f,55i,10105i,5

fi

orvmvmvm

pp

=−

=

Solve for vf,5:

5

i,10105i,5f,5 m

vmvmv

−=

Substitute numerical values and evaluate vf,5:

( )( ) ( )( )

m/s00.2

kg5m/s3kg10m/s4kg5

f,5

−=

−=v

where the minus sign means that the 5-kg object is moving to the left after the collision.


553

(b) Evaluate ∆K for the collision:

( )( ) ( )( )[ ( )( ) ] J0.75m/s3kg10m/s4kg5m/s2kg5 2212

212

21

if −=+−=−=∆ KKK

inelastic. wascollision the0, K Because ≠∆

70 • Picture the Problem The pictorial representation shows the ball and bat just before and just after their collision. Take the direction the bat is moving to be the positive direction. Because the collision is elastic, we can equate the speeds of recession and approach, with the approximation that vi,bat ≈ vf,bat to find vf,ball.

Express the speed of approach of the bat and ball:

( )balli,bati,ballf,batf, vvvv −−=−

Because the mass of the bat is much greater than that of the ball:

batf,bati, vv ≈


( )balli,batf,ballf,batf, vvvv −−=−

Solve for and evaluate vf,ball: ( )

v

vvvvvvvv

3

22 batf,balli,

balli,batf,batf,ballf,

=

+=+−=

−+=

*71 •• Picture the Problem Let the direction the proton is moving before the collision be the positive x direction. We can use both conservation of momentum and conservation of mechanical energy to obtain an expression for velocity of the proton after the collision. (a) Use the expression for the total momentum of a system to find vcm:

( )

( )i

iv

v

vvP

ˆm/s1.23

ˆm/s30012

and

131ip,

cm

cm

=

=+

=

== ∑

mmm

Mmi

ii

rr

rrr

Chapter 8

554

(b) Use conservation of momentum to obtain one relation for the final velocities:

fnuc,nucfp,pip,p vmvmvm += (1)

Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach:

( ) ip,ip,inuc,fp,fnuc, vvvvv =−−=− (2)

To eliminate vnuc,f, solve equation (2) for vnuc,f, and substitute the result in equation (1):

fp,ip,fnuc, vvv +=

( )fp,ip,nucfp,pip,p vvmvmvm ++=

Solve for and evaluate vp,f:

( ) m/s254m/s30013

12

ip,nucp

nucpfp,

−=−

=

+−

=

mmm

vmmmm

v

72 •• Picture the Problem We can use conservation of momentum and the definition of an elastic collision to obtain two equations in v2f and v3f that we can solve simultaneously. Use conservation of momentum to obtain one relation for the final velocities:

2f23f33i3 vmvmvm += (1)


( ) 3i3i2i3f2f vvvvv =−−=− (2)

Solve equation (2) for v3f , substitute in equation (1) to eliminate v3f, and solve for and evaluate v2f:

( )( )

m/s80.4

kg3kg2m/s4kg322

32

3i32f

=

+=

+=

mmvmv

Use equation (2) to find v3f:

m/s0.800

m/s4.00m/s4.803i2f3f

=

−=−= vvv

Evaluate Ki and Kf: ( )( )

J24.0m/s4kg3 2

212

3i321

3ii

==== vmKK


555

and

( )( )( )( )J0.24

m/s4.8kg2

m/s0.8kg32

21

221

22f22

123f32

12f3ff

=+

=

+=+= vmvmKKK

elastic.been havingcollision with theconsistent are and for obtained values that theconcludecan we, Because 3f2ffi vvKK =

73 •• Picture the Problem We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the maximum compression of the spring and express the initial (i.e., before collision) and final (i.e., at separation) velocities. Finally, we’ll transform the velocities from the center of mass frame of reference to the table frame of reference. (a) Use the definition of the total momentum of a system to relate the initial momenta to the velocity of the center of mass:

cmvvPrrr

Mmi

ii == ∑

or ( ) cm211i1 vmmvm +=

Solve for vcm:

21

2i21i1cm mm

vmvmv++

=

Substitute numerical values and evaluate vcm:

( )( ) ( )( )

m/s00.5

kg5kg2m/s3kg5m/s10kg2

cm

=

++

=v

(b) Find the kinetic energy of the system at maximum compression (u1 = u2 = 0):

( )( ) J87.5m/s5kg7 221

2cm2

1cm

==

== MvKK

Use conservation of energy to relate the kinetic energy of the system to the potential energy stored in the spring at maximum compression:

0s =∆+∆ UK

or 0sisfif =−+− UUKK

Because Kf = Kcm and Usi = 0: ( ) 0221

icm =∆+− xkKK

Chapter 8

556

Solve for ∆x: ( )

[ ]

kKvmvm

kKvmvm

kKKx

cm2i22

2i11

cm2i222

12i112

1

cmi

2

2

2

−+=

−+=

−=∆


( )( ) ( )( ) ( ) m250.0N/m1120

J87.52N/m1120

m/s3kg5m/s10kg2 22

=⎥⎦

⎤−

+=∆x

(c) Find u1i, u2i, and u1f for this elastic collision:

m/s5m/s50and

m/s,2m/s5m/s3m/s,5m/s5m/s10

cm1f1f

cm2i2i

cm1i1i

−=−=−=

−=−=−==−=−=

vvu

vvuvvu

Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach and solve for u2f:

( )1i2i1f2f uuuu −−=−

and

( )m/s2

m/s5m/s5m/s21f1i2i2f

=−+−−=

++−= uuuu

Transform u1f and u2f to the table frame of reference:

0m/s5m/s5cm1f1f =+−=+= vuv

and

m/s00.7m/s5m/s2cm2f2f

=+=

+= vuv

*74 •• Picture the Problem Let the system include the earth, the bullet, and the sheet of plywood. Then Wext = 0. Choose the zero of gravitational potential energy to be where the bullet enters the plywood. We can apply both conservation of energy and conservation of momentum to obtain the various physical quantities called for in this problem. (a) Use conservation of mechanical energy after the bullet exits the sheet of plywood to relate its exit speed to the height to which it rises:

0=∆+∆ UK or, because Kf = Ui = 0,

0221 =+− mghmvm


557

Solve for vm: ghvm 2=

Proceed similarly to relate the initial velocity of the plywood to the height to which it rises:

gHvM 2=

(b) Apply conservation of momentum to the collision of the bullet and the sheet of plywood:

fi pp rr=

or Mmm Mvmvmv +=i

Substitute for vm and vM and solve for vmi:

gHmMghvm 22i +=

(c) Express the initial mechanical energy of the system (i.e., just before the collision):

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛++=

=

HmMhH

mMhmg

mvE m

2

2i2

1i

2

Express the final mechanical energy of the system (i.e., when the bullet and block have reached their maximum heights):

( )MHmhgMgHmghE +=+=f

(d) Use the work-energy theorem with Wext = 0 to find the energy dissipated by friction in the inelastic collision:

0frictionif =+− WEE

and

⎥⎦

⎤⎢⎣

⎡−+=

−=

12

fifriction

mM

HhgMH

EEW

75 •• Picture the Problem We can find the velocity of the center of mass from the definition of the total momentum of the system. We’ll use conservation of energy to find the speeds of the particles when their separation is least and when they are far apart. (a) Noting that when the distance between the two particles is least, both move at the same speed, namely vcm, use the definition of the total momentum of a system to relate the initial momenta to the velocity of

cmvvPrrr

Mmi

ii == ∑

or ( ) cmppip vmmvm α+= .

Chapter 8

558

the center of mass: Solve for and evaluate vcm:

0

0

21

ipipcm

200.0

40'

v

mmmv

mmvmvm

vv

=

++

=++

== αα

(b) Use conservation of momentum to obtain one relation for the final velocities:

fpfp0p ααvmvmvm += (1)


( ) piipifpf vvvvv −=−−=− αα (2)

Solve equation (2) for vpf , substitute in equation (1) to eliminate vpf, and solve for vαf:

00

p

0pf 400.0

422

vmm

mvmmvm

v =+

=+

=α

α

76 • Picture the Problem Let the numeral 1 denote the electron and the numeral 2 the hydrogen atom. We can find the final velocity of the electron and, hence, the fraction of its initial kinetic energy that is transferred to the atom, by transforming to the center-of-mass reference frame, calculating the post-collision velocity of the electron, and then transforming back to the laboratory frame of reference. Express f, the fraction of the electron’s initial kinetic energy that is transferred to the atom: 2

1i

f12i112

1

2f112

1

i

f

i

fi

11

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

−=−

=

vv

vmvm

KK

KKKf

(1)

Find the velocity of the center of mass:

21

i11cm mm

vmv+

=

or, because m2 = 1840m1,

1i11

i11cm 1841

11840

vmm

vmv =+

=

Find the initial velocity of the electron in the center-of-mass reference frame:

1i

1i1icm1i1i

184111

18411

v

vvvvu

⎟⎠⎞

⎜⎝⎛ −=

−=−=


559

Find the post-collision velocity of the electron in the center-of-mass reference frame by reversing its velocity:

1i1i1f 11841

1 vuu ⎟⎠⎞

⎜⎝⎛ −=−=

To find the final velocity of the electron in the original frame, add vcm to its final velocity in the center-of-mass reference frame:

1icm1f1f 11841

2 vvuv ⎟⎠⎞

⎜⎝⎛ −=+=


%217.01017.2

11841

211

18412

1

3

2

2

1i

1i

=×=

⎟⎠⎞

⎜⎝⎛ −−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛ −

−=

−

v

vf

77 •• Picture the Problem The pictorial representation shows the bullet about to imbed itself in the bob of the ballistic pendulum and then, later, when the bob plus bullet have risen to their maximum height. We can use conservation of momentum during the collision to relate the speed of the bullet to the initial speed of the bob plus bullet (V). The initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy when they reach their maximum height. Hence we apply conservation of mechanical energy to relate V to the angle through which the bullet plus bob swings and then solve the momentum and energy equations simultaneously for the speed of the bullet.

Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob plus bullet:

( )VMmmv +=b

Solve for the speed of the bullet:

VmMv ⎟

⎠⎞

⎜⎝⎛ += 1b (1)

Use conservation of energy to relate 0=∆+∆ UK

Chapter 8

560

the initial kinetic energy of the bullet to the final potential energy of the system:

or, because Kf = Ui = 0, 0fi =+− UK

Substitute for Ki and Uf and solve for V:

( )( ) ( ) 0cos1

221

=−+++−

θgLMmVMm

and ( )θcos12 −= gLV

Substitute for V in equation (1) to obtain:

( )θcos121b −⎟⎠⎞

⎜⎝⎛ += gL

mMv

Substitute numerical values and evaluate vb:

( )( )( ) m/s450cos601m2.3m/s9.812kg0.016

kg1.51 2b =°−⎟⎟

⎠

⎞⎜⎜⎝

⎛+=v

*78 •• Picture the Problem We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the colliding objects that we can solve for v1f and v2f. Apply conservation of momentum to the elastic collision of the particles to obtain:

2i21i1f22f11 vmvmvmvm +=+ (1)

Relate the initial and final kinetic energies of the particles in an elastic collision:

2i222

12i112

12f222

12f112

1 vmvmvmvm +=+

Rearrange this equation and factor to obtain:

( ) ( )2f1

2i11

2i2

2f22 vvmvvm −=−

or ( )( )

( )( )1fi11fi11

2if22if22

vvvvmvvvvm

+−=+−

(2)

Rearrange equation (1) to obtain:

( ) ( )1f1i12i2f2 vvmvvm −=− (3)

Divide equation (2) by equation (3) to obtain:

1fi12if2 vvvv +=+

Rearrange this equation to obtain equation (4):

1ii2f2f1 vvvv −=− (4)

Multiply equation (4) by m2 and add it to equation (1) to obtain:

( ) ( ) 2i21i211f21 2 vmvmmvmm +−=+


561

Solve for v1f to obtain: iif v

mmmv

mmmmv 2

21

21

21

211

2+

++−

=

Multiply equation (4) by m1 and subtract it from equation (1) to obtain:

( ) ( ) 1i1i212f221 2 vmvmmvmm +−=+

Solve for v2f to obtain: i2

21

12i1

21

1f2

2 vmmmmv

mmmv

+−

++

=

Remarks: Note that the velocities satisfy the condition that ( )1i2i1f2f vvvv −−=− . This verifies that the speed of recession equals the speed of approach. 79 •• Picture the Problem As in this problem, Problem 78 involves an elastic, one-dimensional collision between two objects. Both solutions involve using the conservation of momentum equation 2i21i1f22f11 vmvmvmvm +=+ and the elastic collision equation 1ii2f2f1 vvvv −=− . In part (a) we can simply set the masses equal to each other and substitute in the equations in Problem 78 to show that the particles "swap" velocities. In part (b) we can divide the numerator and denominator of the equations in Problem 78 by m2 and use the condition that m2 >> m1 to show that v1f ≈ −v1i+2v2i and v2f ≈ v2i. (a) From Problem 78 we have: 2i

21

2i1

21

21f1

2 vmm

mvmmmmv

++

+−

= (1)

and

2i21

121i

21

12f

2 vmmmmv

mmmv

+−

++

= (2)

Set m1 = m2 = m to obtain:

i2i2f12 vv

mmmv =+

=

and

1i1if22 vv

mmmv =+

=

(b) Divide the numerator and denominator of both terms in equation (1) by m2 to obtain:

2i

2

1i1

2

1

2

1

f1

1

2

1

1v

mmv

mmmm

v+

++

−=

If m2 >> m1:

2ii1f1 2vvv +−≈

Chapter 8

562

Divide the numerator and denominator of both terms in equation (2) by m2 to obtain:

2i

2

1

2

1

1i

2

1

2

1

2f

1

1

1

2v

mm

mm

v

mm

mm

v+

−+

+=

If m2 >> m1:

2i2f vv ≈

Remarks: Note that, in both parts of this problem, the velocities satisfy the condition that ( )1i2i1f2f vvvv −−=− . This verifies that the speed of recession equals the speed of approach. Perfectly Inelastic Collisions and the Ballistic Pendulum 80 •• Picture the Problem Choose Ug = 0 at the bob’s equilibrium position. Momentum is conserved in the collision of the bullet with bob and the initial kinetic energy of the bob plus bullet is transformed into gravitational potential energy as it swings up to the top of the circle. If the bullet plus bob just makes it to the top of the circle with zero speed, it will swing through a complete circle. Use conservation of momentum to relate the speed of the bullet just before impact to the initial speed of the bob plus bullet:

( )Vmmvm 211 +=

Solve for the speed of the bullet: V

mmv ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

1

21 (1)

Use conservation of energy to relate the initial kinetic energy of the bob plus bullet to their potential energy at the top of the circle:


0fi =+− UK

Substitute for Ki and Uf: ( ) ( ) ( ) 02212

2121 =+++− LgmmVmm

Solve for V:

gLV =

Substitute for V in equation (1) and simplify to obtain: gL

mmv ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

1

21


563

*81 •• Picture the Problem Choose Ug = 0 at the equilibrium position of the ballistic pendulum. Momentum is conserved in the collision of the bullet with the bob and kinetic energy is transformed into gravitational potential energy as the bob swings up to its maximum height. Letting V represent the initial speed of the bob as it begins its upward swing, use conservation of momentum to relate this speed to the speeds of the bullet just before and after its collision with the bob:

( ) Vmvmvm 221

11 +=

Solve for the speed of the bob: vmmV

2

1

2= (1)

Use conservation of energy to relate the initial kinetic energy of the bob to its potential energy at its maximum height:


0fi =+− UK

Substitute for Ki and Uf: 022

221 =+− ghmVm

Solve for h: g

Vh2

2

= (2)

Substitute V from equation (1) in equation (2) and simplify to obtain: 2

2

12

2

2

1

822

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

=mm

gv

g

vmm

h

82 • Picture the Problem Let the mass of the bullet be m, that of the wooden block M, the pre-collision velocity of the bullet v, and the post-collision velocity of the block+bullet be V. We can use conservation of momentum to find the velocity of the block with the bullet imbedded in it just after their perfectly inelastic collision. We can use Newton’s 2nd law to find the acceleration of the sliding block and a constant-acceleration equation to find the distance the block slides.

Chapter 8

564

Using a constant-acceleration equation, relate the velocity of the block+bullet just after their collision to their acceleration and displacement before stopping:

xaV ∆+= 20 2 because the final velocity of the block+bullet is zero.

Solve for the distance the block slides before coming to rest: a

Vx2

2

−=∆ (1)

Use conservation of momentum to relate the pre-collision velocity of the bullet to the post-collision velocity of the block+bullet:

( )VMmmv +=

Solve for V: v

MmmV+

=

Substitute in equation (1) to obtain: 2

21

⎟⎠⎞

⎜⎝⎛

+−=∆ v

Mmm

ax (2)

Apply aF rr

m=∑ to the block+bullet (see the FBD in the diagram):

( )aMmfFx +=−=∑ k (3) and

( ) 0n =+−=∑ gMmFFy (4)

Use the definition of the coefficient of kinetic friction and equation (4) to obtain:

( )gMmFf +== knkk µµ

Substitute in equation (3): ( ) ( )aMmgMm +=+− kµ

Solve for a to obtain: ga kµ−=


k21

⎟⎠⎞

⎜⎝⎛

+=∆ v

Mmm

gx

µ


565


( )( ) ( ) m130.0m/s750kg10.5kg0.0105

kg0.0105m/s9.810.222

12

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛+

=∆x

83 •• Picture the Problem The collision of the ball with the box is perfectly inelastic and we can find the speed of the box-and-ball immediately after their collision by applying conservation of momentum. If we assume that the kinetic friction force is constant, we can use a constant-acceleration equation to find the acceleration of the box and ball combination and the definition of µk to find its value. Using its definition, express the coefficient of kinetic friction of the table:

( )( ) g

agmMamM

Ff

=++

==n

kkµ (1)

Use conservation of momentum to relate the speed of the ball just before the collision to the speed of the ball+box immediately after the collision:

( )vMmMV +=

Solve for v:

MmMVv+

= (2)

Use a constant-acceleration equation to relate the sliding distance of the ball+box to its initial and final velocities and its acceleration:

xavv ∆+= 22i

2f

or, because vf = 0 and vi = v, xav ∆+= 20 2

Solve for a:

xva∆

−=2

2


xgv∆

=2

2

kµ

Use equation (2) to eliminate v:

2

2

k

121

21

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+∆=

⎟⎠⎞

⎜⎝⎛

+∆=

Mm

Vxg

MmMV

xgµ

Chapter 8

566


( )( ) 0529.01

kg0.425kg0.327m/s1.3

m0.52m/s9.8121

2

2k =

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=µ

*84 •• Picture the Problem Jane’s collision with Tarzan is a perfectly inelastic collision. We can find her speed v1 just before she grabs Tarzan from conservation of energy and their speed V just after she grabs him from conservation of momentum. Their kinetic energy just after their collision will be transformed into gravitational potential energy when they have reached their greatest height h.

Use conservation of energy to relate the potential energy of Jane and Tarzan at their highest point (2) to their kinetic energy immediately after Jane grabbed Tarzan:

12 KU = or

2TJ2

1TJ Vmghm ++ =

Solve for h to obtain: g

Vh2

2

= (1)

Use conservation of momentum to relate Jane’s velocity just before she collides with Tarzan to their velocity just after their perfectly inelastic collision:

Vmvm TJ1J +=

Solve for V: 1

TJ

J vmmV

+

= (2)

Apply conservation of energy to relate Jane’s kinetic energy at 1 to her potential energy at 0:

01 UK = or

gLmvm J21J2

1 =


567

Solve for v1: gLv 21 =

Substitute in equation (2) to obtain: gL

mmV 2

TJ

J

+

=

Substitute in equation (1) and simplify:

LmmgL

mm

gh

2

TJ

J

2

TJ

J 221

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

++

Substitute numerical values and evaluate h: ( ) m3.94m25

kg82kg45kg45

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=h

Exploding Objects and Radioactive Decay 85 •• Picture the Problem This nuclear reaction is 4Be → 2α + 1.5×10−14 J. In order to conserve momentum, the alpha particles will have move in opposite directions with the same velocities. We’ll use conservation of energy to find their speeds. Letting E represent the energy released in the reaction, express conservation of energy for this process:

( ) EvmK == 22122 ααα

Solve for vα:

αα m

Ev =

Substitute numerical values and evaluate vα: m/s1050.1

kg106.68J101.5 6

27

14

×=××

= −

−

αv

86 •• Picture the Problem This nuclear reaction is 5Li → α + p + 3.15 × 10−13 J. To conserve momentum, the alpha particle and proton must move in opposite directions. We’ll apply both conservation of energy and conservation of momentum to find the speeds of the proton and alpha particle. Use conservation of momentum in this process to express the alpha particle’s velocity in terms of the proton’s:

0fi == pp

and αα vmvm −= pp0

Chapter 8

568

Solve for vα and substitute for mα to obtain: p4

1p

p

pp

p

4vv

mm

vmm

v ===α

α

Letting E represent the energy released in the reaction, apply conservation of energy to the process:

EKK =+ αp

or Evmvm =+ 2

212

pp21

αα

Substitute for vα: ( ) Evmvm =+ 2

p41

212

pp21

α

Solve for vp and substitute for mα to obtain:

pppp 416

3216

32mm

Emm

Ev+

=+

=α

Substitute numerical values and evaluate vp:

( )( )

m/s1074.1

kg101.6720J103.1532

7

27

13

p

×=

××

= −

−

v

Use the relationship between vp and vα to obtain vα:

( )m/s104.34

m/s101.746

741

p41

×=

×== vvα

87 ••• Picture the Problem The pictorial representation shows the projectile at its maximum elevation and is moving horizontally. It also shows the two fragments resulting from the explosion. We chose the system to include the projectile and the earth so that no external forces act to change the momentum of the system during the explosion. With this choice of system we can also use conservation of energy to determine the elevation of the projectile when it explodes. We’ll also find it useful to use constant-acceleration equations in our description of the motion of the projectile and its fragments.


569

(a) Use conservation of momentum to relate the velocity of the projectile before its explosion to the velocities of its two parts after the explosion:

jjii

vvvpp

ˆˆˆˆ22111133

221133

fi

yyx vmvmvmvm

mmm

−+=

+==

rrr

rr

The only way this equality can hold is if:

2211

1133

and

yy

x

vmvm

vmvm

=

=

Express v3 in terms of v0 and substitute for the masses to obtain: ( ) m/s312cos30m/s1203

cos33 031

=°=== θvvvx

and 21 2 yy vv = (1)

Using a constant-acceleration equation with the downward direction positive, relate vy2 to the time it takes the 2-kg fragment to hit the ground:

( )221

2 tgtvy y ∆+∆=∆

( )t

tgyvy ∆∆−∆

=2

21

2 (2)

With Ug = 0 at the launch site, apply conservation of energy to the climb of the projectile to its maximum elevation:

0=∆+∆ UK Because Kf = Ui = 0, 0fi =+− UK

or 03

2032

1 =∆+− ygmvm y

Solve for ∆y: ( )

gv

gv

y y

230sin

2

20

20 °

==∆

Substitute numerical values and evaluate ∆y:

( )[ ]( ) m183.5

m/s9.812sin30m/s120

2

2

=°

=∆y

Substitute in equation (2) and evaluate vy2:

( )( )

m/s33.3s3.6

s3.6m/s9.81m183.5 2221

2

=

−=yv

Substitute in equation (1) and evaluate vy1:

( ) m/s66.6m/s33.321 ==yv

Chapter 8

570

Express 1vr in vector form:

( ) ( ) ji

jiv

ˆm/s6.66ˆm/s312

ˆˆ111

+=

+= yx vvr

(b) Express the total distance d traveled by the 1-kg fragment:

'xxd ∆+∆= (3)

Relate ∆x to v0 and the time-to-explosion:

( )( )exp0 cos tvx ∆=∆ θ (4)

Using a constant-acceleration equation, express ∆texp: g

vg

vt y θsin00exp ==∆

Substitute numerical values and evaluate ∆texp:

( ) s6.12m/s9.81

sin30m/s1202exp =

°=∆t

Substitute in equation (4) and evaluate ∆x:

( )( )( )m636.5

s6.12cos30m/s120=

°=∆x

Relate the distance traveled by the 1-kg fragment after the explosion to the time it takes it to reach the ground:

t'vx' x ∆=∆ 1

Using a constant-acceleration equation, relate the time ∆t′ for the 1-kg fragment to reach the ground to its initial speed in the y direction and the distance to the ground:

( )221

1 t'gt'vy y ∆−∆=∆

Substitute to obtain the quadratic equation:

( ) ( ) 0s4.37s6.13 22 =−∆−∆ t't'

Solve the quadratic equation to find ∆t′:

∆t′ = 15.9 s

Substitute in equation (3) and evaluate d: ( )( )

km5.61

s15.9m/s312m636.51

=

+=∆+∆=∆+∆= t'vxx'xd x


571

(c) Express the energy released in the explosion:

ifexp KKKE −=∆= (5)

Find the kinetic energy of the fragments after the explosion: ( ) ( ) ( )[ ]

( )( )kJ0.52

m/s33.3kg2

m/s66.6m/s312kg12

21

2221

2222

12112

121f

=+

+=

+=+= vmvmKKK

Find the kinetic energy of the projectile before the explosion:

( )( ) ( )[ ]

kJ2.1630cosm/s201kg3

cos2

21

2032

12332

1i

=

°=

== θvmvmK

Substitute in equation (5) to determine the energy released in the explosion:

kJ35.8

kJ16.2kJ0.52ifexp

=

−=−= KKE

*88 ••• Picture the Problem This nuclear reaction is 9B → 2α + p + 4.4×10−14 J. Assume that the proton moves in the –x direction as shown in the figure. The sum of the kinetic energies of the decay products equals the energy released in the decay. We’ll use conservation of momentum to find the angle between the velocities of the proton and the alpha particles. Note that 'αα vv = .

Express the energy released to the kinetic energies of the decay products:

relp 2 EKK =+ α

or ( ) rel

2212

pp21 2 Evmvm =+ αα

Solve for vα:

αα m

vmEv

2pp2

1rel −

=

Chapter 8

572

Substitute numerical values and evaluate vα:

( )( ) m/s1044.1kg106.68

m/s106kg101.67kg106.68J104.4 6

27

262721

27

14

×=×

××−

××

= −

−

−

−

αv

Given that the boron isotope was at rest prior to the decay, use conservation of momentum to relate the momenta of the decay products:

0if == pp rr ⇒ 0f =xp

( ) 0cos2 pp =−∴ vmvm θαα

or ( ) 0cos42 ppp =− vmvm θα

Solve for θ :

( ) °±=⎥⎦

⎤⎢⎣

⎡×

×=

⎥⎦

⎤⎢⎣

⎡=

−

−

7.58m/s101.448

m/s106cos

8cos

6

61

p1

α

θvv

Let θ′ equal the angle the velocities of the alpha particles make with that of the proton:

( )°±=

°−°±=

121

7.58180'θ

Coefficient of Restitution 89 • Picture the Problem The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which the ball was dropped and the height to which it rebounded by using conservation of mechanical energy. Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

app

rec

vve =

Letting Ug = 0 at the surface of the steel plate, apply conservation of energy to express the velocity of approach:

0=∆+∆ UK Because Ki = Uf = 0,

0or

0

app2app2

1

if

=−

=−

mghmv

UK

Solve for vapp: appapp 2ghv =


573

In like manner, show that: recrec 2ghv =

Substitute in the equation for e to obtain:

app

rec

app

rec

22

hh

ghgh

e ==

Substitute numerical values and evaluate e:

913.0m3m2.5

==e

*90 • Picture the Problem The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which an object was dropped and the height to which it rebounded by using conservation of mechanical energy. Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

app

rec

vve =

Letting Ug = 0 at the surface of the steel plate, apply conservation of energy to express the velocity of approach:


0or

0

app2app2

1

if

=−

=−

mghmv

UK




app

rec

app

rec

22

hh

ghgh

e ==

Find emin: 825.0

cm254cm173

min ==e

Find emax: 849.0

cm254cm183

max ==e

Chapter 8

574

and 849.0825.0 ≤≤ e

91 • Picture the Problem Because the rebound kinetic energy is proportional to the rebound height, the percentage of mechanical energy lost in one bounce can be inferred from knowledge of the rebound height. The coefficient of restitution is defined as the ratio of the velocity of recession to the velocity of approach. These velocities can be determined from the heights from which an object was dropped and the height to which it rebounded by using conservation of mechanical energy. (a) We know, from conservation of energy, that the kinetic energy of an object dropped from a given height h is proportional to h:

K α h.

If, for each bounce of the ball, hrec = 0.8happ:

lost. isenergy mechanical its of %20

(b) Use its definition to relate the coefficient of restitution to the velocities of approach and recession:

app

rec

vve =

Letting Ug = 0 at the surface from which the ball is rebounding, apply conservation of energy to express the velocity of approach:


0or

0

app2app2

1

if

=−

=−

mghmv

UK




app

rec

app

rec

22

hh

ghgh

e ==

Substitute for app

rec

hh

to obtain: 894.08.0 ==e


575

92 •• Picture the Problem Let the numeral 2 refer to the 2-kg object and the numeral 4 to the 4-kg object. Choose a coordinate system in which the direction the 2-kg object is moving before the collision is the positive x direction and let the system consist of the earth, the surface on which the objects slide, and the objects. Then we can use conservation of momentum to find the velocity of the recoiling 4-kg object. We can find the energy transformed in the collision by calculating the difference between the kinetic energies before and after the collision and the coefficient of restitution from its definition. (a) Use conservation of momentum in one dimension to relate the initial and final momenta of the participants in the collision:

f22f44i22

fi

orvmvmvm −=

= pp rr

Solve for and evaluate the final velocity of the 4-kg object:

( )( ) m/s3.50kg4

m/s1m/s6kg24

f22i22f4

=+

=

+=

mvmvmv

(b) Express the energy lost in terms of the kinetic energies before and after the collision:

( )( )[ ]2

f442f2

2i222

1

2f442

12f222

12i222

1

filost

vmvvm

vmvmvm

KKE

−−=

+−=

−=

Substitute numerical values and evaluate Elost:

( ) ( ) ( ){ }( ) ( )( )[ ] J5.10m/s3.5kg4m/s1m/s6kg2 22221

lost =−−=E

(c) Use the definition of the coefficient of restitution:

( ) 0.750m/s6

m/s1m/s3.5

i2

f2f4

app

rec =−−

=−

==v

vvvve

93 •• Picture the Problem Let the numeral 2 refer to the 2-kg block and the numeral 3 to the 3-kg block. Choose a coordinate system in which the direction the blocks are moving before the collision is the positive x direction and let the system consist of the earth, the surface on which the blocks move, and the blocks. Then we can use conservation of momentum find the velocity of the 2-kg block after the collision. We can find the coefficient of restitution from its definition.

Chapter 8

576

(a) Use conservation of momentum in one dimension to relate the initial and final momenta of the participants in the collision:

f33f223i3i22

fi

orvmvmvmvm +=+

= pp rr

Solve for the final velocity of the 2-kg object: 2

f33i33i22f2 m

vmvmvmv −+=

Substitute numerical values and evaluate v2f:

( )( ) ( )( ) m/s70.1kg2

m/s4.2m/s2kg3m/s5kg2f2 =

−+=v

(b) Use the definition of the coefficient of restitution:

0.833

m/s2m/s5m/s7.1m/s2.4

i3i2

f2f3

app

rec

=

−−

=−−

==vvvv

vve

Collisions in Three Dimensions *94 •• Picture the Problem We can use the definition of the magnitude of a vector and the definition of the dot product to establish the result called for in (a). In part (b) we can use the result of part (a), the conservation of momentum, and the definition of an elastic collision (kinetic energy is conserved) to show that the particles separate at right angles. (a) Find the dot product of CB

rr+

with itself: ( ) ( )

CB

CBCBrr

rrrr

⋅++=

+⋅+

222 CB

Because CBA

rrr+= : ( ) ( )CBCBCB

rrrrrr+⋅+=+=

22A

Substitute to obtain: CB

rr⋅++= 2222 CBA

(b) Apply conservation of momentum to the collision of the particles:

Ppprrr

=+ 21

Form the dot product of each side of this equation with itself to obtain:

( ) ( ) PPpppprrrrrr

⋅=+⋅+ 2121 or

221

22

21 2 Ppp =⋅++ pp

rr (1)

Apply the definition of an elastic collision to obtain: m

Pm

pm

p222

222

21 =+


577

or 22

221 Ppp =+ (2)

Subtract equation (1) from equation (2) to obtain:

02 21 =⋅ pp rror 021 =⋅ pp rr

i.e., the particles move apart along paths that are at right angles to each other.

95 •

Picture the Problem Let the initial direction of motion of the cue ball be the positive x direction. We can apply conservation of energy to determine the angle the cue ball makes with the positive x direction and the conservation of momentum to find the final velocities of the cue ball and the eight ball. (a) Use conservation of energy to relate the velocities of the collision participants before and after the collision:

28

2cf

2ci

282

12cf2

12ci2

1

orvvv

mvmvmv

+=

+=

This Pythagorean relationship tells us that 8cfci and,, vvv rrr

form a right

triangle. Hence: °=

°=+

60

and90

cf

8cf

θ

θθ

(b) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision:

88cfcfci

xfxi

coscosor

θθ mvmvmv +=

= pp rr

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision:

88cfcf

yfyi

sinsin0or

θθ mvmv +=

= pprr


m/s33.4

and

m/s50.2

8

cf

=

=

v

v

Chapter 8

578

96 •• Picture the Problem We can find the final velocity of the object whose mass is M1 by using the conservation of momentum. Whether the collision was elastic can be decided by examining the difference between the initial and final kinetic energy of the interacting objects. (a) Use conservation of momentum to relate the initial and final velocities of the two objects:

fi pp rr=

or ( ) ( ) 1f04

102

10

ˆ2ˆ2ˆ viji rmvmvmmv +=+

Simplify to obtain:

1f021

00ˆˆˆ viji r

+=+ vvv

Solve for 1fvr : jiv ˆˆ002

11f vv +=r

(b) Express the difference between the kinetic energy of the system before the collision and its kinetic energy after the collision:

( ) [ ][ ] [ ]

( ) ( )[ ] 2016

12016

1204

5204

1202

1

2f2

2f1

2i2

2i12

12f2

2f1

2i2

2i12

1

2f22

2f11

2i22

2i112

12f1f2i1ifi

22

2222

mvvvvvm

vvvvmmvmvmvmv

vMvMvMvMKKKKKKE

=−−+=

−−+=−−+=

−−+=+−+=−=∆

. iscollision the0, Because inelasticE ≠∆

*97 •• Picture the Problem Let the direction of motion of the puck that is moving before the collision be the positive x direction. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v1 and v2 that we can solve simultaneously. We can decide whether the collision was elastic by either calculating the system’s kinetic energy before and after the collision or by determining whether the angle between the final velocities is 90°. (a) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision: °+°=

°+°=

=

60cos30cosor

60cos30cosor

21

21

xfxi

vvv

mvmvmv

pp


579

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: °−°=

°−°=

=

60sin30sin0or

60sin30sin0or

21

21

yfyi

vv

mvmv

pp


m/s00.1andm/s73.1 21 == vv

(b) . wascollision the,90 is and between angle theBecause 21 elastic°vv rr

98 •• Picture the Problem Let the direction of motion of the object that is moving before the collision be the positive x direction. Applying conservation of momentum to the motion in both the x and y directions will lead us to two equations in the unknowns v2 and θ2 that we can solve simultaneously. We can show that the collision was elastic by showing that the system’s kinetic energy before and after the collision is the same. (a) Use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision:

22100

22100

xfxi

cos2cos53

orcos2cos53

or

θθ

θθ

vvv

mvmvmv

pp

+=

+=

=

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: 2210

2210

yfyi

sin2sin50

orsin2sin50

or

θθ

θθ

vv

mvmv

pp

−=

−=

=

Note that if tanθ1 = 2, then:

52sinand

51cos 11 == θθ

Substitute in the momentum equations to obtain:

220

2200

cosor

cos25

153

θ

θ

vv

vvv

=

+=

and

Chapter 8

580

220

220

sin0or

sin25

250

θ

θ

vv

vv

−=

−=

Solve these equations simultaneously for θ2 :

°== − 0.451tan 12θ

Substitute to find v2:

00

2

02 2

45coscosvvvv =

°==

θ

(b) To show that the collision was elastic, find the before-collision and after-collision kinetic energies:

( )

( ) ( )( )20

2

021

2

021

f

20

202

1i

5.4

225

and5.43

mv

vmvmK

mvvmK

=

+=

==

elastic. iscollision the, Because fi KK =

*99 •• Picture the Problem Let the direction of motion of the ball that is moving before the collision be the positive x direction. Let v represent the velocity of the ball that is moving before the collision, v1 its velocity after the collision and v2 the velocity of the initially-at-rest ball after the collision. We know that because the collision is elastic and the balls have the same mass, v1 and v2 are 90° apart. Applying conservation of momentum to the collision in both the x and y directions will lead us to two equations in the unknowns v1 and v2 that we can solve simultaneously. Noting that the angle of deflection for the recoiling ball is 60°, use conservation of momentum in the x direction to relate the velocities of the collision participants before and after the collision:

°+°=

°+°=

=

60cos30cosor

60cos30cosor

21

21

xfxi

vvv

mvmvmv

pp

Use conservation of momentum in the y direction to obtain a second equation relating the velocities of the collision participants before and after the collision: °−°=

°−°=

=

60sin30sin0or

60sin30sin0or

21

21

yfyi

vv

mvmv

pp


581


m/s00.5andm/s66.8 21 == vv

100 •• Picture the Problem Choose the coordinate system shown in the diagram below with the x-axis the axis of initial approach of the first particle. Call V the speed of the target particle after the collision. In part (a) we can apply conservation of momentum in the x and y directions to obtain two equations that we can solve simultaneously for tanθ. In part (b) we can use conservation of momentum in vector form and the elastic-collision equation to show that v = v0cosφ.

(a) Apply conservation of momentum in the x direction to obtain:

θφ coscos0 Vvv += (1)

Apply conservation of momentum in the y direction to obtain:

θφ sinsin Vv = (2)

Solve equation (1) for Vcosθ :

φθ coscos 0 vvV −= (3)

Divide equation (2) by equation (3) to obtain: φ

φθθ

cossin

cossin

0 vvv

VV

−=

or

φφθ

cossintan

0 vvv−

=

(b) Apply conservation of momentum to obtain:

Vvvrrr

+=0

Draw the vector diagram representing this equation:

Use the definition of an elastic 222

0 Vvv +=

Chapter 8

582

collision to obtain: If this Pythagorean condition is to hold, the third angle of the triangle must be a right angle and, using the definition of the cosine function:

φcos0vv =

Center-of-Mass Frame 101 •• Picture the Problem The total kinetic energy of a system of particles is the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass. The kinetic energy of a particle of mass m is related to momentum according to mpK 22= .

Express the total kinetic energy of the system:

cmrel KKK += (1)

Relate the kinetic energy relative to the center of mass to the momenta of the two particles:

( )21

2121

2

21

1

21

rel 222 mmmmp

mp

mpK +

=+=

Express the kinetic energy of the center of mass of the two particles:

( )( ) 21

21

21

21

cm2

22

mmp

mmpK

+=

+=

Substitute in equation (1) and simplify to obtain:

( )

⎥⎦

⎤⎢⎣

⎡+

++=

++

+=

2212

21

2221

21

21

21

21

21

2121

62

22

mmmmmmmmp

mmp

mmmmpK

In an elastic collision:

⎥⎦

⎤⎢⎣

⎡+

++=

⎥⎦

⎤⎢⎣

⎡+

++=

=

2212

21

2221

21

21

2212

21

2221

21

21

fi

62

62

mmmmmmmmp'

mmmmmmmmp

KK

Simplify to obtain: ( ) ( ) 11

21

21 pppp '' ±=⇒=

and collide.not do particles the, If 11 pp' +=


583

*102 •• Picture the Problem Let the numerals 3 and 1 denote the blocks whose masses are 3 kg and 1 kg respectively. We can use cmvv rr Mm

iii =∑ to find the velocity of the center-of-

mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass:

( ) cm31cm

3311

vv

vvvPrr

rrrr

mmM

mmmi

ii

+==

+== ∑

Solve for cmvr :

13

1133cm mm

mm++

=vvvrr

r


( )( ) ( )( )

( )i

iiv

ˆm/s3.00

kg1kg3


−=

++−

=r

(b) Find the velocity of the 3-kg block in the center of mass reference frame:

( ) ( )( )i

iivvuˆm/s2.00

ˆm/s3ˆm/s5cm33

−=

−−−=−=rrr

Find the velocity of the 1-kg block in the center of mass reference frame:

( ) ( )( )i

iivvuˆm/s00.6

ˆm/s3ˆm/s3cm11

=

−−=−=rrr

(c) Express the after-collision velocities of both blocks in the center of mass reference frame:

( )iu ˆm/s00.23 ='r

and

( )iu ˆm/s00.61 −='r

(d) Transform the after-collision velocity of the 3-kg block from the center of mass reference frame to the original reference frame:

( ) ( )( )i

iivuvˆm/s00.1

ˆm/s3ˆm/s2cm33

−=

−+=+=rrr ''

Transform the after-collision velocity of the 1-kg block from the center of mass reference frame to the original reference frame:

( ) ( )( )i

iivuvˆm/s00.9

ˆm/s3ˆm/s6cm11

−=

−+−=+=rrr ''

(e) Express Ki in the original frame of 2112

12332

1i vmvmK +=

Chapter 8

584

reference:

Substitute numerical values and evaluate Ki:

( )( ) ( )( )[ ]J42.0

m/s3kg1m/s5kg3 2221

i

=

+=K

Express Kf in the original frame of reference:

2112

12332

1f v'mv'mK +=

Substitute numerical values and evaluate Kf:

( )( ) ( )( )[ ]J42.0

m/s9kg1m/s1kg3 2221

f

=

+=K

103 •• Picture the Problem Let the numerals 3 and 1 denote the blocks whose masses are 3 kg and 1 kg respectively. We can use cmvv rr Mm

iii =∑ to find the velocity of the center-of-

mass of the system and simply follow the directions in the problem step by step. (a) Express the total momentum of this two-particle system in terms of the velocity of its center of mass:

( ) cm53cm

5533

vv

vvvPrr

rrrr

mmM

mmmi

ii

+==

+== ∑

Solve for cmvr :

53

5533cm mm

mm++

=vvvrr

r


( )( ) ( )( )

0

kg5kg3


=

++−

=iivr

(b) Find the velocity of the 3-kg block in the center of mass reference frame:

( )( ) i

ivvuˆm/s5

0ˆm/s5cm33

−=

−−=−=rrr

Find the velocity of the 5-kg block in the center of mass reference frame:

( )( )i

ivvuˆm/s3

0ˆm/s3cm55

=

−=−=rrr

(c) Express the after-collision velocities of both blocks in the center of mass reference frame:

( )iu ˆm/s53 ='r

and


585

m/s75.05 ='u

(d) Transform the after-collision velocity of the 3-kg block from the center of mass reference frame to the original reference frame:

( )( )i

ivuvˆm/s5

0ˆm/s5cm33

=

+=+=rrr ''

Transform the after-collision velocity of the 5-kg block from the center of mass reference frame to the original reference frame:

( )( )i

ivuvˆm/s3

0ˆm/s3cm55

−=

+−=+=rrr ''

(e) Express Ki in the original frame of reference:

2552

12332

1i vmvmK +=

Substitute numerical values and evaluate Ki:

( )( ) ( )( )[ ]J60.0

m/s3kg5m/s5kg3 2221

i

=

+=K

Express Kf in the original frame of reference:

2552

12332

1f v'mv'mK +=

Substitute numerical values and evaluate Kf:

( )( )[ ( )( ) ] J60.0m/s3kg5m/s5kg3 2221

f =+=K

Systems With Continuously Varying Mass: Rocket Propulsion 104 •• Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to exth udtdmF = .

Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases:

exth udtdmF =

Substitute numerical values and evaluate Fth:

( )( ) MN20.1km/s6kg/s200th ==F

Chapter 8

586

105 •• Picture the Problem The thrust of a rocket Fth depends on the burn rate of its fuel dm/dt and the relative speed of its exhaust gases uex according to exth udtdmF = . The final

velocity vf of a rocket depends on the relative speed of its exhaust gases uex, its payload to initial mass ratio mf/m0 and its burn time according to ( ) b0fexf ln gtmmuv −−= .

(a) Using its definition, relate the rocket’s thrust to the relative speed of its exhaust gases:

exth udtdmF =

Substitute numerical values and evaluate Fth:

( )( ) kN360km/s8.1kg/s200th ==F

(b) Relate the time to burnout to the mass of the fuel and its burn rate:

dtdmm

dtdmmt

/8.0

/0fuel

b ==

Substitute numerical values and evaluate tb:

( ) s120kg/s200

kg30,0000.8b ==t

(c) Relate the final velocity of a rocket to its initial mass, exhaust velocity, and burn time:

b0

fexf ln gt

mmuv −⎟⎟

⎠

⎞⎜⎜⎝

⎛−=


( ) ( )( ) km/s72.1s120m/s9.8151lnkm/s1.8 2

f =−⎟⎠⎞

⎜⎝⎛−=v

*106 •• Picture the Problem We can use the dimensions of thrust, burn rate, and acceleration to show that the dimension of specific impulse is time. Combining the definitions of rocket thrust and specific impulse will lead us to spex gIu = . (a) Express the dimension of specific impulse in terms of the dimensions of Fth, R, and g:

[ ] [ ][ ][ ] T

TL

TM

TLM

2

2th

sp =⋅

⋅

==gR

FI

(b) From the definition of rocket thrust we have:

exth RuF =

Solve for uex:

RFu th

ex =


587

Substitute for Fth to obtain: sp

spex gI

RRgI

u == (1)

(c) Solve equation (1) for Isp and substitute for uex to obtain: Rg

FI thsp =

From Example 8-21 we have: R = 1.384×104 kg/s and Fth = 3.4×106 N

Substitute numerical values and evaluate Isp: ( )( )

s25.0

m/s81.9kg/s101.384N103.4

24

6

sp

=

××

=I

*107 ••• Picture the Problem We can use the rocket equation and the definition of rocket thrust to show that ga00 1+=τ . In part (b) we can express the burn time tb in terms of the initial and final masses of the rocket and the rate at which the fuel burns, and then use this equation to express the rocket’s final velocity in terms of Isp, τ0, and the mass ratio m0/mf. In part (d) we’ll need to use trial-and-error methods or a graphing calculator to solve the transcendental equation giving vf as a function of m0/mf. (a) Express the rocket equation:

maRumg =+− ex

From the definition of rocket thrust we have:

exth RuF =


maFmg =+− th

Solve for Fth at takeoff: 000th amgmF +=

Divide both sides of this equation by m0g to obtain: g

agm

F 0

0

th 1+=

Because )/( 0th0 gmF=τ :

ga0

0 1+=τ

(b) Use equation 8-42 to express the final speed of a rocket that starts from rest with mass m0:

bf

0exf ln gt

mmuv −= , (1)

where tb is the burn time.

Express the burn time in terms of the burn rate R (assumed constant):

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

0

f0f0b 1

mm

Rm

Rmmt

Chapter 8

588

Multiply tb by one in the form gT/gT and simplify to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

0

f

0

sp

0

fth

th

0

0

f0

th

thb

1

1

1

mmI

mm

gRF

Fgm

mm

Rm

gFgFt

τ

Substitute in equation (1):

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

0

f

0

sp

f

0exf 1ln

mmgI

mmuv

τ

From Problem 32 we have:

spex gIu = , where uex is the exhaust velocity of the propellant.

Substitute and factor to obtain:

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

0

f

0f

0sp

0

f

0

sp

f

0spf

11ln

1ln

mm

mmgI

mmgI

mmgIv

τ

τ

(c) A spreadsheet program to calculate the final velocity of the rocket as a function of the mass ratio m0/mf is shown below. The constants used in the velocity function and the formulas used to calculate the final velocity are as follows:

Cell Content/Formula Algebraic Form B1 250 Isp B2 9.81 g B3 2 τ D9 D8 + 0.25 m0/mf E8 $B$2*$B$1*(LOG(D8) −

(1/$B$3)*(1/D8)) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛

0

f

0f

0sp 11ln

mm

mmgI

τ

A B C D E 1 Isp = 250 s 2 g = 9.81 m/s^2 3 tau = 2 4 5 6 7 mass ratio vf 8 2.00 1.252E+029 2.25 3.187E+02


589

10 2.50 4.854E+0211 2.75 6.316E+0212 3.00 7.614E+02

36 9.00 2.204E+0337 9.25 2.237E+0338 9.50 2.269E+0339 9.75 2.300E+0340 10.00 2.330E+0341 725.00 7.013E+03

A graph of final velocity as a function of mass ratio is shown below.

0

1

2

2 4 6 8 10

m 0/m f

v f (k

m/s

)

(d) Substitute the data given in part (c) in the equation derived in part (b) to obtain:

( )( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

0

f

f

02 121lns250m/s9.81km/s7

mm

mm

or

xx 5.05.0ln854.2 +−= where x = m0/mf.

Use trial-and-error methods or a graphing calculator to solve this transcendental equation for the root greater than 1:

1.28=x ,

a value considerably larger than the practical limit of 10 for single-stage rockets.

108 •• Picture the Problem We can use the velocity-at-burnout equation from Problem 106 to find vf and constant-acceleration equations to approximate the maximum height the rocket will reach and its total flight time. (a) Assuming constant acceleration, relate the maximum height reached

2top2

1 gth = (1)

Chapter 8

590

by the model rocket to its time-to-top-of-trajectory: From Problem 106 we have: ⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

f

f

0spf 11ln

mm

mmgIv

τ

Evaluate the velocity at burnout vf for Isp = 100 s, m0/mf = 1.2, and τ = 5:

( )( )

( )

m/s1462.1

11512.1ln

s100m/s9.81 2f

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−×

=v

Assuming that the time for the fuel to burn up is short compared to the total flight time, find the time to the top of the trajectory:

s14.9m/s9.81m/s146

2f

top ===gvt

Substitute in equation (1) and evaluate h:

( )( ) km1.09s14.9m/s9.81 2221 ==h

(b) Find the total flight time from the time it took the rocket to reach its maximum height:

( ) s29.8s14.922 topflight === tt

(c) Express and evaluate the fuel burn time tb:

s3.331.211

5s1001

0

=

⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

mmI

t fspb τ

burnout. untilon acceleraticonstant assuming m, 243 is timein this movepossibly couldrocket model the

distance maximum theas accuracy, 30%about togood be however, should,It accurate. very be to)(Part in obtained answer we expect the

tcan' wee,flight tim total theof 1/5ely approximat is burn time thisBecause

b21 =vt

b

General Problems 109 • Picture the Problem Let the direction of motion of the 250-g car before the collision be the positive x direction. Let the numeral 1 refer to the 250-kg car, the numeral 2 refer to the 400-kg car, and V represent the velocity of the linked cars. Let the system include the earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their initial and final kinetic energies.


591

Use conservation of momentum to relate the speeds of the cars immediately before and immediately after their collision:

( )Vmmvm

pp

2111

fxix

or+=

=

Solve for V:

21

11

mmvmV

+=

Substitute numerical values and evaluate V:

( )( ) m/s192.0kg0.400kg0.250

m/s0.50kg0.250=

+=V

Find the initial kinetic energy of the cars:

( )( )mJ3.31

m/s0.50kg0.250 2212

1121

i

=

== vmK

Find the final kinetic energy of the coupled cars:

( )( )( )

mJ0.12

m/s0.192kg0.400kg0.250 221

2212

1f

=

+=

+= VmmK

110 • Picture the Problem Let the direction of motion of the 250-g car before the collision be the positive x direction. Let the numeral 1 refer to the 250-kg car and the numeral 2 refer to the 400-g car and the system include the earth and the cars. We can use conservation of momentum to find their speed after they have linked together and the definition of kinetic energy to find their initial and final kinetic energies. (a) Express and evaluate the initial kinetic energy of the cars:

( )( )mJ3.31

m/s0.50kg0.250 2212

1121

i

=

== vmK

(b) Relate the velocity of the center of mass to the total momentum of the system:

cmi

ii vvPrrr

mm == ∑

Solve for vcm:

21

2211cm mm

vmvmv++

=


( )( ) m/s192.0kg0.400kg0.250

m/s0.50kg0.250cm =

+=v

Chapter 8

592

Find the initial velocity of the 250-g car relative to the velocity of the center of mass:

m/s0.308

m/s0.192m/s.500cm11

=

−=−= vvu

Find the initial velocity of the 400-g car relative to the velocity of the center of mass:

m/s192.0

m/s0.192m/s0cm22

−=

−=−= vvu

Express the initial kinetic energy of the system relative to the center of mass:

2222

12112

1reli, umumK +=

Substitute numerical values and evaluate Ki,rel:

( )( )( )( )

mJ19.2

m/s0.192kg0.400

m/s0.308kg0.2502

21

221

reli,

=

−+

=K

(c) Express the kinetic energy of the center of mass:

2cm2

1cm MvK =

Substitute numerical values and evaluate Kcm:

( )( )mJ12.0

m/s0.192kg0.650 221

cm

=

=K

(d) Relate the initial kinetic energy of the system to its initial kinetic energy relative to the center of mass and the kinetic energy of the center of mass:

mJ31.2mJ12.0mJ9.21

cmreli,i

=+=

+= KKK

cmreli,i KKK +=∴

*111 • Picture the Problem Let the direction the 4-kg fish is swimming be the positive x direction and the system include the fish, the water, and the earth. The velocity of the larger fish immediately after its lunch is the velocity of the center of mass in this perfectly inelastic collision. Relate the velocity of the center of mass to the total momentum of the system:

cmi

ii vvPrrr

mm == ∑


593

Solve for vcm:

2.14

2.12.144cm mm

vmvmv+−

=


( )( )

m/s462.0

kg2.1kg4)m/s(3kg)2.1(m/s5.1kg4

cm

=

+−

=v

112 • Picture the Problem Let the direction the 3-kg block is moving be the positive x direction and include both blocks and the earth in the system. The total kinetic energy of the two-block system is the sum of the kinetic energies of the blocks. We can relate the momentum of the system to the velocity of its center of mass and use this relationship to find vcm. Finally, we can use the definition of kinetic energy to find the kinetic energy relative to the center of mass. (a) Express the total kinetic energy of the system in terms of the kinetic energy of the blocks:

2662

12332

1tot vmvmK +=

Substitute numerical values and evaluate Ktot:

( )( ) ( )( )J81.0

m/s3kg6m/s6kg3 2212

21

tot

=

+=K

(b) Relate the velocity of the center of mass to the total momentum of the system:

cmi

ii vvPrrr

mm == ∑

Solve for vcm:

21

6633cm mm

vmvmv++

=


( )( ) ( )( )

m/s00.4

kg6kg3m/s3kg6m/s6kg3

cm

=

++

=v

(c) Find the center of mass kinetic energy from the velocity of the center of mass:

( )( )J72.0

m/s4kg9 2212

cm21

cm

=

== MvK

Chapter 8

594

(d) Relate the initial kinetic energy of the system to its initial kinetic energy relative to the center of mass and the kinetic energy of the center of mass:

J00.9

J0.27J0.81cmtotrel

=

−=−= KKK

113 • Picture the Problem Let east be the positive x direction and north the positive y direction. Include both cars and the earth in the system and let the numeral 1 denote the 1500-kg car and the numeral 2 the 2000-kg car. Because the net external force acting on the system is zero, momentum is conserved in this perfectly inelastic collision. (a) Express the total momentum of the system: ij

vvpppˆˆ

2211

221121

vmvm

mm

−=

+=+=rrrrr

Substitute numerical values and evaluate p

r:

( )( ) ( )( )

( ) ( )ji

ijpˆkm/hkg1005.1ˆkm/hkg1010.1

ˆkm/h55kg2000ˆkm/h70kg150055 ⋅×+⋅×−=

−=r

(b) Express the velocity of the wreckage in terms of the total momentum of the system:

Mpvvr

rr== cmf

Substitute numerical values and evaluate fvr :

( ) ( )

( ) ( ) ji

jiv

ˆkm/h0.30ˆkm/h4.31

kg2000kg1500

ˆkm/hkg101.05kg2000kg1500

ˆkm/hkg101.10 55

f

+−=

+⋅×

++

⋅×−=

r

Find the magnitude of the velocity of the wreckage:

( ) ( )km/h43.4

km/h30.0km/h31.4 22f

=

+=v

Find the direction of the velocity of the wreckage: °−=⎥

⎦

⎤⎢⎣

⎡−

= − 7.43km/h31.4

km/h30.0tan 1θ

north. of west 46.3is wreckage theofdirection The

°


595

*114 •• Picture the Problem Take the origin to be at the initial position of the right-hand end of raft and let the positive x direction be to the left. Let ″w″ denote the woman and ″r″ the raft, d be the distance of the end of the raft from the pier after the woman has walked to its front. The raft moves to the left as the woman moves to the right; with the center of mass of the woman-raft system remaining fixed (because Fext,net = 0). The diagram shows the initial (xw,i) and final (xw,f) positions of the woman as well as the initial (xr_cm,i) and final (xr_cm,f) positions of the center of mass of the raft both before and after the woman has walked to the front of the raft.

x

x

xw,

f

xr_cm,i

xr_cm,f

xw,

i =6 m

xr_cm,i

0

0×

×

CM

CM

xCM

d

0.5 mP I E R

(a) Express the distance of the raft from the pier after the woman has walked to the front of the raft:

wf,m5.0 xd += (1)

Express xcm before the woman has walked to the front of the raft:

rw

i r_cm,riw,wcm mm

xmxmx

+

+=

Express xcm after the woman has walked to the front of the raft:

rw

fr_cm,rfw,wcm mm

xmxmx

+

+=

Because Fext,net = 0, the center of mass remains fixed and we can equate these two expressions for xcm to obtain:

fr_cm,rfw,wir_cm,ri,ww xmxmxmxm +=+

Solve for xw,f: ( )ir_cm,fr_cm,w

riw,fw, xx

mmxx −−=

From the figure it can be seen that xr_cm,f – xr_cm,i = xw,f. Substitute xw,f rw

iw,wfw, mm

xmx

+=

Chapter 8

596

for xr_cm,f – xr_cm,i and to obtain: Substitute numerical values and evaluate xw,f:

( )( ) m00.2kg120kg60

m6kg60fw, =

+=x

Substitute in equation (1) to obtain: m50.2m5.0m00.2 =+=d

(b) Express the total kinetic energy of the system:

2rr2

12ww2

1tot vmvmK +=

Noting that the elapsed time is 2 s, find vw and vr:

m/s2s2

m6m2iw,fw,w −=

−=

∆−

=txx

v

relative to the dock, and

m/s1s2

m0.5m50.2ir,fr,r =

−=

∆−

=txx

v ,

also relative to the dock.

Substitute numerical values and evaluate Ktot:

( )( )( )( )J180

m/s1kg120

m/s2kg602

21

221

tot

=

+

−=K

Evaluate K with the raft tied to the pier:

( )( )J270

m/s3kg60 2212

ww21

tot

=

== vmK

(c) energy. internalher

into ed transformisenergy kinetic thefriction, static viastops she assuming and, woman theofenergy chemical thefrom derivesenergy kinetic theAll

(d)

raft. theoffront at the lands and m 6 of range a has alsoshot theframe,woman -raft in the Thus, land. the

of frame reference in the m 6 of range a hadshot that thedid asvelocity initial same thehasshot theframeIn that frame. reference inertialan

sconstitute systemwoman -raft thehand, s woman' theleavesshot After the


597

115 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the 1-kg block. We can use conservation of energy to find the speed of the bob just before its perfectly elastic collision with the block and conservation of momentum to find the speed of the block immediately after the collision. We’ll apply Newton’s 2nd law to find the acceleration of the sliding block and use a constant-acceleration equation to find how far it slides before coming to rest. (a) Use conservation of energy to find the speed of the bob just before its collision with the block: 0

or0

ifif =−+−

=∆+∆

UUKK

UK

Because Ki = Uf = 0:

hgv

hgmvm

∆=

=∆+

2

and0

ball

ball2ballball2

1

Substitute numerical values and evaluate vball:

( )( ) m/s6.26m2m/s9.812 2ball ==v

Because the collision is perfectly elastic and the ball and block have the same mass:

m/s26.6ballblock == vv

(b) Using a constant-acceleration equation, relate the displacement of the block to its acceleration and initial speed and solve for its displacement: block

2block

block

2i

f

block2i

2f

22

0, Since2

av

avx

vxavv

−=

−=∆

=∆+=

Apply ∑ = aF rr

m to the sliding

block:

∑

∑

=−=

=−=

0and

blockn

blockk

gmFF

mafF

y

x

Using the definition of fk (µkFn) eliminate fk and Fn between the two equations and solve for ablock:

ga kblock µ−=

Substitute for ablock to obtain: g

vg

vxk

2block

k

2block

22 µµ=

−−

=∆

Chapter 8

598


( )( )( ) m0.20

m/s9.810.12m/s6.26

2

2

==∆x

*116 •• Picture the Problem We can use conservation of momentum in the horizontal direction to find the recoil velocity of the car along the track after the firing. Because the shell will neither rise as high nor be moving as fast at the top of its trajectory as it would be in the absence of air friction, we can apply the work-energy theorem to find the amount of thermal energy produced by the air friction.

(a) conserved. benot willsystem theof momentum the

so and force externalan is rails theof forcereaction verticalThe No.

(b) Use conservation of momentum in the horizontal (x) direction to obtain:

0=∆ xp

or 030cos recoil =−° Mvmv

Solve for and evaluate vrecoil:

Mmvv °

=30cos

recoil

Substitute numerical values and evaluate vrecoil:

( )( )

m/s33.4

kg5000cos30m/s125kg200

recoil

=

°=v

(c) Using the work-energy theorem, relate the thermal energy produced by air friction to the change in the energy of the system:

KUEWW ∆+∆=∆== sysfext

Substitute for ∆U and ∆K to obtain:

( ) ( )22f2

1if

2212

f21

ifext

i

i

vvmyymg

mvmvmgymgyW

−+−=

−+−=

Substitute numerical values and evaluate Wext:

( )( )( ) ( ) ( ) ( )[ ] kJ569m/s125m/s80kg200m180m/s81.9kg200 22212

ext −=−+=W


599

117 •• Picture the Problem Because this is a perfectly inelastic collision, the velocity of the block after the collision is the same as the velocity of the center of mass before the collision. The distance the block travels before hitting the floor is the product of its velocity and the time required to fall 0.8 m; which we can find using a constant-acceleration equation. Relate the distance D to the velocity of the center of mass and the time for the block to fall to the floor:

tvD ∆= cm

Relate the velocity of the center of mass to the total momentum of the system and solve for vcm:

cmi

ii vvPrrr

Mm == ∑

and

blockbullet

blockblockbulletbulletcm mm

vmvmv++

=


( )( ) m/s9.20kg0.8kg0.015m/s500kg0.015

cm =+

=v

Using a constant-acceleration equation, find the time for the block to fall to the floor:

( )

gytv

tatvy

∆=∆=

∆+∆=∆

2 0, Because 0

221

0


gyvD ∆

=2

cm

Substitute numerical values and evaluate D: ( ) ( ) m72.3

m/s9.81m0.82m/s20.9 2 ==D

118 •• Picture the Problem Let the direction the particle whose mass is m is moving initially be the positive x direction and the direction the particle whose mass is 4m is moving initially be the negative y direction. We can determine the impulse delivered by F

r and,

hence, the change in the momentum of the system from the change in the momentum of the particle whose mass is m. Knowing p

r∆ , we can express the final momentum of the

particle whose mass is 4m and solve for its final velocity. Express the impulse delivered by the force F

r: ( ) iii

pppFIˆ3ˆˆ4

if

mvmvvm

T

=−=

−=∆==rrrrr

Chapter 8

600

Express m4'pr : ( )ij

ppvpˆ3ˆ4

04 m44m

mvmv

'm'

+−=

∆+==rrrr

Solve for 'vr : jiv ˆˆ

43 vv' −=

r

119 •• Picture the Problem Let the numeral 1 refer to the basketball and the numeral 2 to the baseball. The left-hand side of the diagram shows the balls after the basketball’s elastic collision with the floor and just before they collide. The right-hand side of the diagram shows the balls just after their collision. We can apply conservation of momentum and the definition of an elastic collision to obtain equations relating the initial and final velocities of the masses of the colliding objects that we can solve for v1f and v2f.

(a) Because both balls are in free-fall, and both are in the air for the same amount of time, they have the same velocity just before the basketball rebounds. After the basketball rebounds elastically, its velocity will have the same magnitude, but the opposite direction than just before it hit the ground.

baseball. theof velocity thetodirection in

oppositebut magnitudein equal be willbasketball theof velocity The

(b) Apply conservation of momentum to the collision of the balls to obtain:

2i21i1f22f11 vmvmvmvm +=+ (1)

Relate the initial and final kinetic energies of the balls in their elastic collision:

2i222

12i112

12f222

12f112

1 vmvmvmvm +=+

Rearrange this equation and factor to obtain:

( ) ( )2f1

2i11

2i2

2f22 vvmvvm −=−

or ( )( )

( )( )1fi11fi11

2if22if22

vvvvmvvvvm

+−=+−

(2)

Rearrange equation (1) to obtain:

( ) ( )1f1i12i2f2 vvmvvm −=− (3)

Divide equation (2) by equation (3) to obtain:

1fi12if2 vvvv +=+


601

Rearrange this equation to obtain equation (4):

1ii2f2f1 vvvv −=− (4)

Multiply equation (4) by m2 and add it to equation (1) to obtain:

( ) ( ) 2i21i211f21 2 vmvmmvmm +−=+


21

2i1

21

21f1

2 vmm

mvmmmmv

++

+−

=

or, because v2i = −v1i,

i121

21

i121

2i1

21

21f1

3

2

vmmmm

vmm

mvmmmmv

+−

=

+−

+−

=

For m1 = 3m2 and v1i = v: 0

333

22

22f1 =

+−

= vmmmmv

(c) Multiply equation (4) by m1 and subtract it from equation (1) to obtain:

( ) ( ) 1i1i212f221 2 vmvmmvmm +−=+


21

12i1

21

1f2

2 vmmmmv

mmmv

+−

++

=

or, because v2i = −v1i,

i121

21

i121

12i1

21

1f2

3

2

vmmmm

vmmmmv

mmmv

+−

=

+−

−+

=

For m1 = 3m2 and v1i = v:

( ) vvmm

mmv 2333

22

22f2 =

+−

=

Chapter 8

602

120 ••• Picture the Problem In Problem 119 only two balls are dropped. They collide head on, each moving at speed v, and the collision is elastic. In this problem, as it did in Problem 119, the solution involves using the conservation of momentum equation

2i21i1f22f11 vmvmvmvm +=+ and the elastic collision equation

,1ii2f2f1 vvvv −=− where the numeral 1 refers to the baseball, and the numeral 2 to the top ball. The diagram shows the balls just before and just after their collision. From Problem 119 we know that that v1i = 2v and v2i = −v.

(a) Express the final speed v1f of the baseball as a function of its initial speed v1i and the initial speed of the top ball v2i (see Problem 78):

i221

2i1

21

21f1

2 vmm

mvmmmmv

++

+−

=

Substitute for v1i and , v2i to obtain: ( ) ( )v

mmmv

mmmmv −

++

+−

=21

2

21

21f1

22

Divide the numerator and denominator of each term by m2 to introduce the mass ratio of the upper ball to the lower ball:

( ) ( )v

mmv

mmmm

v −+

++

−=

1

221

1

2

1

2

1

2

1

f1

Set the final speed of the baseball v1f equal to zero, let x represent the mass ratio m1/m2, and solve for x:

( ) ( )vx

vxx

−+

++−

=1

22110

and

21

2

1 ==mmx

(b) Apply the second of the two equations in Problem 78 to the collision between the top ball and the baseball:

2i21

12i1

21

1f2

2 vmmmmv

mmmv

+−

++

=

Substitute v1i = 2v and are given that v2i = −v to obtain: ( ) ( )v

mmmmv

mmmv −

+−

++

=21

12

21

1f2 22


603

In part (a) we showed that m2 = 2m1. Substitute and simplify:

( ) ( )

( )

v

vvvmmv

mm

vmmmmv

mmmv

37

31

38

1

1

1

1

11

11

11

13f

32

34

222

222

=

−=−=

+−

−+

=

*121 •• Picture the Problem Let the direction the probe is moving after its elastic collision with Saturn be the positive direction. The probe gains kinetic energy at the expense of the kinetic energy of Saturn. We’ll relate the velocity of approach relative to the center of mass to urec and then to v. (a) Relate the velocity of recession to the velocity of recession relative to the center of mass:

cmrec vuv +=

Find the velocity of approach: km/s0.20

km/s0.41km/s9.6app

−=

−−=u

Relate the relative velocity of approach to the relative velocity of recession for an elastic collision:

km/s0.20apprec =−= uu

Because Saturn is so much more massive than the space probe:

km/s6.9Saturncm == vv

Substitute and evaluate v:

km/s29.6

km/s9.6km/s02cmrec

=

+=+= vuv

(b) Express the ratio of the final kinetic energy to the initial kinetic energy:

10.8km/s10.4km/s29.6

2

2

i

rec2i2

1

2rec2

1

i

f

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛==

vv

MvMv

KK

Saturn. of slowing smallly immeasuraban from comesenergy The

Chapter 8

604

*122 •• Picture the Problem We can use the relationships mcP ∆= and 2mcE ∆=∆ to show that .cEP ∆= We can then equate this expression with the change in momentum of the flashlight to find the latter’s final velocity. (a) Express the momentum of the mass lost (i.e., carried away by the light) by the flashlight:

mcP ∆=

Relate the energy carried away by the light to the mass lost by the flashlight:

2cEm ∆

=∆

Substitute to obtain: c

EcEcP ∆

=∆

= 2

(b) Relate the final momentum of the flashlight to ∆E:

mvpcE

=∆=∆

because the flashlight is initially at rest.

Solve for v:

mcEv ∆

=

Substitute numerical values and evaluate v: ( )( )

m/s33.3

m/s1033.3m/s102.998kg1.5

J101.5

6

8

3

µ=

×=

××

=

−

v

123 • Picture the Problem We can equate the change in momentum of the block to the momentum of the beam of light and relate the momentum of the beam of light to the mass converted to produce the beam. Combining these expressions will allow us to find the speed attained by the block. Relate the change in momentum of the block to the momentum of the beam:

( ) beamPvmM =− because the block is initially at rest.

Express the momentum of the mass converted into a well-collimated beam of light:

mcP =beam


( ) mcvmM =−

Solve for v:

mMmcv−

=


605


( )( )

m/s1000.3

kg0.001kg1m/s102.998kg0.001

5

8

×=

−×

=v

124 •• Picture the Problem Let the origin of the coordinate system be at the end of the boat at which your friend is sitting prior to changing places. If we let the system include you and your friend, the boat, the water and the earth, then Fext,net = 0 and the center of mass is at the same location after you change places as it was before you shifted. Express the center of mass of the system prior to changing places:

( )mmmmxmmx

mmmmxxmxm

x

++++

=

++++

=

youboat

friendyouboatyou

friendyouboat

friendyouyouboatboatcm

Substitute numerical values and simplify to obtain an expression for xcm in terms of m:

( )( ) ( )

m

mmx

+⋅

=

++++

=

kg140mkg280

kg80kg600kg80kg60m2

cm

Find the center of mass of the system after changing places:

( )( ) ( )mmm

mmmm

mmmmm

mxxmxmx'

++±

+++±+

=++

++=

youboat

you

youboat

boat

friendyouboat

friendyouyouboatboatcm

m0.2m0.2m2

Substitute numerical values and simplify to obtain:

( )( ) ( )( )

( )m

mmmmm

mx'

+⋅±±

+

+⋅±⋅

=++

±+

++±+

=

kg140mkg16m2.0m2

kg140mkg12mkg120

kg80kg60m0.2kg80

kg80kg60m0.2m2kg60

cm

Because Fext,net = 0, cmcm xx' = .

Equate the two expressions and solve for m to obtain:

( )( ) kg

0.2228160

±±

=m

Calculate the largest possible mass for your friend:

( )( ) kg104kg

0.2228160

=−

+=m

Chapter 8

606

Calculate the smallest possible mass for your friend:

( )( ) kg0.60kg

0.2228160

=+

−=m

125 •• Picture the Problem Let the system include the woman, both vehicles, and the earth. Then Fext,net = 0 and acm = 0. Include the mass of the man in the mass of the truck. We can use Newton’s 2nd and 3rd laws to find the acceleration of the truck and net force acting on both the car and the truck. (a) Relate the action and reaction forces acting on the car and truck:

truckcar FF =

or truckwomantruckcarcar amam +=

Solve for the acceleration of the truck:

womantruck

carcartruck

+

=m

ama

Substitute numerical values and evaluate atruck:

( )( ) 22

truck m/s600.0kg1600

m/s1.2kg800==a

(b) Apply Newton’s 2nd law to either vehicle to obtain:

carcarnet amF =

Substitute numerical values and evaluate Fnet:

( )( ) N960m/s1.2kg800 2net ==F

126 •• Picture the Problem Let the system include the block, the putty, and the earth. Then Fext,net = 0 and momentum is conserved in this perfectly inelastic collision. We’ll use conservation of momentum to relate the after-collision velocity of the block plus blob and conservation of energy to find their after-collision velocity. Noting that, because this is a perfectly elastic collision, the final velocity of the block plus blob is the velocity of the center of mass, use conservation of momentum to relate the velocity of the center of mass to the velocity of the glob before the collision:

cmglgl

fi

orMvvm

pp

=

=

where M = mgl + mbl.


607

Solve for vgl to obtain: cm

glgl v

mMv = (1)

Use conservation of energy to find the initial energy of the block plus glob:

0f =+∆+∆ WUK

Because ∆U = Kf = 0, 0k

2cm2

1 =∆+− xfMv

Use fk = µkMg to eliminate fk and solve for vcm:

xgv ∆= kcm 2µ


( )( )( )m/s1.08

m0.15m/s9.810.42 2cm

=

=v

Substitute numerical values in equation (1) and evaluate vgl:

( )

m/s36.2

m/s1.08kg0.4

kg0.4kg13gl

=

+=v

*127 •• Picture the Problem Let the direction the moving car was traveling before the collision be the positive x direction. Let the numeral 1 denote this car and the numeral 2 the car that is stopped at the stop sign and the system include both cars and the earth. We can use conservation of momentum to relate the speed of the initially-moving car to the speed of the meshed cars immediately after their perfectly inelastic collision and conservation of energy to find the initial speed of the meshed cars. Using conservation of momentum, relate the before-collision velocity to the after-collision velocity of the meshed cars:

( )Vmmvm

pp

2111

fi

or+=

=

Solve for v1: VmmV

mmmv ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

+=

1

2

1

211 1

Using conservation of energy, relate the initial kinetic energy of the meshed cars to the work done by friction in bringing them to a stop:

0thermal =∆+∆ EK

or, because Kf = 0 and ∆Ethermal = f∆s, 0ki =∆+− sfK

Substitute for Ki and, using fk = µkFn = µkMg, eliminate fk to

0k2

21 =∆+− xMgMV µ

Chapter 8

608

obtain:

Solve for V: xgV ∆= k2µ


xgmmv ∆⎟⎟

⎠

⎞⎜⎜⎝

⎛+= k

1

21 21 µ


( )( )( ) km/h23.3m/s48.6m0.76m/s9.810.922kg1200kg9001 2

1 ==⎟⎟⎠

⎞⎜⎜⎝

⎛+=v

km/h. 23.3at traveling wasHe truth. thegnot tellin driver was The

128 •• Picture the Problem Let the zero of gravitational potential energy be at the lowest point of the bob’s swing and note that the bob can swing either forward or backward after the collision. We’ll use both conservation of momentum and conservation of energy to relate the velocities of the bob and the block before and after their collision. Express the kinetic energy of the block in terms of its after-collision momentum:

mpK m

2

2

m =

Solve for m to obtain:

m

m

Kpm

2

2

= (1)

Use conservation of energy to relate Km to the change in the potential energy of the bob:

0=∆+∆ UK or, because Ki = 0,

0if =−+ UUKm

Solve for Km:

( ) ( )[ ][ ]ifbob

fibob

if

coscoscos1cos1

θθθθ

−=−−−=

+−=

gLmLLgm

UUKm

Substitute numerical values and evaluate Km:

( )( )( )[ ] J2.47cos53cos5.73m1.6m/s9.81kg0.4 2 =°−°=mK


609

Use conservation of energy to find the velocity of the bob just before its collision with the block:

0=∆+∆ UK or, because Ki = Uf = 0,

0if =−UK

( )

( )i

ibob2

bob21

cos12

or0cos1

θ

θ

−=

=−−∴

gLv

gLmvm


( )( )( )m/s3.544

cos531m1.6m/s9.812 2

=

°−=v

Use conservation of energy to find the velocity of the bob just after its collision with the block:


0fi =+− UK

Substitute for Ki and Uf to obtain: ( ) 0cos1' fbob

2bob2

1 =−+− θgLmvm

Solve for v′: ( )fcos12' θ−= gLv

Substitute numerical values and evaluate v′:

( )( )( )m/s396.0

cos5.731m1.6m/s9.812' 2

=

°−=v

Use conservation of momentum to relate pm after the collision to the momentum of the bob just before and just after the collision:

mpvmvm

pp

±=

=

'or

bobbob

fi

Solve for and evaluate pm:

( )( )m/skg0.158m/skg.4181m/s0.396m/s3.544kg0.4

'bobbob

⋅±⋅=±=

±= vmvmpm

Find the larger value for pm:

m/skg1.576m/skg0.158m/skg.4181

⋅=⋅+⋅=mp

Find the smaller value for pm:

m/skg1.260m/skg0.158m/skg.4181

⋅=⋅−⋅=mp

Substitute in equation (1) to determine the two values for m:

( )( ) kg503.0

J47.22m/skg576.1 2

=⋅

=m

Chapter 8

610

or ( )

( ) kg321.0J47.22m/skg260.1 2

=⋅

=m

129 •• Picture the Problem Choose the zero of gravitational potential energy at the location of the spring’s maximum compression. Let the system include the spring, the blocks, and the earth. Then the net external force is zero as is work done against friction. We can use conservation of energy to relate the energy transformations taking place during the evolution of this system. Apply conservation of energy: 0sg =∆+∆+∆ UUK

Because ∆K = 0:

0sg =∆+∆ UU

Express the change in the gravitational potential energy:

θsing MgxhmgU −∆−=∆

Express the change in the potential energy of the spring:

221

s kxU =∆


0sin 221 =+−∆− kxMgxhmg θ

Solve for M: x

hmgkx

gxhmgkxM ∆

−=°∆−

=2

30sin

221

Relate ∆h to the initial and rebound positions of the block whose mass is m:

( ) m720.030sinm56.2m4 =°−=∆h

Substitute numerical values and evaluate M:

( ) ( ) ( )( ) kg8.85m0.04

m0.72kg12m/s9.81

m0.04N/m10112

3

=−×

=M

*130 •• Picture the Problem By symmetry, xcm = 0. Let σ be the mass per unit area of the disk. The mass of the modified disk is the difference between the mass of the whole disk and the mass that has been removed.


611

Start with the definition of ycm:

hole

holeholediskdisk

hole

iii

cm

mMymym

mM

ymy

−−

=

−=

∑

Express the mass of the complete disk: 2rAM σπσ ==

Express the mass of the material removed:

Mrrm 412

41

2

hole 2==⎟

⎠⎞

⎜⎝⎛= σπσπ

Substitute and simplify to obtain: ( ) ( )( ) r

MMrMMy 6

1

41

21

41

cm0

=−

−−=

131 •• Picture the Problem Let the horizontal axis by the y axis and the vertical axis the z axis. By symmetry, xcm = ycm = 0. Let ρ be the mass per unit volume of the sphere. The mass of the modified sphere is the difference between the mass of the whole sphere and the mass that has been removed. Start with the definition of ycm:

hole

holeholespheresphere

hole

iii

cm

mMymym

mM

ymz

−

−=

−=

∑

Express the mass of the complete sphere:

334 rVM ρπρ ==

Express the mass of the material removed: ( ) Mrrm 8

1334

81

3

34

hole 2==⎟

⎠⎞

⎜⎝⎛= ρπρπ

Substitute and simplify to obtain: ( ) ( )( ) r

MMrMMz 14

1

81

21

81

cm0

=−

−−=

*132 •• Picture the Problem In this elastic head-on collision, the kinetic energy of recoiling nucleus is the difference between the initial and final kinetic energies of the neutron. We can derive the indicated results by using both conservation of energy and conservation of momentum and writing the kinetic energies in terms of the momenta of the particles before and after the collision.

Chapter 8

612

(a) Use conservation of energy to relate the kinetic energies of the particles before and after the collision:

Mp

mp

mp

222

2nucleus

2nf

2ni += (1)

Apply conservation of momentum to obtain a second relationship between the initial and final momenta:

nucleusnfni ppp += (2)

Eliminate pnf in equation (1) using equation (2):

022

ninucleusnucleus =−+mp

mp

Mp

(3)

Use equation (3) to write mp 22

ni in

terms of pnucleus:

( )mM

mMpKm

p2

22nucleus

n

2ni

82+

== (4)

Use equation (4) to express MpK 22

nucleusnucleus = in terms of

Kn:

( ) ⎥⎦

⎤⎢⎣

⎡

+= 2nnucleus

4mM

MmKK (5)

(b) Relate the change in the kinetic energy of the neutron to the after-collision kinetic energy of the nucleus:

nucleusn KK −=∆

Using equation (5), express the fraction of the energy lost in the collision: ( ) 22

n

n

1

44

⎟⎠⎞

⎜⎝⎛ +

=+

=∆−

MmMm

mMMm

KK

133 •• Picture the Problem Problem 132 (b) provides an expression for the fractional loss of energy per collision. (a) Using the result of Problem 132 (b), express the fractional loss of energy per collision:

( )( )2

2

0

nni

ni

nf

mMmM

EKK

KK

+−

=∆−

=

Evaluate this fraction to obtain: ( )

( )716.0

1212

2

2

0

nf =+−

=mmmm

EK

Express the kinetic energy of one

0nf 716.0 EK N=


613

neutron after N collisions: (b) Substitute for Knf and E0 to obtain:

810716.0 −=N

Take the logarithm of both sides of the equation and solve for N:

550.716log

8N ≈−

=

134 •• Picture the Problem We can relate the number of collisions needed to reduce the energy of a neutron from 2 MeV to 0.02 eV to the fractional energy loss per collision and solve the resulting exponential equation for N. (a) Using the result of Problem 132 (b), express the fractional loss of energy per collision: 37.0

63.0

ni

nini

0

nni

ni

nf

=

−=

∆−=

KKK

EKK

KK

Express the kinetic energy of one neutron after N collisions:

0nf 37.0 EK N=

Substitute for Knf and E0 to obtain:

81037.0 −=N


190.37log8

≈−

=N

(b) Proceed as in (a) to obtain:

89.0

11.0

ni

nini

0

nni

ni

nf

=

−=

∆−=

KKK

EKK

KK

Express the kinetic energy of one neutron after N collisions:

0nf 89.0 EK N=

Substitute for Knf and E0 to obtain:

81089.0 −=N


1580.89log8

≈−

=N

Chapter 8

614

135 •• Picture the Problem Let λ = M/L be the mass per unit length of the rope, the subscript 1 refer to the portion of the rope that is being supported by the force F at any given time, and the subscript 2 refer to the rope that is still on the table at any given time. We can find the height hcm of the center of mass as a function of time and then differentiate this expression twice to find the acceleration of the center of mass. (a) Apply the definition of the center of mass to obtain:

Mhmhm

h cm2,2cm1,1cm

+= (1)

From the definition of λ we have:

vtm

LM 1= ⇒ vt

LMm =1

h1,cm and h2,cm are given by : vth

21

cm1, = and h2,cm = 0

Substitute for m1, h1,cm, and h2,cm in equation (1) and simplify to obtain:

( )2

22cm1,

cm 2

0t

Lv

M

mhvtLM

h =+⎟

⎠⎞

⎜⎝⎛

=

(b) Differentiate hcm twice to obtain acm:

Lva

dthd

tLvt

Lv

dtdh

2

cm2cm

2

22cm

and2

2

==

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

(c) Letting N represent the normal force that the table exerts on the rope, apply ∑ = cmmaFy to the

rope to obtain:

cmMaMgNF =−+

Solve for F, substitute for acm and N to obtain: gm

LvMMg

NMaMgF

2

2cm

−+=

−+=

Use the definition of λ again to obtain:

LM

vtLm

=−

2 ⇒ ⎟⎠⎞

⎜⎝⎛ −=

LvtMm 12


615

Substitute for m2 and simplify:

Mggtv

Lvt

Lvt

gLvMgg

Lvtg

LvgMg

LvtM

LvMMgF

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+=⎟

⎠⎞

⎜⎝⎛ −−+=

1

1222

136 •• Picture the Problem The free-body diagram shows the forces acting on the platform when the spring is partially compressed. The scale reading is the force the scale exerts on the platform and is represented on the FBD by Fn. We can use Newton’s 2nd law to determine the scale reading in part (a). We’ll use both conservation of energy and momentum to obtain the scale reading when the ball collides inelastically with the cup.

(a) Apply ∑ = yy maF to the

spring when it is compressed a distance d:

0springonballpn =−− FgmF

Solve for Fn:

( )gmmgmgm

kgm

kgm

kdgm

FgmF

bpbp

bp

p

springonballpn

+=+=

⎟⎠⎞

⎜⎝⎛+=

+=

+=

(b) Letting the zero of gravitational energy be at the initial elevation of the cup and vbi represent the velocity of the ball just before it hits the cup, use conservation of energy to find this velocity:

0 where0 gfig ===∆+∆ UKUK

ghv

mghvm

2

and0

bi

2bib2

1

=

=−∴

Use conservation of momentum to fi pp rr=

Chapter 8

616

find the velocity of the center of mass: ⎥

⎦

⎤⎢⎣

⎡+

=+

=∴cb

b

cb

bibcm 2

mmmgh

mmvmv

Apply conservation of energy to the collision to obtain:

0scm =∆+∆ UK

or, with Kf = Usi = 0, ( ) 02

212

cmcb21 =++− kxvmm

Substitute for vcm and solve for kx2: ( )

( )

cb

2b

2

cb

bcb

2cmcb

2

2

2

mmghm

mmmmmgh

vmmkx

+=

⎥⎦

⎤⎢⎣

⎡+

+=

+=

Solve for x:

( )cbb

2mmk

ghmx+

=

From part (a):

( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛

++=

++=

+=

cbbp

cbbp

pn

2

2

mmgkhmmg

mmkghkmgm

kxgmF

(c) height. original its toreturnsnever ball theinelastic, iscollision theBecause

137 •• Picture the Problem Let the direction that astronaut 1 first throws the ball be the positive direction and let vb be the initial speed of the ball in the laboratory frame. Note that each collision is perfectly inelastic. We can apply conservation of momentum and the definition of the speed of the ball relative to the thrower to each of the perfectly inelastic collisions to express the final speeds of each astronaut after one throw and one catch. Use conservation of momentum to relate the speeds of astronaut 1 and the ball after the first throw:

0bb11 =+ vmvm (1)

Relate the speed of the ball in the laboratory frame to its speed relative

1b vvv −= (2)


617

to astronaut 1: Eliminate vb between equations (1) and (2) and solve for v1:

vmm

mv

b1

b1 +

−= (3)

Substitute equation (3) in equation (2) and solve for vb:

vmm

mvb1

1b +

= (4)

Apply conservation of momentum to express the speed of astronaut 2 and the ball after the first catch:

( ) 2b2bb0 vmmvm +== (5)

Solve for v2: b

b2

b2 v

mmm

v+

= (6)

Express v2 in terms of v by substituting equation (4) in equation (6):

( )( ) vmmmm

mm

vmm

mmm

mv

⎥⎦

⎤⎢⎣

⎡++

=

++=

b1b2

1b

b1

1

b2

b2

(7)

Use conservation of momentum to express the speed of astronaut 2 and the ball after she throws the ball:

( ) 2f2bfb2b2 vmvmvmm +=+ (8)

Relate the speed of the ball in the laboratory frame to its speed relative to astronaut 2:

bf2f vvv −= (9)

Eliminate vbf between equations (8) and (9) and solve for v2f: v

mmm

mmmv ⎥

⎦

⎤⎢⎣

⎡+

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

=b1

1

b2

bf2 1 (10)

Substitute equation (10) in equation (9) and solve for vbf:

vmm

m

mmmv

⎥⎦

⎤⎢⎣

⎡+

+×

⎥⎦

⎤⎢⎣

⎡+

−−=

b1

1

b2

bbf

1

1 (11)

Apply conservation of momentum to express the speed of astronaut 1 and the ball after she catches the ball:

( ) 11bfb1fb1 vmvmvmm +=+ (12)

Chapter 8

618

Using equations (3) and (11), eliminate vbf and v1 in equation (12) and solve for v1f:

( )( ) ( )

vmmmm

mmmmv

b22

b1

b1b21f

2++

+−=

*138 •• Picture the Problem We can use the definition of the center of mass of a system containing multiple objects to locate the center of mass of the earth−moon system. Any object external to the system will exert accelerating forces on the system. (a) Express the center of mass of the earth−moon system relative to the center of the earth:

∑=i

iicm rrrr

mM

or ( )

1

0

m

e

em

me

emm

me

emmecm

+=

+=

++

=

mM

rmM

rmmM

rmMr

Substitute numerical values and evaluate rcm:

km467013.81km1084.3 5

cm =+

×=r

earth. theof surface thebelow is systemmoon earth theof mass ofcenter theofposition theearth, theof radius than theless is distance thisBecause

−

(b) planets.other andsun thee.g.,

system, on the forces exerts systemmoon earth in thenot object Any −

(c) sun. the towardis system theofon accelerati thesystem,

moon earth on the force externaldominant theexertssun theBecause −

(d) Because the center of mass is at a fixed distance from the sun, the distance d moved by the earth in this time interval is:

( ) km9340km467022 em === rd

139 •• Picture the Problem Let the numeral 2 refer to you and the numeral 1 to the water leaving the hose. Apply conservation of momentum to the system consisting of yourself, the water, and the earth and then differentiate this expression to relate your recoil acceleration to your mass, the speed of the water, and the rate at which the water is


619

leaving the hose. Use conservation of momentum to relate your recoil velocity to the velocity of the water leaving the hose:

0or

0

2211

21

=+

=+

vmvm

pp rr

Differentiate this expression with respect to t:

022

22

11

11 =+++

dtdmv

dtdvm

dtdmv

dtdvm

or

0222

1111 =+++

dtdmvma

dtdmvam

Because the acceleration of the water leaving the hose, a1, is zero …

as is dt

dm2 , the rate at which you are

losing mass: dtdm

mv

a

amdt

dmv

1

2

12

221

1

and

0

−=

=+

Substitute numerical values and evaluate a2:

2

2

m/s960.0

)kg/s(2.4kg75m/s30

−=

−=a

*140 ••• Picture the Problem Take the zero of gravitational potential energy to be at the elevation of the pan and let the system include the balance, the beads, and the earth. We can use conservation of energy to find the vertical component of the velocity of the beads as they hit the pan and then calculate the net downward force on the pan from Newton’s 2nd law. Use conservation of energy to relate the y component of the bead’s velocity as it hits the pan to its height of fall:


0221 =− mghmvy

Solve for vy: ghvy 2=

Substitute numerical values and evaluate vy:

( )( ) m/s3.13m0.5m/s9.812 2 ==yv

Express the change in momentum in the y direction per bead:

( ) yyyyyy mvmvmvppp 2if =−−=−=∆

Chapter 8

620

Use Newton’s 2nd law to express the net force in the y direction exerted on the pan by the beads:

tp

NF yy ∆

∆=net,

Letting M represent the mass to be placed on the other pan, equate its weight to the net force exerted by the beads, substitute for ∆py, and solve for M:

tp

NMg y

∆

∆=

and

⎟⎟⎠

⎞⎜⎜⎝

⎛∆

=g

mvt

NM y2

Substitute numerical values and evaluate M:

( ) ( )( )[ ]

g9.31

m/s9.81m/s3.13kg0.00052s/100 2

=

=M

141 ••• Picture the Problem Assume that the connecting rod goes halfway through both balls, i.e., the centers of mass of the balls are separated by L. Let the system include the dumbbell, the wall and floor, and the earth. Let the zero of gravitational potential be at the center of mass of the lower ball and use conservation of energy to relate the speeds of the balls to the potential energy of the system. By symmetry, the speeds will be equal when the angle with the vertical is 45°. Use conservation of energy to express the relationship between the initial and final energies of the system:

fi EE =

Express the initial energy of the system:

mgLE =i

Express the energy of the system when the angle with the vertical is 45°:

( ) 221

f 245sin vmmgLE +°=


21 vgLgL +⎟

⎠

⎞⎜⎝

⎛=

Solve for v:

⎟⎠⎞

⎜⎝⎛ −=

211gLv


621

Substitute numerical values and evaluate v: ( )

( ) L

Lv

/sm70.1

211m/s81.9

21

2

=

⎟⎠⎞

⎜⎝⎛ −=

Chapter 8

622

623

Chapter 9 Rotation Conceptual Problems *1 • Determine the Concept Because r is greater for the point on the rim, it moves the greater distance. Both turn through the same angle. Because r is greater for the point on the rim, it has the greater speed. Both have the same angular velocity. Both have zero tangential acceleration. Both have zero angular acceleration. Because r is greater for the point on the rim, it has the greater centripetal acceleration. 2 •

(a) False. Angular velocity has the dimensions ⎥⎦⎤

⎢⎣⎡T1

whereas linear velocity has

dimensions ⎥⎦⎤

⎢⎣⎡TL

.

(b) True. The angular velocity of all points on the wheel is dθ/dt. (c) True. The angular acceleration of all points on the wheel is dω/dt. 3 •• Picture the Problem The constant-acceleration equation that relates the given variables is θαωω ∆+= 22

02 . We can set up a proportion to determine the number of revolutions

required to double ω and then subtract to find the number of additional revolutions to accelerate the disk to an angular speed of 2ω. Using a constant-acceleration equation, relate the initial and final angular velocities to the angular acceleration:

θαωω ∆+= 220

2

or, because 20ω = 0,

θαω ∆= 22

Let ∆θ10 represent the number of revolutions required to reach an angular velocity ω:

102 2 θαω ∆= (1)

Let ∆θ2ω represent the number of revolutions required to reach an angular velocity ω:

( ) ωθαω 22 22 ∆= (2)

Divide equation (2) by equation (1) and solve for ∆θ2ω:

( )10102

2

2 42 θθωωθ ω ∆=∆=∆

Chapter 9

624

The number of additional revolutions is: ( ) rev30rev10334 101010 ==∆=∆−∆ θθθ

and correct. is )(c

*4 •

Determine the Concept Torque has the dimension ⎥⎦

⎤⎢⎣

⎡2

2

TML

.

(a) Impulse has the dimension ⎥⎦⎤

⎢⎣⎡

TML

.

(b) Energy has the dimension ⎥⎦

⎤⎢⎣

⎡2

2

TML

. correct. is )(b

(c) Momentum has the dimension ⎥⎦⎤

⎢⎣⎡

TML

.

5 • Determine the Concept The moment of inertia of an object is the product of a constant that is characteristic of the object’s distribution of matter, the mass of the object, and the square of the distance from the object’s center of mass to the axis about which the object is rotating. Because both (b) and (c) are correct correct. is )(d

*6 • Determine the Concept Yes. A net torque is required to change the rotational state of an object. In the absence of a net torque an object continues in whatever state of rotational motion it was at the instant the net torque became zero. 7 • Determine the Concept No. A net torque is required to change the rotational state of an object. A net torque may decrease the angular speed of an object. All we can say for sure is that a net torque will change the angular speed of an object. 8 • (a) False. The net torque acting on an object determines the angular acceleration of the object. At any given instant, the angular velocity may have any value including zero. (b) True. The moment of inertia of a body is always dependent on one’s choice of an axis of rotation. (c) False. The moment of inertia of an object is the product of a constant that is characteristic of the object’s distribution of matter, the mass of the object, and the square of the distance from the object’s center of mass to the axis about which the object is

Rotation

625

rotating. 9 • Determine the Concept The angular acceleration of a rotating object is proportional to the net torque acting on it. The net torque is the product of the tangential force and its lever arm. Express the angular acceleration of the disk as a function of the net torque acting on it:

dIF

IFd

I=== netτα

i.e., d∝α

Because d∝α , doubling d will double the angular acceleration.

correct. is )(b

*10 • Determine the Concept From the parallel-axis theorem we know that

,2cm MhII += where Icm is the moment of inertia of the object with respect to an axis

through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Therefore, I is always greater than Icm by Mh2. correct. is )( d

11 • Determine the Concept The power delivered by the constant torque is the product of the torque and the angular velocity of the merry-go-round. Because the constant torque causes the merry-go-round to accelerate, neither the power input nor the angular velocity of the merry-go-round is constant. correct. is )(b

12 • Determine the Concept Let’s make the simplifying assumption that the object and the surface do not deform when they come into contact, i.e., we’ll assume that the system is rigid. A force does no work if and only if it is perpendicular to the velocity of an object, and exerts no torque on an extended object if and only if it’s directed toward the center of the object. Because neither of these conditions is satisfied, the statement is false. 13 • Determine the Concept For a given applied force, this increases the torque about the hinges of the door, which increases the door’s angular acceleration, leading to the door being opened more quickly. It is clear that putting the knob far from the hinges means that the door can be opened with less effort (force). However, it also means that the hand on the knob must move through the greatest distance to open the door, so it may not be the quickest way to open the door. Also, if the knob were at the center of the door, you would have to walk around the door after opening it, assuming the door is opening toward you.

Chapter 9

626

*14 • Determine the Concept If the wheel is rolling without slipping, a point at the top of the wheel moves with a speed twice that of the center of mass of the wheel, but the bottom of the wheel is momentarily at rest. correct. is )(c

15 •• Picture the Problem The kinetic energies of both objects is the sum of their translational and rotational kinetic energies. Their speed dependence will differ due to the differences in their moments of inertia. We can express the total kinetic of both objects and equate them to decide which of their translational speeds is greater. Express the kinetic energy of the cylinder:

( )2cyl4

3

2cyl2

12

2cyl2

21

21

2cyl2

12cylcyl2

1cyl

mv

mvrv

mr

mvIK

=

+=

+= ω

Express the kinetic energy of the sphere:

( )2sph10

7

2sph2

12

2sph2

52

21

2sph2

12sphlsph2

1sph

mv

mvr

vmr

mvIK

=

+=

+= ω

Equate the kinetic energies and simplify to obtain:

sphsph1514

cyl vvv <=

and correct. is )(b

*16 • Determine the Concept You could spin the pipes about their center. The one which is easier to spin has its mass concentrated closer to the center of mass and, hence, has a smaller moment of inertia. 17 •• Picture the Problem Because the coin and the ring begin from the same elevation, they will have the same kinetic energy at the bottom of the incline. The kinetic energies of both objects is the sum of their translational and rotational kinetic energies. Their speed dependence will differ due to the differences in their moments of inertia. We can express the total kinetic of both objects and equate them to their common potential energy loss to decide which of their translational speeds is greater at the bottom of the incline.

Rotation

627

Express the kinetic energy of the coin at the bottom of the incline:

( )2coincoin4

3

2coincoin2

12

2coin2

coin21

21

2coincoin2

12coincyl2

1coin

vm

vmr

vrm

vmIK

=

+=

+= ω

Express the kinetic energy of the ring at the bottom of the incline:

( )2ringring

2ringring2

12

2ring2

ring21

2ringring2

12ringring2

1ring

vm

vmr

vrm

vmIK

=

+=

+= ω

Equate the kinetic of the coin to its change in potential energy as it rolled down the incline and solve for vcoin:

ghv

ghmvm

342

coin

coin2coincoin4

3

and=

=

Equate the kinetic of the ring to its change in potential energy as it rolled down the incline and solve for vring:

ghv

ghmvm

=

=

2ring

ring2ringring

and

correct. is )( and Therefore, ringcoin bvv >

18 •• Picture the Problem We can use the definitions of the translational and rotational kinetic energies of the hoop and the moment of inertia of a hoop (ring) to express and compare the kinetic energies. Express the translational kinetic energy of the hoop:

221

trans mvK =

Express the rotational kinetic energy of the hoop:

( ) 221

2

22

212

hoop21

rot mvrvmrIK === ω

Therefore, the translational and rotational

kinetic energies are the same and correct. is )( c

Chapter 9

628

19 •• Picture the Problem We can use the definitions of the translational and rotational kinetic energies of the disk and the moment of inertia of a disk (cylinder) to express and compare the kinetic energies. Express the translational kinetic energy of the disk:

221

trans mvK =

Express the rotational kinetic energy of the disk:

( ) 241

2

22

21

212

hoop21

rot mvrvmrIK === ω

Therefore, the translational kinetic energy is

greater and correct. is )( a

20 •• Picture the Problem Let us assume that f ≠ 0 and acts along the direction of motion. Now consider the acceleration of the center of mass and the angular acceleration about the point of contact with the plane. Because Fnet ≠ 0, acm ≠ 0. However, τ = 0 because l = 0, so α = 0. But α = 0 is not consistent with acm ≠ 0. Consequently, f = 0. 21 • Determine the Concept True. If the sphere is slipping, then there is kinetic friction which dissipates the mechanical energy of the sphere. 22 • Determine the Concept Because the ball is struck high enough to have topspin, the frictional force is forward; reducing ω until the nonslip condition is satisfied.

correct. is )(a

Estimation and Approximation 23 •• Picture the Problem Assume the wheels are hoops, i.e., neglect the mass of the spokes, and express the total kinetic energy of the bicycle and rider. Let M represent the mass of the rider, m the mass of the bicycle, mw the mass of each bicycle wheel, and r the radius of the wheels. Express the ratio of the kinetic energy associated with the rotation of the wheels to that associated with the total kinetic energy of the bicycle and rider:

rottrans

rot

tot

rot

KKK

KK

+= (1)

Rotation

629

Express the translational kinetic energy of the bicycle and rider: 2

212

21

riderbicycletrans

Mvmv

KKK

+=

+=

Express the rotational kinetic energy of the bicycle wheels:

( )( ) 2

w2

22

w

2w2

1wheel1rot,rot 22

vmrvrm

IKK

==

== ω


w

w21

21

w2

w2

212

21

2w

tot

rot

2

2

mMmmMm

mvmMvmv

vmKK

++

=++

=++

=

Substitute numerical values and evaluate Krot/Ktot:

%3.10

kg3kg38kg142

2

tot

rot =+

+=

KK

24 •• Picture the Problem We can apply the definition of angular velocity to find the angular orientation of the slice of toast when it has fallen a distance of 0.5 m from the edge of the table. We can then interpret the orientation of the toast to decide whether it lands jelly-side up or down. Relate the angular orientation θ of the toast to its initial angular orientation, its angular velocity ω, and time of fall ∆t:

t∆+= ωθθ 0 (1)

Use the equation given in the problem statement to find the angular velocity corresponding to this length of toast:

rad/s9.470.1m

m/s9.81956.02

==ω

Using a constant-acceleration equation, relate the distance the toast falls ∆y to its time of fall ∆t:

( )221

0 tatvy yy ∆+∆=∆ or, because v0y = 0 and ay = g,

( )221 tgy ∆=∆

Solve for ∆t: g

yt ∆=∆

2

Substitute numerical values and evaluate ∆t:

( ) s0.319m/s9.81

m0.522 ==∆t

Chapter 9

630

( )0

2f cos

2' θ+

gLv

Substitute in equation (1) to

find θ :

( )( )

°=°

×=

+=

203rad

180rad54.3

s0.319rad/s9.476

π

πθ

down. side-jelly with thei.e. ground, therespect towith 203 of anglean at be thereforel toast wilof slice theofn orientatio The °

*25 •• Picture the Problem Assume that the mass of an average adult male is about 80 kg, and that we can model his body when he is standing straight up with his arms at his sides as a cylinder. From experience in men’s clothing stores, a man’s average waist circumference seems to be about 34 inches, and the average chest circumference about 42 inches. We’ll also assume that about 20% of the body’s mass is in the two arms, and each has a length L = 1 m, so that each arm has a mass of about m = 8 kg. Letting Iout represent his moment of inertia with his arms straight out and Iin his moment of inertia with his arms at his side, the ratio of these two moments of inertia is:

in

armsbody

in

out

III

II +

= (1)

Express the moment of inertia of the ″man as a cylinder″:

221

in MRI =

Express the moment of inertia of his arms:

( ) 231

arms 2 mLI =

Express the moment of inertia of his body-less-arms:

( ) 221

body RmMI −=


( ) ( )2

21

2312

21

in

out 2MR

mLRmMII +−

=

Assume the circumference of the cylinder to be the average of the average waist circumference and the average chest circumference:

in382

in42in34av =

+=c

Find the radius of a circle whose circumference is 38 in:

m154.02π

cm100m1

incm2.54in38

2av

=

××==

πcR

Substitute numerical values and evaluate Iout/ Iin:

Rotation

631

( )( ) ( )( )( )( )

42.6m0.154kg80

m1kg8m0.154kg16kg802

21

2322

21

in

out =+−

=II

Angular Velocity and Angular Acceleration 26 • Picture the Problem The tangential and angular velocities of a particle moving in a circle are directly proportional. The number of revolutions made by the particle in a given time interval is proportional to both the time interval and its angular speed. (a) Relate the angular velocity of the particle to its speed along the circumference of the circle:

ωrv =

Solve for and evaluate ω: rad/s0.278m90

m/s25===

rvω

(b) Using a constant-acceleration equation, relate the number of revolutions made by the particle in a given time interval to its angular velocity:

( )

rev33.1

rad2rev1s30

srad278.0

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=∆=∆

πωθ t

27 • Picture the Problem Because the angular acceleration is constant; we can find the various physical quantities called for in this problem by using constant-acceleration equations. (a) Using a constant-acceleration equation, relate the angular velocity of the wheel to its angular acceleration and the time it has been accelerating:

t∆+= αωω 0

or, when ω0 = 0, t∆= αω

Evaluate ω when ∆t = 6 s: ( ) rad/s15.6s62rad/s2.6 =⎟⎠⎞⎜

⎝⎛=ω

(b) Using another constant-acceleration equation, relate the angular displacement to the wheel’s angular acceleration and the time it

( )221

0 tt ∆+∆=∆ αωθ

or, when ω0 = 0, ( )2

21 t∆=∆ αθ

Chapter 9

632

has been accelerating: Evaluate θ∆ when ∆t = 6 s: ( ) ( )( ) rad8.46s6rad/s2.6s6 22

21 ==∆θ

(c) Convert ( )s6θ∆ from rad to

revolutions: ( ) rev45.7

rad2rev1rad8.46s6 =×=∆

πθ

(d) Relate the angular velocity of the particle to its tangential speed and evaluate the latter when ∆t = 6 s:

( )( ) m/s4.68rad/s15.6m0.3 === ωrv

Relate the resultant acceleration of the point to its tangential and centripetal accelerations when ∆t = 6 s:

( ) ( )42

2222c

2t

ωα

ωα

+=

+=+=

r

rraaa

Substitute numerical values and evaluate a:

( ) ( ) ( )2

422

m/s73.0

rad/s15.6rad/s2.6m0.3

=

+=a

*28 • Picture the Problem Because we’re assuming constant angular acceleration; we can find the various physical quantities called for in this problem by using constant-acceleration equations. (a) Using its definition, express the angular acceleration of the turntable:

tt ∆−

=∆∆

= 0ωωωα

Substitute numerical values and evaluate α:

2

31

rad/s0.134

s26s60

min1rev

rad2πminrev330

=

××−=α

Rotation

633

(b) Because the angular acceleration is constant, the average angular velocity is the average of its initial and final values:

rad/s75.12

s60min1

revrad2

minrev33

2

31

0av

=

××=

+=

π

ωωω

(c) Using the definition of ωav, find the number or revolutions the turntable makes before stopping:

( )( )

rev24.7rad2

rev1rad5.45

s26rad/s1.75av

=×=

=∆=∆

π

ωθ t

29 • Picture the Problem Because the angular acceleration of the disk is constant, we can use a constant-acceleration equation to relate its angular velocity to its acceleration and the time it has been accelerating. We can find the tangential and centripetal accelerations from their relationships to the angular velocity and angular acceleration of the disk. (a) Using a constant-acceleration equation, relate the angular velocity of the disk to its angular acceleration and time during which it has been accelerating:

t∆+= αωω 0

or, because ω0 = 0, t∆= αω

Evaluate ω when t = 5 s: ( ) ( )( ) rad/s40.0s5rad/s8s5 2 ==ω

(b) Express at in terms of α:

αra =t

Evaluate at when t = 5 s: ( ) ( )( )2

2t

m/s960.0

rad/s8m0.12s5

=

=a

Express ac in terms of ω:

2c ωra =

Evaluate ac when t = 5 s: ( ) ( )( )2

2c

m/s192

rad/s40.0m0.12s5

=

=a

30 • Picture the Problem We can find the angular velocity of the Ferris wheel from its definition and the linear speed and centripetal acceleration of the passenger from the relationships between those quantities and the angular velocity of the Ferris wheel.

Chapter 9

634

(a) Find ω from its definition: rad/s233.0s27

rad2==

∆∆

=πθω

t

(b) Find the linear speed of the passenger from his/her angular speed:

( )( )m/s79.2

rad/s0.233m12

=

== ωrv

Find the passenger’s centripetal acceleration from his/her angular velocity:

( )( )2

22c

m/s651.0

rad/s0.233m12

=

== ωra

31 • Picture the Problem Because the angular acceleration of the wheels is constant, we can use constant-acceleration equations in rotational form to find their angular acceleration and their angular velocity at any given time. (a) Using a constant-acceleration equation, relate the angular displacement of the wheel to its angular acceleration and the time it has been accelerating:

( )221

0 tt ∆+∆=∆ αωθ

or, because ω0 = 0, ( )2

21 t∆=∆ αθ

Solve for α: ( )22

t∆∆

=θα


( )

( )2

2 rad/s589.0s8

revrad2rev32

=⎟⎠⎞

⎜⎝⎛

=

π

α

(b) Using a constant-acceleration equation, relate the angular velocity of the wheel to its angular acceleration and the time it has been accelerating:

t∆+= αωω 0

or, when ω0 = 0, t∆= αω

Evaluate ω when ∆t = 8 s: ( ) ( )( ) rad/s71.4s8rad/s589.0s8 2 ==ω

Rotation

635

32 • Picture the Problem The earth rotates through 2π radians every 24 hours. Find ω using its definition:

rad/s1027.7h

s3600h24

rad2

5−×=

×=

∆∆

≡πθω

t

33 • Picture the Problem When the angular acceleration of a wheel is constant, its average angular velocity is the average of its initial and final angular velocities. We can combine this relationship with the always applicable definition of angular velocity to find the initial angular velocity of the wheel. Express the average angular velocity of the wheel in terms of its initial and final angular speeds:

20

avωω

ω+

=

or, because ω = 0, 02

1av ωω =

Express the definition of the average angular velocity of the wheel:

t∆∆

≡θω av

Equate these two expressions and solve for ω0:

( ) s57.3s2.8

rad5220 ==

∆∆

=tθω and

correct. is )(d

34 • Picture the Problem The tangential and angular accelerations of the wheel are directly proportional to each other with the radius of the wheel as the proportionality constant. Provided there is no slippage, the acceleration of a point on the rim of the wheel is the same as the acceleration of the bicycle. We can use its defining equation to determine the acceleration of the bicycle. Relate the tangential acceleration of a point on the wheel (equal to the acceleration of the bicycle) to the wheel’s angular acceleration and solve for its angular acceleration:

αraa == t

and

ra

=α

Chapter 9

636

Use its definition to express the acceleration of the wheel: t

vvtva

∆−

=∆∆

= 0

or, because v0 = 0,

tva∆

=

Substitute in the expression for α to obtain: tr

v∆

=α


( )( )2rad/s794.0

s14.0m0.6km

m1000s3600

h1h

km24

=

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

=α

*35 •• Picture the Problem The two tapes will have the same tangential and angular velocities when the two reels are the same size, i.e., have the same area. We can calculate the tangential speed of the tape from its length and running time and relate the angular velocity to the constant tangential speed and the radius of the reels when they are turning with the same angular velocity. Relate the angular velocity of the tape to its tangential speed:

rv

=ω (1)

Letting Rf represent the outer radius of the reel when the reels have the same area, express the condition that they have the same speed:

( )222122

f rRrR ππππ −=−

Solve for Rf:

2

22

frRR +

=

Substitute numerical values and evaluate Rf:

( ) ( ) mm32.92

mm12mm45 22

f =+

=R

Find the tangential speed of the tape from its length and running time: cm/s42.3

hs3600h2

mcm100m246

∆=

×

×==

tLv

Rotation

637

Substitute in equation (1) and evaluate ω:

rad/s1.04

mm10cm1mm32.9

cm/s3.42

f

=

×==

Rvω

Convert 1.04 rad/s to rev/min:

rev/min93.9

mins60

rad2rev1

srad04.1rad/s04.1

=

××=π

Torque, Moment of Inertia, and Newton’s Second Law for Rotation 36 • Picture the Problem The force that the woman exerts through her axe, because it does not act at the axis of rotation, produces a net torque that changes (decreases) the angular velocity of the grindstone. (a) From the definition of angular acceleration we have: tt ∆

−=

∆∆

= 0ωωωα

or, because ω = 0,

t∆−

= 0ωα


2rad/s49.8

s9s60

min1rev

rad2minrev730

−=

××−=

π

α

where the minus sign means that the grindstone is slowing down.

(b) Use Newton’s 2nd law in rotational form to relate the angular acceleration of the grindstone to the net torque slowing it:

ατ I=net

Express the moment of inertia of disk with respect to its axis of rotation:

221 MRI =

Chapter 9

638

Substitute to obtain: ατ MR21

net =

Substitute numerical values and evaluate τnet:

( )( ) ( )mN0462.0

rad/s8.49m0.08kg1.7 2221

net

⋅=

=τ

*37 • Picture the Problem We can find the torque exerted by the 17-N force from the definition of torque. The angular acceleration resulting from this torque is related to the torque through Newton’s 2nd law in rotational form. Once we know the angular acceleration, we can find the angular velocity of the cylinder as a function of time. (a) Calculate the torque from its definition:

( )( ) mN1.87m0.11N17 ⋅=== lFτ

(b) Use Newton’s 2nd law in rotational form to relate the acceleration resulting from this torque to the torque:

Iτα =

Express the moment of inertia of the cylinder with respect to its axis of rotation:

221 MRI =


2MR

τα =


( )( )( )

22 rad/s124

m0.11kg2.5mN1.872

=⋅

=α

(c) Using a constant-acceleration equation, express the angular velocity of the cylinder as a function of time:

tαωω += 0

or, because ω0 = 0, tαω =

Evaluate ω (5 s): ( ) ( )( ) rad/s620s5rad/s124s5 2 ==ω

38 •• Picture the Problem We can find the angular acceleration of the wheel from its definition and the moment of inertia of the wheel from Newton’s 2nd law.

Rotation

639

(a) Express the moment of inertia of the wheel in terms of the angular acceleration produced by the applied torque:

ατ

=I

Find the angular acceleration of the wheel:

2rad/s14.3s20

s60min1

revrad2

minrev600

=

××=

∆∆

=

πωαt

Substitute and evaluate I: 2

2 mkg9.15rad/s3.14

mN50⋅=

⋅=I

(b) Because the wheel takes 120 s to slow to a stop (it took 20 s to acquire an angular velocity of 600 rev/min) and its angular acceleration is directly proportional to the accelerating torque:

( ) mN33.8mN5061

61

fr ⋅=⋅== ττ

39 •• Picture the Problem The pendulum and the forces acting on it are shown in the free-body diagram. Note that the tension in the string is radial, and so exerts no tangential force on the ball. We can use Newton’s 2nd law in both translational and rotational form to find the tangential component of the acceleration of the bob. (a) Referring to the FBD, express the component of g

rm that is tangent

to the circular path of the bob:

θsint mgF =

Use Newton’s 2nd law to express the tangential acceleration of the bob:

θsintt g

mFa ==

(b) Noting that, because the line-of-action of the tension passes through the pendulum’s pivot point, its lever arm is zero and the net torque is due

∑ = θτ sinpointpivot mgL

Chapter 9

640

to the weight of the bob, sum the torques about the pivot point to obtain: (c) Use Newton’s 2nd law in rotational form to relate the angular acceleration of the pendulum to the net torque acting on it:

αθτ ImgL == sinnet

Solve for α to obtain: I

mgL θα sin=

Express the moment of inertia of the bob with respect to the pivot point:

2mLI =

Substitute to obtain: L

gmL

mgL θθα sinsin2 ==

Relate α to at: θθα sinsin

t gL

gLra =⎟⎠⎞

⎜⎝⎛==

*40 ••• Picture the Problem We can express the velocity of the center of mass of the rod in terms of its distance from the pivot point and the angular velocity of the rod. We can find the angular velocity of the rod by using Newton’s 2nd law to find its angular acceleration and then a constant-acceleration equation that relates ω to α. We’ll use the impulse-momentum relationship to derive the expression for the force delivered to the rod by the pivot. Finally, the location of the center of percussion of the rod will be verified by setting the force exerted by the pivot to zero. (a) Relate the velocity of the center of mass to its distance from the pivot point:

ω2cmLv = (1)

Express the torque due to F0:

ατ pivot0 IxF ==

Solve for α: pivot

0

IxF

=α

Express the moment of inertia of the rod with respect to an axis through

231

pivot MLI =

Rotation

641

its pivot point: Substitute to obtain:

203

MLxF

=α

Express the angular velocity of the rod in terms of its angular acceleration:

203

MLtxFt ∆

=∆= αω


MLtxF

v2

3 0cm

∆=

(b) Let IP be the impulse exerted by the pivot on the rod. Then the total impulse (equal to the change in momentum of the rod) exerted on the rod is:

cm0P MvtFI =∆+

and tFMvI ∆−= 0cmP

Substitute our result from (a) to obtain:

⎟⎠⎞

⎜⎝⎛ −∆=∆−

∆= 1

23

23

000

P LxtFtF

LtxFI

Because tFI ∆= PP :

⎟⎠⎞

⎜⎝⎛ −= 1

23

0P LxFF

In order for FP to be zero:

0123

=−Lx

⇒3

2Lx =

41 ••• Picture the Problem We’ll first express the torque exerted by the force of friction on the elemental disk and then integrate this expression to find the torque on the entire disk. We’ll use Newton’s 2nd law to relate this torque to the angular acceleration of the disk and then to the stopping time for the disk. (a) Express the torque exerted on the elemental disk in terms of the friction force and the distance to the elemental disk:

kf rdfd =τ (1)

Using the definition of the coefficient of friction, relate the

gdmdf kk µ= (2)

Chapter 9

642

force of friction to µk and the weight of the circular element: Letting σ represent the mass per unit area of the disk, express the mass of the circular element:

drrdm σπ2= (3)

Substitute equations (2) and (3) in (1) to obtain:

drrgd 2kf 2 σµπτ = (4)

Because 2RM

πσ = : drr

RgMd 2

2k

f2µτ =

(b) Integrate fτd to obtain the total

torque on the elemental disk: gMRdrr

RgM R

k32

0

22

kf

2 µµτ == ∫

(c) Relate the disk’s stopping time to its angular velocity and acceleration:

αω

=∆t

Using Newton’s 2nd law, express α in terms of the net torque acting on the disk:

Ifτ

α =

The moment of inertia of the disk, with respect to its axis of rotation, is:

221 MRI =

Substitute and simplify to obtain: g

Rtk4

3µ

ω=∆

Calculating the Moment of Inertia 42 • Picture the Problem One can find the formula for the moment of inertia of a thin spherical shell in Table 9-1. The moment of inertia of a thin spherical shell about its diameter is:

232 MRI =

Rotation

643


( )( )25

232

mkg104.66

m0.035kg0.057

⋅×=

=−

I

*43 • Picture the Problem The moment of inertia of a system of particles with respect to a given axis is the sum of the products of the mass of each particle and the square of its distance from the given axis. Use the definition of the moment of inertia of a system of particles to obtain:

244

233

222

211

i

2ii

rmrmrmrm

rmI

+++=

= ∑


( )( ) ( )( )( )( ) ( )( )

2

22

22

mkg0.56

m2kg30kg4

m22kg4m2kg3

⋅=

++

+=I

44 • Picture the Problem Note, from symmetry considerations, that the center of mass of the system is at the intersection of the diagonals connecting the four masses. Thus the distance of each particle from the axis through the center of mass is m2 . According to the parallel-axis theorem, 2

cm MhII += , where Icm is the moment of inertia of the

object with respect to an axis through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Express the parallel axis theorem:

2cm MhII +=

Solve for Icm and substitute from Problem 44: ( )( )

2

22

2cm

mkg28.0

m2kg14mkg6.05

⋅=

−⋅=

−= MhII

Use the definition of the moment of inertia of a system of particles to express Icm:

244

233

222

211

i

2iicm

rmrmrmrm

rmI

+++=

= ∑

Substitute numerical values and evaluate Icm:

( )( ) ( )( )( )( ) ( )( )

2

22

22

cm

mkg0.28

m2kg3m2kg4

m2kg4m2kg3

⋅=

++

+=I

Chapter 9

644

45 • Picture the Problem The moment of inertia of a system of particles with respect to a given axis is the sum of the products of the mass of each particle and the square of its distance from the given axis. (a) Apply the definition of the moment of inertia of a system of particles to express Ix:

244

233

222

211

i

2ii

rmrmrmrm

rmI x

+++=

= ∑

Substitute numerical values and evaluate Ix:

( )( ) ( )( )( )( ) ( )( )

2

22

mkg0.28

0kg30kg4m2kg4m2kg3

⋅=

+++=xI

(b) Apply the definition of the moment of inertia of a system of particles to express Iy:

244

233

222

211

i

2ii

rmrmrmrm

rmI y

+++=

= ∑

Substitute numerical values and evaluate Iy:

( )( ) ( )( )( )( ) ( )( )

2

2

2

mkg0.28

m2kg30kg4

m2kg40kg3

⋅=

++

+=yI

Remarks: We could also use a symmetry argument to conclude that Iy = Ix . 46 • Picture the Problem According to the parallel-axis theorem, ,2

cm MhII += where Icm

is the moment of inertia of the object with respect to an axis through its center of mass, M is the mass of the object, and h is the distance between the parallel axes. Use Table 9-1 to find the moment of inertia of a sphere with respect to an axis through its center of mass:

252

cm MRI =

Express the parallel axis theorem:

2cm MhII +=

Substitute for Icm and simplify to obtain:

25722

52 MRMRMRI =+=

Rotation

645

47 •• Picture the Problem The moment of inertia of the wagon wheel is the sum of the moments of inertia of the rim and the six spokes. Express the moment of inertia of the wagon wheel as the sum of the moments of inertia of the rim and the spokes:

spokesrimwheel III +=

Using Table 9-1, find formulas for the moments of inertia of the rim and spokes: 2

spoke31

spoke

2rimrim

andLMI

RMI

=

=

Substitute to obtain: ( )

2spoke

2rim

2spoke3

12rimwheel

2

6

LMRM

LMRMI

+=

+=

Substitute numerical values and evaluate Iwheel:

( )( ) ( )( )2

22wheel

mkg60.2

m0.5kg1.22m5.0kg8

⋅=

+=I

*48 •• Picture the Problem The moment of inertia of a system of particles depends on the axis with respect to which it is calculated. Once this choice is made, the moment of inertia is the sum of the products of the mass of each particle and the square of its distance from the chosen axis. (a) Apply the definition of the moment of inertia of a system of particles:

( )22

21

i

2ii xLmxmrmI −+== ∑

(b) Set the derivative of I with respect to x equal to zero in order to identify values for x that correspond to either maxima or minima:

( )( )

( )extremafor 0

2

122

221

21

=−+=

−−+=

Lmxmxm

xLmxmdxdI

If 0=dxdI

, then:

0221 =−+ Lmxmxm

Chapter 9

646

Solve for x:

21

2

mmLmx

+=

Convince yourself that you’ve found

a minimum by showing that 2

2

dxId

is

positive at this point. . from mass ofcenter theof distance

the,definitionby is, 21

2

mmmLmx

+=

49 •• Picture the Problem Let σ be the mass per unit area of the uniform rectangular plate. Then the elemental unit has mass dm = σ dxdy. Let the corner of the plate through which the axis runs be the origin. The distance of the element whose mass is dm from the corner r is related to the coordinates of dm through the Pythagorean relationship r2 = x2 + y2. (a) Express the moment of inertia of the element whose mass is dm with respect to an axis perpendicular to it and passing through one of the corners of the uniform rectangular plate:

( )dxdyyxdI 22 += σ

Integrate this expression to find I: ( )

( ) ( )323133

31

0 0

22

bamabba

dxdyyxIa b

+=+=

+= ∫ ∫σ

σ

(b) Letting d represent the distance from the origin to the center of mass of the plate, use the parallel axis theorem to relate the moment of inertia found in (a) to the moment of inertia with respect to an axis through the center of mass:

( ) 222312

cm

2cm

ormdbammdII

mdII

−+=−=

+=

Using the Pythagorean theorem, relate the distance d to the center of

( ) ( ) ( )22412

212

212 babad +=+=

Rotation

647

mass to the lengths of the sides of the plate: Substitute for d2 in the expression for Icm and simplify to obtain:

( ) ( )( )22

121

2224122

31

cm

bam

bambamI

+=

+−+=

*50 •• Picture the Problem Corey will use the point-particle relationship

222

211

i

2iiapp rmrmrmI +== ∑ for his calculation whereas Tracey’s calculation will take

into account not only the rod but also the fact that the spheres are not point particles. (a) Using the point-mass approximation and the definition of the moment of inertia of a system of particles, express Iapp:

222

211

i

2iiapp rmrmrmI +== ∑

Substitute numerical values and evaluate Iapp:

( )( ) ( )( )2

22app

mkg0.0400

m0.2kg0.5m0.2kg0.5

⋅=

+=I

Express the moment of inertia of the two spheres and connecting rod system:

rodspheres III +=

Use Table 9-1 to find the moments of inertia of a sphere (with respect to its center of mass) and a rod (with respect to an axis through its center of mass):

2rod12

1rod

2sphere5

2sphere

andLMI

RMI

=

=

Because the spheres are not on the axis of rotation, use the parallel axis theorem to express their moment of inertia with respect to the axis of rotation:

rotation. of axis the tosphere a of mass of

center thefrom distance theish where

2sphere

2sphere5

2sphere hMRMI +=

Substitute to obtain: { } 2rod12

12sphere

2sphere5

22 LMhMRMI ++=


Chapter 9

648

( )( ) ( )( ){ } ( )( )2

212122

52

mkg0415.0

m0.3kg0.06m0.2kg0.5m0.05kg0.52

⋅=

++=I

Compare I and Iapp by taking their ratio:

964.0mkg0.0415mkg0.0400

2

2app =

⋅⋅

=I

I

(b) sphere. solid a of an greater th

is sphere hollow a of because increase wouldinertia rotational The

cm

cm

II

51 •• Picture the Problem The axis of rotation passes through the center of the base of the tetrahedron. The carbon atom and the hydrogen atom at the apex of the tetrahedron do not contribute to I because the distance of their nuclei from the axis of rotation is zero. From the geometry, the distance of the three H nuclei from the rotation axis is 3/a , where a is the length of a side of the tetrahedron. Apply the definition of the moment of inertia for a system of particles to obtain: 2

H

2

H

23H

22H

21H

i

2ii

33 amam

rmrmrmrmI

=⎟⎠

⎞⎜⎝

⎛=

++== ∑


( )( )247

2927

mkg1041.5

m1018.0kg101.67

⋅×=

××=−

−−I

52 •• Picture the Problem Let the mass of the element of volume dV be dm = ρdV = 2πρhrdr where h is the height of the cylinder. We’ll begin by expressing the moment of inertia dI for the element of volume and then integrating it between R1 and R2.

Rotation

649

Express the moment of inertia of the element of mass dm:

drhrdmrdI 32 2πρ==

Integrate dI from R1 to R2 to obtain: ( )

( )( )21

22

21

222

1

41

422

132

1

2

RRRRh

RRhdrrhIR

R

+−=

−== ∫πρ

πρπρ

The mass of the hollow cylinder is ( )2

122 RRhm −= ρπ , so:

( )21

22 RRhm

−=

πρ

Substitute for ρ and simplify to obtain:

( ) ( )( ) ( )21

222

121

22

21

222

122

21 RRmRRRRh

RRhmI +=+−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=π

π

53 ••• Picture the Problem We can derive the given expression for the moment of inertia of a spherical shell by following the procedure outlined in the problem statement. Find the moment of inertia of a sphere, with respect to an axis through a diameter, in Table 9-1:

252 mRI =

Express the mass of the sphere as a function of its density and radius:

334 Rm ρπ=


5158 RI ρπ=

Express the differential of this expression:

dRRdI 438 ρπ= (1)

Express the increase in mass dm as the radius of the sphere increases by dR:

dRRdm 24 ρπ= (2)

Eliminate dR between equations (1) and (2) to obtain:

dmRdI 232=

. is mass of shell sphericalthe of inertia ofmoment theTherefore,2

32 mRm

Chapter 9

650

*54 ••• Picture the Problem We can find C in terms of M and R by integrating a spherical shell of mass dm with the given density function to find the mass of the earth as a function of M and then solving for C. In part (b), we’ll start with the moment of inertia of the same spherical shell, substitute the earth’s density function, and integrate from 0 to R. (a) Express the mass of the earth using the given density function:

33

0

3

0

2

0

2

22.13

4

422.14

4

CRCR

drrR

CdrrC

drrdmM

RR

R

ππ

ππ

ρπ

−=

−=

==

∫∫

∫∫

Solve for C as a function of M and R to obtain:

3508.0RMC =

(b) From Problem 9-40 we have: drrdI 438 ρπ=

Integrate to obtain:

( )

2

553

0 0

543

0

438

329.0

61

522.126.4

122.13508.08

MR

RRR

M

drrR

drrR

M

drrI

R R

R

=

⎥⎦⎤

⎢⎣⎡ −=

⎥⎦

⎤⎢⎣

⎡−=

=

∫ ∫

∫π

ρπ

Rotation

651

55 ••• Picture the Problem Let the origin be at the apex of the cone, with the z axis along the cone’s symmetry axis. Then the radius of the elemental ring, at a distance z from the apex, can be obtained from the

proportionHR

zr

= . The mass dm of the

elemental disk is ρdV = ρπr2dz. We’ll integrate r2dm to find the moment of inertia of the disk in terms of R and H and then integrate dm to obtain a second equation in R and H that we can use to eliminate H in our expression for I.

Express the moment of inertia of the cone in terms of the moment of inertia of the elemental disk:

102

4

0

44

4

22

22

02

2

21

221

HRdzzHR

dzzHRz

HR

dmrI

H

H

πρπρ

ρπ

==

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

∫

∫

∫

Express the total mass of the cone in terms of the mass of the elemental disk: HR

dzzHRdzrM

HH

231

0

22

2

0

2

πρ

πρπρ

=

== ∫∫

Divide I by M, simplify, and solve for I to obtain:

2103 MRI =

56 ••• Picture the Problem Let the axis of rotation be the x axis. The radius r of the

elemental area is 22 zR − and its mass,

dm, is dzzRdA 222 −= σσ . We’ll

integrate z2 dm to determine I in terms of σ and then divide this result by M in order to eliminate σ and express I in terms of M and R.

Chapter 9

652

Express the moment of inertia about the x axis:

( )4

41

222

22

2

R

dzzRz

dAzdmzIR

R

σπ

σ

σ

=

−=

==

∫

∫∫

−

The mass of the thin uniform disk is:

2RM σπ=

Divide I by M, simplify, and solve for I to obtain:

241 MRI = , a result in agreement with

the expression given in Table 9-1 for a cylinder of length L = 0.

57 ••• Picture the Problem Let the origin be at the apex of the cone, with the z axis along the cone’s symmetry axis, and the axis of rotation be the x rotation. Then the radius of the elemental disk, at a distance z from the apex, can be obtained from the

proportionHR

zr

= . The mass dm of the

elemental disk is ρdV = ρπr2dz. Each elemental disk rotates about an axis that is parallel to its diameter but removed from it by a distance z. We can use the result from Problem 9-57 for the moment of inertia of the elemental disk with respect to a diameter and then use the parallel axis theorem to express the moment of inertia of the cone with respect to the x axis.

Using the parallel axis theorem, express the moment of inertia of the elemental disk with respect to the x axis:

2disk zdmdIdI x += (1)

where dzrdVdm 2ρπρ ==

In Problem 9-57 it was established that the moment of inertia of a thin uniform disk of mass M and radius R rotating about a diameter is 2

41 MR . Express this result in

( )

dzzHR

rdzrdI2

22

2

41

2241

disk

⎟⎟⎠

⎞⎜⎜⎝

⎛=

=

ρπ

ρπ

Rotation

653

terms of our elemental disk: Substitute in equation (1) to obtain:

22

22

2

2

41

zdzzHR

dzzHRdI x

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

πρ

πρ

Integrate from 0 to H to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛= ∫

520

41

324

0

42

222

2

2

HRHR

dzzHRz

HRI

H

x

πρ

πρ

Express the total mass of the cone in terms of the mass of the elemental disk: HR

dzzHRdzrM

HH

231

0

22

2

0

2

πρ

πρπρ

=

== ∫∫

Divide Ix by M, simplify, and solve for Ix to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

2053

22 RHMI x

Remarks: Because both H and R appear in the numerator, the larger the cones are,

the greater their moment of inertia and the greater the energy consumption required to set them into motion. Rotational Kinetic Energy 58 • Picture the Problem The kinetic energy of this rotating system of particles can be calculated either by finding the tangential velocities of the particles and using these values to find the kinetic energy or by finding the moment of inertia of the system and using the expression for the rotational kinetic energy of a system. (a) Use the relationship between v and ω to find the speed of each particle:

( )( )

( )( ) m/s0.8rad/s2m0.4and

m/s0.4rad/s2m0.2

11

33

===

===

ω

ω

rv

rv

Chapter 9

654

Find the kinetic energy of the system: ( )( ) ( )( )

J1.12

m/s0.8kg1m/s0.4kg3

2222

211

23313

=

+=

+=+= vmvmKKK

(b) Use the definition of the moment of inertia of a system of particles to obtain:

244

233

222

211

2

rmrmrmrm

rmIi

ii

+++=

= ∑


( )( ) ( )( )( )( ) ( )( )

2

22

22

mkg560.0m0.2kg3m0.4kg1

m0.2kg3m0.4kg1

⋅=

++

+=I

Calculate the kinetic energy of the system of particles:

( )( )J1.12

rad/s2mkg0.560 22212

21

=

⋅== ωIK

*59 • Picture the Problem We can find the kinetic energy of this rotating ball from its angular speed and its moment of inertia. We can use the same relationship to find the new angular speed of the ball when it is supplied with additional energy. (a) Express the kinetic energy of the ball:

221 ωIK =

Express the moment of inertia of ball with respect to its diameter:

252 MRI =

Substitute for I: 2251 ωMRK =


( )( )

Jm6.84

s60min1

revrad2

minrev70

m0.075kg1.42

251

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×××

=

π

K

(b) Express the new kinetic energy with K′ = 2.0846 J:

221 '' ωIK =

Express the ratio of K to K′: 2

221

221

'⎟⎠⎞

⎜⎝⎛==

ωω

ωω '

I'I

KK'

Rotation

655

Solve for ω′:

KK'' ωω =

Substitute numerical values and evaluate ω′: ( )

rev/min347

J0.0846J2.0846rev/min70

=

='ω

60 • Picture the Problem The power delivered by an engine is the product of the torque it develops and the angular speed at which it delivers the torque. Express the power delivered by the engine as a function of the torque it develops and the angular speed at which it delivers this torque:

ωτ=P


( ) kW155s60

min1rev

rad2minrev3700mN400 =⎟⎟

⎠

⎞⎜⎜⎝

⎛××⋅=

πP

61 •• Picture the Problem Let r1 and r2 be the distances of m1 and m2 from the center of mass. We can use the definition of rotational kinetic energy and the definition of the center of mass of the two point masses to show that K1/K2 = m2/m1. Use the definition of rotational kinetic energy to express the ratio of the rotational kinetic energies:

222

211

2222

2211

222

1

212

1

2

1

rmrm

rmrm

II

KK

===ωω

ωω

Use the definition of the center of mass to relate m1, m2, r1, and r2:

2211 mrmr =

Solve for 2

1

rr

, substitute and

simplify to obtain: 1

2

2

1

2

2

1

2

1

mm

mm

mm

KK

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

62 •• Picture the Problem The earth’s rotational kinetic energy is given by

221

rot ωIK = where I is its moment of inertia with respect to its axis of rotation. The

Chapter 9

656

center of mass of the earth-sun system is so close to the center of the sun and the earth-sun distance so large that we can use the earth-sun distance as the separation of their centers of mass and assume each to be point mass. Express the rotational kinetic energy of the earth:

221

rot ωIK = (1)

Find the angular speed of the earth’s rotation using the definition of ω:

rad/s1027.7h

s3600h24

rad2

5−×=

×=

∆∆

=πθω

t

From Table 9-1, for the moment of inertia of a homogeneous sphere, we find:

( )( )237

262452

252

mkg109.83

m106.4kg106.0

⋅×=

××=

= MRI

Substitute numerical values in equation (1) to obtain:

( )( )

J102.60

rad/s107.27

mkg109.83

29

25

23721

rot

×=

××

⋅×=−

K

Express the earth’s orbital kinetic energy:

2orb2

1orb ωIK = (2)

Find the angular speed of the center of mass of the earth-sun system:

rad/s1099.1h

s3600dayh24days365.25

rad2

7−×=

××=

∆∆

=

π

θωt

Express and evaluate the orbital moment of inertia of the earth: ( )( )

247

21124

2orbE

mkg101.35m101.50kg106.0

⋅×=

××=

= RMI

Substitute in equation (2) to obtain: ( )

( )J102.67

rad/s101.99

mkg101.35

33

27

24721

orb

×=

××

⋅×=−

K

Rotation

657

Evaluate the ratio rot

orb

KK

: 429

33

rot

orb 10J102.60J102.67

≈××

=KK

*63 •• Picture the Problem Because the load is not being accelerated, the tension in the cable equals the weight of the load. The role of the massless pulley is to change the direction the force (tension) in the cable acts. (a) Because the block is lifted at constant speed:

( )( )kN19.6

m/s9.81kg2000 2

=

== mgT

(b) Apply the definition of torque at the winch drum:

( )( )mkN5.89

m0.30kN19.6

⋅=

== Trτ

(c) Relate the angular speed of the winch drum to the rate at which the load is being lifted (the tangential speed of the cable on the drum):

rad/s0.267m0.30

m/s0.08===

rvω

(d) Express the power developed by the motor in terms of the tension in the cable and the speed with which the load is being lifted:

( )( )kW1.57

m/s0.08kN19.6

=

== TvP

64 •• Picture the Problem Let the zero of gravitational potential energy be at the lowest point of the small particle. We can use conservation of energy to find the angular velocity of the disk when the particle is at its lowest point and Newton’s 2nd law to find the force the disk will have to exert on the particle to keep it from falling off. (a) Use conservation of energy to relate the initial potential energy of the system to its rotational kinetic energy when the small particle is at its lowest point:

0=∆+∆ UK or, because Uf = Ki = 0,

( ) 02fparticledisk2

1 =∆−+ hmgII ω

Solve for ωf:

particlediskf

2II

hmg+

∆=ω

Chapter 9

658

Substitute for Idisk, Iparticle, and ∆h and simplify to obtain:

( )( )MmR

mgmRMRRmg

+=

+=

2822

2221fω

(b) The mass is in uniform circular motion at the bottom of the disk, so the sum of the force F exerted by the disk and the gravitational force must be the centripetal force:

2fωmRmgF =−

Solve for F and simplify to obtain:

( )

⎟⎠⎞

⎜⎝⎛

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

Mmmmg

mMRmgmRmg

mRmgF

281

28

2fω

65 •• Picture the Problem Let the zero of gravitational potential energy be at the center of mass of the ring when it is directly below the point of support. We’ll use conservation of energy to relate the maximum angular velocity and the initial angular velocity required for a complete revolution to the changes in the potential energy of the ring. (a) Use conservation of energy to relate the initial potential energy of the ring to its rotational kinetic energy when its center of mass is directly below the point of support:

0=∆+∆ UK or, because Uf = Ki = 0,

02max2

1 =∆− hmgIPω (1)

Use the parallel axis theorem and Table 9-1 to express the moment of inertia of the ring with respect to its pivot point P:

2cm mRII P +=

Substitute in equation (1) to obtain: ( ) 02max

2221 =−+ mgRmRmR ω

Solve for ωmax:

Rg

=maxω

Substitute numerical values and evaluate ωmax:

rad/s3.62m0.75

m/s9.81 2

max ==ω

Rotation

659

(b) Use conservation of energy to relate the final potential energy of the ring to its initial rotational kinetic energy:

0=∆+∆ UK or, because Ui = Kf = 0,

02i2

1 =∆+− hmgI Pω

Noting that the center of mass must rise a distance R if the ring is to make a complete revolution, substitute for IP and ∆h to obtain:

( ) 02i

2221 =++− mgRmRmR ω

Solve for ωi:

Rg

i =ω

Substitute numerical values and evaluate ωi: rad/s3.62

m0.75m/s9.81 2

==iω

66 •• Picture the Problem We can find the energy that must be stored in the flywheel and relate this energy to the radius of the wheel and use the definition of rotational kinetic energy to find the wheel’s radius. Relate the kinetic energy of the flywheel to the energy it must deliver:

( )( )MJ600

km300MJ/km22cyl2

1rot

=

== ωIK

Express the moment of inertia of the flywheel:

221

cyl MRI =

Substitute for Icyl and solve for ω:

MKR rot2

ω=

Substitute numerical values and evaluate R:

m95.1

kg100MJ

J10MJ600

revrad2

srev400

26

=

×

×= πR

67 •• Picture the Problem We’ll solve this problem for the general case of a ladder of length L, mass M, and person of mass m. Let the zero of gravitational potential energy be at floor level and include you, the ladder, and the earth in the system. We’ll use

Chapter 9

660

conservation of energy to relate your impact speed falling freely to your impact speed riding the ladder to the ground. Use conservation of energy to relate the speed with which a person will strike the ground to the fall distance L:


02f2

1 =− mgLmv

Solve for 2fv : gLv 22

f =

Letting ωr represent the angular velocity of the ladder+person system as it strikes the ground, use conservation of energy to relate the initial and final momenta of the system:


( ) 02

2rladderperson2

1 =⎟⎠⎞

⎜⎝⎛ +−+

LMgmgLII ω

Substitute for the moments of inertia to obtain:

023

1 2f

221 =⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ +

LMgmgLLMm ω

Substitute vr for Lωf and solve for 2rv :

3

22

2r Mm

MmgLv

+

⎟⎠⎞

⎜⎝⎛ +

=

Express the ratio 2f

2r

vv

:

3

22f

2r

Mm

Mm

vv

+

+=

Solve for vr to obtain: Mm

Mmvv2636

fr ++

=

ground. the tofall and golet better to isIt . zero, is ladder, theof mass the, Unless fr vvM >

Rotation

661

Pulleys, Yo-Yos, and Hanging Things *68 •• Picture the Problem We’ll solve this problem for the general case in which the mass of the block on the ledge is M, the mass of the hanging block is m, and the mass of the pulley is Mp, and R is the radius of the pulley. Let the zero of gravitational potential energy be 2.5 m below the initial position of the 2-kg block and R represent the radius of the pulley. Let the system include both blocks, the shelf and pulley, and the earth. The initial potential energy of the 2-kg block will be transformed into the translational kinetic energy of both blocks plus rotational kinetic energy of the pulley. (a) Use energy conservation to relate the speed of the 2 kg block when it has fallen a distance ∆h to its initial potential energy and the kinetic energy of the system:


( ) 02pulley2

1221 =−++ mghIvMm ω

Substitute for Ipulley and ω to obtain: ( ) ( ) 02

22

21

212

21 =−++ mgh

RvMRvMm

Solve for v:

pMmMmghv

21

2++

=


( )( )( )( )

m/s3.95

kg0.6kg2kg4m2.5m/s9.81kg22

21

2

=

++=v

(b) Find the angular velocity of the pulley from its tangential speed:

rad/s49.3m0.08

m/s3.95===

Rvω

69 •• Picture the Problem The diagrams show the forces acting on each of the masses and the pulley. We can apply Newton’s 2nd law to the two blocks and the pulley to obtain three equations in the unknowns T1, T2, and a.

Chapter 9

662

Apply Newton’s 2nd law to the two blocks and the pulley:

∑ == amTFx 41 , (1)

( )∑ =−= ατ pp IrTT 12 , (2)

and

∑ =−= amTgmFx 222 (3)

Eliminate α in equation (2) to obtain:

aMTT p21

12 =− (4)

Eliminate T1 and T2 between equations (1), (3) and (4) and solve for a:

pMmmgma

21

42

2

++=


( )( )( )

2

21

2

m/s3.11kg0.6kg4kg2

m/s9.81kg2=

++=a

Using equation (1), evaluate T1: ( )( ) N12.5m/s3.11kg4 21 ==T

Solve equation (3) for T2: ( )agmT −= 22

Substitute numerical values and evaluate T2:

( )( )N13.4

m/s3.11m/s9.81kg2 222

=

−=T

70 •• Picture the Problem We’ll solve this problem for the general case in which the mass of the block on the ledge is M, the mass of the hanging block is m, the mass of the pulley is Mp, and R is the radius of the pulley. Let the zero of gravitational potential energy be 2.5 m below the initial position of the 2-kg block. The initial potential energy of the 2-kg block will be transformed into the translational kinetic energy of both blocks plus rotational kinetic energy of the pulley plus work done against friction. (a) Use energy conservation to relate the speed of the 2 kg block when it has fallen a distance ∆h to its initial potential energy, the kinetic energy of the system and the work done against friction:

0f =+∆+∆ WUK

or, because Ki = Uf = 0, ( )

0k

2pulley2

1221

=+−

++

Mghmgh

IvMm

µ

ω

Substitute for Ipulley and ω to obtain: ( ) ( )0k

2

2

21

212

21

=+−

++

MghmghRvMvMm p

µ

Rotation

663

Solve for v: ( )pMmM

Mmghv21

k2++−

=µ


( )( ) ( )( )[ ]( ) m/s79.2

kg0.6kg2kg4kg425.0kg2m2.5m/s9.812

21

2

=++

−=v

(b) Find the angular velocity of the pulley from its tangential speed:

rad/s9.43m0.08

m/s79.2===

Rvω

71 •• Picture the Problem Let the zero of gravitational potential energy be at the water’s surface and let the system include the winch, the car, and the earth. We’ll apply energy conservation to relate the car’s speed as it hits the water to its initial potential energy. Note that some of the car’s initial potential energy will be transformed into rotational kinetic energy of the winch and pulley. Use energy conservation to relate the car’s speed as it hits the water to its initial potential energy:


02pp2

12ww2

1221 =∆−++ hmgIImv ωω

Express ωw and ωp in terms of the speed v of the rope, which is the same throughout the system:

2p

2

p2w

2

w andrv

rv

== ωω


p

2

p21

2w

2

w212

21 =∆−++ hmg

rvI

rvImv

Solve for v:

2p

p2

w

w

2

rI

rIm

hmgv++

∆=


( )( )( )

( ) ( )m/s21.8

m0.3mkg4

m0.8mkg320kg1200

m5m/s9.81kg12002

2

2

2

2

2

=

⋅+

⋅+

=v

Chapter 9

664

*72 •• Picture the Problem Let the system include the blocks, the pulley and the earth. Choose the zero of gravitational potential energy to be at the ledge and apply energy conservation to relate the impact speed of the 30-kg block to the initial potential energy of the system. We can use a constant-acceleration equations and Newton’s 2nd law to find the tensions in the strings and the descent time.

(a) Use conservation of energy to relate the impact speed of the 30-kg block to the initial potential energy of the system:


03020

2pp2

12202

12302

1

=∆−∆+

++

hgmhgm

Ivmvm ω

Substitute for ωp and Ip to obtain: ( )

03020

2

22

p21

212

20212

3021

=∆−∆+

⎟⎟⎠

⎞⎜⎜⎝

⎛++

hgmhgmrvrMvmvm

Solve for v: ( )

p21

3020

20302Mmm

mmhgv++−∆

=


( )( )( )( )

m/s73.2

kg5kg30kg20kg20kg30m2m/s9.812

21

2

=

++−

=v

(b) Find the angular speed at impact from the tangential speed at impact and the radius of the pulley:

rad/s27.3m0.1m/s2.73

===rvω

(c) Apply Newton’s 2nd law to the blocks:

∑ =−= amgmTFx 20201 (1)

∑ =−= amTgmFx 30230 (2)

Using a constant-acceleration equation, relate the speed at impact to the fall distance and the

havv ∆+= 220

2

or, because v0 = 0,

Rotation

665

acceleration and solve for and evaluate a:

( )( )

222

m/s1.87m22m/s2.73

2==

∆=

hva

Substitute in equation (1) to find T1: ( )

( )( )N234

m/s1.87m/s9.81kg20 22201

=

+=

+= agmT

Substitute in equation (2) to find T2: ( )

( )( )N238

m/s1.87m/s9.81kg30 22302

=

−=

−= agmT

(d) Noting that the initial speed of the 30-kg block is zero, express the time-of-fall in terms of the fall distance and the block’s average speed:

vh

vh

vht ∆

=∆

=∆

=∆2

21

av


( ) s1.47m/s2.73m22

==∆t

73 •• Picture the Problem The force diagram shows the forces acting on the sphere and the hanging object. The tension in the string is responsible for the angular acceleration of the sphere and the difference between the weight of the object and the tension is the net force acting on the hanging object. We can use Newton’s 2nd law to obtain two equations in a and T that we can solve simultaneously.

(a)Apply Newton’s 2nd law to the sphere and the hanging object:

∑ == ατ sphere0 ITR (1)

and

∑ =−= maTmgFx (2)

Substitute for Isphere and α in equation (1) to obtain:

( )RaMRTR 2

52= (3)

Chapter 9

666

Eliminate T between equations (2) and (3) and solve for a to obtain:

mM

ga

521+

=

(b) Substitute for a in equation (2) and solve for T to obtain: Mm

mMgT25

2+

=

74 •• Picture the Problem The diagram shows the forces acting on both objects and the pulley. By applying Newton’s 2nd law of motion, we can obtain a system of three equations in the unknowns T1, T2, and a that we can solve simultaneously.

(a) Apply Newton’s 2nd law to the pulley and the two objects:

∑ =−= amgmTFx 111 , (1)

( )∑ =−= ατ 0120 IrTT , (2)

and

∑ =−= amTgmFx 222 (3)

Substitute for I0 = Ipulley and α in equation (2) to obtain:

( ) ( )ramrrTT 2

21

12 =− (4)

Eliminate T1 and T2 between equations (1), (3) and (4) and solve for a to obtain:

( )mmm

gmma21

21

12

++−

=


( )( )( )

2

21

2

cm/s9.478

g50g510g500cm/s981g500g510

=

++−

=a

(b) Substitute for a in equation (1) and solve for T1 to obtain:

( )( )( )

N4.9524

m/s0.09478m/s9.81kg0.500 2211

=

+=

+= agmT

Rotation

667

Substitute for a in equation (3) and solve for T2 to obtain:

( )( )( )

N4.9548

m/s0.09478m/s9.81kg0.510 2222

=

−=

−= agmT

Find ∆T:

N0.0024

N4.9524N.9548412

=

−=−=∆ TTT

(c) If we ignore the mass of the pulley, our acceleration equation is:

( )21

12

mmgmma

+−

=


( )( )

2

2

cm/s9.713

g510g500cm/s981g500g510

=

+−

=a

Substitute for a in equation (1) and solve for T1 to obtain:

( )agmT += 11

Substitute numerical values and evaluate T1:

( )( ) N4.9536m/s0.09713m/s9.81kg0.500 221 =+=T

From equation (4), if m = 0:

21 TT =

*75 •• Picture the Problem The diagram shows the forces acting on both objects and the pulley. By applying Newton’s 2nd law of motion, we can obtain a system of three equations in the unknowns T1, T2, and α that we can solve simultaneously.

(a) Express the condition that the system does not accelerate:

02211net =−= gRmgRmτ

Chapter 9

668

Solve for m2:

2

112 R

Rmm =

Substitute numerical values and evaluate m2:

( ) kg72.0m0.4m1.2

kg242 ==m

(b) Apply Newton’s 2nd law to the objects and the pulley:

∑ =−= amTgmFx 111 , (1)

∑ =−= ατ 022110 IRTRT , (2)

and

∑ =−= amgmTFx 222 (3)

Eliminate a in favor of α in equations (1) and (3) and solve for T1 and T2:

( )α111 RgmT −= (4)

and ( )α222 RgmT += (5)

Substitute for T1 and T2 in equation (2) and solve for α to obtain:

( )0

222

211

2211

IRmRmgRmRm

++−

=α


( )( ) ( )( )[ ]( )( )( ) ( )( )

2222

2

rad/s37.1mkg40m0.4kg72m1.2kg36

m/s9.81m0.4kg72m1.2kg36=

⋅++−

=α

Substitute in equation (4) to find T1:

( )[ ( )( )] N294rad/s1.37m1.2m/s9.81kg36 221 =−=T

Substitute in equation (5) to find T2:

( )[ ( )( )] N467rad/s1.37m4.0m/s9.81kg27 222 =+=T

Rotation

669

76 •• Picture the Problem Choose the coordinate system shown in the diagram. By applying Newton’s 2nd law of motion, we can obtain a system of two equations in the unknowns T and a. In (b) we can use the torque equation from (a) and our value for T to findα. In (c) we use the condition that the acceleration of a point on the rim of the cylinder is the same as the acceleration of the hand, together with the angular acceleration of the cylinder, to find the acceleration of the hand.

(a) Apply Newton’s 2nd law to the cylinder about an axis through its center of mass:

∑ ==RaITR 00τ (1)

and

∑ =−= 0TMgFx (2)

Solve for T to obtain:

MgT =

(b) Rewrite equation (1) in terms of α:

α0ITR =

Solve for α:

0ITR

=α

Substitute for T and I0 to obtain:

Rg

MRMgR 2

221

==α

(c) Relate the acceleration a of the hand to the angular acceleration of the cylinder:

αRa =

Substitute for α to obtain: gRgRa 22

=⎟⎠⎞

⎜⎝⎛=

Chapter 9

670

77 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. By applying Newton’s 2nd law to the cylinder and the block we can obtain simultaneous equations in a, T, and α from which we can express a and T. By applying the conservation of energy, we can derive an expression for the speed of the block when it reaches the bottom of the incline.

(a) Apply Newton’s 2nd law to the cylinder and the block:

∑ == ατ 00 ITR (1)

and

∑ =−= amTgmFx 22 sinθ (2)

Substitute for α in equation (1), solve for T, and substitute in equation (2) and solve for a to obtain:

2

1

21

sin

mm

ga+

=θ

(b) Substitute for a in equation (2) and solve for T:

2

1

121

21

sin

mm

gmT

+=

θ

(c) Noting that the block is released from rest, express the total energy of the system when the block is at height h:

ghmKUE 2=+=

(d) Use the fact that this system is conservative to express the total energy at the bottom of the incline:

ghmE 2bottom =

(e) Express the total energy of the system when the block is at the bottom of the incline in terms of its kinetic energies:

202

1222

1

rottranbottom

ωIvm

KKE

+=

+=

Rotation

671

Substitute for ω and I0 to obtain: ( ) ghmrvrmvm 22

22

121

212

221 =+

Solve for v to obtain:

2

1

21

2

mm

ghv+

=

(f) For θ = 0: 0== Ta

For θ = 90°:

2

1

21

mm

ga+

= ,

am

mmgm

T 121

2

1

121

21

=+

= ,

and

2

1

21

2

mm

ghv+

=

For m1 = 0: θsinga = , 0=T , and

ghv 2=

*78 •• Picture the Problem Let r be the radius of the concentric drum (10 cm) and let I0 be the moment of inertia of the drum plus platform. We can use Newton’s 2nd law in both translational and rotational forms to express I0 in terms of a and a constant-acceleration equation to express a and then find I0. We can use the same equation to find the total moment of inertia when the object is placed on the platform and then subtract to find its moment of inertia.

(a) Apply Newton’s 2nd law to the platform and the weight:

∑ == ατ 00 ITr (1)

∑ =−= MaTMgFx (2)

Chapter 9

672

Substitute a/r for α in equation (1) and solve for T:

arIT 2

0=

Substitute for T in equation (2) and solve for a to obtain:

( )a

agMrI −=

2

0 (3)

Using a constant-acceleration equation, relate the distance of fall to the acceleration of the weight and the time of fall and solve for the acceleration:

( )221

0 tatvx ∆+∆=∆

or, because v0 = 0 and ∆x = D,

( )22

tDa

∆=

Substitute for a in equation (3) to obtain:

( )⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆=⎟

⎠⎞

⎜⎝⎛ −= 1

21

222

0 DtgMr

agMrI

Substitute numerical values and evaluate I0:

( )( )( )( )

( )2

22

20

mkg1.177

1m1.82

s4.2m/s9.81

m0.1kg2.5

⋅=

⎥⎦

⎤⎢⎣

⎡−×

=I

(b) Relate the moments of inertia of the platform, drum, shaft, and pulley (I0) to the moment of inertia of the object and the total moment of inertia:

( )⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆=

⎟⎠⎞

⎜⎝⎛ −=+=

12

1

22

20tot

DtgMr

agMrIII

Substitute numerical values and evaluate Itot:

( )( )( )( )

( )2

22

2tot

mkg125.3

1m1.82

s8.6m/s9.81

m0.1kg2.5

⋅=

⎥⎦

⎤⎢⎣

⎡−×

=I

Solve for and evaluate I:

2

2

20tot

mkg1.948

mkg1.177

mkg3.125

⋅=

⋅−

⋅=−= III

Rotation

673

Objects Rolling Without Slipping *79 •• Picture the Problem The forces acting on the yo-yo are shown in the figure. We can use a constant-acceleration equation to relate the velocity of descent at the end of the fall to the yo-yo’s acceleration and Newton’s 2nd law in both translational and rotational form to find the yo-yo’s acceleration.

Using a constant-acceleration equation, relate the yo-yo’s final speed to its acceleration and fall distance:

havv ∆+= 220

2

or, because v0 = 0, hav ∆= 2 (1)

Use Newton’s 2nd law to relate the forces that act on the yo-yo to its acceleration:

∑ =−= maTmgFx (2)

and ατ 00 ITr ==∑ (3)

Use αra = to eliminate α in equation (3) r

aITr 0= (4)

Eliminate T between equations (2) and (4) to obtain:

maarI

mg =− 20 (5)

Substitute 2

21 mR for I0 in equation

(5): maa

rmR

mg =− 2

221

Solve for a:

2

2

21

rR

ga+

=

Substitute numerical values and evaluate a: ( )

( )

2

2

2

2

m/s0.0864

m0.12m1.51

m/s9.81=

+=a

Substitute in equation (1) and evaluate v:

( )( )m/s3.14

m57m/s0.08642 2

=

=v

Chapter 9

674

80 •• Picture the Problem The diagram shows the forces acting on the cylinder. By applying Newton’s 2nd law of motion, we can obtain a system of two equations in the unknowns T, a, and α that we can solve simultaneously.

(a) Apply Newton’s 2nd law to the cylinder:

∑ == ατ 00 ITR (1)

and

∑ =−= MaTMgFx (2)

Substitute for α and I0 in equation (1) to obtain:

( ) ⎟⎠⎞

⎜⎝⎛=

RaMRTR 2

21

Solve for T:

MaT 21= (3)

Substitute for T in equation (2) and solve for a to obtain:

ga 32=

(b) Substitute for a in equation (3) to obtain:

( ) MggMT 31

32

21 ==

81 •• Picture the Problem The forces acting on the yo-yo are shown in the figure. Apply Newton’s 2nd law in both translational and rotational form to obtain simultaneous equations in T, a, and α from which we can eliminate α and solve for T and a.

Apply Newton’s 2nd law to the yo-yo: ∑ =−= maTmgFx (1)

and ατ 00 ITr ==∑ (2)

Use αra = to eliminate α in equation (2) r

aITr 0= (3)

Rotation

675

Eliminate T between equations (1) and (3) to obtain:

maarI

mg =− 20 (4)

Substitute 221 mR for I0 in equation

(4): maa

rmR

mg =− 2

221

Solve for a:

2

2

21

rR

ga+

=

Substitute numerical values and evaluate a: ( )

( )

2

2

2

2

m/s0.192

m0.012m1.01

m/s9.81=

+=a

Use equation (1) to solve for and evaluate T:

( )( )( )

N0.962

m/s0.192m/s9.81kg0.1 22

=

−=

−= agmT

*82 • Picture the Problem We can determine the kinetic energy of the cylinder that is due to its rotation about its center of mass by examining the ratio KK rot .

Express the rotational kinetic energy of the homogeneous solid cylinder:

( ) 241

2

22

21

212

cyl21

rot mvrvmrIK === ω

Express the total kinetic energy of the homogeneous solid cylinder:

2432

212

41

transrot mvmvmvKKK =+=+=

Express the ratio K

K rot : 31

243

241

rot ==mvmv

KK

and correct. is )(b

83 • Picture the Problem Any work done on the cylinder by a net force will change its kinetic energy. Therefore, the work needed to give the cylinder this motion is equal to its kinetic energy. Express the relationship between the work needed to stop the cylinder and its kinetic energy:

2212

21 ωImvKW +=∆=

Chapter 9

676

Because the cylinder is rolling without slipping, its translational and angular speeds are related according to:

ωrv =

Substitute for I (see Table 9-1) and ω and simplify to obtain:

( )2

43

2

22

21

212

21

2212

21

mvrvmrmv

ImvW

=

+=

+= ω

Substitute for m and v to obtain: ( )( ) kJ1.13m/s5kg60 2

43 ==W

84 • Picture the Problem The total kinetic energy of any object that is rolling without slipping is given by rottrans KKK += . We can find the percentages associated with each

motion by expressing the moment of inertia of the objects as kmr2 and deriving a general expression for the ratios of rotational kinetic energy to total kinetic energy and translational kinetic energy to total kinetic energy and substituting the appropriate values of k. Express the total kinetic energy associated with a rotating and translating object: ( )

( )kmvkmvmvrvkmrmv

ImvKKK

+=+=

+=

+=+=

12212

212

21

2

22

212

21

2212

21

rottrans ω

Express the ratio K

K rot : ( )

kk

kkmv

kmvK

K11

1112

21

221

rot

+=

+=

+=

Express the ratio K

K trans : ( ) kkmv

mvK

K+

=+

=1

112

21

221

trans

(a) Substitute k = 2/5 for a uniform sphere to obtain:

%6.28286.0

4.011

1rot ==+

=K

K

and

%4.71714.04.01

1trans ==+

=K

K

Rotation

677

(b) Substitute k = 1/2 for a uniform cylinder to obtain:

%3.33

5.011

1rot =+

=K

K

and

%7.665.01

1trans =+

=K

K

(c) Substitute k = 1 for a hoop to obtain: %0.50

111

1rot =+

=K

K

and

%0.5011

1trans =+

=K

K

85 • Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. As the hoop rolls up the incline its translational and rotational kinetic energies are transformed into gravitational potential energy. We can use energy conservation to relate the distance the hoop rolls up the incline to its total kinetic energy at the bottom of the incline. Using energy conservation, relate the distance the hoop will roll up the incline to its kinetic energy at the bottom of the incline:


0fi =+− UK (1)

Express Ki as the sum of the translational and rotational kinetic energies of the hoop:

2212

21

rottransi ωImvKKK +=+=

When a rolling object moves with speed v, its outer surface turns with a speed v also. Hence ω = v/r. Substitute for I and ω to obtain:

( ) 22

22

212

21

i mvrvmrmvK =+=

Letting ∆h be the change in elevation of the hoop as it rolls up the incline and ∆L the distance it rolls along the incline, express Uf:

θsinf LmghmgU ∆=∆=


0sin2 =∆+− θLmgmv

Chapter 9

678

Solve for ∆L: θsin

2

gvL =∆

Substitute numerical values and evaluate ∆L:

( )( ) m45.9

sin30m/s9.81m/s15

2

2

=°

=∆L

*86 •• Picture the Problem From Newton’s 2nd law, the acceleration of the center of mass equals the net force divided by the mass. The forces acting on the sphere are its weight

grm downward, the normal force nFr

that balances the normal component of the weight,

and the force of friction fr

acting up the incline. As the sphere accelerates down the incline, the angular velocity of rotation must increase to maintain the nonslip condition. We can apply Newton’s 2nd law for rotation about a horizontal axis through the center of mass of the sphere to find α, which is related to the acceleration by the nonslip condition. The only torque about the center of mass is due to f

rbecause both grm and nF

ract through

the center of mass. Choose the positive direction to be down the incline.

Apply aF rr

m=∑ to the sphere: cmsin mafmg =−θ (1)

Apply ατ cmI=∑ to the sphere: αcmIfr =

Use the nonslip condition to eliminate α and solve for f:

raIfr cm

cm=

and

cm2cm a

rIf =

Substitute this result for f in equation (1) to obtain:

cmcm2cmsin maa

rImg =−θ

From Table 9-1 we have, for a solid sphere:

252

cm mrI =

Rotation

679


cmcm52sin maamg =−θ

Solve for and evaluate θ :

( )°=⎥

⎦

⎤⎢⎣

⎡=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

3.165

2.07sin

57sin

1

cm1

gg

gaθ

87 •• Picture the Problem From Newton’s 2nd law, the acceleration of the center of mass equals the net force divided by the mass. The forces acting on the thin spherical shell are its weight grm downward, the normal force nF

rthat balances the normal component of the

weight, and the force of friction fr

acting up the incline. As the spherical shell accelerates down the incline, the angular velocity of rotation must increase to maintain the nonslip condition. We can apply Newton’s 2nd law for rotation about a horizontal axis through the center of mass of the sphere to find α, which is related to the acceleration by the nonslip condition. The only torque about the center of mass is due to f

rbecause both grm and

nFr

act through the center of mass. Choose the positive direction to be down the incline.

Apply aF rr

m=∑ to the thin spherical shell:

cmsin mafmg =−θ (1)

Apply ατ cmI=∑ to the thin spherical shell:

αcmIfr =

Use the nonslip condition to eliminate α and solve for f:

raIfr cm

cm= and cm2cm a

rIf =

Substitute this result for f in equation (1) to obtain:

cmcm2cmsin maa

rImg =−θ

From Table 9-1 we have, for a thin 232

cm mrI =

Chapter 9

680

spherical shell:


cmcm32sin maamg =−θ


( )°==

=

−

−

5.193

2.05sin

35sin

1

cm1

gg

gaθ

Remarks: This larger angle makes sense, as the moment of inertia for a given mass is larger for a hollow sphere than for a solid one. 88 •• Picture the Problem The three forces acting on the basketball are the weight of the ball, the normal force, and the force of friction. Because the weight can be assumed to be acting at the center of mass, and the normal force acts through the center of mass, the only force which exerts a torque about the center of mass is the frictional force. We can use Newton’s 2nd law to find a system of simultaneous equations that we can solve for the quantities called for in the problem statement.

(a) Apply Newton’s 2nd law in both translational and rotational form to the ball:

∑ =−= mafmgFx ssinθ , (1)

∑ =−= 0cosn θmgFFy (2)

and

∑ == ατ 0s0 Irf (3)

Because the basketball is rolling without slipping we know that:

ra

=α

Substitute in equation (3) to obtain: r

aIrf 0s = (4)

From Table 9-1 we have:

232

0 mrI =

Substitute for I0 and α in equation (4) and solve for fs:

( ) maframrrf 3

2s

232

s =⇒= (5)

Rotation

681

Substitute for fs in equation (1) and solve for a:

θsin53 ga =

(b) Find fs using equation (5): ( ) θθ sinsin 5

253

32

s mggmf ==

(c) Solve equation (2) for Fn:

θcosn mgF =

Use the definition of fs,max to obtain:

maxsnsmaxs, cosθµµ mgFf ==

Use the result of part (b) to obtain: maxsmax52 cossin θµθ mgmg =

Solve for θmax: ( )s2

51max tan µθ −=

89 •• Picture the Problem The three forces acting on the cylinder are the weight of the cylinder, the normal force, and the force of friction. Because the weight can be assumed to be acting at the center of mass, and the normal force acts through the center of mass, the only force which exerts a torque about the center of mass is the frictional force. We can use Newton’s 2nd law to find a system of simultaneous equations that we can solve for the quantities called for in the problem statement.

(a) Apply Newton’s 2nd law in both translational and rotational form to the cylinder:

∑ =−= mafmgFx ssinθ , (1)

∑ =−= 0cosn θmgFFy (2)

and

∑ == ατ 0s0 Irf (3)

Because the cylinder is rolling without slipping we know that:

ra

=α

Substitute in equation (3) to obtain: r

aIrf 0s = (4)

From Table 9-1 we have:

221

0 mrI =

Chapter 9

682


( ) maframrrf 2

1s

221

s =⇒= (5)


θsin32 ga =

(b) Find fs using equation (5): ( ) θθ sinsin 3

132

21

s mggmf ==

(c) Solve equation (2) for Fn:

θcosn mgF =

Use the definition of fs,max to obtain:

maxsnsmaxs, cosθµµ mgFf ==

Use the result of part (b) to obtain: maxsmax31 cossin θµθ mgmg =

Solve for θmax: ( )s

1max 3tan µθ −=

*90 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation where the spheres leave the ramp. The distances the spheres will travel are directly proportional to their speeds when they leave the ramp. Express the ratio of the distances traveled by the two spheres in terms of their speeds when they leave the ramp:

vv'

tvtv'

LL'

=∆∆

= (1)

Use conservation of mechanical energy to find the speed of the spheres when they leave the ramp:


0if =−UK (2)

Express Kf for the spheres:

( )

( ) 221

2212

21

2

22

212

21

2cm2

1221

rottransf

1 mvk

kmvmvRvkmRmv

Imv

KKK

+=

+=

+=

+=

+=

ω

where k is 2/3 for the spherical shell and 2/5 for the uniform sphere.

Substitute in equation (2) to obtain: ( ) mgHmvk =+ 2211

Rotation

683

Solve for v:

kgH

v+

=12


09.111

11

5232

=++

=++

=k'k

LL'

or LL' 09.1=

91 •• Picture the Problem Let the subscripts u and h refer to the uniform and thin-walled spheres, respectively. Because the cylinders climb to the same height, their kinetic energies at the bottom of the incline must be equal. Express the total kinetic energy of the thin-walled cylinder at the bottom of the inclined plane: ( ) 2

h2

22

h212

h21

2h2

12h2

1rottransh

vmrvrmvm

IvmKKK

=+=

+=+= ω

Express the total kinetic energy of the solid cylinder at the bottom of the inclined plane: ( ) 2

u43

2

22

u21

212

u21

2u2

12u2

1rottransu '

v'mrv'rmv'm

IvmKKK

=+=

+=+= ω

Because the cylinders climb to the same height:

ghmvm

ghmv'm

h2

h

u2

u43

and=

=

Divide the first of these equations by the second: ghm

ghmvmv'm

h

u2

h

2u4

3=

Simplify to obtain:

143

2

2

=vv'

Solve for v′:

vv'34

=

Chapter 9

684

92 •• Picture the Problem Let the subscripts s and c refer to the solid sphere and thin-walled cylinder, respectively. Because the cylinder and sphere descend from the same height, their kinetic energies at the bottom of the incline must be equal. The force diagram shows the forces acting on the solid sphere. We’ll use Newton’s 2nd law to relate the accelerations to the angle of the incline and use a constant acceleration to relate the accelerations to the distances traveled down the incline.

Apply Newton’s 2nd law to the sphere:

sssin∑ =−= mafmgFx θ , (1)

∑ =−= 0cosn θmgFFy , (2)

and

∑ == ατ 0s0 Irf (3)


( ) s52

s2

52

s maframrrf =⇒=


θsin75

s ga =

Proceed as above for the thin-walled cylinder to obtain:

θsin21

c ga =

Using a constant-acceleration equation, relate the distance traveled down the incline to its acceleration and the elapsed time:

( )221

0 tatvs ∆+∆=∆

or, because v0 = 0, ( )2

21 tas ∆=∆ (4)

Because ∆s is the same for both objects: 2

cc2ss tata =

where ( ) 76.58.44.2 s

2s

2s

2c ++=+= tttt

provided tc and ts are in seconds.

Substitute for as and ac to obtain the quadratic equation:

2s7

10s

2s 76.58.4 ttt =++

Rotation

685

Solve for the positive root to obtain:

s3.12s =t

Substitute in equation (4), simplify, and solve for θ : ⎥

⎦

⎤⎢⎣

⎡ ∆= −

2s

1

514sin

gtsθ


( )( )( )°=

⎥⎦

⎤⎢⎣

⎡= −

324.0

s12.3m/s9.815m314sin 22

1θ

93 ••• Picture the Problem The kinetic energy of the wheel is the sum of its translational and rotational kinetic energies. Because the wheel is a composite object, we can model its moment of inertia by treating the rim as a cylindrical shell and the spokes as rods. Express the kinetic energy of the wheel:

2

2

cm212

tot21

2cm2

12tot2

1

rottrans

RvIvM

IvM

KKK

+=

+=

+=

ω

where Mtot = Mrim + 4Mspoke

Express the moment of inertia of the wheel: ( )

( ) 2spoke3

4rim

2spoke3

12rim

spokesrimcm

4

RMM

RMRM

III

+=

+=

+=

Substitute for Icm in the equation for K:

( )[ ]( )[ ] 2

spoke32

rimtot21

2

22

spoke34

rim212

tot21

vMMMRvRMMvMK

++=

++=


( ) ( )[ ]( )J223

m/s6kg1.2kg3kg7.8 232

21

=

++=K

Chapter 9

686

94 ••• Picture the Problem Let M represent the combined mass of the two disks and their connecting rod and I their moment of inertia. The object’s initial potential energy is transformed into translational and rotational kinetic energy as it rolls down the incline. The force diagram shows the forces acting on this composite object as it rolls down the incline. Application of Newton’s 2nd law will allow us to derive an expression for the acceleration of the object.

(a) Apply Newton’s 2nd law to the disks and rod:

∑ =−= MafMgFx ssinθ , (1)

∑ =−= 0cosn θMgFFy , (2)

and

∑ == ατ Irf s0 (3)

Eliminate fs and α between equations (1) and (3) and solve for a to obtain: 2

sin

rIM

Mga+

=θ

(4)

Express the moment of inertia of the two disks plus connecting rod: ( )

2rod2

12disk

2rod2

12disk2

1

roddisk

2

2

rmRm

rmRm

III

+=

+=

+=


( )( ) ( )( )2

2212

mkg1.80

m0.02kg1m0.3kg20

⋅=

+=I

Substitute in equation (4) and evaluate a:

( )( )

( )2

2

2

2

m/s0.0443

m0.02mkg1.80kg41

sin30m/s9.81kg41

=

⋅+

°=a

(b) Find α from a: 2

2

rad/s2.21m0.02m/s0.0443

===raα

Rotation

687

(c) Express the kinetic energy of translation of the disks-plus-rod when it has rolled a distance ∆s down the incline:

221

trans MvK =

Using a constant-acceleration equation, relate the speed of the disks-plus-rod to their acceleration and the distance moved:

savv ∆+= 220

2

or, because v0 = 0, sav ∆= 22

Substitute to obtain: ( )( )( )

J3.63

m2m/s0.0443kg41 2trans

=

=

∆= sMaK

(d) Express the rotational kinetic energy of the disks after rolling 2 m in terms of their initial potential energy and their translational kinetic energy:

transtransirot KMghKUK −=−=

Substitute numerical values and evaluate Krot:

( )( )( )

J399

J3.63sin30m2m/s9.81kg41 2

rot

=

−°=K

95 ••• Picture the Problem We can express the coordinates of point P as the sum of the coordinates of the center of the wheel and the coordinates, relative to the center of the wheel, of the tip of the vector 0r

r. Differentiation of these expressions with respect to time

will give us the x and y components of the velocity of point P. (a) Express the coordinates of point P relative to the center of the wheel:

θ

θ

sinand

cos

0

0

ry

rx

=

=

Because the coordinates of the center of the circle are X and R:

( ) ( )θθ sin,cos, 00 rRrXyx PP ++=

Chapter 9

688

(b) Differentiate xP to obtain: ( )

dtdr

dtdX

rXdtdvPx

θθ

θ

⋅−=

+=

sin

cos

0

0

Note that

RV

dtdV

dtdX

−=−== ωθand so: θsin0

RVr

VvPx +=

Differentiate yP to obtain: ( )

dtdrrR

dtdvPy

θθθ ⋅=+= cossin 00

BecauseRV

dtd

−=−= ωθ: θcos0

RVrvPy −=

(c) Calculate rv rr

⋅ :

( )

( )

0

sincos

cossin

00

00

=

+⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛ +=

+=⋅

θθ

θθ

rRRVr

rRVrV

rvrv yPyxPxrv rr

(d) Express v in terms of its components:

2

200

20

20

22

sin21

cossin

Rr

Rr

V

RVr

RVr

V

vvv yx

++=

⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛ +=

+=

θ

θθ

Express r in terms of its components:

( ) ( )

2

200

20

20

22

sin21

sincos

Rr

Rr

R

rRr

rrr yx

++=

++=

+=

θ

θθ

Divide v by r to obtain:

RV

rv

==ω

Rotation

689

*96 ••• Picture the Problem Let the letter B identify the block and the letter C the cylinder. We can find the accelerations of the block and cylinder by applying Newton’s 2nd law and solving the resulting equations simultaneously. Apply xx maF =∑ to the block: B' mafF =− (1)

Apply xx maF =∑ to the cylinder: CMaf = , (2)

Apply ατ CMCM I=∑ to the cylinder:

αCMIfR = (3)

Substitute for ICM in equation (3) and solve for f = f ′ to obtain:

αMRf 21= (4)

Relate the acceleration of the block to the acceleration of the cylinder:

CBBC aaa +=

or, because aCB = −Rα is the acceleration of the cylinder relative to the block,

αRaa −= BC

and CB aaR −=α (5)

Equate equations (2) and (4) and substitute from (5) to obtain:

CB 3aa =

Substitute equation (4) in equation (1) and substitute for aC to obtain:

BB31 maMaF =−

Solve for aB: mM

Fa3

3B +

=

97 ••• Picture the Problem Let the letter B identify the block and the letter C the cylinder. In this problem, as in Problem 97, we can find the accelerations of the block and cylinder by applying Newton’s 2nd law and solving the resulting equations simultaneously.

Chapter 9

690

Apply xx maF =∑ to the block: BmafF =− (1)


Apply ατ CMCM I=∑ to the cylinder:

αCMIfR = (3)

Substitute for ICM in equation (3) and solve for f:

αMRf 21= (4)


CBBC aaa +=

or, because aCB = −Rα, αRaa −= BC


(a) Solve for α and substitute for aB to obtain:

( )mMRF

Ra

Raa

Raa

32

23 CCCCB

+=

=−

=−

=α

direction. ckwisecounterclo thein is , thereforeand, torquethat the

evident isit diagram force theFromα

(b) Equate equations (2) and (4) and substitute (5) to obtain:

CB 3aa =

From equations (1) and (4) we obtain:

BB31 maMaF =−

Solve for aB: mM

Fa3

3B +

=

Substitute to obtain the linear acceleration of the cylinder relative to the table:

mMFaa B 33

1C +

==

Rotation

691

(c) Express the acceleration of the cylinder relative to the block:

mMF

aaaaaa

32

23 CCCBCCB

+−=

−=−=−=

98 ••• Picture the Problem Let the system include the earth, the cylinder, and the block. Then F

ris an external force that

changes the energy of the system by doing work on it. We can find the kinetic energy of the block from its speed when it has traveled a distance d. We can find the kinetic energy of the cylinder from the sum of its translational and rotational kinetic energies. In part (c) we can add the kinetic energies of the block and the cylinder to show that their sum is the work done by Fr

in displacing the system a distance d.

(a) Express the kinetic energy of the block: 2

B21

blockonB mvWK ==

Using a constant-acceleration equation, relate the velocity of the block to its acceleration and the distance traveled:

davv B20

2B 2+=

or, because the block starts from rest, dav B

2B 2=


( ) dmadamK BB21

B 2 == (1)

Apply xx maF =∑ to the block: BmafF =− (2)


Apply ατ CMCM I=∑ to the

cylinder:

αCMIfR = (4)

Substitute for ICM in equation (4) and solve for f:

αMRf 21= (5)


CBBC aaa +=

or, because aCB = −Rα,

Chapter 9

692

αRaa −= BC


Equate equations (3) and (5) and substitute in (6) to obtain:

CB 3aa =

Substitute equation (5) in equation (2) and use CB 3aa = to obtain:

BC maMaF =−

or BB3

1 maMaF =−

Solve for aB:

MmFa

31B +

=

Substitute in equation (1) to obtain: Mm

mFdK31B +

=

(b) Express the total kinetic energy of the cylinder:

2

2CB

CM212

C21

2CM2

12C2

1rottranscyl

RvIMv

IMvKKK

+=

+=+= ω(7)

where BCCB vvv −= .

In part (a) it was established that: CB 3aa =

Integrate both sides of the equation with respect to time to obtain:

constant3 CB += vv

where the constant of integration is determined by the initial conditions that vC = 0 when vB = 0.

Substitute the initial conditions to obtain: 0constant = and

CB 3vv =

Substitute in our expression for vCB to obtain:

CCCBCCB 23 vvvvvv −=−=−=

Substitute for ICM and vCB in equation (7) to obtain: ( )( )

2C2

3

2

2C2

21

212

C21

cyl2

MvRvMRMvK

=

−+=

(8)

Rotation

693

Because B31

C vv = : 2B9

12C vv =

It part (a) it was established that: dav B

2B 2=

and

MmFa

31B +

=

Substitute to obtain: ( )

( )MmFd

dMm

Fdav

31

319

2B9

12C

92

2

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==


( )

( )MmMFd

MmFdMK

31

312

3cyl

3

92

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

(c) Express the total kinetic energy of the system and simplify to obtain:

( )( )( ) FdFd

MmMm

MmMFd

MmmFd

KKK

=++

=

++

+=

+=

31

31

31

cylBtot

33

3

99 •• Picture the Problem The forces responsible for the rotation of the gears are shown in the diagram to the right. The forces acting through the centers of mass of the two gears have been omitted because they produce no torque. We can apply Newton’s 2nd law in rotational form to obtain the equations of motion of the gears and the not slipping condition to relate their angular accelerations.

(a) Apply ατ I=∑ to the gears to obtain their equations of motion:

111mN 2 αIFR =−⋅ (1) and

222 αIFR = (2) where F is the force keeping the gears from slipping with respect to each other.

Because the gears do not slip 2211 αα RR =

Chapter 9

694

relative to each other, the tangential accelerations of the points where they are in contact must be the same:

or

121

12

12 ααα ==

RR

(3)

Divide equation (1) by R1 to obtain:

11

1

1

mN 2 αRIF

R=−

⋅

Divide equation (2) by R2 to obtain:

22

2 αRIF =

Add these equations to obtain:

22

21

1

1

1

mN 2 ααRI

RI

R+=

⋅

Use equation (3) to eliminate α2:

12

21

1

1

1 2mN 2 αα

RI

RI

R+=

⋅

Solve for α1 to obtain:

22

11

1

2

mN2

IRRI +

⋅=α

Substitute numerical values and evaluate α1:

( ) ( )2

221

rad/s400.0

mkg16m12m0.5mkg1

mN2

=

⋅+⋅

⋅=α

Use equation (3) to evaluate α2: ( ) 22

21

2 rad/s0.200rad/s0.400 ==α

(b) To counterbalance the 2-N·m torque, a counter torque of 2 N·m must be applied to the first gear. Use equation (2) with α1 = 0 to find F:

0mN 2 1 =−⋅ FR and

N4.00m0.5mN2mN2

1

=⋅

=⋅

=R

F

Rotation

695

*100 •• Picture the Problem Let r be the radius of the marble, m its mass, R the radius of the large sphere, and v the speed of the marble when it breaks contact with the sphere. The numeral 1 denotes the initial configuration of the sphere-marble system and the numeral 2 is configuration as the marble separates from the sphere. We can use conservation of energy to relate the initial potential energy of the marble to the sum of its translational and rotational kinetic energies as it leaves the sphere. Our choice of the zero of potential energy is shown on the diagram.

(a) Apply conservation of energy:

0=∆+∆ KU or

01212 =−+− KKUU

Because U2 = K1 = 0: ( )[ ]0

cos2

212

21 =++

+−+−

ω

θ

ImvrRrRmg

or ( )( )[ ]

0cos1

2212

21 =++

−+−

ω

θ

ImvrRmg

Use the rolling-without-slipping condition to eliminate ω:

( )( )[ ]

0

cos1

2

2

212

21 =++

−+−

rvImv

rRmg θ

From Table 9-1 we have: 2

52 mrI =

Substitute to obtain: ( )( )[ ]

( ) 0

cos1

2

22

52

212

21 =++

−+−

rvmrmv

rRmg θ

or ( )( )[ ]

0cos1

2512

21 =++

−+−

mvmvrRmg θ

Solve for v2 to obtain: ( )( )θcos1

7102 −+= rRgv

Apply rr maF =∑ to the marble as it separates from the sphere:

rRvmmg+

=2

cosθ

or

Chapter 9

696

( )rRgv

+=

2

cosθ

Substitute for v2:

( ) ( )( )

( )⎥⎦⎤

⎢⎣⎡ −=

⎥⎦⎤

⎢⎣⎡ −+

+=

θ

θθ

cos17

10

cos17

101cos rRgrRg


°=⎟⎠⎞

⎜⎝⎛= − 0.541710cos 1θ

(b)

sphere. theleavesit before slippingwithout rolling ball thekeep toneeded force than theless bemust friction

of force that themeaning sphere, theleaves ball theepoint wher at the 0 todecreases force normal theHowever, marble. on the force

normal by the multiplied than less always isfriction of force The sµ

Rolling With Slipping 101 • Picture the Problem Part (a) of this problem is identical to Example 9-16. In part (b) we can use the definitions of translational and rotational kinetic energy to find the ratio of the final and initial kinetic energies. (a) From Example 9-16:

gv

sk

20

1 4912

µ= ,

gv

tk

01 7

2µ

= , and

01k25

1 75 vgtv == µ

(b) When the ball rolls without slipping, v1 = rω. Express the final kinetic energy of the ball:

( )2014

52110

7

2

212

52

212

121

2212

121

rottransf

MvMvrvMrMv

IMv

KKK

==

+=

+=

+=

ω

Rotation

697

Express the ratio of the final and initial kinetic energies: 7

5202

1

2014

5

i

f ==MvMv

KK

(c) Substitute in the expressions in (a) to obtain:

( )( )( ) m6.26

m/s9.810.06m/s8

4912

2

2

1 ==s

( )( ) s3.88m/s9.810.06

m/s872

21 ==t

( ) m/s5.71m/s8

75

1 ==v

*102 •• Picture the Problem The cue stick’s blow delivers a rotational impulse as well as a translational impulse to the cue ball. The rotational impulse changes the angular momentum of the ball and the translational impulse changes its linear momentum. Express the rotational impulse Prot as the product of the average torque and the time during which the rotational impulse acts:

tP ∆= avrot τ

Express the average torque it produces about an axis through the center of the ball:

( ) ( )rhPrhP −=−= 00av sinθτ

where θ (= 90°) is the angle between F and the lever arm h − r.

Substitute in the expression for Prot to obtain:

( ) ( )( )( ) 0trams

00rot

ωILrhPrhtPtrhPP

=∆=−=−∆=∆−=

The translational impulse is also given by:

00trans mvptPP =∆=∆=

Substitute to obtain: ( ) 02

52

0 ωmrrhmv =−

Solve for ω0: ( )

20

0 25

rrhv −

=ω

Chapter 9

698

103 •• Picture the Problem The angular velocity of the rotating sphere will decrease until the condition for rolling without slipping is satisfied and then it will begin to roll. The force diagram shows the forces acting on the sphere. We can apply Newton’s 2nd law to the sphere and use the condition for rolling without slipping to find the speed of the center of mass when the sphere begins to roll without slipping. Relate the velocity of the sphere when it begins to roll to its acceleration and the elapsed time:

tav ∆= (1)

Apply Newton’s 2nd law to the sphere:

∑ == mafFx k , (2)

∑ =−= 0n mgFFy , (3)

and

∑ == ατ 0k0 Irf (4)

Using the definition of fk and Fn from equation (3), substitute in equation (2) and solve for a:

ga kµ=

Substitute in equation (1) to obtain: tgtav ∆=∆= kµ (5)

Solve for α in equation (4):

rg

mrmar

Irf k

252

0

k

25 µα ===

Express the angular speed of the sphere when it has been moving for a time ∆t:

trgt ∆−=∆−=

25 k

00µωαωω (6)

Express the condition that the sphere rolls without slipping:

ωrv =

Substitute from equations (5) and (6) and solve for the elapsed time until the sphere begins to roll:

gr

tk

0

72

µω

=∆

Rotation

699

Use equation (4) to find v when the sphere begins to roll: 7

272 0

k

k0k

ωµ

µωµ rg

grtgv ==∆=

104 •• Picture the Problem The sharp force delivers a rotational impulse as well as a translational impulse to the ball. The rotational impulse changes the angular momentum of the ball and the translational impulse changes its linear momentum. In parts (c) and (d) we can apply Newton’s 2nd law to the ball to obtain equations describing both the translational and rotational motion of the ball. We can then solve these equations to find the constant accelerations that allow us to apply constant-acceleration equations to find the velocity of the ball when it begins to roll and its sliding time.

(a) Relate the translational impulse delivered to the ball to its change in its momentum:

0avtrans mvptFP =∆=∆=

Solve for v0: m

tFv ∆= av

0


( )( ) m/s200kg0.02

s102kN20 4

0 =×

=−

v

(b) Express the rotational impulse Prot as the product of the average torque and the time during which the rotational impulse acts:

tP ∆= avrot τ

Letting h be the height at which the impulsive force is delivered, express the average torque it produces about an axis through the center of the ball:

θτ sinav lF=

where θ is the angle between F and the lever arm l .

Substitute h − r for l and 90° for θ ( )rhF −=avτ

Chapter 9

700

to obtain: Substitute in the expression for Prot to obtain:

( ) trhFP ∆−=rot

Because Ptrans = F∆t:

( )0

252

0transrot

ω

ω

mr

ILrhPP

=

=∆=−=

Express the translational impulse delivered to the cue ball:

00trans mvptPP =∆=∆=

Substitute for Ptrans to obtain:

002

52 mvmr =ω

Solve for ω0: ( )2

00 2

5r

rhv −=ω

Substitute numerical values and evaluate ω0:

( )( )( )

rad/s8000

m.052m0.05m0.09m/s2005

20

=

−=ω

(c) and (d) Relate the velocity of the ball when it begins to roll to its acceleration and the elapsed time:

tav ∆= (1)

Apply Newton’s 2nd law to the ball: ∑ == mafFx k , (2)

∑ =−= 0n mgFFy , (3)

and

∑ == ατ 0k0 Irf (4)


ga kµ=

Substitute in equation (1) to obtain: tgtav ∆=∆= kµ (5)


rg

mrmar

Irf k

252

0

k

25 µα ===

Rotation

701

Express the angular speed of the ball when it has been moving for a time ∆t:

trgt ∆−=∆−=

25 k

00µωαωω (6)

Express the speed of the ball when it has been moving for a time ∆t:

tgvv ∆+= k0 µ (7)

Express the condition that the ball rolls without slipping:

ωrv =

Substitute from equations (6) and (7) and solve for the elapsed time until the ball begins to roll:

gvrt

k

00

72

µω −

=∆


( )( )( )( )

s11.6

m/s9.810.5m/s200rad/s8000m0.05

72

2

=

⎥⎦

⎤⎢⎣

⎡ −=∆t

Use equation (4) to express v when the ball begins to roll:

tgvv ∆+= k0 µ


( )( )( )m/s572

s11.6m/s9.810.5m/s200 2

=

+=v

105 •• Picture the Problem Because the impulse is applied through the center of mass, ω0 = 0. We can use the results of Example 9-16 to find the rolling time without slipping, the distance traveled to rolling without slipping, and the velocity of the ball once it begins to roll without slipping. (a) From Example 9-16 we have:

gvtk

01 7

2µ

=

Substitute numerical values and evaluate t1:

( )( ) s0.194m/s9.810.6

m/s472

21 ==t

(b) From Example 9-16 we have: g

vsk

20

1 4912

µ=

Chapter 9

702

Substitute numerical values and evaluate s1:

( )( )( ) m0.666

m/s9.810.6m/s4

4912

2

2

1 ==s

(c) From Example 9-16 we have: 01 7

5 vv =


( ) m/s2.86m/s475

1 ==v

106 •• Picture the Problem Because the impulsive force is applied below the center line, the spin is backward, i.e., the ball will slow down. We’ll use the impulse-momentum theorem and Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constant-acceleration equations to relate these quantities to each other and to the elapsed time to rolling without slipping. (a) Express the rotational impulse delivered to the ball:

( ) 02

52

0cm00rot 32

ω

ω

mR

IRmvrmvP

=

===

Solve for ω0:

Rv0

0 35

=ω

(b) Apply Newton’s 2nd law to the ball to obtain:

∑ == ατ cmk0 IRf , (1)

∑ =−= 0n mgFFy , (2)

and

∑ =−= mafFx k (3)

Using the definition of fk and Fn

from equation (2), solve for α: Rg

mRmgR

ImgR

25 k

252

k

cm

k µµµα ===

Using a constant-acceleration equation, relate the angular speed of the ball to its acceleration:

tR

gt ∆+=∆+=2

5 k00

µωαωω

Rotation

703

Using the definition of fk and Fn

from equation (2), solve equation (3) for a:

ga kµ−=

Using a constant-acceleration equation, relate the speed of the ball to its acceleration:

tgvtavv ∆−=∆+= k00 µ (4)

Impose the condition for rolling without slipping to obtain:

tgvtR

gR ∆−=⎟⎠⎞

⎜⎝⎛ ∆+ k0

k0 2

5µ

µω

Solve for ∆t:

gv

tk

0

2116

µ=∆


0

0k

0k0

238.0

215

2116

v

vgvgvv

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

µµ

(c) Express the initial kinetic energy of the ball:

( )20

20

202

52

212

021

202

1202

1rottransi

056.1

1819

35

mv

mvRvmRmv

ImvKKK

=

=⎟⎠⎞

⎜⎝⎛+=

+=+= ω

(d) Express the work done by friction in terms of the initial and final kinetic energies of the ball:

fifr KKW −=

Express the final kinetic energy of the ball:

( )( ) 2

02

0107

2107

2

22

52

212

21

2cm2

1221

f

0397.0238.0 mvvm

mvRvmRmv

ImvK

==

=+=

+= ω

Substitute to find Wfr:

20

20

20fr

016.1

0397.0056.1

mv

mvmvW

=

−=

Chapter 9

704

107 •• Picture the Problem The figure shows the forces acting on the bowling during the sliding phase of its motion. Because the ball has a forward spin, the friction force is in the direction of motion and will cause the ball’s translational speed to increase. We’ll apply Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constant-acceleration equations to relate these quantities to each other and to the elapsed time to rolling without slipping.

(a) and (b) Relate the velocity of the ball when it begins to roll to its acceleration and the elapsed time:

tavv ∆+= 0 (1)

Apply Newton’s 2nd law to the ball: ∑ == mafFx k , (2)

∑ =−= 0n mgFFy , (3)

and

∑ == ατ 0k0 IRf (4)


ga kµ=

Substitute in equation (1) to obtain: tgvtavv ∆+=∆+= k00 µ (5)


Rg

mRmaR

IRf k

252

0

k

25 µα ===

Relate the angular speed of the ball to its acceleration:

tR

g∆−= k

0 25 µ

ωω

Apply the condition for rolling without slipping:

⎟⎠⎞

⎜⎝⎛ ∆−=

⎟⎠⎞

⎜⎝⎛ ∆−==

tR

gRvR

tR

gRRv

k0

k0

253

25

µ

µωω

Rotation

705

∴ tgvv ∆−= k0 253 µ (6)

Equate equations (5) and (6) and solve ∆t: g

vt

k

0

74

µ=∆

Substitute for ∆t in equation (6) to obtain: 00 57.1

711 vvv ==

(c) Relate ∆x to the average speed of the ball and the time it moves before beginning to roll without slipping:

( )

gv

gv

gvvv

tvvtvx

k

20

k

20

k

0002

1

021

av

735.04936

74

711

µµ

µ

==

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +=

∆+=∆=∆

*108 •• Picture the Problem The figure shows the forces acting on the cylinder during the sliding phase of its motion. The friction force will cause the cylinder’s translational speed to decrease and eventually satisfy the condition for rolling without slipping. We’ll use Newton’s 2nd law to find the linear and rotational velocities and accelerations of the ball and constant-acceleration equations to relate these quantities to each other and to the distance traveled and the elapsed time until the satisfaction of the condition for rolling without slipping.

(a) Apply Newton’s 2nd law to the cylinder:

∑ =−= MafFx k , (1)

∑ =−= 0n MgFFy , (2)

and

∑ == ατ 0k0 IRf (3)

Use fk = µkFn to eliminate Fn between equations (1) and (2) and solve for a:

ga kµ−=

Chapter 9

706

Using a constant-acceleration equation, relate the speed of the cylinder to its acceleration and the elapsed time:

tgvtavv ∆−=∆+= k00 µ

Similarly, eliminate fk between equations (2) and (3) and solve for α:

Rgk2µ

α =

Using a constant-acceleration equation, relate the angular speed of the cylinder to its acceleration and the elapsed time:

tR

gt ∆=∆+= k0

2µαωω

Apply the condition for rolling without slipping:

tg

tR

gRRtgvv

∆=

⎟⎠⎞

⎜⎝⎛ ∆==∆−=

k

kk0

2

2

µ

µωµ

Solve for ∆t:

gv

tk

0

3µ=∆

Substitute for ∆t in the expression for v: 0

k

0k0 3

23

vg

vgvv =−=µ

µ

(b) Relate the distance the cylinder travels to its average speed and the elapsed time:

( )

gv

gvvvtvx

k

20

k

003

202

1av

185

3

µ

µ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=∆=∆

(c) Express the ratio of the energy dissipated in friction to the cylinder’s initial mechanical energy:

i

fi

i

fr

KKK

KW −

=

Express the kinetic energy of the cylinder as it begins to roll without slipping: ( )

20

2

02

2

22

21

212

21

2cm2

1221

f

31

32

43

43 MvvMMv

RvMRMv

IMvK

=⎟⎠⎞

⎜⎝⎛==

+=

+= ω

Rotation

707

Substitute for Ki and Kf and simplify to obtain: 3

1202

1

203

1202

1

i

fr =−

=Mv

MvMvKW

109 •• Picture the Problem The forces acting on the ball as it slides across the floor are its weight ,mg

r the normal force nF

rexerted by

the floor, and the friction force .fv

Because the weight and normal force act through the center of mass of the ball and are equal in magnitude, the friction force is the net (decelerating) force. We can apply Newton’s 2nd law in both translational and rotational form to obtain a set of equations that we can solve for the acceleration of the ball. Once we have determined the ball’s acceleration, we can use constant-acceleration equations to obtain its velocity when it begins to roll without slipping.

(a) Apply aF rr

m=∑ to the ball: ∑ =−= mafFx (1) and

∑ =−= 0n mgFFy (2)

From the definition of the coefficient of kinetic friction we have:

nk Ff µ= (3)

Solve equation (2) for Fn: mgF =n

Substitute in equation (3) to obtain: mgf kµ=

Substitute in equation (1) to obtain: mamg =− kµ or

ga kµ−=

Apply ατ I=∑ to the ball: αIfr =

Solve for α to obtain: Imgr

Ifr kµα ==

Assuming that the coefficient of kinetic friction is constant*, we can use constant-acceleration equations to describe how long it will take the ball to begin

tgtavv k ∆−=∆=− µf (4) and

tIgmrk ∆=

µωf (5)

Chapter 9

708

rolling without slipping:

Once rolling without slipping has been established, we also have: r

vff =ω (6)

Equate equations (5) and (6):

tIgmr

rv k ∆=

µf

Solve for ∆t:

2f

gmrIvt

kµ=∆


f2

2f

f

vmr

Igmr

Ivgvvk

k

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

µµ

Solve for vf:

v

mrIv f

21

1

+=

(b) Express the total kinetic energy of the ball:

2f

2f 2

121 ωImvK +=

Because the ball is now rolling without slipping, fωrv = and:

( )

⎟⎠⎞

⎜⎝⎛

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

++=⎟

⎠⎞

⎜⎝⎛

++⎟

⎠⎞

⎜⎝⎛

+=

22

2

222

2

22

22

2

2

/11

21

/11/1

21

/11

21

/11

21

mrImv

mrImrImv

rv

mrIIv

mrImK

* Remarks: This assumption is not necessary. One can use the impulse-momentum theorem and the related theorem for torque and change in angular momentum to prove that the result holds for an arbitrary frictional force acting on the ball, so long as the ball moves along a straight line and the force is directed opposite to the direction of motion of the ball. General Problems *110 • Picture the Problem The angular velocity of an object is the ratio of the number of revolutions it makes in a given period of time to the elapsed time.

Rotation

709

The moon’s angular velocity is:

rad/s102.66

s3600h1

h24day1

revrad2

days27.3rev1days27.3

rev1

6−×=

×××=

=

π

ω

111 • Picture the Problem The moment of inertia of the hoop, about an axis perpendicular to the plane of the hoop and through its edge, is related to its moment of inertia with respect to an axis through its center of mass by the parallel axis theorem. Apply the parallel axis theorem: 2222

cm 2mRMRMRMhII =+=+=

112 •• Picture the Problem The force you exert on the rope results in a net torque that accelerates the merry-go-round. The moment of inertia of the merry-go-round, its angular acceleration, and the torque you apply are related through Newton’s 2nd law. (a) Using a constant-acceleration equation, relate the angular displacement of the merry-go-round to its angular acceleration and acceleration time:

( )221

0 tt ∆+∆=∆ αωθ


21 t∆=∆ αθ

Solve for and evaluate α: ( )

( )( )

222 rad/s0873.0

s12rad222

==∆∆

=πθα

t

(b) Use the definition of torque to obtain: ( )( ) mN572m2.2N260 ⋅=== Frτ

(c) Use Newton’s 2nd law to find the moment of inertia of the merry-go-round: 23

2net

mkg106.55

rad/s0.0873mN572

⋅×=

⋅==

ατI

Chapter 9

710

113 • Picture the Problem Because there are no horizontal forces acting on the stick, the center of mass of the stick will not move in the horizontal direction. Choose a coordinate system in which the origin is at the horizontal position of the center of mass. The diagram shows the stick in its initial raised position and when it has fallen to the ice. Express the displacement of the right end of the stick ∆x as the difference between the position coordinates x2 and x2:

12 xxx −=∆

Using trigonometry, find the initial coordinate of the right end of the stick:

( ) m0.866cos30m1cos1 =°== θlx

Because the center of mass has not moved horizontally:

m12 == lx

Substitute to find the displacement of the right end of the stick:

m0.134m0.866m1 =−=∆x

114 •• Picture the Problem The force applied to the string results in a torque about the center of mass of the disk that accelerates it. We can relate these quantities to the moment of inertia of the disk through Newton’s 2nd law and then use constant-acceleration equations to find the disk’s angular velocity the angle through which it has rotated in a given period of time. The disk’s rotational kinetic energy can be found from its definition. (a) Use the definition of torque to obtain:

( )( ) mN2.40m0.12N20 ⋅==≡ FRτ

(b) Use Newton’s 2nd law to express the angular acceleration of the disk in terms of the net torque acting on it and its moment of inertia:

221

netnet

MRIττα ==


( )( )( )

22 rad/s66.7

m0.12kg5mN2.402

=⋅

=α

(c) Using a constant-acceleration equation, relate the angular velocity of the disk to its angular

t∆+= αωω 0

or, because ω0 = 0, t∆= αω

Rotation

711

acceleration and the elapsed time:

Substitute numerical values and evaluate ω:

( )( ) rad/s333s5rad/s66.7 2 ==ω

(d) Use the definition of rotational kinetic energy to obtain:

( ) 2221

212

21

rot ωω MRIK ==

Substitute numerical values and evaluate Krot:

( )( ) ( )kJ2.00

rad/s333m0.12kg5 2241

rot

=

=K

(e) Using a constant-acceleration equation, relate the angle through which the disk turns to its angular acceleration and the elapsed time:

( )221

0 tt ∆+∆=∆ αωθ


21 t∆=∆ αθ

Substitute numerical values and evaluate ∆θ :

( )( ) rad834s5rad/s66.7 2221 ==∆θ

(f) Express K in terms of τ and θ : ( ) ( )

θτ

αταατω

∆=

∆=∆⎟⎠⎞

⎜⎝⎛== 2

212

212

21 ttIK

115 •• Picture the Problem The diagram shows the rod in its initial horizontal position and then, later, as it swings through its vertical position. The center of mass is denoted by the numerals 0 and 1. Let the length of the rod be represented by L and its mass by m. We can use Newton’s 2nd law in rotational form to find, first, the angular acceleration of the rod and then, from α, the acceleration of any point on the rod. We can use conservation of energy to find the angular velocity of the center of mass of the rod when it is vertical and then use this value to find its linear velocity.

(a) Relate the acceleration of the center of the rod to the angular

αα2La == l

Chapter 9

712

acceleration of the rod: Use Newton’s 2nd law to relate the torque about the suspension point of the rod (exerted by the weight of the rod) to the rod’s angular acceleration:

Lg

ML

LMg

I 232

231

===τα


( )( )

22

rad/s18.4m0.82

m/s9.813==α


( )( ) 2221 m/s7.36rad/s18.4m0.8 ==a

(b) Relate the acceleration of the end of the rod to α:

( )( )2

2end

m/s14.7

rad/s18.4m0.8

=

== αLa

(c) Relate the linear velocity of the center of mass of the rod to its angular velocity as it passes through the vertical:

Lhv ωω 21=∆=

Use conservation of energy to relate the changes in the kinetic and potential energies of the rod as it swings from its initial horizontal orientation through its vertical orientation:

00101 =−+−=∆+∆ UUKKUK

or, because K0 = U1 = 0, 001 =−UK


hmgI P ∆=221 ω

Substitute for ∆h and solve for ω:

Lg3

=ω

Substitute to obtain: gL

LgLv 33

21

21 ==

Substitute numerical values and evaluate v: ( )( ) m/s2.43m0.8m/s9.813 2

21 ==v

Rotation

713

116 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the track. The initial potential energy of the marble is transformed into translational and rotational kinetic energy as it rolls down the track to its lowest point and then, because the portion of the track to the right is frictionless, into translational kinetic energy and, eventually, into gravitational potential energy. Using conservation of energy, relate h2 to the kinetic energy of the marble at the bottom of the track:


0fi =+− UK

Substitute for Ki and Uf to obtain: 022

21 =−− MghMv

Solve for h2:

gvh2

2

2 = (1)

Using conservation of energy, relate h1 to the kinetic energy of the marble at the bottom of the track:


0if =−UK

Substitute for Kf and Ui to obtain: 01

2212

21 =−+ MghIMv ω

Substitute for I and solve for v2 to obtain:

17102 ghv =


17517

10

2 2h

ggh

h ==

*117 •• Picture the Problem To stop the wheel, the tangential force will have to do an amount of work equal to the initial rotational kinetic energy of the wheel. We can find the stopping torque and the force from the average power delivered by the force during the slowing of the wheel. The number of revolutions made by the wheel as it stops can be found from a constant-acceleration equation. (a) Relate the work that must be done to stop the wheel to its kinetic energy:

( ) 224122

21

212

21 ωωω mrmrIW ===

Chapter 9

714


( )( )

kJ780

s60min1

revrad2

minrev1100

m1.4kg1202

241

=

⎥⎦

⎤⎢⎣

⎡×××

=

π

W

(b) Express the stopping torque is terms of the average power required:

avav τω=P

Solve for τ :

av

av

ωτ P

=

Substitute numerical values and evaluate τ : ( )( )

( )( )( )

mN3.902

smin/601rad/rev2rev/min1100s/min60min2.5

kJ780

⋅=

= πτ

Relate the stopping torque to the magnitude of the required force and solve for F:

N151m0.6

mN90.3=

⋅==

RF τ

(c) Using a constant-acceleration equation, relate the angular displacement of the wheel to its average angular velocity and the stopping time:

t∆=∆ avωθ

Substitute numerical values and evaluate ∆θ:

( )

rev1380

min2.52

rev/min1100

=

⎟⎠⎞

⎜⎝⎛=∆θ

118 •• Picture the Problem The work done by the four children on the merry-go-round will change its kinetic energy. We can use the work-energy theorem to relate the work done by the children to the distance they ran and Newton’s 2nd law to find the angular acceleration of the merry-go-round.

Rotation

715

(a) Use the work-kinetic energy theorem to relate the work done by the children to the kinetic energy of the merry-go-round:

f

forcenet

K

KW

=

∆=

or 2

214 ωIsF =∆

Substitute for I and solve for ∆s to obtain:

Fmr

Fmr

FIs

1688

2222212 ωωω

===∆

Substitute numerical values and evaluate ∆s: ( )( )

( )m6.11

N2616rev

rad2s2.8

rev1m2kg2402

2

=

⎥⎦

⎤⎢⎣

⎡×

=∆

π

s

(b) Apply Newton’s 2nd law to express the angular acceleration of the merry-go-round:

mrF

mrFr

I84

221

net ===τα


( )( )( )

2rad/s0.433m2kg240

N268==α

(c) Use the definition of work to relate the force exerted by each child to the distance over which that force is exerted:

( )( ) J302m11.6N26 ==∆= sFW

(d) Relate the kinetic energy of the merry-go-round to the work that was done on it:

sFKKW ∆=−=∆= 40fforcenet

Substitute numerical values and evaluate Wnet force:

( )( ) kJ1.21m11.6N264forcenet ==W

119 •• Picture the Problem Because the center of mass of the hoop is at its center, we can use Newton’s second law to relate the acceleration of the hoop to the net force acting on it. The distance moved by the center of the hoop can be determined using a constant-acceleration equation, as can the angular velocity of the hoop. (a) Using a constant-acceleration equation, relate the distance the

( )2cm2

1 tas ∆=∆

Chapter 9

716

center of the travels in 3 s to the acceleration of its center of mass: Relate the acceleration of the center of mass of the hoop to the net force acting on it:

mF

a netcm =

Substitute to obtain: ( )mtFs

2

2∆=∆

Substitute numerical values and evaluate ∆s:

( )( )( ) m15.0

kg1.52s3N5 2

==∆s

(b) Relate the angular velocity of the hoop to its angular acceleration and the elapsed time:

t∆= αω

Use Newton’s 2nd law to relate the angular acceleration of the hoop to the net torque acting on it:

mRF

mRFR

I=== 2

netτα

Substitute to obtain: mR

tF∆=ω


( )( )( )( ) rad/s15.4

m0.65kg1.5s3N5

==ω

120 •• Picture the Problem Let R represent the radius of the grinding wheel, M its mass, r the radius of the handle, and m the mass of the load attached to the handle. In the absence of information to the contrary, we’ll treat the 25-kg load as though it were concentrated at a point. Let the zero of gravitational potential energy be where the 25-kg load is at its lowest point. We’ll apply Newton’s 2nd law and the conservation of mechanical energy to determine the initial angular acceleration and the maximum angular velocity of the wheel. (a) Use Newton’s 2nd law to relate the acceleration of the wheel to the net torque acting on it:

2221

net

mrMRmgr

I +==

τα

Rotation

717


( )( )( )( )( ) ( )( )

2

2221

2

rad/s9.58

m0.65kg25m0.45kg60m0.65m/s9.81kg25

=

+=α

(b) Use the conservation of mechanical energy to relate the initial potential energy of the load to its kinetic energy and the rotational kinetic energy of the wheel when the load is directly below the center of mass of the wheel:


0irotf,transf, =−+ UKK .

Substitute and solve for ω: ( ) 02221

212

21 =−+ mgrMRmv ω ,

0224122

21 =−+ mgrMRmr ωω ,

and

2224

MRmrmgr+

=ω


( )( )( )( )( ) ( )( )

rad/s38.4

m0.45kg60m0.65kg252m0.65m/s9.81kg254

22

2

=

+=ω

*121 •• Picture the Problem Let the smaller block have the dimensions shown in the diagram. Then the length, height, and width of the larger block are ,lS h,S and w,S respectively. Let the numeral 1 denote the smaller block and the numeral 2 the larger block and express the ratios of the surface areas, masses, and moments of inertia of the two blocks.

(a) Express the ratio of the surface areas of the two blocks:

( )( ) ( )( ) ( )( )

( )

2

21

2

S222222S

222SS2SS2SS2

=

++++

=

++++

=

whhwwhhw

whhwhwhw

AA

ll

ll

ll

ll

Chapter 9

718

(b) Express the ratio of the masses of the two blocks:

( )( )( )

( ) 33

1

2

1

2

1

2

SS

SSS

==

===

hwhw

hwhw

VV

VV

MM

l

l

l

l

ρρ

(c) Express the ratio of the moments of inertia, about the axis shown in the diagram, of the two blocks:

( ) ( )[ ][ ]

[ ][ ] ( )2

1

222

222

1

2

22112

1

22212

1

1

2

SS

SS

⎟⎟⎠

⎞⎜⎜⎝

⎛=

++

=

++

=

MM

hh

MM

hMhM

II

l

l

l

l

In part (b) we showed that: 3

1

2 S=MM

Substitute to obtain: ( )( ) 523

1

2 SSS ==II

122 •• Picture the Problem We can derive the perpendicular-axis theorem for planar objects by following the step-by-step procedure outlined in the problem. (a) and (b) ( )

yx

z

II

dmydmx

dmyxdmrI

+=

+=

+==

∫ ∫∫∫

22

222

(c) Let the z axis be the axis of rotation of the disk. By symmetry:

yx II =

Express Iz in terms of Ix:

xz II 2=

Letting M represent the mass of the disk, solve for Ix:

( ) 2412

21

21

21 MRMRII zx ===

123 •• Picture the Problem Let the zero of gravitational potential energy be at the center of the disk when it is directly below the pivot. The initial gravitational potential energy of the disk is transformed into rotational kinetic energy when its center of mass is directly below the pivot. We can use Newton’s 2nd law to relate the force exerted by the pivot to the weight of the disk and the centripetal force acting on it at its lowest point.

Rotation

719

(a) Use the conservation of mechanical energy to relate the initial potential energy of the disk to its kinetic energy when its center of mass is directly below the pivot:


0irotf, =−UK

Substitute for rotf,K and iU :

0221 =− MgrIω (1)

Use the parallel-axis theorem to relate the moment of inertia of the disk about the pivot to its moment of inertia with respect to an axis through its center of mass:

2cm MhII +=

or 2

2322

21 MrMrMrI =+=

Solve equation (1) for ω and substitute for I to obtain: r

g34

=ω

(b) Letting F represent the force exerted by the pivot, use Newton’s 2nd law to express the net force acting on the swinging disk as it passes through its lowest point:

2net ωMrMgFF =−=

Solve for F and simplify to obtain:

MgMgMgrgMrMgMrMgF

37

34

2

34

=+=

+=+= ω

124 •• Picture the Problem The diagram shows a vertical cross-piece. Because we’ll need to take moments about the point of rotation (point P), we’ll need to use the parallel-axis theorem to find the moments of inertia of the two parts of this composite structure. Let the numeral 1 denote the vertical member and the numeral 2 the horizontal member. We can apply Newton’s 2nd law in rotational form to the structure to express its angular acceleration in terms of the net torque causing it to fall and its moment of inertia with respect to point P.

Chapter 9

720

(a) Taking clockwise rotation to be positive (this is the direction the structure is going to rotate), apply

ατ PI=∑ :

αPIwgmgm =⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛

22 12

2l

Solve for α to obtain:

PIgwmgm

2122 −

=lα

or ( )

( )PP IIwmmg

21

122

2 +−

=lα (1)

Convert wand,, 21 ll to SI units:

m3.66ft3.281

m1ft121 =×=l ,

m83.1ft3.281

m1ft62 =×=l , and

m610.0ft3.281

m1ft2 =×=w

Using Table 9-1 and the parallel-axis theorem, express the moment of inertia of the vertical member about an axis through point P:

( )2412

131

1

2

12113

11 2

wm

wmmI P

+=

⎟⎠⎞

⎜⎝⎛+=

l

l

Substitute numerical values and evaluate I1P:

( ) ( ) ( )[ ]23

2412

31

1

mkg1060.1

m0.610m3.66kg350

⋅×=

+=PI

Using the parallel-axis theorem, express the moment of inertia of the horizontal member about an axis through point P:

22cm,22 dmII P += (2)

where ( ) ( )2

2212

21

12 wwd −++= ll

Solve for d: ( ) ( )2

2212

21

1 wwd −++= ll


( )[ ] ( )[ ] m86.3m0.610m1.83m0.610m3.66 2212

21 =−++=d

From Table 9-1 we have: 2

22121

cm,2 lmI =

Substitute in equation (2) to obtain: ( )22

2121

2

22

22212

12

dm

dmmI P

+=

+=

l

l

Rotation

721

Evaluate I2P: ( ) ( ) ( )[ ]23

22121

2

mkg1066.2

m3.86m1.83kg175

⋅×=

+=PI

Substitute in equation (1) and evaluate α:

( ) ( )( ) ( )( )[ ]( )

223

2

rad/s123.0mkg102.661.602

m0.61kg350m1.83kg175m/s9.81=

⋅×+−

=α

(b) Express the magnitude of the acceleration of the sparrow:

Ra α= where R is the distance of the sparrow from the point of rotation and

( ) ( )22

21

2 wwR −++= ll

Solve for R: ( ) ( )22

21 wwR −++= ll

Substitute numerical values and evaluate R:

( ) ( ) m4.44m0.610m1.83m0.610m3.66 22 =−++=R


( )( )2

2

m/s0.546

m4.44rad/s0.123

=

=a

(c) Refer to the diagram to express ax in terms of a: R

waaax+

== 1cos lθ

Substitute numerical values and evaluate ax: ( )

2

2

m/s0.525

m4.44m0.61m3.66m/s0.546

=

+=xa

Chapter 9

722

125 •• Picture the Problem Let the zero of gravitational potential energy be at the bottom of the incline. The initial potential energy of the spool is transformed into rotational and translational kinetic energy when the spool reaches the bottom of the incline. We can apply the conservation of mechanical energy to find an expression for its speed at that location. The force diagram shows the forces acting on the spool when there is enough friction to keep it from slipping. We’ll use Newton’s 2nd law in both translational and rotational form to derive an expression for the static friction force.

(a) In the absence of friction, the forces acting on the spool will be its weight, the normal force exerted by the incline, and the tension in the string. A component of its weight will cause the spool to accelerate down the incline and the tension in the string will exert a torque that will cause counterclockwise rotation of the spool.

unwinds. string asdirection ckwisecounterclo a

in spinning on,acceleraticonstant at plane down the move willspool The

Use the conservation of mechanical energy to relate the speed of the center of mass of the spool at the bottom of the slope to its initial potential energy:


0irotf,transf, =−+ UKK .

Substitute for transf,K , rotf,K and iU :

0sin2212

21 =−+ θω MgDIMv (1)

Substitute for ω and solve for v to obtain:

0sin2

2

212

21 =−+ θMgD

rvIMv

and

2

sin2

rIM

MgDv+

=θ

Rotation

723

(b) Apply Newton’s 2nd law to the spool: ∑ =−−= 0sin sfTMgFx θ

∑ =−= 0s0 RfTrτ

Eliminate T between these equations to obtain:

rR

Mgf+

=1

sins

θ, up the incline.

126 •• Picture the Problem While the angular acceleration of the rod is the same at each point along its length, the linear acceleration and, hence, the force exerted on each coin by the rod, varies along its length. We can relate this force the linear acceleration of the rod through Newton’s 2nd law and the angular acceleration of the rod. Letting x be the distance from the pivot, use Newton’s 2nd law to express the force F acting on a coin:

( ) ( )xmaxFmgF =−=net

or ( ) ( )( )xagmxF −= (1)

Use Newton’s 2nd law to relate the angular acceleration of the system to the net torque acting on it: L

gML

LMg

I 232

231

net ===τα

Relate a(x) and α: ( ) ( ) gxgxxxa ===

m5.123α


( ) ( ) ( )xmggxgmxF −=−= 1

Evaluate F(0.25 m): ( ) ( ) mgmgF 75.0m25.01m25.0 =−=



Evaluate F(1 m): ( ) ( ) ( ) 0m5.1m25.1m1 === FFF

*127 •• Picture the Problem The diagram shows the force the hand supporting the meterstick exerts at the pivot point and the force the earth exerts on the meterstick acting at the center of mass. We can relate the angular acceleration to the acceleration of the end of the meterstick using αLa = and use Newton’s 2nd law in rotational form to relate α to the moment of inertia of the meterstick.

Chapter 9

724

(a) Relate the acceleration of the far end of the meterstick to the angular acceleration of the meterstick:

αLa = (1)

Apply ατ PP I=∑ to the meterstick:

αPILMg =⎟⎠⎞

⎜⎝⎛

2

Solve for α:

PIMgL2

=α

From Table 9-1, for a rod pivoted at one end, we have:

2

31 MLIP =


gMLMgL

23

23

2 ==α


3ga =


( ) 22

m/s14.72m/s9.813

==a

(b) Express the acceleration of a point on the meterstick a distance x from the pivot point:

xLgxa

23

== α

Express the condition that the meterstick leaves the penny behind:

ga >

Substitute to obtain: gx

Lg

>23

Solve for and evaluate x: ( ) cm7.66

3m12

32

==>Lx

Rotation

725

128 •• Picture the Problem Let m represent the 0.2-kg mass, M the 0.8-kg mass of the cylinder, L the 1.8-m length, and x + ∆x the distance from the center of the objects whose mass is m. We can use Newton’s 2nd law to relate the radial forces on the masses to the spring’s stiffness constant and use the work-energy theorem to find the work done as the system accelerates to its final angular speed. (a) Express the net inward force acting on each of the 0.2-kg masses:

( )∑ ∆+=∆= 2radial ωxxmxkF

Solve for k: ( ) 2

xxxmk

∆∆+

=ω

Substitute numerical values and evaluate k:

( )( )( )

N/m230

m0.4rad/s24m0.8kg0.2 2

=

=k

(b) Using the work-energy theorem, relate the work done to the change in energy of the system:

( )2212

21

springrot

xkI

UKW

∆+=

∆+=

ω (1)

Express I as the sum of the moments of inertia of the cylinder and the masses:

m

mM

IMLMr

III

221212

21

2

++=

+=

From Table 9-1 we have, for a solid cylinder about a diameter through its center:

21212

41 mLmrI +=

where L is the length of the cylinder.

For a disk (thin cylinder), L is small and:

241 mrI =

Apply the parallel-axis theorem to obtain:

2241 mxmrIm +=

Substitute to obtain: ( )( )22

412

1212

21

22412

1212

21

2

2

xrmMLMr

mxmrMLMrI

+++=

+++=


Chapter 9

726

( )( ) ( )( ) ( ) ( ) ( )[ ]2

22412

1212

21

mN492.0

m0.8m0.2kg0.22m1.8kg0.8m0.2kg0.8

⋅=

+++=I


( )( ) ( )( ) J160m0.4N/m230rad/s24mN0.492 22122

21 =+⋅=W

129 •• Picture the Problem Let m represent the 0.2-kg mass, M the 0.8-kg mass of the cylinder, L the 1.8-m length, and x + ∆x the distance from the center of the objects whose mass is m. We can use Newton’s 2nd law to relate the radial forces on the masses to the spring’s stiffness constant and use the work-energy theorem to find the work done as the system accelerates to its final angular speed. Using the work-energy theorem, relate the work done to the change in energy of the system:

( )2212

21

springrot

xkI

UKW

∆+=

∆+=

ω (1)

Express I as the sum of the moments of inertia of the cylinder and the masses:

m

mM

IMLMr

III

221212

21

2

++=

+=

From Table 9-1 we have, for a solid cylinder about a diameter through its center:

21212

41 mLmrI +=

where L is the length of the cylinder.

For a disk (thin cylinder), L is small and:

241 mrI =

Apply the parallel-axis theorem to obtain:

2241 mxmrIm +=

Substitute to obtain: ( )( )22

412

1212

21

22412

1212

21

2

2

xrmMLMr

mxmrMLMrI

+++=

+++=


( )( ) ( )( ) ( ) ( ) ( )[ ]2

22412

1212

21

mN492.0

m0.8m0.2kg0.22m1.8kg0.8m0.2kg0.8

⋅=

+++=I

Rotation

727

Express the net inward force acting on each of the 0.2-kg masses:

( )∑ ∆+=∆= 2radial ωxxmxkF

Solve for ω:

( )xxmxk∆+

∆=ω


( )( )( )( ) rad/s12.2

m0.8kg0.2m0.4N/m60

==ω

Substitute numerical values in equation (1) to obtain:

( )( )( )( )J4.14

m0.4N/m06

rad/s2.21mN0.4922

21

2221

=

+

⋅=W

130 •• Picture the Problem The force diagram shows the forces acting on the cylinder. Because F causes the cylinder to rotate clockwise, f, which opposes this motion, is to the right. We can use Newton’s 2nd law in both translational and rotational forms to relate the linear and angular accelerations to the forces acting on the cylinder. (a) Use Newton’s 2nd law to relate the angular acceleration of the center of mass of the cylinder to F:

MRF

MRFR

I2

221

net ===τα

Use Newton’s 2nd law to relate the acceleration of the center of mass of the cylinder to F:

MF

MFa == net

cm

Express the rolling-without-slipping condition to the accelerations:

αα 2cm ===MRF

Ra'

(b) Take the point of contact with the floor as the ″pivot″ point, express the net torque about that point, and solve for α:

ατ IFR == 2net

and

IFR2

=α

Chapter 9

728

Express the moment of inertia of the cylinder with respect to the pivot point:

22322

21 MRMRMRI =+=

Substitute to obtain: MRF

MRFR

342

223

==α

Express the linear acceleration of the cylinder: M

FRa34

cm == α

Apply Newton’s 2nd law to the forces acting on the cylinder:

∑ =+= cmMafFFx

Solve for f:

direction. positive in the 3

4

31

cm

xF

FFFMaf

=

−=−=

131 •• Picture the Problem As the load falls, mechanical energy is conserved. As in Example 9-7, choose the initial potential energy to be zero. Apply conservation of mechanical energy to obtain an expression for the speed of the bucket as a function of its position and use the given expression for t to determine the time required for the bucket to travel a distance y. Apply conservation of mechanical energy:

000iiff =+=+=+ KUKU (1)

Express the total potential energy when the bucket has fallen a distance y:

⎟⎠⎞

⎜⎝⎛−−=

++=

2c

wfcfbff

y'gmmgy

UUUU

where 'mc is the mass of the hanging part of the cable.

Assume the cable is uniform and express 'mc in terms of mc, y, and L: L

my'm cc = or y

Lm'm c

c =


gymmgyU

2

2c

f −−=

Rotation

729

Noting that bucket, cable, and rim of the winch have the same speed v, express the total kinetic energy when the bucket is falling with speed v: ( )

2412

c212

21

2

22

21

212

c212

21

2f2

12c2

1221

wfcfbff

MvvmmvRvMRvmmv

Ivmmv

KKKK

++=

++=

++=

++=

ω


02

2412

c21

221

2c

=++

+−−

Mvvm

mvLgymmgy

Solve for v:

c

c

mmMLgymmgy

v22

242

++

+=

A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic Form D9 0 y0

D10 D9+$B$8 y + ∆y E9 0 v0

E10 ((4*$B$3*$B$7*D10+2*$B$7*D10^2/(2*$B$5))/ ($B$1+2*$B$3+2*$B$4))^0.5

c

c

mmMLgymmgy

22

242

++

+

F10 F9+$B$8/((E10+E9)/2) yvvt nn

n ∆⎟⎠⎞

⎜⎝⎛ +

+ −− 2

11

J9 0.5*$B$7*H9^2 2

21 gt

A B C D E F G H I J

1 M= 10 kg 2 R= 0.5 m 3 m= 5 kg 4 mc= 3.5 kg 5 L= 10 m 6 7 g= 9.81 m/s^2 8 dy= 0.1 m y v(y) t(y) t(y) y 1/2gt^29 0.0 0.00 0.00 0.00 0.0 0.00

10 0.1 0.85 0.23 0.23 0.1 0.27 11 0.2 1.21 0.33 0.33 0.2 0.54 12 0.3 1.48 0.41 0.41 0.3 0.81 13 0.4 1.71 0.47 0.47 0.4 1.08 15 0.5 1.91 0.52 0.52 0.5 1.35

Chapter 9

730

105 9.6 9.03 2.24 2.24 9.6 24.61 106 9.7 9.08 2.25 2.25 9.7 24.85 107 9.8 9.13 2.26 2.26 9.8 25.09 108 9.9 9.19 2.27 2.27 9.9 25.34 109 10.0 9.24 2.28 2.28 10.0 25.58

The solid line on the graph shown below shows the position y of the bucket when it is in free fall and the dashed line shows y under the conditions modeled in this problem.

0

2

4

6

8

10

12

14

16

18

20

0.0 0.4 0.8 1.2 1.6 2.0

t (s)

y (m

)

y'free fall

132 •• Picture the Problem The pictorial representation shows the forces acting on the cylinder when it is stationary. First, we note that if the tension is small, then there can be no slipping, and the system must roll. Now consider the point of contact of the cylinder with the surface as the “pivot” point. If τ about that point is zero, the system will not roll. This will occur if the line of action of the tension passes through the pivot point. From the diagram we see that:

⎟⎠⎞

⎜⎝⎛= −

Rr1cosθ

Rotation

731

*133 •• Picture the Problem Free-body diagrams for the pulley and the two blocks are shown to the right. Choose a coordinate system in which the direction of motion of the block whose mass is M (downward) is the positive y direction. We can use the given relationship θµ ∆= s

max' TeT to relate the tensions in the rope on either side of the pulley and apply Newton’s 2nd law in both rotational form (to the pulley) and translational form (to the blocks) to obtain a system of equations that we can solve simultaneously for a, T1, T2, and M. (a) Use θµ ∆= s

max' TeT to evaluate the maximum tension required to prevent the rope from slipping on the pulley:

( ) ( ) N7.25N10' 3.0max == πeT

(c) Given that the angle of wrap is π radians, express T2 in terms of T1:

13.0

12 57.2 TeTT == π (1)

Because the rope doesn’t slip, we can relate the angular acceleration, α, of the pulley to the acceleration, a, of the hanging masses by:

ra

=α

Apply yy maF =∑ to the two blocks to obtain:

mamgT =−1 (2) and

MaTMg =− 2 (3)

Apply ∑ = ατ I to the pulley to obtain:

( )raIrTT =− 12 (4)

Substitute for T2 from equation (1) in equation (4) to obtain:

( )raIrTT =− 1157.2

Solve for T1 and substitute numerical values to obtain: ( )

( )a

aar

IT

kg91.9m0.151.57mkg0.35

57.1 2

2

21

=

⋅==

(5)


( ) mamga =−kg91.9

Chapter 9

732

Solve for and evaluate a:

22

m/s10.11

kg1kg9.91m/s9.81

1kg91.9kg91.9

=−

=

−=

−=

m

gm

mga

(b) Solve equation (3) for M:

agTM−

= 2

Substitute in equation (5) to find T1: ( )( ) N10.9m/s1.10kg91.9 2

1 ==T

Substitute in equation (1) to find T2: ( )( ) N28.0N10.9.5722 ==T

Evaluate M: kg21.3

m/s1.10m/s9.81N28.0

22 =−

=M

134 ••• Picture the Problem When the tension is horizontal, the cylinder will roll forward and the friction force will be in the direction of .T

r We can use Newton’s 2nd law to obtain

equations that we can solve simultaneously for a and f. (a) Apply Newton’s 2nd law to the cylinder:

∑ =+= mafTFx (1)

and

∑ =−= ατ IfRTr (2)

Substitute for I and α in equation (2) to obtain:

mRaRamRfRTr 2

1221 ==− (3)

Solve equation (3) for f: ma

RTrf 2

1−= (4)

Substitute equation (4) in equation (1) and solve for a:

⎟⎠⎞

⎜⎝⎛ +=

Rr

mTa 1

32

(5)

Substitute equation (5) in equation (4) to obtain: ⎟

⎠⎞

⎜⎝⎛ −= 12

3 RrTf

(b) Equation (4) gives the acceleration of the center of mass: ⎟

⎠⎞

⎜⎝⎛ +=

Rr

mTa 1

32

Rotation

733

(c) Express the condition that mTa > : 11

321

32

>⎟⎠⎞

⎜⎝⎛ +⇒>⎟

⎠⎞

⎜⎝⎛ +

Rr

mT

Rr

mT

or Rr 2

1>

(d) If Rr 2

1> : . ofdirection in the i.e., ,0 Tr

>f

135 ••• Picture the Problem The system is shown in the drawing in two positions, with angles θ0 and θ with the vertical. The drawing also shows all the forces that act on the stick. These forces result in a rotation of the stick—and its center of mass—about the pivot, and a tangential acceleration of the center of mass. We’ll apply the conservation of mechanical energy and Newton’s 2nd law to relate the radial and tangential forces acting on the stick. Use the conservation of mechanical energy to relate the kinetic energy of the stick when it makes an angle θ with the vertical and its initial potential energy:

0ifif =−+− UUKK

or, because Kf = 0,

0cos2

cos2 0

221 =−+− θθω LMgLMgI

Substitute for I and solve for ω2: ( )02 coscos3 θθω −=

Lg

Express the centripetal force acting on the center of mass:

( )

( )0

0

2c

coscos2

3

coscos32

2

θθ

θθ

ω

−=

−=

=

MgLgLM

LMF

Express the radial component of g

rM :

( ) θcosradial MgMg =

Express the total radial force at the hinge:

F|| = Fc + (Mg)radial

Chapter 9

734

( )

( )021

0

cos3cos5

coscoscos2

3

θθ

θθθ

−=

+−=

Mg

MgMg

Relate the tangential acceleration of the center of mass to its angular acceleration:

a⊥= 21 Lα

Use Newton’s 2nd law to relate the angular acceleration of the stick to the net torque acting on it: L

gML

LMg

I 2sin3sin

22

31

net θθτα ===

Express a⊥ in terms of α: a⊥= 2

1 Lα = 43 gsinθ = gsinθ + F⊥/M

Solve for F⊥ to obtain: F⊥ θsin4

1 Mg−= where the minus sign

indicates that the force is directed oppositely to the tangential component of

.gr

M

735

Chapter 10 Conservation of Angular Momentum Conceptual Problems *1 • (a) True. The cross product of the vectors A

rand B

ris defined to be .ˆsin nBA φAB=×

rr

If Ar

and Br

are parallel, sinφ = 0. (b) True. By definition, ωr is along the axis. (c) True. The direction of a torque exerted by a force is determined by the definition of the cross product.

2 • Determine the Concept The cross product of the vectors A

rand B

ris defined to be

.ˆsin nBA φAB=×rr

Hence, the cross product is a maximum when sinφ = 1. This

condition is satisfied provided Ar

and Br

are perpendicular. correct. is )(c

3 • Determine the Concept L

rand p

r are related according to .prL

rrr×= From this

definition of the cross product, Lr

and pr

are perpendicular; i.e., the angle between them

is 90°.

4 • Determine the Concept L

rand p


rrr×= Because the

motion is along a line that passes through point P, r = 0 and so is L. correct. is )(b

*5 •• Determine the Concept L

rand p


rrr×=

(a) Because L

r is directly proportional

to :pr

. doubles Doubling Lprr

(b) Because Lr

is directly proportional to :rr

. doubles Doubling Lrrr

Chapter 10

736

6 •• Determine the Concept The figure shows a particle moving with constant speed in a straight line (i.e., with constant velocity and constant linear momentum). The magnitude of L is given by rpsinφ = mv(rsinφ).

Referring to the diagram, note that the distance rsinφ from P to the line along which the

particle is moving is constant. Hence, mv(rsinφ) is constant and so constant. is Lr

7 • False. The net torque acting on a rotating system equals the change in the system’s angular momentum; i.e., dtdL=netτ , where L = Iω. Hence, if netτ is zero, all we can say

for sure is that the angular momentum (the product of I and ω) is constant. If I changes, so mustω. *8 •• Determine the Concept Yes, you can. Imagine rotating the top half of your body with arms flat at sides through a (roughly) 90° angle. Because the net angular momentum of the system is 0, the bottom half of your body rotates in the opposite direction. Now extend your arms out and rotate the top half of your body back. Because the moment of inertia of the top half of your body is larger than it was previously, the angle which the bottom half of your body rotates through will be smaller, leading to a net rotation. You can repeat this process as necessary to rotate through any arbitrary angle. 9 • Determine the Concept If L is constant, we know that the net torque acting on the system is zero. There may be multiple constant or time-dependent torques acting on the system as long as the net torque is zero. correct. is )(e

10 •• Determine the Concept No. In order to do work, a force must act over some distance. In each ″inelastic collision″ the force of static friction does not act through any distance. 11 •• Determine the Concept It is easier to crawl radially outward. In fact, a radially inward force is required just to prevent you from sliding outward. *12 •• Determine the Concept The pull that the student exerts on the block is at right angles to its motion and exerts no torque (recall that Frτ rrr

×= and θτ sinrF= ). Therefore, we

Conservation of Angular Momentum

737

can conclude that the angular momentum of the block is conserved. The student does, however, do work in displacing the block in the direction of the radial force and so the block’s energy increases. correct. is )(b

*13 •• Determine the Concept The hardboiled egg is solid inside, so everything rotates with a uniform velocity. By contrast, it is difficult to get the viscous fluid inside a raw egg to start rotating; however, once it is rotating, stopping the shell will not stop the motion of the interior fluid, and the egg may start rotating again after momentarily stopping for this reason. 14 • False. The relationship dtdLτ

rr= describes the motion of a gyroscope independently of

whether it is spinning.

15 • Picture the Problem We can divide the expression for the kinetic energy of the object by the expression for its angular momentum to obtain an expression for K as a function of I and L. Express the rotational kinetic energy of the object:

221 ωIK =

Relate the angular momentum of the object to its moment of inertia and angular velocity:

ωIL =

Divide the first of these equations by the second and solve for K to obtain:

ILK2

2

= and so correct. is )(b

16 • Determine the Concept The purpose of the second smaller rotor is to prevent the body of the helicopter from rotating. If the rear rotor fails, the body of the helicopter will tend to rotate on the main axis due to angular momentum being conserved. 17 •• Determine the Concept One can use a right-hand rule to determine the direction of the torque required to turn the angular momentum vector from east to south. Letting the fingers of your right hand point east, rotate your wrist until your fingers point south. Note that your thumb points downward. correct. is )(b

Chapter 10

738

18 •• Determine the Concept In turning east, the man redirects the angular momentum vector from north to east by exerting a clockwise torque (viewed from above) on the gyroscope. As a consequence of this torque, the front end of the suitcase will dip downward.

correct. is )(d

19 •• (a) The lifting of the nose of the plane rotates the angular momentum vector upward. It veers to the right in response to the torque associated with the lifting of the nose. (b) The angular momentum vector is rotated to the right when the plane turns to the right. In turning to the right, the torque points down. The nose will move downward. 20 •• Determine the Concept If L

r points up and the car travels over a hill or through a

valley, the force on the wheels on one side (or the other) will increase and car will tend to tip. If L

r points forward and car turns left or right, the front (or rear) of the car will tend

to lift. These problems can be averted by having two identical flywheels that rotate on the same shaft in opposite directions. 21 •• Determine the Concept The rotational kinetic energy of the woman-plus-stool system is given by .222

21

rot ILIK == ω Because L is constant (angular momentum is conserved)

and her moment of inertia is greater with her arms extended, correct. is )(b

*22 •• Determine the Concept Consider the overhead view of a tether pole and ball shown in the adjoining figure. The ball rotates counterclockwise. The torque about the center of the pole is clockwise and of magnitude RT, where R is the pole’s radius and T is the tension. So L must decrease and correct. is )(e

23 •• Determine the Concept The center of mass of the rod-and-putty system moves in a straight line, and the system rotates about its center of mass.


739

24 • (a) True. The net external torque acting a system equals the rate of change of the angular

momentum of the system; i.e.,dtdLτr

r=∑

iexti, .

(b) False. If the net torque on a body is zero, its angular momentum is constant but not necessarily zero. Estimation and Approximation *25 •• Picture the Problem Because we have no information regarding the mass of the skater, we’ll assume that her body mass (not including her arms) is 50 kg and that each arm has a mass of 4 kg. Let’s also assume that her arms are 1 m long and that her body is cylindrical with a radius of 20 cm. Because the net external torque acting on her is zero, her angular momentum will remain constant during her pirouette. Express the conservation of her angular momentum during her pirouette:

fi LL =

or inarmsinarmsoutarmsoutarms ωω II = (1)

Express her total moment of inertia with her arms out:

armsbodyoutarms III +=

Treating her body as though it is cylindrical, calculate its moment of inertia of her body, minus her arms:

( )( )2

2212

21

body

mkg00.1

m0.2kg50

⋅=

== mrI

Modeling her arms as though they are rods, calculate their moment of inertia when she has them out:

( )( )[ ]2

231

arms

mkg67.2

m1kg42

⋅=

=I

Substitute to determine her total moment of inertia with her arms out: 2

22outarms

mkg67.3

mkg67.2mkg00.1

⋅=

⋅+⋅=I

Express her total moment of inertia with her arms in: ( )( )[ ]

2

22

armsbodyiarms

mkg32.1m0.2kg42mkg00.1

⋅=

+⋅=

+= III n

Chapter 10

740

Solve equation (1) for inarmsω and

substitute to obtain:

( )

rev/s17.4

rev/s5.1mkg32.1mkg3.67

2

2

outarmsinarms

outarmsinarms

=

⋅⋅

=

= ωωII

26 •• Picture the Problem We can express the period of the earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to I and then use differentials to approximate the changes in I and T. Express the period of the earth’s rotation in terms of its angular velocity of rotation:

ωπ2

=T

Relate the earth’s angular velocity of rotation to its angular momentum and moment of inertia:

IL

=ω


IT π2=

Find dT/dI:

IT

LdIdT

==π2

Solve for dT/T and approximate ∆T:

IdI

TdT

= or TIIT ∆

≈∆

Substitute for ∆I and I to obtain:

TMmT

RMmrT

E2EE5

2

232

35

=≈∆

Substitute numerical values and evaluate ∆T:

( )( ) ( )

s552.0h

s3600d

h24d1039.6

d1039.6

d1kg1063kg102.35

6

6

24

19

=

×××=

×=

××

=∆

−

−

T


741

27 • Picture the Problem We can use L = mvr to find the angular momentum of the particle. In (b) we can solve the equation ( )hll 1+=L for ( )1+ll and the approximate value of

l . (a) Use the definition of angular momentum to obtain: ( )( )( )

/smkg102.40

m104m/s103kg10228

333

⋅×=

×××=

=

−

−−−

mvrL

(b) Solve the equation

( )hll 1+=L for ( )1+ll : ( ) 2

2

1h

llL

=+

Substitute numerical values and evaluate ( )1+ll : ( )

52

2

34

28

1022.5

sJ011.05/smkg102.401

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅×

⋅×=+ −

−

ll

Because l >>1, approximate its value with the square root of

( )1+ll :

261029.2 ×≈l

(c)

.1102 and102between atedifferentican experiment no because physics cmacroscopiin noticednot is momentumangular ofon quantizati The

26

26

+×=

×=

l

l

*28 •• Picture the Problem We can use conservation of angular momentum in part (a) to relate the before-and-after collapse rotation rates of the sun. In part (b), we can express the fractional change in the rotational kinetic energy of the sun as it collapses into a neutron star to decide whether its rotational kinetic energy is greater initially or after the collapse. (a) Use conservation of angular momentum to relate the angular momenta of the sun before and after its collapse:

aabb ωω II = (1)

Using the given formula, approximate the moment of inertia Ib of the sun before collapse: ( ) ( )

246

2530

2sunb

mkg1069.5km106.96kg1099.1059.0

059.0

⋅×=

××=

= MRI

Chapter 10

742

Find the moment of inertia Ia of the sun when it has collapsed into a spherical neutron star of radius 10 km and uniform mass distribution:

( )( )237

23052

252

a

mkg1096.7

km10kg1099.1

⋅×=

×=

= MRI

Substitute in equation (1) and solve for ωa to obtain:

b8

b237

246

ba

ba

1015.7

mkg1096.7mkg1069.5

ω

ωωω

×=

⋅×⋅×

==II

Given that ωb = 1 rev/25 d, evaluate ωa:

rev/d1086.2

d25rev11015.7

7

8a

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×=ω

smaller. getssun theas decreases which energy, potential nalgravitatioof expense at the comesenergy kinetic rotational additional The

Note that the rotational period decreases by the same factor of Ib/Ia and becomes:

s1002.3

s3600h1

h24d1

revrad2

drev1086.2

22 3

7aa

−×=××××

== πT πωπ

(b) Express the fractional change in the sun’s rotational kinetic energy as a consequence of its collapse and simplify to obtain:

1

1

1

2bb

2aa

2bb2

1

2aa2

1

b

a

b

ba

b

−=

−=

−=−

=∆

ωω

ωω

II

II

KK

KKK

KK

Substitute numerical values and evaluate ∆K/Kb:

827

8b

1015.71drev/251

rev/d102.861015.7

1×=−⎟⎟

⎠

⎞⎜⎜⎝

⎛ ×⎟⎠⎞

⎜⎝⎛

×=

∆KK

(i.e., the rotational kinetic

energy increases by a factor of approximately 7×108.) 29 •• Picture the Problem We can solve 2CMRI = for C and substitute numerical values in order to determine an experimental value of C for the earth. We can then compare this value to those for a spherical shell and a sphere in which the mass is uniformly distributed to decide whether the earth’s mass density is greatest near its core or near its crust.


743

(a) Express the moment of inertia of the earth in terms of the constant C:

2CMRI =

Solve for C to obtain: 2MR

IC =

Substitute numerical values and evaluate C: ( )( )

331.0

km6370kg105.98mkg108.03

224

237

=

×⋅×

=C

(b) If all of the mass were in the crust, the moment of inertia of the earth would be that of a thin spherical shell:

232

shell spherical MRI =

If the mass of the earth were uniformly distributed throughout its volume, its moment of inertia would be:

252

sphere solid MRI =

earth. theofcenter near thegreater bemust density mass the0.4, 2/5 ally experiment Because =<C

*30 •• Picture the Problem Let’s estimate that the diver with arms extended over head is about 2.5 m long and has a mass M = 80 kg. We’ll also assume that it is reasonable to model the diver as a uniform stick rotating about its center of mass. From the photo, it appears that he sprang about 3 m in the air, and that the diving board was about 3 m high. We can use these assumptions and estimated quantities, together with their definitions, to estimate ω and L. Express the diver’s angular velocity ω and angular momentum L:

t∆∆

=θω (1)

and ωIL = (2)

Using a constant-acceleration equation, express his time in the air:

gy

gy

ttt

downup

m 6 fallm 3 rise

22 ∆+

∆=

∆+∆=∆


( ) ( ) s89.1m/s9.81m62

m/s9.81m32

22 =+=∆t

Estimate the angle through which he rotated in 1.89 s:

radrev5.0 πθ =≈∆

Chapter 10

744

Substitute in equation (1) and evaluate ω: rad/s66.1

s89.1rad

==πω

Use the ″stick rotating about an axis through its center of mass″ model to approximate the moment of inertia of the diver:

2121 MLI =


ω2121 MLL =


( )( ) ( )/smkg70/smkg2.69

rad/s1.66m2.5kg8022

2121

⋅≈⋅=

=L

Remarks: We can check the reasonableness of this estimation in another way. Because he rose about 3 m in the air, the initial impulse acting on him must be about 600 kg⋅m/s (i.e., I = ∆p = Mvi). If we estimate that the lever arm of the force is roughly l = 1.5 m, and the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°, we obtain °= sin5IL l ≈ 78 kg⋅m2/s, which is not too bad considering the approximations made here. 31 •• Picture the Problem First we assume a spherical diver whose mass M = 80 kg and whose diameter, when curled into a ball, is 1 m. We can estimate his angular velocity when he has curled himself into a ball from the ratio of his angular momentum to his moment of inertia. To estimate his angular momentum, we’ll guess that the lever arml of the force that launches him from the diving board is about 1.5 m and that the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°. Express the diver’s angular velocity ω when he curls himself into a ball in mid-dive:

IL

=ω (1)

Using a constant-acceleration equation, relate the speed with which he left the diving board v0 to his maximum height ∆y and our estimate of his angle with the vertical direction:

yav yy ∆+= 20 20

where °= 5cos00 vv y

Solve for v0:

°∆

=5cos

220

ygv


( )( )m/s7.70

5cosm3m/s9.812 2

0 =°

=v


745

Approximate the impulse acting on the diver to launch him with the speed v0:

0MvpI =∆=

Letting l represent the lever arm of the force acting on the diver as he leaves the diving board, express his angular momentum:

°=°= 5sin5sin 0ll MvIL

Use the ″uniform sphere″ model to approximate the moment of inertia of the diver:

252 MRI =


02

520

25sin55sin

Rv

MRMv °

=°

=llω


( )( )( )

rad/s10.1

m0.525sinm1.5m/s7.705

2

=

°=ω

*32 •• Picture the Problem We’ll assume that he launches himself at an angle of 45° with the horizontal with his arms spread wide, and then pulls them in to increase his rotational speed during the jump. We’ll also assume that we can model him as a 2-m long cylinder with an average radius of 0.15 m and a mass of 60 kg. We can then find his take-off speed and ″air time″ using constant-acceleration equations, and use the latter, together with the definition of rotational velocity, to find his initial rotational velocity. Finally, we can apply conservation of angular momentum to find his initial angular momentum. Using a constant-acceleration equation, relate his takeoff speed v0 to his maximum elevation ∆y:

yavv yy ∆+= 220

2

or, because v0y = v0sin45°, v = 0, and ay = − g,

ygv ∆−°= 245sin0 220

Solve for v0 to obtain:

°∆

=°

∆=

45sin2

45sin2

20ygygv


( )( )m/s4.85

sin45m0.6m/s9.812 2

0 =°

=v

Use its definition to express Goebel’s angular velocity: t∆

∆=

θω

Use a constant-acceleration equation to express Goebel’s ″air time″ ∆t: g

ytt ∆=∆=∆

222 m 0.6 rise

Chapter 10

746


( ) s699.0m/s9.81

m6.022 2 ==∆t

Substitute numerical values and evaluate ω: rad/s0.36

revrad2π

s0.699rev4

=×=ω

Use conservation of angular momentum to relate his take-off angular velocity ω0 to his average angular velocity ω as he performs a quadruple Lutz:

ωω II =00

Assuming that he can change his angular momentum by a factor of 2 by pulling his arms in, solve for and evaluate ω0:

( ) rad/s18.0rad/s3621

00 === ωω

II

Express his take-off angular momentum:

000 ωIL =

Assuming that we can model him as a solid cylinder of length l with an average radius r and mass m, express his moment of inertia with arms drawn in (his take-off configuration):

( ) 2221

0 2 mrmrI == where the factor of 2 represents our assumption that he can double his moment of inertia by extending his arms.


20 ωmrL =

Substitute numerical values and evaluate L0:

( )( ) ( )/smkg3.24

rad/s18m0.15kg602

20

⋅=

=L

Vector Nature of Rotation 33 • Picture the Problem We can express F

rand r

rin terms of the unit vectors i and j and

then use the definition of the cross product to find .τr Express F

rin terms of F and the unit

vector :i

iF ˆF−=r

Express rr

in terms of R and the unit vector :j

jr ˆR=r


747

Calculate the cross product of rr

and :F

r

( )( ) kji

ijFrτˆˆˆ

ˆˆ

FRFR

FR

=×=

−×=×=rrr

34 • Picture the Problem We can find the torque is the cross product of r

rand .F

r

Compute the cross product of rr and :F

r ( )( )

( ) ( )k

jjji

jjiFrτ

ˆ

ˆˆˆˆ

ˆˆˆ

mgx

mgymgx

mgyx

−=

×−×−=

−+=×=rrr

35 • Picture the Problem The cross product of the vectors jiA ˆˆ

yx AA +=r

and jiB ˆˆyx BB +=

ris given by

( ) ( ) ( ) ( )jjijjiiiBA ˆˆˆˆˆˆˆˆ ×+×+×+×=× yyxyyxxx BABABABArr

( ) ( ) ( ) ( )0ˆˆ0 yyxyyxxx BABABABA +−++= kk

( ) ( )kk ˆˆ −+= xyyx BABA

(a) Find A

r× Br

for Ar

= 4 i and Br

= 6 i + j6 : ( )

( ) ( )( ) kk

jiii

jiiBA

ˆ24ˆ24024

ˆˆ24ˆˆ24

ˆ6ˆ6ˆ4

=+=

×+×=

+×=×rr

(b) Find A

r× Br

for Ar

= 4 i and Br

= 6 i + 6 k : ( )

( ) ( )( ) ( ) jj

kiii

kiiBA

ˆ24ˆ24024

ˆˆ24ˆˆ24

ˆ6ˆ6ˆ4

−=−+=

×+×=

+×=×rr

(c) Find Ar

× Br

for Ar

= 2 i + j3

and Br

=3 i + j2 :

( ) ( )( ) ( ) ( )

( )( ) ( ) ( ) ( )

k

kk

jj

ijjiii

jijiBA

ˆ5

06ˆ9ˆ406

ˆˆ6

ˆˆ9ˆˆ4ˆˆ6

ˆ2ˆ3ˆ3ˆ2

−=

+−++=

×+

×+×+×=

+×+=×rr

Chapter 10

748

*36 • Picture the Problem The magnitude of A

r× Br

is given by θsinAB .

Equate the magnitudes of A

r× Br

and BA

rr⋅ :

θθ cossin ABAB =

θθ cossin =∴

or 1tan ±=θ

Solve for θ to obtain: °±°±=±= − 135or451tan 1θ

37 •• Picture the Problem Let rr be in the xy plane. Then ωr points in the positive z direction. We can establish the results called for in this problem by forming the appropriate cross products and by differentiating .vr

(a) Express ωr using unit vectors: kω ˆω=r

Express r

rusing unit vectors: ir ˆr=

r

Form the cross product of ω

rand :rr ( )

j

jikikrωˆ

ˆˆˆˆˆ

v

rrr

=

=×=×=× ωωωrr

rωv

rrr×=∴

(b) Differentiate v

rwith respect to t to

express ar

: ( )

( )ct

t

aarωωa

vωrω

rωrω

rωva

rr

rrrr

rrrr

rrr

r

rrr

r

+=××+=

×+×=

×+×=

×==

dtd

dtd

dtd

dtd

dtd

where ( )rωωa rrrr××=c

and ct and aa rrare the tangential and


749

centripetal accelerations, respectively.

38 •• Picture the Problem Because Bz = 0, we can express B

ras jiB ˆˆ

yx BB +=r

and form its

cross product with Ar

to determine Bx and By. Express B

rin terms of its components: jiB ˆˆ

yx BB +=r

(1)

Express A

r× Br

: ( ) kkjiiBA ˆ12ˆ4ˆˆˆ4 ==+×=× yyx BBBrr

Solve for By: 3=yB

Relate B to Bx and By: 222

yx BBB +=

Solve for and evaluate Bx: 435 2222 =−=−= yx BBB

Substitute in equation (1): jiB ˆ3ˆ4 +=

r

39 • Picture the Problem We can write B

rin the form kjiB ˆˆˆ

zyx BBB ++=r

and use the dot

product of Ar

and Br

to find By and their cross product to find Bx and Bz.

Express Br

in terms of its components: kjiB ˆˆˆzyx BBB ++=

r (1)

Evaluate A

r⋅ :Br

123 ==⋅ yBBArr

and By = 4

Evaluate Ar

× :Br

( )ik

kjijBAˆ3ˆ3

ˆˆ4ˆˆ3

zx

zx

BB

BB

+−=

++×=×rr

Because A

r× Br

= 9 :i Bx = 0 and Bz = 3.

Substitute in equation (1) to obtain: kjB ˆ3ˆ4 +=r

Chapter 10

750

40 •• Picture the Problem The dot product of A

rwith the cross product of B

rand C

ris a scalar

quantity and can be expressed in determinant form as

zyx

zyx

zyx

cccbbbaaa

. We can expand this

determinant by minors to show that it is equivalent to )( CBArrr

×⋅ , )( BACrrr

×⋅ , and

)( ACBrrr

×⋅ . The dot product of A

rwith the cross

product of Br

and Cr

is a scalar quantity and can be expressed in determinant form as:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( CBArrr

Expand the determinant by minors to obtain:

xyzyxz

zxyxzy

yzxzyx

zyx

zyx

zyx

cbacba

cbacba

cbacbacccbbbaaa

−+

−+

−=

(1)

Evaluate the cross product of B

rand

Cr

to obtain:

( )( ) ( )kj

iCBˆˆ

ˆ

xyyxzxxz

yzzy

cbcbcbcb

cbcb

−+−+

−=×rr

Form the dot product of A

rwith

Br

× Cr

to obtain:

( )

xyzyxz

zxyxzy

yzxzyx

cbacba

cbacba

cbacba

−+

−+

−=×⋅ CBArrr

(2)

Because (1) and (2) are the same, we can conclude that:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( CBArrr

Proceed as above to establish that:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( BACrrr

and


751

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( ACBrrr

41 •• Picture the Problem Let, without loss of generality, the vector C

rlie along the x axis and

the vector Br

lie in the xy plane as shown below to the left. The diagram to the right shows the parallelepiped spanned by the three vectors. We can apply the definitions of the cross- and dot-products to show that ( )CBA

rrr×⋅ is the volume of the parallelepiped.

Express the cross-product of B

rand :C

r ( )( )kCB ˆsin −=× θBC

rr

and ( )

ramparallelog theof area

sin

=

=× CB θCBrr

Form the dot-product of A

rwith the

cross-product of Br

and Cr

to obtain: ( ) ( )

( )( )( )( )

ipedparallelep

heightbase of areacossin

cossin

V

ABCCBA

=

===×⋅

φθφθCBA

rrr

*42 •• Picture the Problem Draw the triangle using the three vectors as shown below. Note that .CBA

rrr=+ We can find the

magnitude of the cross product of Ar

and Br

and of Ar

and Cr

and then use the cross product of A

rand ,C

r using ,CBA

rrr=+ to

show that cABbAC sinsin = or B/sin b = C/ sin c. Proceeding similarly, we can extend the law of sines to the third side of the triangle and the angle opposite it.

Chapter 10

752

Express the magnitude of the cross product of A

rand :B

r

cABsin=× BArr

Express the magnitude of the cross product of A

rand :C

r

bAC sin=×CArr

Form the cross product of Ar

with Cr

to obtain:

( )

BA

BAAA

BAACA

rr

rrrr

rrrrr

×=

×+×=

+×=×

because 0=× AArr

.

Because :BACArrrr

×=× BACArrrr

×=×

and cABbAC sinsin =

Simplify and rewrite this expression to obtain: c

Cb

Bsinsin

=

Proceed similarly to extend this result to the law of sines: c

Cb

Ba

Asinsinsin

==

Angular Momentum 43 • Picture the Problem L

rand p


rrr×= If L

r= 0, then

examination of the magnitude of pr rr× will allow us to conclude that 0sin =φ and that

the particle is moving either directly toward the point, directly away from the point, or through the point.

Because L

r= 0: 0=×=×=× vrvrpr

rrrrrrmm

or 0=× vr rr

Express the magnitude of :vr rr× 0sin ==× φrvvr rr

Because neither r nor v is zero: 0sin =φ

where φ is the angle between rr

and .vr

Solve for φ: °°== − 180or00sin 1φ


753

44 • Picture the Problem We can use their definitions to calculate the angular momentum and moment of inertia of the particle and the relationship between L, I, and ω to determine its angular speed.

(a) Express and evaluate the magnitude of :L

r

( )( )( )/smkg28.0

m4m/s3.5kg22⋅=

== mvrL

(b) Express the moment of inertia of the particle with respect to an axis through the center of the circle in which it is moving:

( )( ) 222 mkg32m4kg2 ⋅=== mrI

(c) Relate the angular speed of the particle to its angular momentum and solve for and evaluate ω:

22

2

rad/s0.875mkg32

/smkg28.0=

⋅⋅

==ILω

45 • Picture the Problem We can use the definition of angular momentum to calculate the angular momentum of this particle and the relationship between its angular momentum and angular speed to describe the variation in its angular speed with time.

(a) Express the angular momentum of the particle as a function of its mass, speed, and distance of its path from the reference point:

( )( )( )/smkg54.0

sin90m/s4.5kg2m6sin

2⋅=

°== θrmvL

(b) Because L = mr2ω: 1

2r∝ω and

recedes.it asdecreases andpoint theapproaches

particle theas increases ω

*46 •• Picture the Problem We can use the formula for the area of a triangle to find the area swept out at t = t1, add this area to the area swept out in time dt, and then differentiate this expression with respect to time to obtain the given expression for dA/dt. Express the area swept out at t = t1: 12

1112

11 cos bxbrA == θ

where θl is the angle between 1rr

and vr

and

Chapter 10

754

x1 is the component of 1rr

in the direction of vr .

Express the area swept out at t = t1 + dt:

( )( )vdtxb

dxxbdAAA+=

+=+=

121

121

1

Differentiate with respect to t: constant2

121 === bv

dtdxb

dtdA

Because rsinθ = b: ( ) ( )

mL

rpm

vrbv

2

sin21sin2

121

=

== θθ

47 •• Picture the Problem We can find the total angular momentum of the coin from the sum of its spin and orbital angular momenta. (a) Express the spin angular momentum of the coin:

spincmspin ωIL =

From Problem 9-44: 241 MRI =

Substitute for I to obtain: spin

241

spin ωMRL =

Substitute numerical values and evaluate Lspin:

( )( )

/smkg1033.1

revrad2

srev10

m0.0075kg0.015

25

241

spin

⋅×=

⎟⎠⎞

⎜⎝⎛ ××

=

−

π

L

(b) Express and evaluate the total angular momentum of the coin: /smkg1033.1

025

spinspinorbit

⋅×=

+=+=−

LLLL

(c) From Problem 10-14: 0orbit =L

and /smkg1033.1 25 ⋅×= −L

(d) Express the total angular momentum of the coin:

spinorbit LLL +=


755

Find the orbital momentum of the coin: ( )( )( )

/smkg107.50m0.1m/s0.05kg0.015

25

orbit

⋅×±=

±=±=

−

MvRL

where the ± is a consequence of the fact that the coin’s direction is not specified.


/smkg1033.1/smkg1050.7

25

25

⋅×+

⋅×±=−

−L

The possible values for L are: /smkg1083.8 25 ⋅×= −L

or /smkg1017.6 25 ⋅×−= −L

48 •• Picture the Problem Both the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum.

Express the net torque about an axis perpendicular to the page and through point O: ( ) 121

2211i

inet

Frr

FrFrττrrr

rrrrrr

×−=

×+×== ∑

because 12 FFrr

−=

Because 21 rr rr

− points along 1Fr

− : ( ) 0121 =×− Frrrrr

Torque and Angular Momentum 49 • Picture the Problem The angular momentum of the particle changes because a net torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s 2nd law to relate the angular speed of the particle to its angular acceleration.

(a) Relate the magnitude of the torque acting on the particle to the rate at which its angular momentum changes:

( )[ ]

mN00.4

mN4net

⋅=

⋅== tdtd

dtdLτ

Chapter 10

756

(b) Using a constant-acceleration equation, relate the angular speed of the particle to its acceleration and time-in-motion:

tαωω += 0

where ω0 = 0

Use Newton’s 2nd law to relate the angular acceleration of the particle to the net torque acting on it:

2netnet

mrIττα ==

Substitute to obtain: tmr 2

netτω =


( )( )( )( ) rad/s0.192

m3.4kg8.1mN4

2

2

t

t

=

⋅=ω

provided t is in seconds. 50 •• Picture the Problem The angular momentum of the cylinder changes because a net torque acts on it. We can find the angular momentum at t = 25 s from its definition and the net torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque. (a) Use its definition to express the angular momentum of the cylinder:

ωω 221 mrIL ==


( )( )

/smkg377

s60min1

revrad2

minrev500

m0.4kg90

2

221

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×××

=

π

L

(b) Express and evaluate dtdL

: ( )

22

2

/smkg15.1

s25/smkg377

⋅=

⋅=

dtdL

(c) Because the torque acting on the uniform cylinder is constant, the rate

22/smkg15.1 ⋅==dtdLτ


757

of change of the angular momentum is constant and hence the instantaneous rate of change of the angular momentum at any instant is equal to the average rate of change over the time during which the torque acts: (d) Using the definition of torque that relates the applied force to its lever arm, express the magnitude of the frictional force f acting on the rim:

N37.7m0.4

/smkg15.1 22

=⋅

==l

τf

*51 •• Picture the Problem Let the system include the pulley, string, and the blocks and assume that the mass of the string is negligible. The angular momentum of this system changes because a net torque acts on it.

(a) Express the net torque about the center of mass of the pulley: ( )12

12net

sin

sin

mmRg

gRmgRm

−=

−=

θ

θτ

where we have taken clockwise to be positive to be consistent with a positive upward velocity of the block whose mass is m1 as indicated in the figure.

(b) Express the total angular momentum of the system about an axis through the center of the pulley:

⎟⎠⎞

⎜⎝⎛ ++=

++=

212

21

mmRIvR

vRmvRmIL ω

(c) Express τ as the time derivative of the angular momentum:

⎟⎠⎞

⎜⎝⎛ ++=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++==

212

212

mmRIaR

mmRIvR

dtd

dtdLτ

Equate this result to that of part (a) and solve for a to obtain:

( )

212

12 sin

mmRI

mmga++

−=

θ

Chapter 10

758

52 •• Picture the Problem The forces resulting from the release of gas from the jets will exert a torque on the spaceship that will slow and eventually stop its rotation. We can relate this net torque to the angular momentum of the spaceship and to the time the jets must fire.

Relate the firing time of the jets to the desired change in angular momentum:

netnet τω

τ∆

=∆

=∆ILt

Express the magnitude of the net torque exerted by the jets:

FR2net =τ

Letting ∆m/∆t′ represent the mass of gas per unit time exhausted from the jets, relate the force exerted by the gas on the spaceship to the rate at which the gas escapes:

vtmF'∆

∆=

Substitute and solve for ∆t to obtain:

vRtm

It

'2

∆∆∆

=∆ω


( )( )( )( ) s52.4

m3m/s800kg/s102s60

min1rev

rad2minrev6mkg4000

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛××⋅

=∆ −

π

t

53 •• Picture the Problem We can use constant-acceleration equations to express the projectile’s position and velocity coordinates as functions of time. We can use these coordinates to express the particle’s position and velocity vectors r

rand .vr Using its

definition, we can express the projectile’s angular momentum Lr

as a function of time and then differentiate this expression to obtain .dtdL

r Finally, we can use the definition of

the torque, relative to an origin located at the launch position, the gravitational force exerts on the projectile to express τr and complete the demonstration that .τL rr

=dtd Using its definition, express the angular momentum vector L

r of the

projectile:

vrL rrrm×= (1)

Using constant-acceleration ( )tVtvx x θcos0 ==


759

equations, express the position coordinates of the projectile as a function of time:

and

( ) 221

221

00

sin gttV

tatvyy yy

−=

++=

θ

Express the projectile’s position vector :rr

( )[ ] ( )[ ] jir ˆsinˆcos 221 gttVtV −+= θθ

r

Using constant-acceleration equations, express the velocity of the projectile as a function of time:

θcos0 Vvv xx == and

gtVtavv yyy

−=

+=

θsin0

Express the projectile’s velocity vector :vr

[ ] [ ] jiv ˆsinˆcos gtVV −+= θθr

Substitute in equation (1) to obtain: ( )[ ] ( )[ ]{ }[ ] [ ]{ }

( )kji

jiL

ˆcos

ˆsinˆcos

ˆsinˆcos

221

221

θ

θθ

θθ

Vmgt

gtVVm

gttVtV

−=

−+×

−+=r

Differentiate L

rwith respect to t to obtain: ( )

( )k

kL

ˆcos

ˆcos221

θ

θ

mgtV

Vmgtdtd

dtd

−=

−=r

(2)

Using its definition, express the torque acting on the projectile:

( )( )[ ] ( )[ ]

( ) j

ji

jrτ

ˆ

ˆsinˆcos

ˆ2

21

mg

gttVtV

mg

−×

−+=

−×=

θθ

rr

or ( )kτ ˆcosθmgtV−=

r (3)

Comparing equations (2) and (3) we see that: τL rr

=dtd

Conservation of Angular Momentum *54 • Picture the Problem Let m represent the mass of the planet and apply the definition of torque to find the torque produced by the gravitational force of attraction. We can use Newton’s 2nd law of motion in the form dtdLτ

rr= to show that L

ris constant and apply

conservation of angular momentum to the motion of the planet at points A and B.

Chapter 10

760

(a) Express the torque produced by the gravitational force of attraction of the sun for the planet:

.ofdirection the

along acts because 0

r

FFrτr

rrrr=×=

(b) Because 0=τr : constant0 =×=⇒= vrLL rrr

r

mdtd

Noting that at points A and B

rv=× vr rr, express the

relationship between the distances from the sun and the speeds of the planets:

2211 vrvr =

or

1

2

2

1

rr

vv

=

55 •• Picture the Problem Let the system consist of you, the extended weights, and the platform. Because the net external torque acting on this system is zero, its angular momentum remains constant during the pulling in of the weights. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the system to its initial and final moments of inertia:

ffii ωω II =

Solve for fω : i

f

if ωω

II

=

Substitute numerical values and evaluate fω : ( ) rev/s5.00rev/s1.5

mkg1.8mkg6

2

2

f =⋅

⋅=ω

(b) Express the change in the kinetic energy of the system:

2ii2

12ff2

1if ωω IIKKK −=−=∆

Substitute numerical values and evaluate ∆K: ( )

( )

J622

revrad2

srev1.5mkg6

revrad2

srev5mkg1.8

22

21

22

21

=

⎟⎠⎞

⎜⎝⎛ ×⋅−

⎟⎠⎞

⎜⎝⎛ ×⋅=∆

π

πK


761

(c) man. theofenergy internal thefrom

comesenergy thesystem, on the work doesagent external no Because

*56 •• Picture the Problem Let the system consist of the blob of putty and the turntable. Because the net external torque acting on this system is zero, its angular momentum remains constant when the blob of putty falls onto the turntable. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the turntable to its initial and final moments of inertia and solve for ωf:

ffi0 ωω II =

and

if

0f ωω

II

=

Express the final rotational inertia of the turntable-plus-blob:

20blob0f mRIIII +=+=

Substitute and simplify to obtain: i

0

2i20

0f

1

1 ωωω

ImRmRI

I

+=

+=

(b) If the blob flies off tangentially to the turntable, its angular momentum doesn’t change (with respect to an axis through the center of turntable). Because there is no external torque acting on the blob-turntable system, the total angular momentum of the system will remain constant and the angular momentum of the turntable will not change. Because the moment of inertia of the table hasn’t changed either, the turntable will continue to spin at f ωω =' .

57 •• Picture the Problem Because the net external torque acting on the Lazy Susan-cockroach system is zero, the net angular momentum of the system is constant (equal to zero because the Lazy Susan is initially at rest) and we can use conservation of angular momentum to find the angular velocity ω of the Lazy Susan. The speed of the cockroach relative to the floor vf is the difference between its speed with respect to the Lazy Susan and the speed of the Lazy Susan at the location of the cockroach with respect to the floor. Relate the speed of the cockroach with respect to the floor vf to the speed of the Lazy Susan at the location of the cockroach:

rvv ω−=f (1)

Use conservation of angular momentum to obtain:

0CLS =− LL

Chapter 10

762

Express the angular momentum of the Lazy Susan:

ωω 221

LSLS MRIL ==

Express the angular momentum of the cockroach:

⎟⎠⎞

⎜⎝⎛ −== ωω

rvmrIL 2

CCC


21 =⎟

⎠⎞

⎜⎝⎛ −− ωω

rvmrMR

Solve for ω to obtain: 22 2

2mrMR

mrv+

=ω


22

2

f 22

mrMRvmrvv

+−=


( )( ) ( )( )( ) ( )( )

mm/s67.9m08.0kg015.02m15.0m25.0

m/s01.0m08.0kg0.0152m/s01.0 22

2

f =+

−=v

*58 •• Picture the Problem The net external torque acting on this system is zero and so we know that angular momentum is conserved as these disks are brought together. Let the numeral 1 refer to the disk to the left and the numeral 2 to the disk to the right. Let the angular momentum of the disk with the larger radius be positive. Using conservation of angular momentum, relate the initial angular speeds of the disks to their common final speed and to their moments of inertia:

ffii ωω II =

or ( ) f210201 ωωω IIII +=−

Solve for ωf: 0

21

21f ωω

IIII

+−

=

Express I1 and I2: ( ) 22

21

1 22 mrrmI ==

and 2

21

2 mrI =

Substitute and simplify to obtain:

053

02212

2212

f 22

ωωω =+−

=mrmrmrmr


763

59 •• Picture the Problem We can express the angular momentum and kinetic energy of the block directly from their definitions. The tension in the string provides the centripetal force required for the uniform circular motion and can be expressed using Newton’s 2nd law. Finally, we can use the work-kinetic energy theorem to express the work required to reduce the radius of the circle by a factor of two. (a) Express the initial angular momentum of the block:

000 mvrL =

(b) Express the initial kinetic energy of the block:

202

10 mvK =

(c) Using Newton’s 2nd law, relate the tension in the string to the centripetal force required for the circular motion:

0

20

c rvmFT ==

Use the work-kinetic energy theorem to relate the required work to the change in the kinetic energy of the block:

( ) 20

20

20

202

1

20

0f

20

0

20

f

20

0

20

f

2f

0f

321

2

1222

22

mrL

mrrmL

IIL

IL

IL

IL

ILKKKW

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−=

−=−=∆=

Substitute the result from part (a) and simplify to obtain:

203

2 mvW −=

*60 •• Picture the Problem Because the force exerted by the rubber band is parallel to the position vector of the point mass, the net external torque acting on it is zero and we can use the conservation of angular momentum to determine the speeds of the ball at points B and C. We’ll use mechanical energy conservation to find b by relating the kinetic and elastic potential energies at A and B. (a) Use conservation of momentum to relate the angular momenta at points A, B and C:

CBA LLL ==

or CCBBAA rmvrmvrmv ==

Solve for vB in terms of vA:

B

AAB r

rvv =

Chapter 10

764

Substitute numerical values and evaluate vB:

( ) m/s2.40m1m0.6m/s4 ==Bv

Solve for vC in terms of vA:

C

AAC r

rvv =

Substitute numerical values and evaluate vC:

( ) m/s00.4m6.0m0.6m/s4 ==Cv

(b) Use conservation of mechanical energy between points A and B to relate the kinetic energy of the point mass and the energy stored in the stretched rubber band:

BA EE =

or 2

212

212

212

21

BBAA brmvbrmv +=+

Solve for b: ( )22

22

BA

AB

rrvvmb

−−

=

Substitute numerical values and evaluate b: ( ) ( ) ( )[ ]

( ) ( )N/m20.3

m1m0.6m/s4m/s2.4kg2.0

22

22

=

−−

=b

Quantization of Angular Momentum *61 • Picture the Problem The electron’s spin angular momentum vector is related to its z component as shown in the diagram.

Using trigonometry, relate the magnitude of sr to its z component:

°== − 7.5475.0

cos 21

1

h

hθ

62 •• Picture the Problem Equation 10-27a describes the quantization of rotational energy. We can show that the energy difference between a given state and the next higher state is proportional to 1+l by using Equation 10-27a to express the energy difference.


765

From Equation 10-27a we have: ( ) r01 EK += lll

Using this equation, express the difference between one rotational state and the next higher state:

( )( ) ( )( ) r0

r0r0

12

121

E

EEE

+=

+−++=∆

l

llll

63 •• Picture the Problem The rotational energies of HBr molecule are related to l and

r0E according to ( ) r01 EK += lll where .22r0 IE h=

(a) Express and evaluate the moment of inertia of the H atom: ( ) ( )

247

2927

2p

mkg103.46

m100.144kg101.67

⋅×=

××=

=

−

−−

rmI

(b) Relate the rotational energies to l and r0E :

( ) r01 EK += lll

Evaluate r0E : ( )( )

meV0.996J101.60

eV1J101.59

mkg103.462sJ101.05

2

1922

247

2342

r0

=×

××=

⋅×⋅×

==

−−

−

−

IE h

Evaluate E for l = 1: ( )( ) meV1.99meV0.996111 =+=E

Evaluate E for l = 2: ( )( )

meV98.5

meV0.9961222

=

+=E

Evaluate E for l = 3: ( )( )

meV0.21

meV0.9961333

=

+=E

64 •• Picture the Problem We can use the definition of the moment of inertia of point particles to calculate the rotational inertia of the nitrogen molecule. The rotational energies of nitrogen molecule are related to l and r0E according

to ( ) r01 EK += lll where .22r0 IE h=

Chapter 10

766

(a) Using a rigid dumbbell model, express and evaluate the moment of inertia of the nitrogen molecule about its center of mass:

2N

2N

2N

i

2ii

2 rm

rmrmrmI

=

+== ∑

Substitute numerical values and evaluate I: ( )( )( )246

21127

mkg101.41

m105.5kg101.66142

⋅×=

××=−

−−I

(b) Relate the rotational energies to l and r0E :

( ) r01 EE += lll

Evaluate r0E : ( )( )

meV0.244J101.60

eV1J1091.3

mkg1041.12sJ101.05

2

1923

246

2342

r0

=×

××=

⋅×⋅×

==

−−

−

−

IE h

Substitute to obtain: ( ) meV1244.0 += lllE

*65 •• Picture the Problem We can obtain an expression for the speed of the nitrogen molecule by equating its translational and rotational kinetic energies and solving for v. Because this expression includes the moment of inertia I of the nitrogen molecule, we can use the definition of the moment of inertia to express I for a dumbbell model of the nitrogen molecule. The rotational energies of a nitrogen molecule depend on the quantum number l according to ( ) .2/12/ 22 IILE hlll +==

Equate the rotational kinetic energy of the nitrogen molecule in its l = 1 quantum state and its translational kinetic energy:

2N2

11 vmE = (1)

Express the rotational energy levels of the nitrogen molecule:

( )II

LE2

12

22 hlll

+==

For l = 1:

( )II

E22

1 2111 hh

=+

=


767


2N2

12

vmI

=h

Solve for v to obtain: Im

vN

22h= (2)

Using a rigid dumbbell model, express the moment of inertia of the nitrogen molecule about its center of mass:

2N

2N

2N

i

2ii 2 rmrmrmrmI =+== ∑

and 22

NN 2 rmIm =


rmrmv

N22

N

2

22 hh

==


( ) ( )m/s5.82

m105.5kg101.6641sJ10055.1

1127

34

=

××⋅×

= −−

−

v

Collision Problems 66 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the rod. Because the net external torque acting on this system is zero, we know that angular momentum is conserved in the collision. We’ll use the definition of angular momentum to express the angular momentum just after the collision and conservation of mechanical energy to determine the speed of the ball just before it makes its perfectly inelastic collision with the rod. Use conservation of angular momentum to relate the angular momentum before the collision to the angular momentum just after the perfectly inelastic collision:

mvrLL

== if

Use conservation of mechanical energy to relate the kinetic energy of the ball just before impact to its initial potential energy:

0ifif =−+− UUKK

or, because Ki = Uf = 0, 0if =−UK

Letting h represent the distance the ghv 2=

Chapter 10

768

ball falls, substitute for if and UK and solve for v to

obtain: Substitute for v to obtain: ghmrL 2f =

Substitute numerical values and evaluate Lf:

( )( ) ( )( )sJ14.0

m1.2m/s9.812m0.9kg3.2 2f

⋅=

=L

*67 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the problem statement, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Let the direction the blob of putty is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction. Using its definition, express the location of the center of mass relative to the center of the bar:

cm mMmdy

+= below the center of the bar.

Using its definition, express the velocity of the center of mass:

mMmvv+

=cm

Using the definition of L in terms of I and ω, express ω:

cm

cm

IL

=ω (1)

Express the angular momentum about the center of mass:

( )

mMmMvd

mMmddmv

ydmvL

+=⎟

⎠⎞

⎜⎝⎛

+−=

−= cmcm

Using the parallel axis theorem, express the moment of inertia of the system relative to its center of mass:

( )2cm

2cm

2121

cm ydmMyMLI −++=

Substitute for ycm and simplify to obtain:


769

( )( )

( ) ( )( )

( )

mMmMd

ML

mM

mMdmMML

mM

dmM

mM

dMmML

mMmdmMd

mmM

dMmML

mMmd

dmmM

mdMMLI

++=

+

++=

++

++=

+−+

++

+=

+−+

++=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

22

2

22

2

22

2

222

2

2

222

222

121

121

121

121

121

cm

Substitute for Icm and Lcm in equation (1) and simplify to obtain: ( ) 22

121 MmdmMML

mMvd++

=ω

Remarks: You can verify the expression for Icm by letting m → 0 to obtain

2cm MLI 12

1= and letting M → 0 to obtain Icm = 0. 68 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the statement of Problem 67, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Kinetic energy is also conserved as the collision of the hard sphere with the bar is elastic. Let the direction the sphere is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction and v′ and V′ be the final velocities of the objects whose masses are m and M, respectively. Apply conservation of linear momentum to obtain:

fi pp =

or '' MVmvmv += (1)

Apply conservation of angular momentum to obtain:

fi LL =

or ω2

121' MLdmvmvd += (2)

Set v′ = 0 in equation (1) and solve for V ′: M

mvV' = (3)

Use conservation of mechanical energy to relate the kinetic energies of translation and rotation before

fi KK =

or ( ) 22121

212

212

21 ' ωMLMVmv += (4)

Chapter 10

770

and after the elastic collision: Substitute (2) and (3) in (4) and simplify to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 2

2121Ld

Mm

Mm

Solve for d:

mmMLd

12−

=

69 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ:


0fi =+− UK

and

( )θω cos12

221 −⎟

⎠⎞

⎜⎝⎛ += mgxdMgI (1)

Apply conservation of momentum to the collision:

fi LL =

or ( )[ ]ωω mdMdIdmv 22

31 8.08.0 +==

Solve for ω to obtain:

2231 64.0

8.0mdMd

dmv+

=ω (2)

Express the moment of inertia of the system about the pivot:

( )2

312

2312

64.0

8.0

Mdmd

MddmI

+=

+= (3)


771

Substitute equations (2) and (3) in equation (1) and simplify to obtain:

( )

( )22

31

2

64.032.0

cos12

mdMddmv

mgddMg

+=

−⎟⎠⎞

⎜⎝⎛ + θ

Solve for v:

( ) ( ) ( )2

2231

32.0cos164.08.05.0

dmgmdMdmMv θ−++

=

Evaluate v for θ = 90° to obtain:

( )( )2

2231

32.064.08.05.0

dmgmdMdmMv ++

=

70 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing. Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ :

0ifif =−+− UUKK

or, because Kf = Ui = 0, 0fi =+− UK

and

( )θω cos12

221 −⎟

⎠⎞

⎜⎝⎛ += mgxdMgI (1)

Apply conservation of momentum to the collision:

fi LL =

or

( )[ ]ωω

mdMd

Idmv22

31 8.0

8.0

+=

=

Chapter 10

772

Solve for ω to obtain: 22

21 64.0

8.0mdMd

dmv+

=ω (2)

Express the moment of inertia of the system about the pivot:

( )( )

( ) ( )[ ]( )2

231

231

2312

mkg0.660

m1.2kg0.8kg0.30.64

64.0

8.0

⋅=

+=

+=

+=

dMm

MddmI

Substitute equation (2) in equation (1) and simplify to obtain:

( )

( )Idmv

dmgdMg

232.0

cos18.02

=

−⎟⎠⎞

⎜⎝⎛ + θ

Solve for v: ( )( )

232.0cos18.05.0

dmImMgv θ−+

=

Substitute numerical values and evaluate v for θ = 60° to obtain:

( ) ( ) ( )[ ]( )( )( )( )

m/s74.7kg0.3m1.20.32

mkg0.6605.0kg0.30.8kg0.80.5m/s9.812

22

=⋅+

=v

71 •• Picture the Problem Let the length of the uniform stick be l. We can use the impulse-change in momentum theorem to express the velocity of the center of mass of the stick. By expressing the velocity V of the end of the stick in terms of the velocity of the center of mass and applying the angular impulse-change in angular momentum theorem we can find the angular velocity of the stick and, hence, the velocity of the end of the stick. (a) Apply the impulse-change in momentum theorem to obtain:

ppppK =−=∆= 0 or, because p0 = 0 and p = Mvcm,

cmMvK =

Solve for vcm to obtain: M

Kv =cm

(b) Relate the velocity V of the end of the stick to the velocity of the center of mass vcm:

( )l21

cmm of c torelcm ω+=+= vvvV (1)

Relate the angular impulse to the change in the angular momentum of the stick:

( ) ωcm021 ILLLK =−=∆=l

or, because L0 = 0, ( ) ωcm2

1 IK =l


773

Refer to Table 9-1 to find the moment of inertia of the stick with respect to its center of mass:

2121

cm lMI =

Substitute to obtain: ( ) ω2121

21 ll MK =

Solve for ω:

lMK6

=ω


MK

MK

MKV 4

26

=⎟⎠⎞

⎜⎝⎛+=

l

l

(c) Relate the velocity V′ of the other end of the stick to the velocity of the center of mass vcm:

( )

MK

MK

MK

vvvV

22

621

cmm of c torelcm

−=⎟⎠⎞

⎜⎝⎛−=

−=−=

l

l

lω

(d) Letting x be the distance from the center of mass toward the end not struck, express the condition that the point at x is at rest:

0cm =− xv ω

Solve for x to obtain: 06

=− xM

KMK

l

Solve for x to obtain:

l

l

61

6 ==

MK

MK

x

Note that for a meter stick struck at the

100-cm mark, the stationary point would be at the 33.3-cm mark.

Remarks: You can easily check this result by placing a meterstick on the floor and giving it a sharp blow at the 100-cm mark. 72 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. (a) Use its definition to express the total angular momentum of the disk and projectile just before impact:

bvmL 0p0 =

Chapter 10

774

(b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision:

ωILL ==0 and I

L0=ω

Express the moment of inertia of the disk + projectile:

2p

221 bmMRI +=

Substitute for I in the expression for ω to obtain: 2

p2

0p

22

bmMRbvm

+=ω

(c) Express the kinetic energy of the system after impact in terms of its angular momentum:

( )( )

( )2

p2

20p

2p

221

20p

2

f

2

22

bmMRbvm

bmMRbvm

ILK

+=

+==

(d) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain:

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=

+−=

−=∆

2p

2

2p2

0p21

2p

2

20p2

0p21

fi

21

2

bmMRbm

vm

bmMRbvm

vm

KKE

*73 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the masses M and m. Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle:

0ifif =−+− UUKK

or, because Ki = 0, 0iff =−+ UUK


775

Substitute for Kf, Uf, and Ui to obtain:

( ) 02 1

12213

121 =−+ MgLLMgML ω

Solve for ω:

1

3Lg

=ω

Letting ω′ represent the angular speed of the rod-and-particle system just after impact, use conservation of angular momentum to relate the angular momenta before and after the collision:

fi LL =

or ( ) ( ) '2

2213

1213

1 ωω mLMLML +=

Solve for ω′: ωω 2

2213

1

213

1

'mLML

ML+

=

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

0ifif =−+− UUKK

or, because Kf = 0, ( )( )

( ) 0cos1cos1'

max2

max1212

21

=−+

−+−

θθω

mgLLMgI

(1)

Express the moment of inertia of the system with respect to the pivot:

22

213

1 mLMLI +=

Substitute for θmax, I and ω′ in equation (1):

( )( ) 212

122

213

1

2213

1

1

3mgLLMg

mLML

MLLg

+=+

Simplify to obtain: 3

21222

21

31 632 L

MmLLLL

MmL ++= (2)

Simplify equation (2) by letting α = m/M and β = L2/L1 to obtain:

01236 232 =−++ αβββα

Substitute for α and simplify to obtain the cubic equation in β:

034912 23 =−++ βββ

Use the solver function* of your calculator to find the only real value

349.0=β

Chapter 10

776

of β:

*Remarks: Most graphing calculators have a ″solver″ feature. One can solve the cubic equation using either the ″graph″ and ″trace″ capabilities or the ″solver″ feature. The root given above was found using SOLVER on a TI-85. 74 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m. (a) Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain:

( ) 02 1

12213


Solve for ω:

1

3Lg

=ω

Letting ω′ represent the angular speed of the system after impact, use conservation of angular momentum to relate the angular momenta before and after the collision:

fi LL =

or ( ) ( ) '2

2213

1213

1 ωω mLMLML += (1)

Solve for ω′:

122

213

1

213

1

22

213

1

213

1

3

'

Lg

mLMLML

mLMLML

+=

+= ωω


777

Substitute numerical values to obtain: ( )( )( )( ) ( )

( )

( )

m

m

m

64.0kg0.960s/kg75.4

m64.0mkg0.960s/mkg75.4

m2.1m/s81.93

m8.0m2.1kg2m2.1kg2'

22

2

2

2231

231

+=

+⋅⋅

=

×

+=ω

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

0ifif =−+− UUKK

or, because Kf = 0, 0ifi =−+− UUK

Substitute for Ki, Uf, and Ui to obtain:

( )( )( ) 0cos1

cos1'

max2

max1212

21

=−+−+−

θθω

mgLLMgI

Express the moment of inertia of the system with respect to the pivot:

22

213

1 mLMLI +=

Substitute for θmax, I and ω′ in equation (1) and simplify to obtain:

( ) ( )21

221

2.064.0kg960.0

kg/s75.4 mLMLgm

+=+

Substitute for M, L1 and L2 and simplify to obtain:

0901.800.32 =−+ mm

Solve the quadratic equation for its positive root:

kg84.1=m

(b) The energy dissipated in the inelastic collision is:

fi UUE −=∆ (2)

Express Ui: 2

1i

LMgU =

Express Uf: ( ) ⎟

⎠⎞

⎜⎝⎛ +−= 2

1maxf 2

cos1 mLLMgU θ

Chapter 10

778


( ) ⎟⎠⎞

⎜⎝⎛ +−−

=∆

21

max

1

2cos1

2

mLLMg

LMgE

θ


( )( )( )

( )( ) ( )( ) ( )( )

J51.6

m0.8kg1.852

m1.2kg2m/s9.81cos371

2m1.2m/s9.81kg2

2

2

f

=

⎟⎠⎞

⎜⎝⎛ +°−−

=U

75 •• Picture the Problem Let ωi and ωf be the angular velocities of the rod immediately before and immediately after the inelastic collision with the mass m. Let ω0 be the initial angular velocity of the rod. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. We apply energy conservation to determine ωf and conservation of angular momentum to determine ωi. We’ll apply energy conservation to determine ω0. Finally, we’ll find the energies of the system immediately before and after the collision and the energy dissipated. Express the energy dissipated in the inelastic collision:

fi UUE −=∆ (1)

Use energy conservation to relate the kinetic energy of the system immediately after the collision to its potential energy after a 180° rotation:

0ifif =−+− UUKK

or, because Kf = Ktop = 0 and Ki = Kbottom, 0bottomtopbottom =−+− UUK

Substitute for Kbottom, Utop, and Ubottom to obtain:

( )( ) 02112

1

211232

f21

=−−−+++−LLmgMgLLLmgMgLIω

Simplify to obtain:

02 212f2

1 =++− mgLMgLIω (2)

Express I: 22

213

1 mLMLI +=

Substitute for I in equation (2) and solve for ωf to obtain:

( )22

213

121

f22mLML

mLMLg++

=ω


779

Substitute numerical values and evaluate ωf:

( ) ( )( ) ( )( )[ ]( )( ) ( )( )

rad/s00.7m0.8kg0.4m1.2kg0.75

m0.8kg0.42m1.2kg0.75m/s9.81222

31

2

f =+

+=ω

Use conservation of angular momentum to relate the angular momentum of the system just before the collision to its angular momentum just after the collision:

fi LL =

or ffii ωω II =

Substitute for Ii and If and solve for ωi:

( ) ( ) f22

213

1i

213

1 ωω mLMLML +=

and

f

2

1

2i

31 ωω⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

LL

Mm

Substitute numerical values and evaluate ωi:

( ) ( )

rad/s0.12

rad/s7.00m1.2m0.8

kg0.75kg0.431

2

i

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=ω

Apply conservation of mechanical energy to relate the initial rotational kinetic energy of the rod to its rotational kinetic energy just before its collision with the particle:

0ifif =−+− UUKK

Substitute to obtain: ( ) ( )0

21

120

213

1212

i213

121

=−

+−

MgL

LMgMLML ωω

Solve for ω0:

1

2i0

3Lg

−= ωω

Substitute numerical values and evaluate ω0: ( ) ( )

rad/s10.9

m1.2m/s9.813

rad/s122

20

=

−=ω

Chapter 10

780

Substitute in equation (1) to express the energy dissipated in the collision:

( ) 212i

213

121 2mgLMgLMLE +−=∆ ω


( )( ) ( ) ( ) ( )( ) ( )( )[ ]J8.10

m0.8kg0.42m1.2kg0.75m/s9.81rad/s12m1.2kg0.75 22261

=

+−=∆E

76 ••• Picture the Problem Let v be the speed of the particle immediately after the collision and ωi and ωf be the angular velocities of the rod immediately before and immediately after the elastic collision with the mass m. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this elastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the elastic collision with the particle and the kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m. Use energy conservation to relate the energies of the system immediately before and after the elastic collision:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain: ( ) 02

cos12

1max

1221 =−−+

LMg

LMgmv θ

Solve for mv2: max1

2 cosθMgLmv = (1)

Apply conservation of energy to express the angular speed of the rod just before the collision:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain: ( ) 0

2 112

i213


Solve for ωi:

1i

3Lg

=ω


781

Apply conservation of energy to the rod after the collision:

( ) ( ) 0cos12 max

12f

213

121 =−− θω LMgML

Solve for ωf:

1f

6.0L

g=ω

Apply conservation of angular momentum to the collision:

fi LL =

or ( ) ( ) 2f

213

1i

213

1 mvLMLML += ωω

Solve for mv: ( )

2

fi213

1

LMLmv ωω −

=

Substitute for ωf and ωI to obtain:

2

11

21

3

6.03

LL

gLgML

mv⎟⎟⎠

⎞⎜⎜⎝

⎛−

= (2)

Divide equation (1) by equation (2) to eliminate m and solve for v:

11

max2

2

11

21

max1

6.03cos3

3

6.03cos

gLgLgL

LL

gLgML

MgLv

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

θ

θ


( ) ( )( ) ( ) ( )( )

m/s72.5m2.1m/s81.96.0m2.1m/s81.93

37cosm8.0m/s81.9322

2

=−

°=v

Solve equation (1) for m:

2max1 cos

vMgLm θ

=

Substitute for v in the expression for mv and solve for m:

( )( )( )( )

kg575.0

m/s72.537cosm2.1m/s81.9kg2

2

2

=

°=m

Because the collision was elastic: 0=∆E

Chapter 10

782

77 •• Picture the Problem We can determine the angular momentum of the wheel and the angular velocity of its precession from their definitions. The period of the precessional motion can be found from its angular velocity and the angular momentum associated with the motion of the center of mass from its definition. (a) Using the definition of angular momentum, express the angular momentum of the spinning wheel:

ωωω 22 RgwMRIL ===


( )

sJ1.18

revrad2

srev12

m0.28m/s9.81N30 2

2

⋅=

⎟⎠⎞

⎜⎝⎛ ××

⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

L

(b) Using its definition, express the angular velocity of precession: L

MgDdtd

==φωp

Substitute numerical values and evaluate ωp:

( )( ) rad/s0.414sJ18.1

m0.25N30p =

⋅=ω

(c) Express the period of the precessional motion as a function of the angular velocity of precession:

s2.15rad/s414.0

22

p

===π

ωπT

(d) Express the angular momentum of the center of mass due to the precession:

p2

pcmp ωω MDIL ==

Substitute numerical values and evaluate Lp:

( ) ( )

sJ0791.0

rad/s414.0m0.25m/s9.81N30 2

2p

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=L

The direction of Lp is either up or down, depending on the direction of L.

*78 •• Picture the Problem The angular velocity of precession can be found from its definition. Both the speed and acceleration of the center of mass during precession are related to the angular velocity of precession. We can use Newton’s 2nd law to find the vertical and


783

horizontal components of the force exerted by the pivot. (a) Using its definition, express the angular velocity of precession:

s2

s2

21

ssp

2ωωω

φωR

gDMRMgD

IMgD

dtd

====

Substitute numerical values and evaluate ωp:

( ) ( )

( )rad/s27.3

s60min1

revrad2π

minrev700m0.064

m0.05m/s9.8122

2

p =

⎟⎟⎠

⎞⎜⎜⎝

⎛××

=ω

(b) Express the speed of the center of mass in terms of its angular velocity of precession:

( )( )m/s0.164

rad/s3.27m0.05pcm

=

== ωDv

(c) Relate the acceleration of the center of mass to its angular velocity of precession:

( )( )2

22pcm

m/s0.535

rad/s3.27m0.05

=

== ωDa

(d) Use Newton’s 2nd law to relate the vertical component of the force exerted by the pivot to the weight of the disk:

( )( )N24.5

m/s9.81kg2.5 2v

=

== MgF

Relate the horizontal component of the force exerted by the pivot to the acceleration of the center of mass:

( )( )N34.1

m/s535.0kg2.5 2cmv

=

== MaF

General Problems 79 • Picture the Problem While the 3-kg particle is moving in a straight line, it has angular momentum given by prL

rrr×= where r

ris its position vector and p

ris its linear

momentum. The torque due to the applied force is given by .Frτrrr

×= (a) Express the angular momentum of the particle:

prLrrr

×=

Express the vectors rr and pr : ( ) ( ) jir ˆm3.5ˆm12 +=r

Chapter 10

784

and ( )( )

( )i

iipˆm/skg9

ˆm/s3kg3ˆ

⋅=

== mvr

Substitute and simplify to find L

r: ( ) ( )[ ] ( )

( )( )( )k

ij

ijiL

ˆ/smkg7.47

ˆˆ/smkg7.47

ˆm/skg9ˆm3.5ˆm12

2

2

⋅−=

×⋅=

⋅×+=r

(b) Using its definition, express the torque due to the force:

Frτrrr

×=

Substitute and simplify to find τr : ( ) ( )[ ] ( )( )( )( )k

ij

ijiτ

ˆmN9.15

ˆˆmN9.15

ˆN3ˆm3.5ˆm12

⋅=

×⋅−=

−×+=r

80 • Picture the Problem The angular momentum of the particle is given by

prLrrr

×= where rr

is its position vector and pr

is its linear momentum. The torque acting

on the particle is given by .dtdLτrr

=

Express the angular momentum of the particle:

dtdm

mmrr

vrvrprLr

r

rrrrrrr

×=

×=×=×=

Evaluate dtdrr

: jr ˆ6tdtd

=r

Substitute and simplify to find L

r: ( ) ( ) ( ){ }[ ]

( )( )k

j

jiL

ˆsJ0.72

ˆm/s6

ˆm/s3ˆm4kg3 22

⋅=

×

+=

t

t

tr

Find the torque due to the force: ( )[ ]

( )k

kLτ

ˆmN0.72

ˆsJ0.72

⋅=

⋅== tdtd

dtdr

r


785

81 •• Picture the Problem The ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by ωcmIL = and its kinetic energy can be found

from .2 cm2 ILK =

(a) Express the angular momentum of the system about the center of mass of the skaters:

ωcmIL =

Using its definition, locate the center of mass, relative to the 85-kg skater, of the system:

( )( ) ( )( )

m0.668kg85kg55

0kg85m1.7kg55cm

=+

+=x

Calculate cmI : ( )( )( )( )

2

2

2cm

mkg5.96m0.668kg85

m0.668m1.7kg55

⋅=

+

−=I

Substitute to determine L: ( )

sJ243

revrad2π

s2.5rev1mkg96.5 2

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×⋅=L

(b) Relate the total kinetic energy of the system to its angular momentum and evaluate K:

cm

2

2ILK =


( )( ) J306

mkg96.52sJ243

2

2

=⋅

⋅=K

Chapter 10

786

*82 •• Picture the Problem Let the origin of the coordinate system be at the pivot (point P). The diagram shows the forces acting on the ball. We’ll apply Newton’s 2nd law to the ball to determine its speed. We’ll then use the derivative of its position vector to express its velocity and the definition of angular momentum to show that L

rhas

both horizontal and vertical components. We can use the derivative of L

rwith

respect to time to show that the rate at which the angular momentum of the ball changes is equal to the torque, relative to the pivot point, acting on it. (a) Express the angular momentum of the ball about the point of support:

vrprLrrrrr

×=×= m (1)

Apply Newton’s 2nd law to the ball: ∑ ==θ

θsin

sin2

rvmTFx

and

∑ =−= 0cos mgTFz θ

Eliminate T between these equations and solve for v:

θθ tansinrgv =


( )( )m/s2.06

tan30sin30m/s9.81m1.5 2

=

°°=v

Express the position vector of the ball:

( ) ( )( ) k

jirˆ30cosm5.1

ˆsinˆcos30sinm5.1

°−

+°= tt ωωr

where .kωω =

Find the velocity of the ball:

( )( )ji

rv

ˆcosˆsinm/s75.0 ttdtd

ωωω +−=

=r

r

Evaluate ω:

( ) rad/s75.230sinm5.1

m/s06.2=

°=ω


787

Substitute for ω to obtain:

( )( )jiv ˆcosˆsinm/s06.2 tt ωω +−=r

Substitute in equation (1) and evaluate Lr

:

( ) ( ) ( ) ( )[ ]( )[ ( )]

( )[ ] sJˆ09.3ˆsinˆcos36.5

ˆcosˆsinm/s06.2

ˆ30cosm5.1ˆsinˆcos30sinm5.1kg2

⋅++=

+−×

°−+°=

kji

ji

kjiL

tt

tt

tt

ωω

ωω

ωωr

The horizontal component of L

ris:

( ) sJˆsinˆcos36.5 ⋅+ ji tt ωω

The vertical component of Lr

is:

sJˆ09.3 ⋅k

(b) Evaluate dtdLr

: ( )[ ] Jˆcosˆsin36.5 jiL ttdtd ωωω +−=r

Evaluate the magnitude of dtdLr

: ( )( )

mN7.14

rad/s75.2smN36.5

⋅=

⋅⋅=dtdLr

Express the magnitude of the torque exerted by gravity about the point of support:

θτ sinmgr=

Substitute numerical values and evaluate τ :

( )( )( )mN7.14

30sinm1.5m/s9.81kg2 2

⋅=

°=τ

83 •• Picture the Problem In part (a) we need to decide whether a net torque acts on the object. In part (b) the issue is whether any external forces act on the object. In part (c) we can apply the definition of kinetic energy to find the speed of the object when the unwrapped length has shortened to r/2. (a) Consider the overhead view of the cylindrical post and the object shown in the adjoining figure. The object rotates counterclockwise. The torque about the center of the cylinder is clockwise and of magnitude RT, where R is the radius of the cylinder and T is the tension. So

Chapter 10

788

L must decrease.

decreases. No, L

(b) Because, in this frictionless environment, no net external forces act on the object:

constant. isenergy kinetic Its

(c) Express the kinetic energy of the object as it spirals inward:

( ) 221

2

22

212

21 mv

rvmrIK === ω

constant.)remainsenergy kinetic (The .0v

84 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for ωi: ωωωf

ii

f

if I

III

==

Express Ii: ( )2

412

101

i 2 lmMLI +=

Express If: ( )2

412

101

f 2 mLMLI +=

Substitute to express fω in terms of ω : ( )

( )

ω

ωω

mML

mM

mLMLmML

5

5

22

2

2

2412

101

2412

101

f

+

+=

++

=

l

l

Express the initial kinetic energy of the system:

( )[ ]( ) 222

201

22412

101

212

i21

i

5

2

ω

ωω

l

l

mML

mMLIK

+=

+==


789

Express the final kinetic energy of the system and simplify to obtain:

( )[ ] ( )

( )

( ) 222

222

201

2

22

201

2

2

2

22201

2f

222012

f2

412

101

212

ff21

f

55

5

5

5

55

52

ω

ωω

ωωω

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

++=

+=+==

mLMLmML

mML

mML

mML

mMmLML

mLMLmLMLIK

l

ll

85 •• Determine the Concept Yes. The net external torque is zero and angular momentum is conserved as the system evolves from its initial to its final state. Because the disks come to the same final position, the initial and final configurations are the same as in Problem 84. Therefore, the answers are the same as for Problem 84. 86 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for ωf: ωωωf

ii

f

if I

III

== (1)

Relate the tension in the string to the angular speed of the system and solve for and evaluate ω:

22

2ωω lmmrT ==

and ( )

( )( )rad/s30.0

m0.6kg0.4N10822

=

==lm

Tω

Chapter 10

790

Express and evaluate Ii: ( )( )( ) ( )( )

2

2212

101

2412

101

i

mkg0.392

m0.6kg0.4m2kg0.8

2

⋅=

+=

+= lmMLI

Express and evaluate If: ( )

( )( ) ( )( )2

2212

101

2412

101

f

mkg12.1

m2kg0.4m2kg0.8

2

⋅=

+=

+= mLMLI

Substitute in equation (1) and solve for fω : ( )

rad/s5.10

rad/s0.30mkg1.12mkg392.0

2

2

f

if

=

⋅⋅

== ωωII

Express and evaluate the initial kinetic energy of the system: ( )( )

J176

rad/s0.30mkg392.0 2221

2i2

1i

=

⋅=

= ωIK

Express and evaluate the final kinetic energy of the system: ( )( )

J7.61

rad/s10.5mkg1.12 2221

2ff2

1f

=

⋅=

= ωIK

87 •• Picture the Problem Until the inelastic collision of the cylindrical objects at the ends of the cylinder, both angular momentum and energy are conserved. Let K’ represent the kinetic energy of the system just before the disks reach the end of the cylinder and use conservation of energy to relate the initial and final kinetic energies to the final radial velocity. Using conservation of mechanical energy, relate the initial and final kinetic energies of the disks:

'i KK =

or ( )2

r212

ff212

i21 2mvII += ωω

Solve for vr:

mII

v2

2ff

2i

rωω −

= (1)

Using conservation of angular momentum, relate the initial and final angular velocities to the initial

fi LL =

or


791

and final moments of inertia:

ffii ωω II =

Solve for fω : ωωωf

i

f

if I

III

==

Express Ii: ( )2

412

101

i 2 lmMLI +=

Express If: ( )2

412

101

f 2 mLMLI +=

Substitute to obtain fω in terms of ω : ( )

( )ω

ωω

22

22

2412

101

2412

101

f

55

22

mLMLmML

mLMLmML

++

=

++

=

l

l


( )22r 2

ll

−= LL

v ω

88 •• Picture the Problem Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities and the initial and final kinetic energy of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for fω : ωωωf

ii

f

if I

III

== (1)

Relate the tension in the string to the angular speed of the system:

22

2ωω lmmrT ==

Solve for ω:

lmT2

=ω


( )( )( ) rad/s30.0

m0.6kg0.4N1082

==ω

Chapter 10

792

Express and evaluate Ii: ( )( )( ) ( )( )

2

2212

101

2412

101

i

mkg0.392

m0.6kg0.4m2kg0.8

2

⋅=

+=

+= lmMLI

Letting L′ represent the final separation of the disks, express and evaluate If:

( )( )( ) ( )( )

2

2212

101

2412

101

f

mkg832.0

m6.1kg0.4m2kg0.8

'2

⋅=

+=

+= mLMLI

Substitute in equation (1) and solve for fω : ( )

rad/s1.14

rad/s0.30mkg832.0mkg392.0

2

2

f

if

=⋅⋅

== ωωII

Express and evaluate the initial kinetic energy of the system: ( )( )

J176

rad/s0.30mkg392.0 2221

2i2

1i

=

⋅=

= ωIK

Express and evaluate the final kinetic energy of the system: ( )( )

J7.82

rad/s1.41mkg832.0 2221

2ff2

1f

=

⋅=

= ωIK

The energy dissipated in friction is:

J93.3

J82.7J176fi

=

−=−=∆ KKE

*89 •• Picture the Problem The drawing shows an elliptical orbit. The triangular element of the area is ( ) .2

21

21 θθ drrdrdA ==

Differentiate dA with respect to t to obtain:

ωθ 2212

21 r

dtdr

dtdA

==

Because the gravitational force acts along the line joining the two objects, τ = 0 and:

constant

2

== ωmrL


793

Eliminate r2ω between the two equations to obtain:

constant2

==mL

dtdA

90 •• Picture the Problem Let x be the radial distance each disk moves outward. Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities to the initial and final moments of inertia. We’ll assume that the disks are thin enough so that we can ignore their lengths in expressing their moments of inertia. Use conservation of angular momentum to relate the initial and final angular velocities of the disks:

fi LL =

or ffii ωω II =

Solve for ωf: i

f

if ωω

II

= (1)

Express the initial moment of inertia of the system:

diskcyli 2III +=

Express the moment of inertia of the cylinder: ( )

( ) ( ) ( )[ ]2

22121

22121

2212

121

cyl

mkg0.232

m0.26m1.8kg0.8

6

⋅=

+=

+=

+=

RLM

MRMLI

Letting l represent the distance of the clamped disks from the center of rotation and ignoring the thickness of each disk (we’re told they are thin), use the parallel-axis theorem to express the moment of inertia of each disk:

( )( ) ( ) ( )[ ]

2

2241

2241

2241

disk

mkg0340.0

m4.04m2.0kg2.0

4

⋅=

+=

+=

+=

l

l

rm

mmrI

With the disks clamped:

( )2

22

diskcyli

mkg300.0mkg0340.02mkg232.0

2

⋅=

⋅+⋅=

+= III

Chapter 10

794

With the disks unclamped, l = 0.6 m and:

( )( ) ( ) ( )[ ]

2

2241

2241

disk

mkg0740.0

m6.04m2.0kg2.0

4

⋅=

+=

+= lrmI

Express and evaluate the final moment of inertia of the system: ( )

2

22

diskcylf

mkg380.0mkg0740.02mkg232.0

2

⋅=

⋅+⋅=

+= III

Substitute in equation (1) to determine ωf:

( )

rad/s32.6

rad/s8mkg380.0mkg300.0

2

2

f

=

⋅⋅

=ω

Express the energy dissipated in friction: ( )2

212

ff212

ii21

fi

kxII

EEE

+−=

−=∆

ωω

Apply Newton’s 2nd law to each disk when they are in their final positions:

∑ == 2radial ωmrkxF

Solve for k: x

mrk2ω

=

Substitute numerical values and evaluate k:

( )( )( )

N/m24.0m0.2

rad/s6.32m0.6kg0.2 2

=

=k

Express the energy dissipated in friction:

( )2212

ff212

ii21

fifr

kxII

EEW

+−=

−=

ωω

Substitute numerical values and evaluate Wfr:

( )( ) ( )( ) ( )( )J1.53

m0.2N/m24rad/s32.6mkg380.0rad/s8mkg0.300 22122

2122

21

fr

=

−⋅−⋅=W

91 •• Picture the Problem Let the letters d, m, and r denote the disk and the letters t, M, and R the turntable. We can use conservation of angular momentum to relate the final angular speed of the turntable to the initial angular speed of the Euler disk and the moments of inertia of the turntable and the disk. In part (b) we’ll need to use the parallel-axis theorem


795

to express the moment of inertia of the disk with respect to the rotational axis of the turntable. You can find the moments of inertia of the disk in its two orientations and that of the turntable in Table 9-1. (a) Use conservation of angular momentum to relate the initial and final angular momenta of the system:

tftfdfdfdidi ωωω III +=

Because ωtf = ωdf:

tftftfdfdidi ωωω III +=

Solve for ωtf: di

tfdf

ditf ωω

III+

= (1)

Ignoring the negligible thickness of the disk, express its initial moment of inertia:

241

di mrI =

Express the final moment of inertia of the disk:

221

df mrI =

Express the final moment of inertia of the turntable:

221

tf MRI =


di

2

2

di2212

21

241

tf

22

1 ω

ωω

mrMR

MRmrmr

+=

+=

(2)

Express ωdi in rad/s:

rad/ss60

min1rev

rad2minrev30di ππω =××=

Substitute numerical values in equation (2) and evaluate ωtf: ( )( )

( )( )rad/s228.0

m0.125kg0.5m0.25kg0.73522

rad/s

2

2tf

=

+=

πω

(b) Use the parallel-axis theorem to express the final moment of inertia of the disk when it is a distance L from the center of the turntable:

( )222122

21

df LrmmLmrI +=+=

Chapter 10

796


( )di

2

2

2

2

di22122

21

241

tf

242

1 ω

ωω

mrMR

rL

MRLrmmr

++=

++=

Substitute numerical values and evaluate ωtf:

( )( )

( )( )( )( )

rad/s192.0

m0.125kg0.5m0.25kg0.7352

m0.125m0.142

rad/s

2

2

2

2tf =++

=πω

92 •• Picture the Problem We can express the period of the earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to T and then use differentials to approximate the changes in r and T. (a) Express the period of the earth’s rotation in terms of its angular velocity of rotation:

ωπ2

=T

Relate the earth’s angular velocity of rotation to its angular momentum and moment of inertia:

252 mr

LIL

==ω

Substitute and simplify to obtain: ( ) 22

52

542

rLm

Lmr

T ππ==

(b) Find dT/dr:

rTr

rTr

Lm

drdT 22

542 2 =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

π

Solve for dT/T:

rdr

TdT 2= or

rr

TT ∆

≈∆ 2

(c) Using the equation we just derived, substitute for the change in the period of the earth:

rr

TT ∆

==×=∆ 2

14601

d365.24y1

yd4

1


797

Solve for and evaluate ∆r: ( ) ( )

km18.2

14602km1037.6

14602

3

=

×==∆

rr

*93 •• Picture the Problem Let ωP be the angular velocity of precession of the earth-as-gyroscope, ωs its angular velocity about its spin axis, and I its moment of inertia with respect to an axis through its poles, and relate ωP to ωs and I using its definition. Use its definition to express the precession rate of the earth as a giant gyroscope:

Lτω =P

Substitute for I and solve for τ: PP ωωωτ IL ==

Express the angular velocity ωs of the earth about its spin axis: T

πω 2= where T is the period of rotation of

the earth.

Substitute to obtain: T

I P2 ωπτ =

Substitute numerical values and evaluateτ:

( ) ( ) mN1047.4

hs3600

dh24d1

s1066.7mkg1003.82 22112237

⋅×=××

×⋅×=

−−πτ

94 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Express the moment of inertia of the system (see Table 9-1):

( )( )( ) ( )

( ) 22

22121

22cylcyl12

1mcyl

kg0.8mkg256.0

kg4.02m1.6kg1.2

22

x

x

mxLMxIIII

+⋅=

+=

+==+=

Chapter 10

798

Evaluate If = I(0.4 m): ( )( )( )

2

22

f

mkg384.0m4.0kg0.8mkg256.0

m4.0

⋅=

+⋅=

= II

Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity:

∑ == 2fradial ωmrTF

where T is the tension in the string at which it breaks.

Solve for ωf:

mrT

=fω

Substitute numerical values and evaluate ωf: ( )( ) rad/s0.25

m0.4kg0.4N100

f ==ω

Substitute in equation (1) to express the work done before the string breaks:

2ff2

1 ωIW =


( )( )J120

rad/s25mkg0.384 2221

=

⋅=W

With the applied torque removed, angular momentum is conserved and we can express the angular momentum as a function of x:

( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xIIx ffωω =

Substitute numerical values to obtain: ( ) ( )( )

( )

( ) 22

22

2

kg0.8mkg0.256sJ60.9

kg0.8mkg0.256rad/s25mkg0.384

x

xx

+⋅⋅

=

+⋅⋅

=ω

95 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied


799

torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Express the moment of inertia of the system (see Table 9-1):

( ) 22cylcyl12

1mcyl 22 mxLMxIIII +==+=


( )( ) ( )( ) 22

22121

kg0.8mkg256.0

kg4.02m1.6kg1.2

x

xI

+⋅=

+=

Evaluate If = I(0.4 m): ( )

( )( )2

22

f

mkg384.0m4.0kg0.8mkg256.0

m4.0

⋅=

+⋅=

= II


∑ == 2frad ωmrTF


Solve for ωf:

mrT

=fω

Substitute numerical values and evaluate ωf:

( )( ) rad/s0.25m0.4kg0.4

N100f ==ω


( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xII

x ff ωω =

Substitute numerical values and simplify to obtain:

( ) ( )( )( ) ( ) 2222

2

kg0.8mkg0.256sJ60.9


xxx

+⋅⋅

=+⋅

⋅=ω

Evaluate ( )m8.0ω :

Chapter 10

800

( )( )( )

rad/s5.12m0.8kg0.8mkg0.256

sJ9.60m8.0 22 =+⋅

⋅=ω

Remarks: Note that this is the angular velocity in both instances. Because the disks leave the cylinder with a tangential velocity of Lω2

1 , the angular momentum of the

system remains constant. 96 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Using the parallel axis theorem and treating the disks as thin disks, express the moment of inertia of the system (see Table 9-1):

( )( )

( ) ( )224122

121

22412

212

121

mcyl

26

2

2

xRmRLM

mxmRMRML

IIxI

+++=

+++=

+=


( ) ( ) ( ) ( )[ ]( ) ( )[ ]

( ) 22

2241

22121

kg0.8mkg384.0

m4.0kg4.02

m4.06m1.6kg1.2

x

x

xI

+⋅=

++

+=

Evaluate If = I(0.4 m): ( )

( )( )2

22

f

mkg512.0m4.0kg0.8mkg384.0

m4.0

⋅=

+⋅=

= II


∑ == 2frad ωmrTF


Solve for ωf:

mrT

=fω

Substitute numerical values and evaluate ωf: ( )( ) rad/s0.25

m0.4kg0.4N100

f ==ω


801

Substitute in equation (1) to express the work done before the string breaks:

2ff2

1 ωIW =


( )( )J160

rad/s25mkg0.512 2221

=

⋅=W


( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xIIx ffωω =

Substitute numerical values to obtain: ( ) ( )( )

( )

( ) 22

22

2

kg0.8mkg0.384sJ8.12


x

xx

+⋅⋅

=

+⋅⋅

=ω

*97 ••• Picture the Problem Let the origin of the coordinate system be at the center of the pulley with the upward direction positive. Let λ be the linear density (mass per unit length) of the rope and L1 and L2 the lengths of the hanging parts of the rope. We can use conservation of mechanical energy to find the angular velocity of the pulley when the difference in height between the two ends of the rope is 7.2 m. (a) Apply conservation of energy to relate the final kinetic energy of the system to the change in potential energy:

0=∆+∆ UK or, because Ki = 0,

0=∆+ UK (1)

Express the change in potential energy of the system: ( ) ( )

( ) ( )[ ]( ) ( )

( ) ( )[ ]22i

21i

22f

21f2

1

22i

21i2

122f

21f2

1

2i2i21

1i1i21

2f2f21

1f1f21

if

LLLLg

gLLgLL

gLLgLLgLLgLL

UUU

+−+−=

+++−=

−−−−−=

−=∆

λ

λλ

λλλλ

Chapter 10

802

Because L1 + L2 = 7.4 m, L2i – L1i = 0.6 m, and L2f – L1f = 7.2 m, we obtain:

L1i = 3.4 m, L2i = 4.0 m, L1f = 0.1 m, and L2f = 7.3 m.

Substitute numerical values and evaluate ∆U:

( )( )( ) ( )[

( ) ( ) ]J75.75

m4m3.4

m7.3m0.1

m/s9.81kg/m0.6

22

22

221

−=−−

+×

−=∆U

Express the kinetic energy of the system when the difference in height between the two ends of the rope is 7.2 m:

( )( ) 22

p21

21

222122

p21

21

2212

p21

ω

ωω

ω

RMM

MRRM

MvIK

+=

+=

+=

Substitute numerical values and simplify: ( )[ ]

( ) 22

22

21

21

mkg1076.02

m2.1kg8.4kg2.2

ω

ωπ

⋅=

⎟⎠⎞

⎜⎝⎛+=K

Substitute in equation (1) and solve for ω:

( ) 0J75.75mkg1076.0 22 =−⋅ ω

and

rad/s5.26mkg1076.0

J75.752 =

⋅=ω

(b) Noting that the moment arm of each portion of the rope is the same, express the total angular momentum of the system:

( )( ) ω

ω

ωω

2rp2

1

2r

2p2

1

2rprp

RMM

RMRM

RMILLL

+=

+=

+=+=

(2)

Letting θ be the angle through which the pulley has turned, express U(θ):

( ) ( ) ( )[ ] gRLRLU λθθθ 22i

21i2

1 ++−−=

Express ∆U and simplify to obtain: ( ) ( )( ) ( )[ ]

( )( ) gRLLgR

gLL

gRLRL

UUUUU

θλλθ

λ

λθθ

θ

2ii122

2i2

21i2

1

22i

21i2

1

if 0

−+−=

++

++−−=

−=−=∆

Assuming that, at t = 0, L1i ≈ L2i: gRU λθ 22−≈∆


803

Substitute for K and ∆U in equation (1) to obtain:

( ) 0mkg1076.0 2222 =−⋅ gR λθω

Solve for ω: 2

22

mkg1076.0 ⋅=

gR λθω


( )( )

( )θ

θπω

1-

2

22

s41.1

mkg1076.0

m/s9.81kg/m0.62

m2.1

=

⋅

⎟⎠⎞

⎜⎝⎛

=

Express ω as the rate of change of θ :

( )θθ 1s41.1 −=dtd

⇒ ( )dtd 1s41.1 −=θθ

Integrate θ from 0 to θ to obtain: ( )t1s41.1ln −=θ

Transform from logarithmic to exponential form to obtain:

( ) ( )tet1s41.1 −

=θ

Differentiate to express ω as a function of time:

( ) ( ) ( )tedtdt

1s41.11s41.1−−==

θω

Substitute for ω in equation (2) to obtain:

( ) ( ) ( )teRMML1s41.112

rp21 s41.1

−−+=


( ) ( )[ ] ( ) ( )[ ] ( ) ( )tt eeL11 s41.12s41.11

2

21 s/mkg303.0s41.1

2m2.1kg4.8kg2.2

−−

⋅=⎟⎠⎞

⎜⎝⎛+= −

π

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