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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
Substitute to obtain:
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 =∆
Conservation of Energy
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
Conservation of Energy
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
Conservation of Energy
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 =
Substitute numerical values and evaluate d:
( )( ) 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
Conservation of Energy
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=
Substitute to obtain:
⎟⎟⎠
⎞⎜⎜⎝
⎛−= −
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+=
Conservation of Energy
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
Substitute to obtain:
( ) 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 =−
Conservation of Energy
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
Substitute to obtain:
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 −=
Conservation of Energy
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
Substitute to obtain:
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
Substitute to obtain:
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
Substitute numerical values and evaluate v:
( )( ) 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
Conservation of Energy
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
Conservation of Energy
457
Solve for θ : ( )⎥⎦
⎤⎢⎣
⎡+= −
0
2f1 cos
2'cos θθ
gLv
Substitute numerical values and evaluate θ :
( )( )( )
°=
⎥⎦
⎤⎢⎣
⎡°+= −
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.
Conservation of Energy
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
Substitute numerical values and evaluate v2:
( ) ( )( )( ) 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
Conservation of Energy
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
Substitute numerical values and evaluate v3:
( ) ( )( )[ ] 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=−
Conservation of Energy
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
Substitute numerical values and evaluate v3:
( ) ( )( )[ ] 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
Conservation of Energy
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
Conservation of Energy
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
Conservation of Energy
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
Substitute numerical values and evaluate vf:
( ) ( )( )( )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
Conservation of Energy
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
Substitute numerical values and evaluate v2:
( )( ) 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
−∆=−=∆
=∆+−=
Substitute numerical values and evaluate ∆Etherm:
( )( )( ) ( )( ) 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:
Conservation of Energy
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=∆=∆
Substitute numerical values and evaluate ∆Etherm:
( )( )( )( )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)
Substitute numerical values and evaluate v:
( )( )( ) ( )( )[ ] 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 =∆
Conservation of Energy
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
Substitute numerical values and evaluate L:
( )( ) ( )[ ]
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
Substitute numerical values and evaluate vf:
( )( ) ( )[ ]] 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
−=−=−=∆=∆
Substitute in equation (1) to obtain:
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 =
Conservation of Energy
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
×=
× −
Conservation of Energy
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
Substitute to obtain:
( ) 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
Conservation of Energy
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 ==∆
Substitute to obtain:
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:
0thermext =∆+∆+∆= EUKW
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
Conservation of Energy
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=
Substitute to obtain:
IRL 24π=
Substitute numerical values and evaluate L:
( ) ( )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
Substitute numerical values and evaluate µk:
( )( )[ ]( )( ) 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:
Conservation of Energy
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
Conservation of Energy
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:
0thermext =∆+∆+∆= EUKW
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
Substitute numerical values and evaluate v1:
( )( ) ( )[ ] 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 +=
Substitute numerical values and evaluate P:
( )( )( )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 −+=
Substitute numerical values and evaluate P:
( )( )( )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:
0thermext =∆+∆+∆= EUKW
or, because ∆K = 0, 0thermis,fs,ig,fg, =∆+−+− EUUUU
Because 0fs,ig, == UU :
0thermis,fg, =∆+− EUU
Conservation of Energy
489
Substitute for Ug,i and Us,f and solve for ∆Etherm:
mghkxUUE −=−=∆ 221
fg,is,therm
Substitute numerical values and evaluate ∆Etherm:
( )( )( )( )( )
J0.602
m24m/s9.81kg0.007
m0.03N/m50002
221
therm
=
−
=∆E
Apply conservation of energy during the dart’s descent:
0thermext =∆+∆+∆= EUKW
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 ∆−
=
Substitute numerical values and evaluate vf:
( )( )( )[ ] 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
Substitute numerical values and evaluate ∆Etherm:
( )( )( )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 ∆
−=
Substitute numerical values and evaluate vf:
( )( ) ( )
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
Conservation of Energy
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
Substitute numerical values and evaluate ∆E:
( ) ( ) ( ) ( )( )[ ] 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.
Conservation of Energy
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:
0therm =∆+∆+∆ EUK
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:
0therm =∆+∆+∆ EUK
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
Conservation of Energy
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
Conservation of Energy
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
Conservation of Energy
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
Substitute numerical values and evaluate θ :
( )( )
°=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ −= −
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
Conservation of Energy
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
Conservation of Energy
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
Substitute numerical values and evaluate P:
( ) 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
Conservation of Energy
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
Substitute to obtain: 02212
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=
Conservation of Energy
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
Substitute to obtain:
( ) 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
Substitute numerical values and evaluate ∆Etherm:
( )( )( )( )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∆−
=
Substitute numerical values and evaluate v2:
( )( ) ( )
m/s0.492
kg5J0.2942cm3N/cm20 2
2
=
−=v
Conservation of Energy
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
Substitute numerical values and evaluate d:
( )( )( )( )( )
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 −+=
Conservation of Energy
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
Conservation of Energy
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)
Conservation of Energy
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
−+=
−+=
∆=
Substitute numerical values and evaluate v:
( ) ( ) ( ) 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 =⇒=
Systems of Particles and Conservation of Momentum
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.
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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 .
Systems of Particles and Conservation of Momentum
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 =
Substitute to obtain:
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
+
⎟⎠⎞
⎜⎝⎛+
=
+
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
=
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 ==
Systems of Particles and Conservation of Momentum
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
=
== ∫∫∫
Systems of Particles and Conservation of Momentum
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
Substitute numerical values and evaluate cmvr :
( )( ) ( )( ) ( ) 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
=
Substitute to obtain:
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,
Systems of Particles and Conservation of Momentum
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+−
=
Substitute to obtain:
( )( )( ) 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+
=
Systems of Particles and Conservation of Momentum
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
Substitute numerical values to obtain:
( )( ) ( ) 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
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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−=
Systems of Particles and Conservation of Momentum
543
Substitute numerical values to obtain:
( ) ( )( )
( ) ( )( )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 +=+=
Substitute numerical values and evaluate K:
( )( ) ( )( )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
Substitute numerical values and evaluate cmvr :
( )( ) ( )( )
( )i
iiv
ˆm/s75.3
kg5kg3
ˆm/s3kg5ˆm/s5kg3cm
=
++
=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
Systems of Particles and Conservation of Momentum
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=
Substitute numerical values and evaluate I:
( ) ( )( )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
Systems of Particles and Conservation of Momentum
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 ===∆
Systems of Particles and Conservation of Momentum
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∆
=
Substitute numerical values and evaluate Fav:
( )( )
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
Substitute numerical values and evaluate Fav:
( )( )N1095.4
m/s9.90kg103
s60min1
mindroplets10
5
5
av
−
−
×=
××
⎟⎟⎠
⎞⎜⎜⎝
⎛×=
∆⎟⎠⎞
⎜⎝⎛
∆= vm
tNF
Systems of Particles and Conservation of Momentum
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.
Systems of Particles and Conservation of Momentum
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 ≈
Substitute to obtain:
( )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)
Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach:
( ) 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
Systems of Particles and Conservation of Momentum
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
−+=
−+=
−=∆
Substitute numerical values and evaluate ∆x:
( )( ) ( )( ) ( ) 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
Systems of Particles and Conservation of Momentum
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)
Use conservation of mechanical energy to set the velocity of recession equal to the negative of the velocity of approach:
( ) 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
⎟⎠⎞
⎜⎝⎛ −=
−=−=
Systems of Particles and Conservation of Momentum
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 ⎟⎠⎞
⎜⎝⎛ −=+=
Substitute in equation (1) to obtain:
%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 +−=+
Systems of Particles and Conservation of Momentum
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:
0=∆+∆ UK or, because Kf = Ui = 0,
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
Systems of Particles and Conservation of Momentum
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:
0=∆+∆ UK or, because Kf = Ui = 0,
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−=
Substitute in equation (2) to obtain: 2
k21
⎟⎠⎞
⎜⎝⎛
+=∆ v
Mmm
gx
µ
Systems of Particles and Conservation of Momentum
565
Substitute numerical values and evaluate ∆x:
( )( ) ( ) 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
Substitute in equation (1) to obtain:
xgv∆
=2
2
kµ
Use equation (2) to eliminate v:
2
2
k
121
21
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+∆=
⎟⎠⎞
⎜⎝⎛
+∆=
Mm
Vxg
MmMV
xgµ
Chapter 8
566
Substitute numerical values and evaluate µk:
( )( ) 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 =
Systems of Particles and Conservation of Momentum
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.
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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 =
Systems of Particles and Conservation of Momentum
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:
0=∆+∆ UK Because Ki = Uf = 0,
0or
0
app2app2
1
if
=−
=−
mghmv
UK
Solve for vapp: appapp 2ghv =
In like manner, show that: recrec 2ghv =
Substitute in the equation for e to obtain:
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:
0=∆+∆ UK Because Ki = Uf = 0,
0or
0
app2app2
1
if
=−
=−
mghmv
UK
Solve for vapp: appapp 2ghv =
In like manner, show that: recrec 2ghv =
Substitute in the equation for e to obtain:
app
rec
app
rec
22
hh
ghgh
e ==
Substitute for app
rec
hh
to obtain: 894.08.0 ==e
Systems of Particles and Conservation of Momentum
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 =+
Systems of Particles and Conservation of Momentum
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
Solve these equations simultaneously to obtain:
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
Systems of Particles and Conservation of Momentum
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
Solve these equations simultaneously to obtain:
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
Systems of Particles and Conservation of Momentum
581
Solve these equations simultaneously to obtain:
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' +=
Systems of Particles and Conservation of Momentum
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
Substitute numerical values and evaluate cmvr :
( )( ) ( )( )
( )i
iiv
ˆm/s3.00
kg1kg3
ˆm/s3kg1ˆm/s5kg3cm
−=
++−
=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
Substitute numerical values and evaluate cmvr :
( )( ) ( )( )
0
kg5kg3
ˆm/s3kg5ˆm/s5kg3cm
=
++−
=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
Systems of Particles and Conservation of Momentum
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 −⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
Substitute numerical values and evaluate vf:
( ) ( )( ) 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 =
Systems of Particles and Conservation of Momentum
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 =
Substitute to obtain:
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
Systems of Particles and Conservation of Momentum
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.
Systems of Particles and Conservation of Momentum
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++
=
Substitute numerical values and evaluate vcm:
( )( ) 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 == ∑
Systems of Particles and Conservation of Momentum
593
Solve for vcm:
2.14
2.12.144cm mm
vmvmv+−
=
Substitute numerical values and evaluate vcm:
( )( )
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++
=
Substitute numerical values and evaluate vcm:
( )( ) ( )( )
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
°
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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
Substitute numerical values and evaluate ∆x:
( )( )( ) 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
Systems of Particles and Conservation of Momentum
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++
=
Substitute numerical values and evaluate vcm:
( )( ) 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
Substitute to obtain:
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 +=+
Systems of Particles and Conservation of Momentum
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 +−=+
Solve for v1f to obtain: i2
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 +−=+
Solve for v2f to obtain: i2
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
Systems of Particles and Conservation of Momentum
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
Substitute to obtain:
( ) mcvmM =−
Solve for v:
mMmcv−
=
Systems of Particles and Conservation of Momentum
605
Substitute numerical values and evaluate v:
( )( )
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.
Systems of Particles and Conservation of Momentum
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
Substitute numerical values and evaluate vcm:
( )( )( )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
Substitute to obtain:
xgmmv ∆⎟⎟
⎠
⎞⎜⎜⎝
⎛+= k
1
21 21 µ
Substitute numerical values and evaluate v1:
( )( )( ) 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
Systems of Particles and Conservation of Momentum
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
Substitute numerical values and evaluate v:
( )( )( )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:
0=∆+∆ UK or, because Kf = Ui = 0,
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 =∆
Substitute to obtain:
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.
Systems of Particles and Conservation of Momentum
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=
Systems of Particles and Conservation of Momentum
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
Take the logarithm of both sides of the equation and solve for 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
Take the logarithm of both sides of the equation and solve for 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
Systems of Particles and Conservation of Momentum
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)
Systems of Particles and Conservation of Momentum
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
Systems of Particles and Conservation of Momentum
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:
0=∆+∆ UK or, because Ki = Uf = 0,
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 +°=
Substitute to obtain: 2
21 vgLgL +⎟
⎠
⎞⎜⎝
⎛=
Solve for v:
⎟⎠⎞
⎜⎝⎛ −=
211gLv
Systems of Particles and Conservation of Momentum
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
Substitute in equation (1) to obtain:
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 −=
Substitute in equation (1) to obtain:
( ) ( )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
Substitute numerical values and evaluate α:
( )
( )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
Substitute numerical values and evaluate 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
Substitute numerical values and evaluate α:
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 =
Substitute to obtain: 2
2MR
τα =
Substitute numerical values and evaluate α:
( )( )( )
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
Substitute in equation (1) to obtain:
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
Substitute numerical values and evaluate I:
( )( )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
+++=
= ∑
Substitute numerical values and evaluate I:
( )( ) ( )( )( )( ) ( )( )
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 ++=
Substitute numerical values and evaluate I:
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
=⎟⎠
⎞⎜⎝
⎛=
++== ∑
Substitute numerical values and evaluate I:
( )( )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=
Substitute to obtain:
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
+++=
= ∑
Substitute numerical values and evaluate I:
( )( ) ( )( )( )( ) ( )( )
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 =
Substitute numerical values and evaluate K:
( )( )
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
Substitute numerical values and evaluate 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:
0=∆+∆ UK or, because Ki = Uf = 0,
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:
0=∆+∆ UK or, because Ki = Uf = 0,
( ) 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:
0=∆+∆ UK or, because Ki = Uf = 0,
( ) 02pulley2
1221 =−++ mghIvMm ω
Substitute for Ipulley and ω to obtain: ( ) ( ) 02
22
21
212
21 =−++ mgh
RvMRvMm
Solve for v:
pMmMmghv
21
2++
=
Substitute numerical values and evaluate v:
( )( )( )( )
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
++=
Substitute numerical values and evaluate a:
( )( )( )
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
Substitute numerical values and evaluate v:
( )( ) ( )( )[ ]( ) 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:
0=∆+∆ UK or, because Ki = Uf = 0,
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
Substitute to obtain: 02
p
2
p21
2w
2
w212
21 =∆−++ hmg
rvI
rvImv
Solve for v:
2p
p2
w
w
2
rI
rIm
hmgv++
∆=
Substitute numerical values and evaluate v:
( )( )( )
( ) ( )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:
0=∆+∆ UK or, because Ki = Uf = 0,
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++−∆
=
Substitute numerical values and evaluate v:
( )( )( )( )
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
Substitute numerical values and evaluate ∆t:
( ) 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
++−
=
Substitute numerical values and evaluate a:
( )( )( )
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
+−
=
Substitute numerical values and evaluate a:
( )( )
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
Substitute numerical values and evaluate 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:
0=∆+∆ UK or, because Kf = Ui = 0,
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 ∆=∆=
Substitute in equation (1) to obtain:
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
Substitute in equation (1) and simplify to obtain:
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:
Substitute in equation (1) and simplify to obtain:
cmcm32sin maamg =−θ
Solve for and evaluate 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
Substitute for I0 and α in equation (4) and solve for fs:
( ) maframrrf 2
1s
221
s =⇒= (5)
Substitute for fs in equation (1) and solve for a:
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:
0=∆+∆ UK or, because Ki = Uf = 0,
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
Substitute in equation (1) to obtain:
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)
Substitute for I0 and α in equation (3) and solve for fs:
( ) s52
s2
52
s maframrrf =⇒=
Substitute for fs in equation (1) and solve for a:
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
Substitute numerical values and evaluate θ :
( )( )( )°=
⎥⎦
⎤⎢⎣
⎡= −
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
++=
++=
Substitute numerical values and evaluate K:
( ) ( )[ ]( )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
+=
+=
+=
Substitute numerical values and evaluate I:
( )( ) ( )( )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 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:
αMRf 21= (4)
Relate the acceleration of the block to the acceleration of the cylinder:
CBBC aaa +=
or, because aCB = −Rα, αRaa −= BC
and CB aaR −=α (5)
(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=
Substitute to obtain:
( ) dmadamK BB21
B 2 == (1)
Apply xx maF =∑ to the block: BmafF =− (2)
Apply xx maF =∑ to the cylinder: CMaf = , (3)
Apply ατ CMCM I=∑ to the
cylinder:
αCMIfR = (4)
Substitute for ICM in equation (4) and solve for f:
αMRf 21= (5)
Relate the acceleration of the block to the acceleration of the cylinder:
CBBC aaa +=
or, because aCB = −Rα,
Chapter 9
692
αRaa −= BC
and CB aaR −=α (6)
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
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
==
Substitute in equation (8) to obtain:
( )
( )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
Solve for and evaluate θ :
°=⎟⎠⎞
⎜⎝⎛= − 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
Substitute numerical values and evaluate v0:
( )( ) 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)
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 ===
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 −
=∆
Substitute numerical values and evaluate ∆t:
( )( )( )( )
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
Substitute numerical values and evaluate v:
( )( )( )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 =
Substitute numerical values and evaluate v1:
( ) 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=∆
Substitute in equation (4) to obtain:
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)
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: tgvtavv ∆+=∆+= k00 µ (5)
Solve for α in equation (4):
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=∆
Substitute in equation (4) to obtain:
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
or, because ω0 = 0, ( )2
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 ==
Substitute numerical values and evaluate α:
( )( )( )
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
or, because ω0 = 0, ( )2
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
Substitute numerical values and evaluate α:
( )( )
22
rad/s18.4m0.82
m/s9.813==α
Substitute numerical values and evaluate a:
( )( ) 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
Substitute to obtain:
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:
0=∆+∆ UK or, because Kf = Ui = 0,
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:
0=∆+∆ UK or, because Ki = Uf = 0,
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 =
Substitute in equation (1) to obtain:
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
Substitute numerical values and evaluate W:
( )( )
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
Substitute numerical values and evaluate α:
( )( )( )
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
Substitute numerical values and evaluate 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
Substitute numerical values and evaluate α:
( )( )( )( )( ) ( )( )
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:
0=∆+∆ UK or, because Ki = Uf = 0,
0irotf,transf, =−+ UKK .
Substitute and solve for ω: ( ) 02221
212
21 =−+ mgrMRmv ω ,
0224122
21 =−+ mgrMRmr ωω ,
and
2224
MRmrmgr+
=ω
Substitute numerical values and evaluate 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:
0=∆+∆ UK or, because Ki = Uf = 0,
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
Substitute numerical values and evaluate d:
( )[ ] ( )[ ] 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
Substitute numerical values and evaluate a:
( )( )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:
0=∆+∆ UK or, because Ki = Uf = 0,
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
Substitute in equation (1) to obtain:
( ) ( ) ( )xmggxgmxF −=−= 1
Evaluate F(0.25 m): ( ) ( ) mgmgF 75.0m25.01m25.0 =−=
Evaluate F(0.5 m): ( ) ( ) mgmgF 5.0m5.01m5.0 =−=
Evaluate F(0.75 m): ( ) ( ) mgmgF 25.0m75.01m75.0 =−=
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 =
Substitute to obtain: L
gMLMgL
23
23
2 ==α
Substitute in equation (1) to obtain: 2
3ga =
Substitute numerical values and evaluate a:
( ) 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
+++=
+++=
Substitute numerical values and evaluate I:
Chapter 9
726
( )( ) ( )( ) ( ) ( ) ( )[ ]2
22412
1212
21
mN492.0
m0.8m0.2kg0.22m1.8kg0.8m0.2kg0.8
⋅=
+++=I
Substitute in equation (1) to obtain:
( )( ) ( )( ) 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
+++=
+++=
Substitute numerical values and evaluate I:
( )( ) ( )( ) ( ) ( ) ( )[ ]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
Substitute numerical values and evaluate 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 =
Substitute to obtain: L
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
Substitute in equation (1) to obtain:
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)
Substitute in equation (2) to obtain:
( ) 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
r are related according to .prL
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
r are related according to .prL
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.
Conservation of Angular Momentum
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
Substitute to obtain: L
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
Conservation of Angular Momentum
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.
Conservation of Angular Momentum
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 ∆+
∆=
∆+∆=∆
Substitute numerical values and evaluate ∆t:
( ) ( ) 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 =
Substitute in equation (2) to obtain:
ω2121 MLL =
Substitute numerical values and evaluate L:
( )( ) ( )/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
Substitute numerical values and evaluate v0:
( )( )m/s7.70
5cosm3m/s9.812 2
0 =°
=v
Conservation of Angular Momentum
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 =
Substitute in equation (1) to obtain: 2
02
520
25sin55sin
Rv
MRMv °
=°
=llω
Substitute numerical values and evaluate ω:
( )( )( )
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
Substitute numerical values and evaluate v0:
( )( )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
Substitute numerical values and evaluate ∆t:
( ) 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.
Substitute to obtain: 0
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
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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
r are related according to .prL
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
Conservation of Angular Momentum
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 +=
Conservation of Angular Momentum
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.
Substitute to obtain:
/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 =
Substitute numerical values and evaluate ω:
( )( )( )( ) 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 ==
Substitute numerical values and evaluate L:
( )( )
/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
Conservation of Angular Momentum
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
Substitute numerical values and evaluate ∆t:
( )( )( )( ) 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 ==
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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
Substitute to obtain: 022
21 =⎟
⎠⎞
⎜⎝⎛ −− ωω
rvmrMR
Solve for ω to obtain: 22 2
2mrMR
mrv+
=ω
Substitute in equation (1):
22
2
f 22
mrMRvmrvv
+−=
Substitute numerical values and evaluate vf:
( )( ) ( )( )( ) ( )( )
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
Conservation of Angular Momentum
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.
Conservation of Angular Momentum
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
=+
=
Conservation of Angular Momentum
767
Substitute in equation (1):
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 =
Substitute in equation (2):
rmrmv
N22
N
2
22 hh
==
Substitute numerical values and evaluate v:
( ) ( )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:
Conservation of Angular Momentum
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:
0=∆+∆ UK or, because Kf = Ui = 0,
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)
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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
Substitute in equation (1) to obtain:
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
Conservation of Angular Momentum
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
or, because Ki = 0, 0iff =−+ UUK
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 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
Conservation of Angular Momentum
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
Substitute in equation (2) to obtain:
( ) ⎟⎠⎞
⎜⎝⎛ +−−
=∆
21
max
1
2cos1
2
mLLMg
LMgE
θ
Substitute numerical values and evaluate ∆E:
( )( )( )
( )( ) ( )( ) ( )( )
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
Conservation of Angular Momentum
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
Substitute numerical values and evaluate ∆E:
( )( ) ( ) ( ) ( )( ) ( )( )[ ]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
or, because Ki = 0, 0iff =−+ UUK
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
or, because Ki = 0, 0iff =−+ UUK
Substitute for Kf, Uf, and Ui to obtain: ( ) 0
2 112
i213
121 =−+ MgLLMgML ω
Solve for ωi:
1i
3Lg
=ω
Conservation of Angular Momentum
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
Substitute numerical values and evaluate v:
( ) ( )( ) ( ) ( )( )
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 ===
Substitute numerical values and evaluate L:
( )
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
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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 =
Substitute numerical values and evaluate K:
( )( ) 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 =
Substitute numerical values and evaluate v:
( )( )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
Conservation of Angular Momentum
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
+=
+==
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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
Substitute in equation (1) and simplify to obtain:
( )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
Substitute numerical values and evaluate 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
Conservation of Angular Momentum
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
Conservation of Angular Momentum
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 =
Substitute in equation (1) to obtain:
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
Substitute in equation (1) to obtain:
( )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
Conservation of Angular Momentum
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 =
Substitute numerical values and evaluate W:
( )( )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
Conservation of Angular Momentum
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 +==+=
Substitute numerical values to obtain:
( )( ) ( )( ) 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
Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity:
∑ == 2frad 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/s0.25m0.4kg0.4
N100f ==ω
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 : ( ) ( )xII
x ff ωω =
Substitute numerical values and simplify to obtain:
( ) ( )( )( ) ( ) 2222
2
kg0.8mkg0.256sJ60.9
kg0.8mkg0.256rad/s25mkg0.384
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
+++=
+++=
+=
Substitute numerical values to obtain:
( ) ( ) ( ) ( )[ ]( ) ( )[ ]
( ) 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
Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity:
∑ == 2frad 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
Conservation of Angular Momentum
801
Substitute in equation (1) to express the work done before the string breaks:
2ff2
1 ωIW =
Substitute numerical values and evaluate W:
( )( )J160
rad/s25mkg0.512 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
kg0.8mkg0.384sJ8.12
kg0.8mkg0.384rad/s25mkg0.512
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−≈∆
Conservation of Angular Momentum
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
Substitute numerical values to obtain:
( )( )
( )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
−−+=
Substitute numerical values and evaluate L:
( ) ( )[ ] ( ) ( )[ ] ( ) ( )tt eeL11 s41.12s41.11
2
21 s/mkg303.0s41.1
2m2.1kg4.8kg2.2
−−
⋅=⎟⎠⎞
⎜⎝⎛+= −
π