OpenStax College Physics
Instructor Solutions Manual
Chapter 10
CHAPTER 10: ROTATIONAL MOTION AND
ANGULAR MOMENTUM
10.1 ANGULAR ACCELERATION
1.
At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its
angular velocity in revolutions per second?
Solution
500 km
1h
1000 m


 138.9 m/s.
1h
3600 s 1 km
v 138.9 m/s
 
 4.630 rad/s.
r
30.0 m
1 rev
  4.630 rad/s. 
 0.737 rev/s
2 rad
2.
Integrated Concepts An ultracentrifuge accelerates from rest to 100,000 rpm in 2.00
2
min. (a) What is its angular acceleration in rad/s ? (b) What is the tangential
acceleration of a point 9.50 cm from the axis of rotation? (c) What is the radial
2
acceleration in m/s and multiples of g of this point at full rpm?
Solution
v
2 rad 1 min

 1.047  10 4 rad/s
1 rev
60 s
4
1.047  10 rad/s
   /t 
 87.27 rad/s 2  87.3 rad/s 2
120 s
5
(a)   10 rev/min 
(b) at  r  (87.27 rad/s 2 )(0.0950 m)  8.29 m/s 2
v 2 r 2 2

 r 2
r
r
 (0.0950 m)(1.047  10 4 rad/s) 2  1.042  10 7 m/s 2  1.04  10 7 m/s 2
(c) a 
1.042  107 m/s 2
 1.06  106 gs
9.80 m/s 2
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
3.
Integrated Concepts You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m
radius, and is turning at 90.0 rpm, and you press a steel axe against it with a radial
force of 20.0 N. (a) Assuming the kinetic coefficient of friction between steel and
stone is 0.20, calculate the angular acceleration of the grindstone. (b) How many
turns will the stone make before coming to rest?
Solution
(a) f   k N  (0.20)(20. 0 N)  4.0 N .   fR  I  1 MR 2 , so that
2
2f
 2(4.0 N)


 0.261 rad/s 2   0.26 rad/s 2
MR (90.0 kg)(0.340 m)
(2 sig. figs due to  k ).
90.0 rev 2 rad 1 min


 9.425 rad/s.  2   02  2 , so that
min
rev
60 s
2
2
2
   0 (0 rad/s)  (9.425 rad/s) 2


2
2(0.261 rad/s 2 )
1 rev
 170.2 rad 
 27.1 rev  27 rev
2 rad
(b)  0 
4.
Unreasonable Results You are told that a basketball player spins the ball with an
angular acceleration of 100 rad/s 2 . (a) What is the ball’s final angular velocity if the
ball starts from rest and the acceleration lasts 2.00 s? (b) What is unreasonable about
the result? (c) Which premises are unreasonable or inconsistent?
Solution
(a) Given:   100 rad/s 2 ,   0  t  t  (100 rad/s 2 )(2.00 s)  200 rad/s
(b) This is approximately 32 revolutions of the ball each second. This is unreasonably
high.
(c) It would be difficult for a human being to spin a ball with such a large angular
acceleration and especially for this length of time. Thus, the angular acceleration
and the time interval are both too large.
10.2 KINEMATICS OF ROTATIONAL MOTION
5.
With the aid of a string, a gyroscope is accelerated from rest to 32 rad/s in 0.40 s. (a)
What is its angular acceleration in rad/s2? (b) How many revolutions does it go
through in the process?
OpenStax College Physics
Solution
Instructor Solutions Manual
(a)   0  t  0  t , so that  

t

Chapter 10
32 rad/s
 80 rad/s 2
0.40 s
1
1 rev
(b)   t 2  0.580 rad/s 2 0.40 s 2  6.4 rad  6.4 rad 
 1.0 rev
2
2π rad
6.
Solution
Suppose a piece of dust finds itself on a CD. If the spin rate of the CD is 500 rpm, and
the piece of dust is 4.3 cm from the center, what is the total distance traveled by the
dust in 3 minutes? (Ignore accelerations due to getting the CD rotating.)
   0  t  0 rad  2 rad/rev  500 rev/min (3.0min)  3000  rad
s  r  (0.043m)(3000rad) = 405 m
7.
A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s 2 .(a) How
long does it take to come to rest? (b) How many revolutions does it make before
stopping?
Solution
(a)   0  t  t 
  0
0  32 rad/s

 45.7 s

 0.700 rad/s 2
1
1
2
(b)    0 t  t 2  32 rad/s 45.7 s    0.700 rad/s 2 45.7 s 
2
2
1 rev
 731 rad 
 116 rev
2π rad
8.
During a very quick stop, a car decelerates at 7.00 m/s 2 . (a) What is the angular
acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement?
(b) How many revolutions do the tires make before coming to rest, given their initial
angular velocity is 95.0 rad/s ? (c) How long does the car take to stop completely? (d)
What distance does the car travel in this time? (e) What was the car’s initial velocity?
(f) Do the values obtained seem reasonable, considering that this stop happens very
quickly?
Solution
(a)  
at  7.00 m/s 2

  25.0 rad/s 2
r
0.280 m
 2  02 0 rad/s 2  95.0 rad/s 2
1 rev

 181 rad 
 28.7 rev
(b)  
2
2
2π rad
2 25.0 rad/s 
OpenStax College Physics
(c) t 
Instructor Solutions Manual
Chapter 10
  0 0  95.0 rad/s

 3.80 s.

 25.0 rad/s 2
(d) x  r  0.280 m 181 rad  50.7 m
(e) v0  r0  0.280 m  95.0 rad/s  26.6 m/s
(f) The car is initial travelling at 59.5 mph (26.6 m/s), so a stopping distance of 166 ft
(50.7 m) seems quite reasonable for a panic stop.
9.
Everyday application: Suppose a yo-yo has a center shaft that has a 0.250 cm radius
and that its string is being pulled. (a) If the string is stationary and the yo-yo
accelerates away from it at a rate of 1.50 m/s 2 , what is the angular acceleration of
the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The
outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point
on its edge?
Solution
(a)  
at
1.50 m/s 2

 600 rad/s 2
r
0.00250 m
(b)    0  t  0 rad/s  600 rad/s 2  0.750 s   450 rad/s
(c) a t  ar  600 rad/s 2  0.0350 m  21.0 m/s
10.3 DYNAMICS OF ROTATIONAL MOTION: ROTATIONAL INERTIA
10.
This problem considers additional aspects of example Calculating the Effect of Mass
Distribution on a Merry-Go-Round. (a) How long does it take the father to give the
merry-go-round and child an angular velocity of 1.50 rad/s? (b) How many
revolutions must he go through to generate this velocity? (c) If he exerts a slowing
force of 300 N at a radius of 1.35 m, how long would it take him to stop them?
Solution
(a)   0  t , t 
  0 1.50 rad/s   0 rad/s 

 0.338 s.

4.44 rad/s 2
2
2
(b)    0  2 
 2  02 1.50 rad/s 2  0 rad/s 2
1 rev


 0.253 rad 
 0.0403 rev
2
2
2π rad
2 4.44 rad/s


OpenStax College Physics
Instructor Solutions Manual
Chapter 10
net  rF

I
I
   0    0 I 0 rad/s  1.50 rad/s  84.38 kg  m 2
t


 0.313 s
1.35 m  300 N 

rF
(c)  


11.
Calculate the moment of inertia of a skater given the following information. (a) The
60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater
with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius,
and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from
the cylinder like rods rotated about their ends.
Solution
MR 2 60.0 kg 0.110 m 

 0.363 kg  m 2
(a) I 
2
2
2
(b)
12.
Solution
MR 2 52.5 kg 0.110 m 
I1 

 0.318 kg  m 2
2
2
2
2  Ml
2
2
I2 
 3.75 kg 0.900 m   2.025 kg  m 2
3
3
I  I 1  I 2  2.343 kg.m 2  2.34 kg  m 2
2
The triceps muscle in the back of the upper arm extends the forearm. This muscle in a
professional boxer exerts a force of 2.00  103 N with an effective perpendicular lever
arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s 2 .
What is the moment of inertia of the boxer’s forearm?
I
net 


rF


0.0300 m2.00 103 N   0.500 kg  m 2
120 rad/s 2
13.
A soccer player extends her lower leg in a kicking motion by exerting a force with the
muscle above the knee in the front of her leg. She produces an angular acceleration of
30.00 rad/s 2 and her lower leg has a moment of inertia of 0.750 kg  m2 . What is the
force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm?
Solution
net   rF  I  F 



I
0.750 kg  m 2 30.0 rad/s 2

 1.18  10 3 N
r
0.0190 m
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
14.
Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone
(a solid disk). (a)What torque is exerted? (b) What is the angular acceleration
assuming negligible opposing friction? (c) What is the angular acceleration if there is
an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Solution
(a)   RF  0.280 m 180 N  50.4 N  m
(b) net   I   
net 
50.4 N  m

 17.1 rad/s 2
I
1 275.0 kg 0.280 m2
(c)  f  rf  0.0150 m  20.0 N  0.300 N  m
 net    -  f  50.4 N  m  0.300 N  m  50.1 N  m

net 
50.1 N  m

 17.0 rad/s 2
2
I
0.575.0 kg 0.280 m 
15.
Consider the 12.0 kg motorcycle wheel shown in Figure 10.38. Assume it to be
approximately an annular ring with an inner radius of 0.280 m and an outer radius of
0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If
the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular
acceleration of the wheel? (b) What is the tangential acceleration of a point on the
outer edge of the tire? (c) How long, starting from rest, does it take to reach an
angular velocity of 80.0 rad/s?
Solution
(a)   rF  0.0500 m  2200 N  110 N  m

1
  ; I  M R12  R22 
I
2
2
2
I  0.512.0 kg 0.280 m   0.330 m   1.124 kg  m 2



110 N  m
 97.9 rad/s 2
2
1.124 kg  m
(b) a t  R2  97.9 rad/s 2  0.330 m  32.3 m/s
(c)    0  t  t ( since  0  0)  t 
 80.0 rad/s

 0.817 s.
 97.9 rad/s 2
OpenStax College Physics
16.
Instructor Solutions Manual
Chapter 10
Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h
by exerting an opposing force at and parallel to the equator. Superman is not
immediately concerned, because he knows Zorch can only exert a force of
4.00 107 N (a little greater than a Saturn V rocket’s thrust). How long must Zorch
push with this force to accomplish his goal? (This period gives Superman time to
devote to other villains.) Explicitly show how you follow the steps found in ProblemSolving Strategy for Rotational Dynamics.
Solution
1 rev 2π rad
1h


 7.272  10 5 rad/s, and
24.0 h 1 rev 3600 s
1 rev 2π rad
1h



 6.233  10 5 rad/s.
28.0 h 1 rev 3600 s
net 
rF
5F



, so that    0  t
2
I
2 Mr / 5 2Mr
   0    0 2Mr
t 


5F
5
2 6.233  10 rad/s  7.272  10 5 rad/s 5.979  10 24 kg 6.376  10 6 m
t
5  4.00  10 7 N
0 






 3.96  1018 s or 1.25  1011 y
17.
An automobile engine can produce 200 N·m of torque. Calculate the angular
acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and
rear wheels of a car, given the following information. The car is suspended so that the
wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius.
The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m
and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of
radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg
drive shaft acts like a rod that has a 3.20-cm radius.
Solution
(a) 200 N  m  
1 lb
1 ft

 147.5 ft  lb  148 ft  lb
4.448 N 0.3048 m
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(b)
Instructor Solutions Manual
Chapter 10
1

2
I 1  2 M 1 R12   15.0 kg 0.180 m   0.486 kg  m 2
2

M 
2
2
I 2  2 2  R22  R' 22  2.00 kg 0.180 m   0.320 m   0.270 kg  m 2
 2 






I 3  2 M 3 R32  210.0 kg 0.330 m   2.178 kg  m 2
2
1
2
M 4 R42  0.514.0 kg 0.0200 m   2.80  10 3 kg  m 2
2
1
2
I 5  M 5 R52  0.530.0 kg 0.0320 m   1.54  10  2 kg  m 2
2
I  I 1  I 2  I 3  I 4  I 5  2.95 kg  m 2
I4 
95%  0.950 / I   0.950200 N  m
18.
Solution

2.95 kg  m 2  64.4 rad/s 2
Starting with the formula for the moment of inertia of a rod rotated around an axis
2
through one end perpendicular to its length ( I  M / 3 ), prove that the moment of
inertia of a rod rotated about an axis through its center perpendicular to its length is
I  M 2 / 12 . You will find the graphics in Figure 10.12 useful in visualizing these
rotations.
2
I end  I center
1
1
1
1
l
 m   I center  I end  ml 2  ml 2  ml 2  ml 2
4
3
4
12
2
19.
Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a
500-N·m torque to slow and then reverse her angular velocity. Her initial angular
velocity is 10.0 rad/s, and her moment of inertia is 0.050 kg  m2 . (a) What time is
required for her to exactly reverse her spin? (b) What is unreasonable about the
result? (c) Which premises are unreasonable or inconsistent?
Solution
Given:   500 N  m;  0  10.0 rad/s; I  0.050 kg  m 2
(a)   I       500 N  m  10,000 rad/s 2
I 0.050 kg  m 2
  0
 10.0 rad/s
   0  t  t 

 0.0010 s  1.0 ms

 10,000 rad/s 2
(b) The time interval is too short.
(c) The moment of inertia is much too small, by one to two orders of magnitude. A
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
torque of 500 N  m is reasonable.
20.
Unreasonable Results An advertisement claims that an 800-kg car is aided by its
20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The
flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the
flywheel must have if 95.0% of its rotational energy is used to get the car up to speed.
(b) What is unreasonable about the result? (c) Which premise is unreasonable or
which premises are inconsistent?
Solution
Given: M car  800 kg; M fly  20.0 kg; Rfly  0.150 m; v  30.0 m/s
1 2 11
2  2
I   M fly Rfly

2
22

1
KE t  kinetic energy of car  M car v 2
2
0.950KE fly  KE t
(a) KE fly 


11
1
2  2
2
0.950  M fly Rfly
  M car v
22
2


M car v 2

2
 (0.950) (1 / 2) M fly Rfly

1/ 2




1/2


(800 kg)(30.0 m/s) 2

2 
 (0.950)(0. 5)(20.0 kg)(0.150 m) 
 1835 rad/s  1.84  10 3 rad/s  292 rev/s  17,500 rpm
(b) This angular velocity is very high for a disk of this size and mass. The radial
acceleration at the edge of the disk is > 50,000 gs.
(c) Flywheel mass and radius should both be much greater, allowing for a lower spin
rate (angular velocity).
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Instructor Solutions Manual
Chapter 10
10.4 ROTATIONAL KINETIC ENERGY: WORK AND ENERGY REVISITED
21.
This problem considers energy and work aspects of Example 10.7—use data from that
example as needed. (a) Calculate the rotational kinetic energy in the merry-go-round
plus child when they have an angular velocity of 20.0 rpm. (b) Using energy
considerations, find the number of revolutions the father will have to push to achieve
this angular velocity starting from rest. (c) Again, using energy considerations,
calculate the force the father must exert to stop the merry-go-round in two
revolutions.
Solution
  20.0 rev/min 
(a) KE rot 
1 min 2π rad

 2.09 rad/s
60 s
1 rev


1 2
2
I  0.5 84.38 kg  m 2 2.09 rad/s   185 J
2
(b) W      W  KE 
185J


375N  m
 1 rev 
 0.493 rad  0.493 rad
  0.0785 rev
 2π rad 
(c) W    rF  F  W r  
185J
 9.81 N
1.50 m4π rad 
22.
What is the final velocity of a hoop that rolls without slipping down a 5.00-m-high hill,
starting from rest?
Solution
PE grav  KE trans  KE rot
mgh 
 v2  1
1 2 1 2 1 2 1
1
mv  I  mv  mR 2  2   mv 2  mv 2
2
2
2
2
2
R  2
mgh  mv 2  gh  v 2



v  gh  9.80 m/s 2 5.00 m 
23.
12
 7.00 m/s
(a) Calculate the rotational kinetic energy of Earth on its axis. (b) What is the
rotational kinetic energy of Earth in its orbit around the Sun?
OpenStax College Physics
Solution
(a)
Instructor Solutions Manual


(b) KE rot 
Solution

1 2 1
I  1.338  10 47 kg  m 2 1.991  10 7 rad/s
2
2

2
 2.65  10 33 J






A baseball pitcher throws the ball in a motion where there is rotation of the forearm
about the elbow joint as well as other movements. If the linear velocity of the ball
relative to the elbow joint is 20.0 m/s at a distance of 0.480 m from the joint and the
moment of inertia of the forearm is 0.500 kg  m2 , what is the rotational kinetic
energy of the forearm?
2
KE rot
Solution


M 2
12.0 kg
0.280 m 2  0.330 m 2  1.124 kg  m 2
R1  R22 
2
2
1
1
2
 I 2  1.124 kg  m 2 120 rad/s   8.09  10 3 J
2
2
I
Solution
26.

Calculate the rotational kinetic energy in the motorcycle wheel (Figure 10.38) if its
angular velocity is 120 rad/s.
KE rot
25.

2
2
I    MRE2  0.4 5.98  10 24 kg 6.38  10 6 m  9.736  10 37 kg  m 2
5
1 rev
1h
2π rad



 7.272  10 5 rad/s
24 h 3600 s 1 rev
2
1 2
KE  I  0.5 9.736  10 37 kg  m 2 7.272  10 5 rad/s  2.57  10 29 J
2

24.
Chapter 10

1
1 v
1
 I 2  I    0.500 kg  m 2
2
2 r
2

2
 20.0 m/s 

  434 J
 0.480 m 
While punting a football, a kicker rotates his leg about the hip joint. The moment of
inertia of the leg is 3.75 kg  m2 and its rotational kinetic energy is 175 J. (a) What is
the angular velocity of the leg? (b) What is the velocity of tip of the punter’s shoe if it
is 1.05 m from the hip joint? (c) Explain how the football can be given a velocity
greater than the tip of the shoe (necessary for a decent kick distance).
(a)  
2KE rot
2175 J 

 9.66 rad/s
I
3.75 kg  m 2
(b) v  r  1.05 m9.66 rad/s   10.1 m/s
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
(c) The tip of the shoe kicks, or collides with, the football. This collision conserves
momentum, not velocity. Since the football’s mass is much smaller than the leg of
the kicker, the football’s speed after the collision is greater than that of the tip of
the shoe.
27.
A bus contains a 1500 kg flywheel (a disk that has a 0.600 m radius) and has a total
mass of 10,000 kg. (a) Calculate the angular velocity the flywheel must have to
contain enough energy to take the bus from rest to a speed of 20.0 m/s, assuming
90.0% of the rotational kinetic energy can be transformed into translational energy.
(b) How high a hill can the bus climb with this stored energy and still have a speed
of 3.00 m/s at the top of the hill? Explicitly show how you follow the steps in the
Problem-Solving Strategy for Rotational Energy.
Solution
90.0%KE rot  KE trans so that (0.900) 1 I 2  1 Mv 2 .
2
2
Let M be the mass of the bus and m be the mass of the flywheel.
Mv 2
, and since the flywheel is a disc :
0.900
1
1
Mv 2
I  mr 2  mr 2 2 
2
2
0.900
(a) I 2 
 210,000 kg 20.0 m/s 2 
2Mv 2



0.900mr 2  0.9001500 kg 0.600 m 2 
12
 128 rad/s
1
1
(b) KE trans  PE' grav  KE' trans so that (0.900) I 2  Mgh' Mv ' 2 ,
2
2
1
1
or (0.900) mr 2 2  Mgh' Mv ' 2 , so that
4
2
h' 
0.900 mr 2 2 Mv ' 2 0.900mr 2 2  2Mv ' 2


4Mg
2g
4Mg
0.9001500 kg 0.600 m 128.3 rad/s 
h' 

 210,000 kg 3.00 m/s 
 19.9 m
410,000 kg  9.80 m/s 2
2

28.
2
2

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. Treating the
ball as a spherical shell, calculate the vertical height it reaches. (b) Repeat the
calculation for the same ball if it slides up the hill without rolling.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 10
KEi  PEf or, more specifically, KE rot  PE grav
2
(a) 1 I 2  mgh   1  2 mr 2  2  mgh   1 r 2  v
 

   2
2
 2  3

3  r
1 v
 1  8.00 m/s 
h   
 2.18 m
2
 3  g  3  9.80 m/s

  gh 

2
2
(b) Here the initial KE is KE trans .
8.00 m/s   3.27 m
1 2
mv  mgh  h  v 2 2 g 
2
2 9.80 m/s 2
2


29.
While exercising in a fitness center, a man lies face down on a bench and lifts a
weight with one lower leg by contacting the muscles in the back of the upper leg. (a)
Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance
of 28.0 cm from the knee joint, the moment of inertia of the lower leg is
0.900 kg  m2 , the muscle force is 1500 N, and its effective perpendicular lever arm is
3.00 cm. (b) How much work is done if the leg rotates through an angle of 20.0 with
a constant force exerted by the muscle?
Solution
(a)  

net  RFM  rmg

I
I leg  mr 2
0.0300 m 1500 N   0.280 m 10.0 kg 9.80 m/s 2   10.4 rad/s 2
2
0.900 kg  m 2  10.0 kg 0.280 m 
(b) net W 


1 2 1 2 1
1
I  I 0  I  2   02  I 2 
2
2
2
2



2π rad 

2
 0.900 kg  m 2  10.0 kg 0.280 m  10.4 rad/s 2  20.0 
  6.11 J
360 

30.
To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her
biceps muscle to flex the lower arm through an angle of 60.0 . (a) What is the
angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a
moment of inertia of 0.250 kg  m2 , and the muscle force is 750 N at an effective
perpendicular lever arm of 2.00 cm? (b) How much work does she do?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 10
(a) Assuming her arm starts extended vertically downward, we can calculate the
initial angular acceleration.
Given :   60 


2π rad
 1.047 rad,
360
FM r
net 

so that
2
I  m w rw I  m w rw2
750 N 0.0200 m 
2
0.250 kg  m 2  2.00 kg 0.240 m 
 41.07 rad/s 2  41.1 rad/s 2
(b) net W  net    FM r  750 N0.0200 m1.047 rad   15.7 J
31.
Solution
Consider two cylinders that start down identical inclines from rest except that one is
frictionless. Thus one cylinder rolls without slipping, while the other slides
frictionlessly without rolling. They both travel a short distance at the bottom and then
start up another incline. (a) Show that they both reach the same height on the other
incline, and that this height is equal to their original height. (b) Find the ratio of the
time the rolling cylinder takes to reach the height on the second incline to the time
the sliding cylinder takes to reach the height on the second incline. (c) Explain why the
time for the rolling motion is greater than that for the sliding motion.
(a)
d
x
h
h
2
1
x  r
Assuming no energy is lost due to friction and other dissipative forces, then by
conservation of energy: Ei  Ef or mghi  mghf , so that hi  hf
Since both cylinders start from the same hi , they both reach the same hf .
(b) Assuming that the time spent travelling across the bottom is negligible, then the
ratio of the times for the entire trip will just be the ratio of the times it takes to
get down the first incline.
For the sliding cylinder:
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
1 2
mv  v1  2 gh
2
d
d
t1 

v1
2 gh
mgh 
For the rolling cylinder:
mgh 
t2 
1 2 1 2 1 2 1  1 2  v2 3 2
mv  I  mv   mr  2  mv  v 2 
2
2
2
22
4
r
d

v2
t2

t1
d
4 gh / 3

4
gh
3
3
d

, so that
2
2 gh
3
 1.22
2
(c) The rolling cylinder takes longer because some of the initial potential energy is
converted into rotational energy, which therefore decreases the linear speed (by
conservation of energy).
32.
What is the moment of inertia of an object that rolls without slipping down a 2.00-mhigh incline starting from rest, and has a final velocity of 6.00 m/s? Express the
moment of inertia as a multiple of MR 2 , where M is the mass of the object and R is
its radius.
Solution
PE grav  KE' trans  KE' rot
Mgh 
1
1
Mv 2  I 2
2
2
2 Mgh  (1 / 2) Mv 2

  2M  gh  1 v

2
2
 2 MR gh  (1 / 2v )


2 
2
2 
v2
(9.80 m/s 2 )(2.00 m)  (0.5)(6.00 m/s) 2
I  2 MR 2
(6.00 m/s) 2
I

2


 1.60 m/s 2   4 
 2 MR 
   MR 2 or 0.0889 MR 2
2 
 36.0 m/s   45 
2
33.
Suppose a 200-kg motorcycle has two wheels like the one described in Problem 10.15
and is heading toward a hill at a speed of 30.0 m/s. (a) How high can it coast up the
hill, if you neglect friction? (b) How much energy is lost to friction if the motorcycle
only gains an altitude of 35.0 m before coming to rest?
OpenStax College Physics
Solution
(a)
Instructor Solutions Manual
Chapter 10
PE f  KE i
PE' grav  KE trans  KE rot
Mgh 
h
 v2 
M
1
I 
1
 1
Mv 2  2 I 2   Mv 2  I  2   v 2   2 
2
2
 2
 R2 
 2 R2 
v2  M
I
  2
Mg  2 R2





(30.0 m/s) 2
1.124 kg  m 2


100
kg

2 
0.330 m 2
 (200 kg)(9.80 m/s ) 

  50.66 m  50.7 m

(b) E  Mgh  (200 kg)(9.80 m/s 2 )(15.66 m)  3.07  10 4 J
34.
In softball, the pitcher throws with the arm fully extended (straight at the elbow). In a
fast pitch the ball leaves the hand with a speed of 139 km/h. (a) Find the rotational
kinetic energy of the pitcher’s arm given its moment of inertia is 0.720 kg  m 2 and the
ball leaves the hand at a distance of 0.600 m from the pivot at the shoulder. (b) What
force did the muscles exert to cause the arm to rotate if their effective perpendicular
lever arm is 4.00 cm and the ball is 0.156 kg?
Solution
(a) KE  1 I 2  1 I  v 
rot
2
2 r
2
2

1
2  25.0 m/s 
(0.720 kg  m 2  0.156 kg 0.600 m  )
  674 J
2
 0.600 m 
(b) Assume the ball starts horizontally behind the pitcher and is released at the
bottom of the swing. Approximate the arm to be a thin rod, so that:
Ml 2
3I 3(0.720 kg  m 2 )
I
M  2 
 6.00 kg, then
3
l
(0.600 m) 2
net W  net ( )  Fr  KE rot  (mgh  MgH ), so that
l
KE rot  (ml  M ) g
2
F
r
674 J  (0.650 kg)(0.600 m)  (6.00kg)(0 .300 m) (9.80 m/s 2 )
F
(0.0400 m)(  /2 rad)
 1.04 10 4 N
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
10.5 ANGULAR MOMENTUM AND ITS CONSERVATION
36.
(a) Calculate the angular momentum of the Earth in its orbit around the Sun. (b)
Compare this angular momentum with the angular momentum of Earth on its axis.
Solution
(a)
I  MR 2 ,
Lorb  I  MR 2
 2 rad
 (5.979  10 24 kg)(1.496  1011 m) 2 
7
 3.16  10
(b)

40
2
  2.66  10 kg  m /s
s
2MR 2
,
5
2
2

 2 rad 
  MR 2   (5.979 10 24 kg)(6.376 10 6 m) 2 

5
5

 24  3600 s 
 7.07 10 33 kg  m 2 /s
I
Lorb
The angular momentum of the earth in its orbit around the sun is 3.76 106 times
larger than the angular momentum of the earth around its axis.
37.
(a) What is the angular momentum of the Moon in its orbit around Earth? (b) How
does this angular momentum compare with the angular momentum of the Moon
on its axis? Remember that the Moon keeps one side toward Earth at all times. (c)
Discuss whether the values found in parts (a) and (b) seem consistent with the fact
that tidal effects with Earth have caused the Moon to rotate with one side always
facing Earth.
Solution
2
(a) Lorb  I  Mr 
 2 rad 
 (7.35  10 22 kg)(3.84  10 8 m) 2 
 2.89  10 34 kg  m 2 /s
6 
 2.36  10 s 
2
2
(b) Lorb   MR 
5

 2 rad 
 0.4(7.35  10 22 kg)(1.74  10 6 m) 2 
 2.37  10 29 kg  m 2 /s
6 
 2.36  10 s 
The orbital angular momentum is 1.22  10 5 times larger than the rotational
angular momentum for the Moon.
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
(c) Since the tidal effects with Earth have caused the Moon to rotate with one side
always facing Earth, its period of rotation is the same as its period of orbit.
Therefore, the rotation angular momentum is smaller by a factor of
approximately (102 ) 2 , which is consistent with the results of parts (a) and (b).
38.
Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s.
What angular momentum is given to the engine if the handle of the crank is 0.300 m
from the pivot and the force is exerted to create maximum torque the entire time?
Solution
L  t  rFt. Since L0  0, L  L, and
L  (0.300 m)(300 N)(0.250 s)  22.5 kg  m 2 /s
39.
Solution
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is
rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a
22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.
L  L'  I  I '  '
I '  I  I c , where I c  mr 2
' 
I
I


I'
I  Ic
 120 kg 
(1 / 2) Mr 2
 M 
(0.500 rev/s)
 
  
2
(1 / 2)( M  2m)r
164
kg
 M  2m 


 '  0.366 rev/s  2.30 rad/s

40.
Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and
33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-goround, what is the new angular velocity in rpm?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 10
I
I '  '  I   '   
 I'
I  I M  I c , where I c  m1  m2  m3 r 2
M

M

I    m1  m2  m3 r 2 and I '    m1  m3 r 2
 2

 2

( M / 2  m1  m2  m3 )
' 

( M / 2  m1  m3 )

(50.0 kg  22.0 kg  28.0 kg  33.0 kg)
(20.0 rpm)  25.3 rpm
(50.0 kg  22.0 kg  33.0 kg)
41.
(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his
moment of inertia is 0.400 kg  m2 . (b) He reduces his rate of spin (his angular velocity)
by extending his arms and increasing his moment of inertia. Find the value of his
moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead
he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What
average torque was exerted if this takes 15.0 s?
Solution
2 rad 

2
(a) L  I  (0.400 kg  m 2 ) 6.00 rev/s 
  15.1 kg  m /s
rev 

(b) L  L' , so that I  I '  '  I ' 
I (0.400 kg  m 2 )(6.00 rev/s)

 1.92 kg  m 2
'
1.25 rev/s
   0 

(c)   I  I 
 t 
 3.00 rev/s  6.00 rev/s 2 rad 
 (0.400 kg  m 2 )

   0.503 N  m
15.0 s
rev 

10.6 COLLISIONS OF EXTENDED BODIES IN TWO DIMENSIONS
43.
Solution
Repeat Example 10.15 in which the disk strikes and adheres to the stick 0.100 m from
the nail.
(a) I '  mr 2 
MR 2
 (0.0500 kg)(0.100 m) 2  (0.667 kg)(1.20 m) 2  0.961 kg  m 2
3
OpenStax College Physics
' 
Instructor Solutions Manual
Chapter 10
mvr (0.0500 kg)(30.0 m/s)(0.100 m)

 0.156 rad/s
I'
0.961 kg  m 2
(b) KE  22.5 J
1
1
KE'  I '  2'  (0.961 kg  m 2 )(0.156 rad/s) 2  1.17  10 2 J
2
2
(c) p  1.50 kg  m/s
M
M 

R '  mr 
R  ' , so that :
2
2 

p '  (0.0500 kg)(0.100 m)  (1.00 kg)(1.20 m)   (0.156 rad/s)  0.188 kg  m/s
p '  mr '
44.
Repeat Example 10.15 in which the disk originally spins clockwise at 1000 rpm and
has a radius of 1.50 cm.
Solution
(a) As in Example 10.15, angular momentum is conserved, but here the initial angular
momentum includes the spin of the disk. Thus, Lafter  L1  L2 , where L1 is the
angular momentum of the incoming disk and L2 is the angular momentum of the
disk spinning on its axis.
L1  mvR  (0.0500 kg)(30.0 m/s)(1.20 m)  1.80 kg  m 2 /s
1

 2 rad  1 min
L2  I   mr 2 (1000 rev/min) 

2

 1 rev  60 s
 0.5(0.0500 kg)(1.50  10  2 m) 2 (104.72 rad/s)



 5.89  10  4 kg  m 2 /s
Thus, Lafter  1.800589 kg  m 2 /s  I' ω'
We know I ' from Example 10.15 to be 1.032 kg  m2 ; thus
' 
Lafter 1.800589 kg  m 2 /s

 1.74 rad/s
I'
1.032 kg  m 2
(b) The total initial kinetic energy is
1
1
1
11

KE linear  KE rotational  mv 2  I 2  mv 2   mr 2  2
2
2
2
22

1
1
 0.0500 kg (30.0 m/s) 2  (0.0500 kg)(1.50  10  2 m) 2  (104.72 rad/s) 2  22.5 J
2
4
The final kinetic energy is
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
KE'  (1 / 2) I '  ' 2  0.5(1.032 kg  m 2 )(1.7448 rad/s) 2  1.57 J
(c) The linear momentum before the collision is the same as in Example 10.15:
p  mv  (0.0500 kg)(30.0 m/s)  1.50 kg  m/s
After the collision (as shown in Example 10.15):
M

p'   m 
r '  (1.050 kg)(1.20 m)(1.7448 rad/s)  2.20 kg  m/s
2 

One implication of this problem is that, although the disk is spinning rapidly on its
axis, its angular momentum and energy have little effect because they are
relatively small.
45.
Twin skaters approach one another as shown in Figure 10.39 and lock hands. (a)
Calculate their final angular velocity, given each had an initial speed of 2.50 m/s
relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass
located 0.800 m from their locked hands. You may approximate their moments of
inertia to be that of point masses at this radius. (b) Compare the initial kinetic
energy and final kinetic energy.
Solution
(a) L  2mvr  I , so that
2mvr 2mvr v 2.50 m/s


 
 3.125 rad/s  3.13 rad/s
I
2mr 2 r 0.800 m
1
(b) KE  2 mv 2   mv 2  (70.0 kg)(2.50 m/s) 2  437.5 J  438 J
2

2
KE' 
1
1
v
I '  ' 2  (2mr 2 )   mv 2  437.5 J  438 J.
2
2
r
They are the same.
46.
Suppose a 0.250-kg ball is thrown at 15.0 m/s to a motionless person standing on ice
who catches it with an outstretched arm as shown in Figure 10.40. (a) Calculate the
final linear velocity of the person, given his mass is 70.0 kg. (b) What is his angular
velocity if each arm is 5.00 kg? You may treat his arms as uniform rods (each has a
length of 0.900 m) and the rest of his body as a uniform cylinder of radius 0.180 m.
Neglect the effect of the ball on his center of mass so that his center of mass remains
in his geometrical center. (c) Compare the initial and final total kinetic energies.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 10
(a) Linear momentum is conserved:
mv  (m  M )vcm 
vcm 
mv
(0.250 kg)(15.0 m/s)

 5.338  10  2 m/s  5.34  10  2 m/s
mM
0.250 kg  70.0 kg
ml 2 (5.00 kg)(0.900 m) 2

 1.35 kg  m 2
3
3
1
I body  Mr 2  0.5(60.0 kg)(0.180 m) 2  0.972 kg  m 2
2
I  2(1.35 kg  m 2 )  0.972 kg  m 2  3.672 kg  m 2
(b) I arm 
Angular momentum is conserved in the CM frame:
v'  15.0 m/s  5.338  10 2 m/s  14.95 m/s
mv' r
I
(0.250 kg)(14.95 m/s)(0.900 m)

 0.916 rad/s
3.672 kg  m 2
L  mv' r  I   
1 2
mv  0.5(0.250 kg)(15.0 m/s) 2  28.125 J  28.1 J
2
1
1
2
KE'  (m  M )vcm
 I 2
2
2
1
1
 (70.25 kg)(5.338  10  2 m/s) 2  (3.672 kg  m 2 )(0.916 rad/s) 2  1.64 J
2
2
(c) KE 
The final KE is much less than the initial KE due to inelastic nature of the
collision, i.e. catching the ball.
47.
Repeat Example 10.15 in which the stick is free to have translational motion as well
as rotational motion.
Solution
Linear momentum is conserved:
mv  (m  M )vcm  vcm 
mv
 (0.0500 kg)(30.0 m/s)/(2.05 0 kg)  0.7317 m/s
mM
(a) Angular momentum is conserved in the CM frame:
OpenStax College Physics
Instructor Solutions Manual
Chapter 10
v'  30 m/s  0.7317 m/s  29.268 m/s
' 
mv' r (0.0500 kg)(29.268 m/s)(1.20 m)

 1.7016 rad/s  1.70 rad/s
I'
1.032 kg  m 2
(b) KE  22.5 J
1
1
2
KE'  (m  M )v cm
 I ' '2
2
2
KE'  0.5(2.050 kg)(0.7317 m/s) 2  0.5(1.032 kg  m 2 )(1.7016 rad/s) 2  2.04 J
(c) p  p'  mv  (0.0500 kg)(30.0 m/s)  1.50 kg  m/s
10.7 GYROSCOPIC EFFECTS: VECTOR ASPECTS OF ANGULAR MOMENTUM
48.
Integrated Concepts The axis of Earth makes a 23.5 angle with a direction
perpendicular to the plane of Earth’s orbit. As shown in Figure 10.41, this axis
precesses, making one complete rotation in 25,780 y. (a) Calculate the change in
angular momentum in half this time. (b) What is the average torque producing this
change in angular momentum? (c) If this torque were created by a single force (it is
not) acting at the most effective point on the equator, what would its magnitude be?
Solution
(a) L  2L sin 23.5
The value of L can be obtained from Example 10.11:
L  7.07  10 33 kg  m 2 /s .


L  2(7.07  10 kg  m 2 /s)sin23.5   5.64  10 33 kg  m 2 /s
33
(b) 25,780 y  25,780 y 
3.16  10 7 s
 8.15  1011 s
1y
1
of this time  4.07  1011 s
2
L 5.64  10 33 kg  m 2 /s


 1.38  10 22 N  m
t
4.07  1011 s

(c)   RF  F 

R

1.38 10 22 N  m
 2.17 1015 N
6.38 10 6 m
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