ROTATIONAL MOTION
10
EXERCISES
Section 10.1 Angular Velocity and Acceleration
14.
The angular speed is   /t.
(a)  E  1 rev/1 d  2 /86, 400 s  7.27  105 s 1 . (b)  min  1 rev/1 h  2 /3600 s  1.75 103 s 1 .
(c) hr  1 rev/12 h  121 min  1.45 104 s1. (d)   300 rev/min  300  2 /60 s  31.4 s 1.
(Note: Radians are a dimensionless angular measure, i.e., pure numbers; therefore angular speed can be expressed
in units of inverse seconds.)
15.
INTERPRET We are asked to compute the linear speed at some location on Earth. The problem involves the
rotational motion of the Earth.
DEVELOP We first calculate the angular speed of the Earth using Equation 10.1:
E 
 1 rev
2 rad


 7.27 105 s 1
t
1d
86, 400 s
The linear speed can then be computed using Equation 10.3: v   r.
EVALUATE (a) On the equator,
v   E RE  (7.27 105 s 1 )(6.37 106 m)  463 m/s
(b) At latitude , r  RE cos  so v  E r  (463 m/s) cos.
ASSESS The angle   0 corresponds to the Equator. So the result found in (b) agrees with (a). In addition, if we
take   90, then we are at the poles, and the linear speed is zero there.
16.
(a) (720 rev/min)(2 /rev)(min/60s)  24 s 1  75.4 s 1. (b) (50/h)( /180)(h/3600 s)  2.42  104 s 1 .
(c) (1000 rev/s)(2/rev)  2000 s 1 < 6.28 103 s 1. (d) (1 rev/y) < 2 /( 107 s)  2 107 s 1.
(See note in solution to Exercise 14. The approximate value for 1 y used in part (d) is often handy for estimates,
and is fairly accurate; see Chapter 1, Problem 20.)
17.
INTERPRET The problem asks about the linear speed of the straight wood saw, but it’s equivalent to finding the
linear speed of the circular saw.
DEVELOP We first convert the angular speed to rad/s:

 3500 rev 2 (3500) rad


 367 rad/s
t
1 min
60 s
The linear speed can then be computed using Equation 10.3: v   r.
EVALUATE The radius of the circular saw is r  12.5 cm  0.125 m. Therefore, its linear speed is
v   r  (367 rad/s)(0.125 m)  45.8 m/s
ASSESS The linear speed of the saw is more than 100 mi/h!
18.
From Equation 10.4 (before the limit is taken), av  /t  (500  200) rpm/(74  60 s) 
6.76 102 rpm/s  7.08 103 s 2 . (Recall that 1 rpm  2/60 s.)
19.
INTERPRET In this problem we are asked to find the angular speed and number of revolutions of a turbine, given
its angular acceleration. The key to this type of rotational problem is to identify the analogous situation for linear
motion. The analogies are summarized in Table 10.1.
10.1
10.2
Chapter 10
DEVELOP Given a constant angular acceleration , the angular velocity and angular position at a later time t can
be found using Equations 10.7 and 10.8:
  0  t
1
2
(a) The initial and final angular velocities are 0  0 and
  0   0 t   t 2
EVALUATE
  3600 rpm 
3600 rev 2 (3600) rad

 377 rad/s
1 min
60 s
Therefore, the amount of time it takes to reach this angular speed is
t
  0 377 rad/s  0

 725 s  12.1 min

0.52 rad/s 2
(b) Using Equation 10.8, we find the number of turns made during this time interval to be
1
2
1
2
1
2
  0  0t  t 2  t 2  (0.52 rad/s2 )(725 s)2  1.37 105 rad  2.17 104 rev
ASSESS The turbine turns very fast. After 12.1 min, it has reached an angular speed of 377 rad/s, or 60 rev/s!
20.
The kinematics of motion with constant angular acceleration are summarized in Table 10.1, where the initial
angular velocity at t  0 is zero if the merry-go-round starts at rest. (a) Equation 10.8 gives the angular
displacement:     0  12  t 2  12 (0.01 s2 )(14 s)2  (0.98 radians)  (1 rev/2 )  0.156 rev  56.1.
(b) Equations 10.6 and 10.9 give the average angular speed: av  12  f  12  t  12 (0.01 s2)(14 s)  0.07 s1 
0.668 rpm. (Note that the average angular speed is always equal to /t , in this case 12  t 2/t , but that
Equations 10.6 to 10.9 apply only to constant .)
Section 10.2 Torque
21.
INTERPRET In this problem we are asked to find the torque produced by the frictional force about the wheel’s axis.
DEVELOP The torque produced by a force is given by Equation 10.10:
  rF sin 
where r  r sin  is the perpendicular distance between the rotation axis and the line of action of the force F. The
frictional force acts tangent to the circumference of the wheel and hence perpendicular (  90) to the radius at
the point of contact.
EVALUATE Using Equation 10.10, we find the torque to be
  rf  (0.5 m)(320 N)  160 N  m
opposite to the direction of rotation.
ASSESS Since frictional force opposes motion, the torque it produces tends to slow down the rotation.
22.
If the force is applied perpendicular to the door, the radial distance should be r  /F  110 N  m/90 N  1.22 m
from the axis. (See Equation 10.10 with   90.)
23.
INTERPRET In this problem we want to find the force required to produce a specific torque.
DEVELOP The torque produced by a force is given by Equation 10.10:
  rF sin   r F  rF

where r  r sin  is the perpendicular distance (lever arm) between the rotation axis and the line of action of the
force F. Alternatively, one can think of F  F sin  as the effective force.
EVALUATE
Using Equation 10.10, the magnitude of the applied force may be obtained as F 
are (a)
F
and
(b)

r sin 

35.0 N  m
 146 N
(0.24 m) sin 90

r sin 
. The forces
Rotational Motion
F

r sin 

10.3
35.0 N  m
 155 N.
(0.24 m) sin 110
ASSESS To produce a specific torque most effectively, the applied force should be at the right angle to r , the
position vector from the axis of rotation to the point where the force is applied. This would yield an effective
force F  F sin   F .
24.
The angle between the minute hand (for the clock in upright position) and the weight of the mouse (vertically
downward) is 120 at ten past the hour, so the magnitude of the torque exerted is   rmg sin   (17 cm)(55 g)
(9.8 m/s 2 )sin 120  7.94  102 N  m. The direction of the torque is clockwise.
25.
INTERPRET We find the torque on a wheel, given that it has a slight off-center mass at a given radius and angle
to the horizontal. We use the definition of torque,   rF sin .
DEVELOP We start with a sketch of the system, shown in the figure below. From this, we see that the angle
between the force and the position vector R is   90  . The valve stem is located a distance R  0.32 m from
the center, and has mass m  0.025 kg.
EVALUATE   rF sin   R(mg ) cos   0.072 Nm
ASSESS We see from our equation that the torque would be zero if the angle  were 90°.
Section 10.3 Rotional Inertia and the Analog of Newton’s Law
26.
With distances and axes as shown in the sketch, Equation 10.12 for rotational inertia gives
2
2
2
(a) I a  mL  mL  0  0  2mL ; and (b) Ib  4  m( 12 L)2  mL2.
27.
INTERPRET We want to find the moment of inertia of a shaft that has a shape of a solid cylinder.
DEVELOP The rotational inertia of a solid cylinder or disk about its axis is I  12 MR2 (see Table 10.2).
EVALUATE The rotational inertia of the shaft is
1
1
MR 2  (6.8 103 kg)(0.85 m/2)2  6.14 102 kg  m 2
2
2
The numerical value is reasonable, given its mass and radius, and the units (kg  m 2 ) are correct.
I
ASSESS
28.
(a) I  I cyl  2Idisk  M cyl R2  2( 12 M disk R2)  (65  22)  103 kg  (7.1  102 m)2  4.39  104 kg  m2 .
(b)   I   (4.39  104 kg  m 2 )(3.4 s 2 )  1.49  10 3 N  m.
10.4
29.
Chapter 10
INTERPRET In this problem we are asked to find the minimum mass of a wheel, given its diameter and
rotational inertia.
DEVELOP Any part of the wheel has a distance from the center less than or equal to the maximum radius.
Therefore, using Equation 10.12, we obtain the following inequality:
2
I   mi ri 2    mi  rmax
EVALUATE
(a) The above equation implies that
M   mi 
I
2
max
r
Therefore, the minimum mass is
M min 
I
2
max
r

7.8 kg  m 2
 36.9 kg
(0.92 m/2) 2
(b) If not all the mass of the wheel is concentrated at the rim, the total mass is greater than this minimum.
ASSESS To have the same rotational inertia I, we can have some of the mass of wheel concentrated near the axis
of rotation. Their contribution to I would be small since I   mi ri 2.
30.
INTERPRET We are asked to find the rotational inertia I of three point masses located at the corners of an
equilateral triangle. We find I for the axis through the center perpendicular to the plane, and for the axis along the
line through one corner and the midpoint of the opposite side, as shown in the figure below. We use the equation
for rotational inertia of a collection of point masses: I   mi ri 2 .
DEVELOP
In each part of this problem, we will need the distances of the particles from the appropriate axis.
In part (a), we see from the figure below that the distance of each particle from the center of the triangle is
ra 
L
2
cos(30 )

. In part (b), we see that the distance of two of the masses from the axis is rb  L2 , and the third
L
2 cos(30)
is at a distance of 0.
The masses are each m  1.2 kg, and L  0.88 m.
EVALUATE
2
 
 
L
2
(a) I   mi ri 2  3  mra2   3  m 
   0.93 kg m
2
cos(30

)
 
 
  L 2 
   
(b) I   mi ri 2  2  mrb2   m(0)  2 m     0.46 kg m2
2
ASSESS As we would expect, the rotational inertia is less for the second axis.
31.
INTERPRET By assuming Earth to be a solid sphere with uniform mass distribution, we want to estimate its
rotational inertia, and the torque needed to change the length of the day by one second every century.
DEVELOP From Table 10.2, the rotational inertia of a solid sphere of radius R and mass M is
I
2
MR 2
5
Once I is known, the torque needed to slow down the rotation can be found by using Equation 10.11:   I.
EVALUATE (a) For a uniform solid sphere, and an axis through the center,
Rotational Motion
IE 
10.5
2
2
M E RE2  (5.97  1024 kg)(6.37 106 m) 2  9.69 1037 kg  m 2
5
5
(b) The angular speed of rotation of Earth is  
change by 1s per century, i.e.,
2
T
, where the period is T  1 d  86, 400 s. If the period were to
dT
1s

 3.16  1010
dt 100  3.16  107 s
this would correspond to an angular acceleration of

d d  2 
2 dT
   2
dt dt  T 
T dt
Therefore, a torque of magnitude
| |  I | |  
2 I dT 2 (9.69 1037 kg  m 2 )
10
19

(3.16  10 )  2.58 10 N  m
2
2
T dt
(86, 400 s)
would be required.
ASSESS The torque in (b) is actually generated by tidal friction between the Moon and the Earth. Note that the
Earth has a core of denser material, so its actual rotational inertia is less than that obtained in (a).
32.
(a) For a uniform sphere about an axis through its center, I  52 MR . Since the density is   M /( 34  R3), the radius
can be eliminated to yield I  52 M (3M /4)2 / 3  0.4M 5/ 3 (0.75/)2 / 3  0.4[(1.8)(1.99 1030 kg)]5/ 3 
[0.75/ (1018 kg/m3 )]2 / 3  1.29  1038 kg  m2. (b) The torque responsible for the spin-down rate (i.e., the angular
deceleration) is   I   (1.29  1038 kg  m 2 )(5  105 s 2 )   6.45  1033 N  m, the negative value indicating
a direction opposite to the angular velocity.
33.
INTERPRET We are asked about the rotational inertia of a Frisbee, given its mass distribution, and the torque
required to generate the rotation.
DEVELOP The rotational inertia (about a perpendicular axis through the center) of the Frisbee is the sum of that
for a disk ( I d  12 M d R2 ) and a ring ( I r  M r R 2 ):
2
I  Id  Ir 
1
M d R2  M r R2
2
Next, using Equation 10.11,   I  , the torque needed to rotate the Frisbee can be calculated.
EVALUATE Since half of the mass is spread uniformly in a disk, and the other half concentrated in the rim, we
have M d  M r  12 (108 g)  54 g. Thus, the rotational inertia is
1
1
1
M d R 2  M r R 2  ( M d  2M r ) R 2  [0.054 kg  2(0.054 kg)](0.12 m) 2
2
2
2
3
2
 1.17  10 kg  m
I
(b) Using Equation 10.9, the angular acceleration is found to be

 2   02 (550 rpm) 2  0 (57.6 rad/s) 2


 1056 rad/s 2
2(  0 )
2(0.25 rev)
 rad
Thus, the torque required is
  I  (1.17 103 kg  m 2 )(1056 rad/s 2 )  1.24 N  m
ASSESS The numerical values all look reasonable, and the units are correct:  in rad/s2, and torque  in N  m.
34.
The frictional torque decelerates the flywheel, so Equations 10.7 and 10.11 yield  f  0   0   t , and
   f k Rs  I  , or t  0 /  0 I / f k Rs , where Rs is the radius of the shaft. Since the radius of the flywheel, Rw,
is much larger than Rs, the rotational inertia of the wheel is approximately that of a solid disk, 12 MRw2 ; therefore
t  (360 rpm)( s1/30 rpm)  12 (7.7 104 kg)(2.4 m)2 /(34 kN)  (0.205 m)  1.20 103 s  20.0 min.
(Note: The exact I for the flywheel is 12 M ( Rs2  Rw2 ), which is just 0.7% different from 12 MRw2 for the given radii).
Section 10.4 Rotional Energy
10.6
35.
Chapter 10
INTERPRET The problem asks about the rotational kinetic energy of a rotating circular saw. In addition, we also
want to find the power required for the saw to start from rest and reach a given angular speed.
DEVELOP The rotational kinetic energy of the saw can be found by using Equation 10.18:
K
1 2
I
2
where I  12 MR2 for a disk. The required power is simply equal to the rate of the work done on the saw.
EVALUATE
(a) With   3500 rpm 
K
2 (3500) rad
60 s
 367 rad/s, the rotational kinetic energy is
1 2 11
1

I   MR2   2  (0.85 kg)(0.125 m)2 (367 rad/s)2  446 J
2
2 2
4

(b) The power required is Pav  Wt 
K
t

446 J
3.2 s
 139 W.
ASSESS To check the answer, we present an alternative approach to computing K. In this problem, the angular
acceleration is

  0
t

367 rad/s
 115 rad/s 2
3.2
and the angular displacement is (Using Equation 10.9)
  0 
 2   02 (367 rad/s) 2

 586 rad
2
2(115 rad/s 2 )
With these quantities, the rotational energy may be calculated as
1
1
MR 2  (0.85 kg)(0.125 m)2 (115 rad/s 2 )(586 rad)  446 J
2
2
K  W     I  
which is same as before.
36.
If kinetic energy of rotation were extracted at a constant rate (power P  1013 W) for a time t, until the period of
rotation increased by 1 minute, then 12 I E (02   2f )  Pt , where 0  2 /T0, and  f  2 /(T0  1 min). Since
T0  1 d >>1 min, we can approximate the difference of angular velocities squared as

1 2
1  2  
2
0   2f      


2
2  T0   T0  1 min 

2
2
 1  2 2   1min  2   2 2  1 min 
    1  1 
 <   

T0    T0   T0 
 2  T0   

and moreover, take IE from Exercise 33(a).
Then t < (2 ) 2 I E (1 min)/PT03  (2 ) 2 (9.69 1037 kg  m 2 )(60 s)/(86, 400 s)3 (1013 W)  3.56 1013 s
or about 1.13 My.
37.
INTERPRET The kinetic energy of the baseball consists of two parts: the kinetic energy of the center of
mass, K cm , and the rotational kinetic energy, K rot . We want to find the fraction of total kinetic energy that’s K rot .
DEVELOP The total kinetic energy has center-of-mass energy and internal rotational energy associated with spin
about the center of mass (see Equation 10.20):
K tot  K cm  K rot 
EVALUATE
1
1
Mv 2  I cm 2
2
2
For a solid sphere, I cm  2 MR 2/5. Therefore, the fraction of the total kinetic energy that’s rotational is
K rot
I cm 2
(2 MR 2 /5) 2
2 R 2 2



K tot Mv 2  I cm 2 Mv 2  (2MR 2 /5) 2 5v 2  2 R 2 2

2(0.037 m/s) 2 (42 rad/s) 2
5(33 m/s) 2  2(0.037 m/s) 2 (42 rad/s) 2
 8.86  104  0.0886%
ASSESS Rotational kinetic energy constitutes a very small fraction of the total kinetic energy. This is
reasonable because the linear speed at a point on the surface of the baseball due to rotation is only
 R  (42 rad/s)(0.037 m)  1.55 m/s.
Rotational Motion
38.
10.7
The rotational kinetic energy of the Earth is Krot  12 I 2, where I is Earth’s moment of inertia and  is its
rotational speed. For  we take the total time to rotate 2 radians as 1 day or 24 hours:


2 rad
2 rad


 7.27 105 rad/s
3600 s
t 24 h  1 h
86400 s
We treat Earth as a uniform solid sphere with mass M  5.97 1024 kg and radius R  6.37  106 m to compute I:
I
2
2
MR 2  (5.97  1024 kg)(6.37 106 m) 2  9.69 1037 kg  m 2
5
5
Then Krot  12 I 2  12 (9.69 1037 kg  m2 )(7.27 105 rad/s)2  2.6 1029 J.
39.
INTERPRET We need to find the energy stored in a massive flywheel. We are given the size and mass of the
flywheel, so we can calculate the rotational inertia. Given the speed, we can calculate the kinetic energy. We also
need to find the power output of a generator if the speed of the flywheel changes a given amount in a certain time.
DEVELOP The energy stored in the flywheel is K  12 I 2 . We will need to convert the angular speed in rpm to
rad/s, and calculate the rotational inertia of the flywheel disk using I  12 mR2 . The power output in the second part
is P  tK .
The mass of the flywheel is m  7.7  104 kg, the radius is R  2.4 m, and the initial rotation rate is 360 rpm. In
the second part, the rotation rate goes to 300 rpm in t  3 s.
EVALUATE
(a) I  12 mR2  222 103 kg m2 .
rotation 1 min 2 radians


 37.7 rad/s
min
60 s 1 rotation
K  1 I 2  1.58 108 J
2
8
(b)  2  300 rpm  31.4 rad/s  K 2  1.09 10 J
  360
K  4.82  107 J  P 
K
 16.1 MW
t
ASSESS This is a good way of generating enormous power pulses.
Section 10.5 Rolling Motion
40.
INTERPRET We find the translational and rotational kinetic energy of a rolling sphere. We are given the
translational speed and the mass of the sphere.
DEVELOP Translational kinetic energy is Kt  12 mv2 . Rotational kinetic energy is Kr  12 I 2 . The rotational
inertia of a sphere is I  52 MR2, and v  r  5.0 m/s. The mass of the sphere is M  2.4 kg.
EVALUATE
1 2
mv  30 J.
2
2
1
12
Mv 2
 v 
(b) K r  I 2   MR 2    
 12 J.
2
25
5
 R 
(a) Kt 
ASSESS Note that the rotational kinetic energy is 2/5 the translational kinetic energy, in this case. Does the 2/5
look familiar? How would the two answers be related if instead of a solid sphere it was a hollow sphere?
41.
INTERPRET What fraction of a disk’s kinetic energy is rotational if it’s rolling? We use the relationship
between  and v.
DEVELOP We use the total kinetic energy K  12 Mv2  12 I 2 , with v   R and I  12 MR2 . The rotational fraction
K
is f  Krot .
EVALUATE
f 
1
2
1
2
I 2
Mv  I
2
1
2
2


1
2
MR 2   2
M ( R)   MR  
2
1
2
2
2

1
2
1
1
2

1
3
ASSESS This is consistent with what we noted in the previous problem: the rotational inertia is ( 12 )MR2, so the
rotational kinetic energy is ( 12 ) the translational kinetic energy when it rolls without slipping.
10.8
42.
Chapter 10
INTERPRET
DEVELOP
From the fraction of kinetic energy due to rotation, we determine whether a ball is solid or hollow.
The fraction of energy that is rotational is f 
K rot
K total
40
 100
 52 . We will find this fraction for the hollow
sphere, I  32 MR2 , and for the solid sphere, I  52 MR2 , and see which gives an answer closer to 52 .
EVALUATE For the solid sphere,
f 
1
2
1
2
I 2
Mv  12 I 
2
2


MR 2   2
2
5
M ( R)   MR  
2
2
5
2
2

2
5
1  52
 2.
7
For the hollow sphere,
f 
1
2
1
2
I 2
Mv  12 I
2
2


2
3
MR 2   2
M ( R )   32 MR  
2
2
2

2
3
1  32
2
5
ASSESS The sphere is hollow.
PROBLEMS
43.
INTERPRET The problem is about the rotational motion of the wheel. The key to this type of rotational problem is
to identify the analogous situation for linear motion. The analogies are summarized in Table 10.1.
DEVELOP For constant angular acceleration, the final angular speed is given by Equation 10.9:
 2   02  2 (  0 )
Similarly, to find the time it takes for the wheel to make 2 turns, we solve Equation 10.8:
1
2
(a) The angular acceleration is   18 rpm/s  1080 rev/min 2 . Using Equation 10.7, we find the final
  0  0t  at 2
EVALUATE
angular speed to be
  02  2 (  0 )  0  2(1080 rev/min 2 )(2 rev)  65.7 rpm
(b) Equation 10.8 gives   0  0  12  t 2 or
t
2


2(2 rev)
 0.061 min  3.65 s
1080 rev/min 2
ASSESS Another way to answer (b) is to use Equation 10.8:
  0  t
 t
  0
65.7 rpm

 0.061 min  3.65 s

1080 rev/min 2
44.
At constant angular acceleration, Equations 10.1 and 10.6 give
  12 (0   f )t  12 (3600  1800)(rev/60 s)(1.4 s)  63.0 rev.
45.
INTERPRET The problem is about the angular velocity (rotation rate) of the CD, with the linear speed specified at
some point on the CD.
DEVELOP Equation 10.3 gives the relation between linear speed and angular speed,
v
r
where r is the distance from the center of rotation. With v  130 cm/s being a constant, the angular speed can be

found once r is specified.
EVALUATE (a) With r  6.0 cm, we have

v 130 cm/s

 21.7 rad/s  207 rpm
r
6.0 cm

v 130 cm/s

 34.7 rad/s  331 rpm
r 3.75 cm
for a point on the CD’s outer edge.
(b) With r  3.75 cm, we have
Rotational Motion
10.9
ASSESS Note that radians are a dimensionless angular measure, i.e., pure numbers; therefore angular speed can be
expressed in units of inverse seconds.
46.
(a)  av   /t  (5500  1200) rpm/2.7 s  1590 rpm/s  167 s 2   , if the angular acceleration is assumed to
be constant (see Equation 10.4). (b) at   r  (167 s 2 )(3.5 cm/2)  2.92 m/s 2 (see Equation 10.5).
(c)   12 (0   f ) t  12 (1200  5500) rpm(2.7/60) min  151 rev (see Equation 10.6).
47.
INTERPRET We are asked to find the angular deceleration of the machinery, given its initial and final angular
speeds and the angular displacement.
DEVELOP The three quantities, the initial and final angular speeds and the angular displacement, are related by
Equation 10.9:
 2   02  2 (  0 )
This is the equation we shall use to solve for a.
EVALUATE Using Equation 10.9, we find the angular deceleration to be

 2   02 (440 rpm) 2  (680 rpm) 2

 747 rev/min 2  1.30 rad/s 2
2(  0 )
2(180 rev)
ASSESS A negative acceleration is a positive deceleration. The amount of time it takes for this deceleration is
  0  t
 t
  0 440 rpm  680 rpm

 0.32 min  19 s

747 rev/min 2
During this time interval, the angular displacement is
1
2
1
2
  0  0t  at 2  0  (680 rpm)(0.32 min)  (747rpm/min 2 )(0.32 min)2
 180 rev
This is indeed the value given in the problem statement.
48.
(b) From Equation 10.6,   0  t  12 (0  )t , or 0  2(  0 )/t , since   0 when the blade is stopped.
Thus, 0  2(1200 rev)/40 s  60 rev/s  3600 rpm. (a) From Equation 10.7,   (  0 )/t  0 /t 
3600 rpm/40 s  90 rpm/s. ( is negative for a deceleration; rpm/s is an appropriate practical unit.)
49.
INTERPRET This problem is about the angular motion of the pulley. We want to find the applied torque that
keeps the pulley from rotating.
DEVELOP If the pulley and string are not moving (no rotation and no slipping) the net torque on the pulley is zero,
and the tensions in the string on either side are equal to the weights tied on either end. The tensions are perpendicular
to the radii to each point of application. Therefore, the torques due to the tensions have a magnitude
TR  mgR,
but are in opposite directions. Thus, the condition for no rotation is
0   app  m1 gR  m2 gR
where we chose positive torques in the direction of the applied torque, which is opposite to the torque produced by the
greater mass.
EVALUATE Using the values given in the problem statement, we find
 app  (m1  m2) gR  (0.47 kg  0.22 kg)(9.8 m/s2)(0.06 m)  0.147 N  m
ASSESS The torque  app equals the torque which would be produced by balancing the pulley with 250 g added to
the side with lesser mass.
10.10
50.
Chapter 10
(a) The rotational inertia of each of the rods perpendicular to the axis is 121 mL2, and that of each rod parallel to the
axis is m( 12 L)2  14 mL2. The total is I a  [2(121 )  2( 14 )]mL2  32 mL2. (b) The rotational inertia of a uniform thin rod,
about an axis through one end, making an angle  with the length of the rod, is
I   y dm  
2
L cos 
0
Mdx
M tan 2  x3
( x tan  )

L cos 
L cos  3
L cos 
2
0
1
 ML2 sin 2 
3
Since there are four rods, all making the same angle,   45, with the axis, the
total Ib is 4( 13 mL2 sin 2 45)  32 mL2. (c) The rotational inertia of one rod (from the parallel axis theorem)
is I1  121 mL2  m( 12 L)2  13 mL2 , and there are four rods, so Ic  34 mL2.
51.
INTERPRET We are asked to find the rotational inertia of a thick ring with inner and outer radii R1 and R2. The
mass distribution is continuous, so we need to do an integral.
DEVELOP For a thick ring, the ring-shaped mass elements used in Example 10.7 have mass
dm =  dA 
M
  R  R12 
2
2
2 rdr
where   M /A is the mass density (units: kg/m2). Note that the ring only extends in radius from R1 to R2 . The
rotational inertia can then be obtained by integrating over
R2
I   r 2 dm
R1
EVALUATE Upon carrying out the integration, the rotational inertia about an axis perpendicular to the ring and
through its center is
R2
I   r dm  M 
R1
2
R2
R1
2 r 3 dr
 R  R
2
2
2
1


M  R24  R14 
2R  R
2
2
2
1


M 2
 R1  R22 
2
ASSESS To see that the result makes sense, let’s consider the following limits: (i) R1  0. In this case, we have a
disk with radius R2 and I  12 MR22. (ii) R1  R2. In this limit, we have a thin ring with I  MR22 .
52.
Table 10.2 lists the rotational inertia of a flat plate about a central axis (in the plane of the plate and through its
center of mass) as I cm  121 Ma 2, when the axis is parallel to a side of length b. The CM is a perpendicular
distance a2 from this side, so the parallel-axis theorem gives Iside b  121 Ma2  M ( 12 a)2  13 Ma2.
53.
INTERPRET We are asked about the rotational inertia of a propeller, treating it as a uniform thin rod. In addition,
we want to find the time it takes to change its angular speed, given the torque.
DEVELOP The rotational inertia of a thin rod of length L is (from Table 10.2)
1
I  ML2
3
Rotational Motion
10.11
where the axis of the rotation passes through one of the endpoints of the rod. For part (b), we note that the average
torque is related to the average rate of change of angular speed (angular acceleration) as
  I  I
EVALUATE

t
(a) The rotational inertia of one blade is 13 ML2 (see Table 10.2). The propeller has three such blades, so
1 2 
I  3  ML   ML2  (10 kg)(1.25 m)2  15.6 kg  m2
3

(b) The engine torque is the only one considered. Therefore, with 0  1400 rpm  147 rad/s and
  1900 rpm  199 rad/s, we have
t 
I 


(15.6 kg  m 2 )(199 rad/s  147 rad/s)
 0.303 s
2700 N  m
ASSESS Our result shows that t is inversely proportional to the applied torque. Increasing the torque reduces the
time required to speed up the propeller.
54.
Divide the plate into thin strips parallel to the central, or x axis. For a uniform plate dm/M  bdy/ab, so
M
I   y dm 
a
2
M y3
y
dy

a /2
a 3
a /2
a /2

2
 a /2
Ma 2
.
12
Alternatively, divide into thin strips perpendicular to the axis. The rotational inertia of each strip is dI  121 (dm)a 2,
where dm/M  adx/ab. Therefore,
I   dI  
55.
a2
Ma 2
dm 
12
12

b
0
dx Ma 2

b
12
INTERPRET We are asked to find the time it takes for the space station to start from rest and reach a certain
angular speed, with a given thrust.
DEVELOP Suppose that any difference in the distance to the axis from the center to the rockets, or any part of the
ring, is negligible compared to the radius of the ring, R  11 m. The angular acceleration (about the axis perpendicular
to the ring and through its center) of the ring is


I

2 FR 2 F

MR 2 MR
Starting from rest (0  0), it would take time t   for the ring to reach a final angular speed  . The angular
displacement during this time interval may be obtained by using Equation 10.8:
1
2
  0  0t  at 2
EVALUATE
Since g   2 R specifies the desired rate of rotation, we find
t

g  MR 
9.8 m/s 2 (5 105 kg)(11 m)

 2.60 104 s  7.21 h



R  2F 
11 m
2(100 N)
10.12
Chapter 10
(b) From either Equation 10.8, we find the number of revolutions to be
1
2
1
2
  0   t 2   t 
1 g
1 9.8 m/s 2
t
(2.60 104 s)  1.23 10 4 rad  1953 rev
2 R
2
11 m
ASSESS Since t 1/F, a larger thrust will enable the space station to reach the desirable spin rate more rapidly. On
the other hand, larger mass M and larger R will give a greater rotational inertia, and hence more time would be
required to attain the desired spin rate.
56.
Choose positive torques in the direction the drum turns, and positive forces on the weight upward. In the absence
of friction, the net torque on the drum is the (vector) sum of the torque applied by the motor and the torque due
to the tension in the rope,  net   app  RT  I , since the rope is perpendicular to the radius of the drum (see
Example 10.9 and Figure 10.19). The net force on the weight is Fnet  T  mg  ma, and if the rope does not slip,
the upward acceleration of the weight equals the tangential acceleration of the drum, a  R. Therefore,
 app  I  RT  ( 12 MR2 )(a = R)  R(mg  ma)  (1.2 m/2)[ 12 (51 kg)(1.1 m/s2 )  (38 kg)(9.8  1.1) m/s2 ] 
265 N  m.
57.
INTERPRET We are asked to find the coefficient of friction between block and slope, given the acceleration of the
block. The rotational motion of the solid drum that is connected to the block by a string must be considered.
DEVELOP The equations of motion for the block and drum are
mg sin   k n  T  ma
n  mg cos   0
TR  I 
2
1
where I  2 MR , and a  R (non-slip string). These equations allow us to determine k .
EVALUATE The drum’s equation gives the tension,
I ( MR 2/2)(a /R) 1

 Ma
R
R
2
so that k can be found from the block’s equations,
T
k 
mg sin   ma  Ma/2 (2.4 kg)(9.8 m/s 2 ) sin 30  (2.4 kg  0.425 kg)(1.6 m/s 2 )

 0.36
mg cos 
(2.4 kg)(9.8 m/s 2 ) cos 30
ASSESS To see that our expression for k makes sense, let’s check some limits: (i) If a  0, then k  tan . This
is precisely the equation we obtained in Chapter 5 (see Example 5.10). (ii) M  0 and k  0. The situation
corresponds to a block of mass m sliding down a frictionless slope with acceleration a  g sin .
58.
Since the wheel was spinning freely, the frictional torque of the wrench is the only one acting about its axis of
rotation, and  f  I , where I  MR 2 is the rotational inertia of the wheel (whose mass is concentrated at its rim).
The frictional force is, of course, tangential, so  f   f k R   k nR, where n is the normal force, and we chose
Rotational Motion
10.13
the positive sense of rotation in the direction of the wheel’s motion. Thus,    k nR/MR 2   k n /MR, and
Equation 10.7 gives the final angular speed,  f   0  ( k n /MR)t  230 rpm  [(0.46)(2.7 N)/(1.9 kg)
(0.33 m)](3.1 s)(60 rpm  2 s 1)  230 rpm  58.6 rpm  171 rpm.
59.
INTERPRET In this problem we want to find the angular speed of the potter’s wheel after some force has been
exerted. The force produces a torque that causes the wheel to rotate.
DEVELOP The work-energy theorem for constant torque (Equation 10.19) gives
W     K 
1 2
I
2
since the wheel starts from rest. The equation allows us to determine the angular velocity  .
EVALUATE Since the force acting on the wheel is circumferential, the resulting torque is   FR. In addition, the
rotational inertia of a disk is I  12 MR2, and   18 rev  4 rad. Thus, we have
2 
2  2 FR 4 F 


I
MR
MR 2 /2
or

4F 
4(75 N)( rad/4)

 2.09 rad/s
MR
(120 kg)(0.45 m)
ASSESS The greater the force exerted on the wheel, the larger the angular speed. On the other hand, larger M and
R result in a larger rotational inertia, and smaller  if the same force is applied.
60.
The only significant energy changes in this problem are the change in the anchor’s gravitational potential
energy, mg y  (5000 N)(16 m)  80 kJ, and the changes in the drum’s and anchor’s kinetic energies,
1
2
I 2 and 12 mv2 respectively. (This is because we are neglecting the mass of the roller and cable, and friction.)
If the cable doesn’t slip over the drum, the drum’s tangential speed equals the anchor’s speed, so R  v. For a
hollow cylinder, I  MR 2 , so the total kinetic energy change is 12 (MR2 ) 2  12 m( R)2  12 R2 2 (M  m), which
equals the 80 kJ change in potential energy. Therefore   [2(80 kJ)/(380 kg  5000 N/(9.8 m/s 2 ))]1/ 2 /
(1.1 m)  12.2 s 1.
61.
INTERPRET The problem is about the rotational motion of a hollow basketball rolling down an incline. We want
to know its speed after it has traveled a certain distance.
DEVELOP The ball rolls through a vertical height h  L sin , where L is the distance along the incline, which
makes an angle  with the horizontal. For a hollow ball, the rotational inertia is I cm  32 MR2 (see Table 10.2).
Next, conservation of mechanical energy, applied to the hollow basketball starting from rest and rolling without
slipping down an incline, yields
1
1
2
I cm 2  Mvcm
2
2
where vcm   R. The equation allows us to determine vcm.
EVALUATE Upon substituting I cm  32 MR2, we obtain
Mgh 
2
12
1
5
 v 
2
2
Mgh   MR 2  cm   Mvcm
 Mvcm
23
2
6
 R 
or
6 gh
6 gL sin 
6(9.8 m/s)(8.4 m) sin 30


 7.03 m/s
5
5
5
In general, for an object having a rotational inertia of I cm  cMR 2 , one can go through a similar
vcm 
ASSESS
derivation and show that the center-of-mass velocity is given by
vcm 
2 gh
1 c
10.14
Chapter 10
If we ignore the rolling motion of the basketball, the velocity then becomes vcm  2gh , which is what we expect
from considering only the linear motion.
62.
If the ball rolls without slipping and we define potential energy to be zero at the incline’s base, then the purely
2
kinetic initial energy is 12 Icm 2  12 Mvcm
. The final energy is all potential at the highest point on the incline with
2
a value of Mgh. Setting the two equal gives: Mgh  12 I cm 2  12 Mvcm
. With I cm  32 MR2 and   vcm /R, this
becomes
Mgh 
1
1
12
1
5

2
2
2
I cm 2  Mvcm
  MR 2  (vcm /R)2  Mvcm
 Mvcm
2
2
2 3
2
6

With vcm  3.7 m/s, h  1.16 m.
63.
INTERPRET The kinetic energy of the wheel consists of two parts: the kinetic energy of the center of
mass, K cm , and the rotational kinetic energy, K rot . We want to find out how changing the moment of inertia and
mass of the wheel affects the total kinetic energy.
DEVELOP The total kinetic energy of the wheel has center-of-mass energy and internal rotational energy
associated with spin about the center of mass (see Equation 10.20):
K tot  K cm  K rot 
1
1
2
Mvcm
 I cm 2
2
2
With the condition for rolling without slipping, v   R, the total kinetic energy can be rewritten as
2
K tot  Kcm  K rot 
I 
1
1
1
v 
2
2 
Mvcm
 I cm  cm   Mvcm
1  cm 2 

2
2
2
MR 
 R 

The initial condition is MRcm2  0.40  40%. After the redesign,
I

I cm
0.9 I cm
I

 1.125 cm 2  (1.125)(0.40)  0.45
2
2

MR
(0.8M ) R
MR
EVALUATE
The fractional decrease in kinetic energy is
 /M R 2 )
M (1  I cm
K  K
K
0.8M (1  0.45)
 1
 1
 1
 0.171  17.1%
2
K
K
M (1  0.40)
M (1  I cm /MR )
1
ASSESS Initially, Kcm accounts for 1.4
 71.4% of the total kinetic energy, while K rot account for the remaining
0.4
  0.8M , the translational kinetic energy decreases by 20%, while the
After
the
redesign,

28.6%.
M

M
1.4
  0.9I cm ) Therefore, the total kinetic energy is now
rotational kinetic energy goes down by 10% ( I cm  I cm
(0.8)
1
0.4
 (0.9)
 0.829  82.9%
1.4
1.4
of the original. This is a 17.1% decrease.
64.
Take gravitational potential energy to be zero at the bottom. The ball starts on the left side with purely potential
2
energy of magnitude Mgh. At the bottom, the ball’s energy is purely kinetic with magnitude 12 Icm 2  12 Mvcm
.
Going up the right side, the ball’s translational kinetic energy will be converted to potential energy, but because the
right side is frictionless, the rotational kinetic energy will be unchanged.
Energy conservation between the top and bottom gives:
Mgh 
1
1
12
1
7

2
2
2
I cm 2  Mvcm
  MR2  (vcm /R)2  Mvcm
 Mvcm
2
2
25
2
10

2
2
So at the bottom, vcm
 107 gh, and the translational kinetic energy is 12 Mvcm
 75 Mgh. This translational kinetic
2
energy is converted entirely to potential energy at the height h . Therefore we have Mgh  12 Mvcm
 75 Mgh, or
5
5
h  7 h. At the height of h  7 h on the right side, the ball still has the same rotational kinetic energy of 72 Mgh
that it had at the bottom.
Rotational Motion
65.
10.15
INTERPRET In this problem we want to find the new rotational inertia of a circular disk after a hole has been
drilled off its center.
DEVELOP Equation 10.17 shows that the rotational inertia of an object is the sum of the rotational inertias of the
pieces, so
I disk  I hole  I remainder
The hint expresses this fact as I remainder  I disk  Ihole. Here I disk is the rotational inertia of the whole disk about an
axis perpendicular to the disk and through its center, which is 12 MR2 (see Example 10.7). On the other hand, I hole is
the rotational inertia of the material removed to form the hole, which the parallel axis theorem gives as
I hole  M hole h 2  I cm
Here, h  R/4 is the distance of the hole’s CM from the axis of the disk, and I cm  12 M hole ( R/4)2 is the rotational
inertia of the hole material about a parallel axis through its CM. With these equations, we can readily
determine I remainder .
EVALUATE Since the planar mass density of the disk (assumed to be uniform) is   M / R 2, the mass of the hole
material is
M hole   Ahole 
M
M
 ( R/4)2 
2
16
R
Therefore, the rotational inertia of the hole is
2
I hole  M hole h 2  I cm 
2
M R
1 M R
3
MR 2
  
  
16  4 
2 16  4 
512
and
I remainder  I disk  I hole 
1
3
253
MR 2 
MR 2 
MR 2  0.494MR 2
2
512
512
ASSESS If the hole drilled were concentric with the disk, we would have (h  0)
2
  I cm 
I hole
1 M R
1
MR 2
  
2 16  4 
512
and

 
I remainder
 I disk  I hole
1
1
255
MR 2 
MR 2 
MR 2  0.498MR 2
2
512
512
The same result is obtained if we use the formula 12 M ( R12  R22 ) derived in Problem 51, with
M    R 2   ( R/4) 2  (15/16) R 2  (15/16) M , R1  R and R2  R/4.
66.
The situation is similar to Example 10.9. (a) The equation of motion (Newton’s second law) of the falling mass is
mg  T  ma (positive “a” downward), so T  m( g  a)  (50 kg)(9.8  3.7)m/s 2  305 N. (b) The equation of
motion of the rotating drum (rotational analog of Newton’s second law) is   RT  I  ( 12 MR2 )(a/R),
so M  2T/a  2(350 N)/(3.7 m/s2 )  165 kg (since   RT and a   R as explained in Example 10.9).
67.
INTERPRET In this problem we want to find out how high up the hill the motorist can go with the total
mechanical energy the system has while traveling on the flat road.
DEVELOP If possible losses are neglected, the mechanical energy of the motorcycle and rider is conserved as it
coasts uphill, so the total kinetic energy at the bottom equals the gain in potential energy at the highest point,
Ktrans  Krot  M tot gh
10.16
Chapter 10
The translation kinetic energy of the cycle and rider (including the wheels) and the rotational kinetic energy of the
wheels (about their CM) are, assuming rolling without slipping,
K trans 
1
M tot v 2 ,
2
1

v
K rot  2  I  2   I  
2

R
2
These expressions can be combined to solve for h.
EVALUATE Substituting the second equation into the first, and using v  85 km/h  23.6 m/s, we find the
maximum vertical height reached to be
h
v2 
2 I  (23.6 m/s) 2 
2(2.1 kg  m 2 ) 

1

1 


  32.9 m
2 g  M tot R 2  2(9.8 m/s 2 )  (395 kg)(0.26 m) 2 
ASSESS If the rolling motion is ignored, the result would be h 
v2
2g
, which is what we expect from considering
only the linear motion.
68.
The CM of the marble travels in a circle of radius R  r inside the loop-the-loop, so at the top, mg  N  mv 2 /
( R  r ). To remain in contact with the track, n  0, or v 2  g ( R  r ). If we assume that energy is conserved between
points A (the start) and B (the top of the loop), and use K A  0 and KB  (1  52 ) 12 mvB2 , we find U A  K A 
mg (h  r )  U B  KB  mg (2R  r )  107 mvB2 , or h  2(R  r )  107 (vB2 /g )  2( R  r )  107 ( R  r )  2.7( R  r ).
69.
INTERPRET In this problem we are given a disk with non-uniform mass density, and asked to find its total mass
and rotational inertia.
DEVELOP As mass elements, choose thin rings of width dr and radius r (as in Example 10.7) so that
20 w 2
 r
dm  (r )dV   0  2 rw dr 
r dr
R
R


The total mass is M  0R dm and the rotational inertia about the disk axis is I  0R r 2 dm.
EVALUATE (a) The disk’s total mass is
R
M   dm 
0
20 w R 2
20 wR 2
r dr 

0
R
3
(b) The disk’s rotational inertia about a perpendicular axis through its center is
R
I   r 2 dm 
0
20 w R 4
20 wR 4 3  20 wR 2
r
dr

 
R 0
5
5
3
 2 3
2
 R  MR
5

ASSESS Our result for I is intermediary between a disk of uniform density and a ring, i.e., 12 MR2  I  MR2, if
expressed in terms of the total mass M, but is less than a disk of uniform density 0 , i.e., I  12 0 R4 w, since 0 is
the maximum density.
70.
Since the axis is frictionless, the conservation of mechanical energy relates  min (when the hole is at the bottom)
to  max (when the hole is at the
2
2
top), K bot  U bot  K top  U top , or 12 I min
 12 Imax
 (Ubot  U top ). I  (253/512) MR 2 is the rotational inertia of
the disk about a perpendicular axis through its center (see Problem 65), and M is the mass of the original complete
disk in Problem 65 (i.e., without the hole). The mass of the material removed to form the hole is M hole  M /16, and
the mass of the remaining wheel is M rem  15M /16. Therefore, the CM of the wheel, at a
distance ycm from the center, opposite to the center of the hole, is given by (15/16) Mycm  (1/16) M ( R/4)  0,
or ycm  R/60. The difference in height of the CM between positions with the hole at the bottom and at the top
Rotational Motion
10.17
is ycm  R/30, so the change in potential energy is U bot  U top  M rem g ycm  (15/16) Mg ( R/30)  MgR/32.
2
2
2
2
  max
 2U /I   max
 ( MgR/16)/(253MR 2/512)   max
 32 g/253 R.
Finally,  min
71.
INTERPRET We are asked to show that the rotational inertia of a planar object around an axis perpendicular to the
plane of the object is equal to the sum of the rotational inertias around two perpendicular axes within the plane of
the object. We will use the integral form of rotational inertia, since this is a continuous distribution.
DEVELOP The rotational inertia is I   r 2 dm. We will set up our coordinate system such that the two rotational
axes within the plane of the object are the x and y coordinate axes.
The rotational inertia around the x axis is I x   r 2 dm   y 2 dm. The rotational inertia around the y
axis is I y   r 2 dm   x 2 dm. The sum of the two is I x  I y   x2 dm   y 2 dm   ( x2  y 2 )dm, but ( x 2  y 2 ) is just
the distance r from the perpendicular z axis, so I x  I y   r 2 dm  I z .
ASSESS We have proven what was requested.
EVALUATE
72.
INTERPRET We use the perpendicular-axis theorem (see Problem 10.71) to check the rotational inertia for a flat
rectangular plate around a perpendicular axis given in Table 10.2. We also use it to find the rotational inertia of a
thin disk around an axis along the diameter.
DEVELOP The rotational inertia of a flat plate around a central axis in the plane of the plate is given in Table 10.2
as I  121 Ma 2. The rotational inertia of a flat plate around a central axis perpendicular to the plate is given in the
same table as I  121 M (a2  b2 ).
We will see that the first, taken along two perpendicular axes, adds to give the second. In the second part, we will
work backward from I  12 MR2 with the perpendicular-axis theorem to find the rotational inertia of a thin disk
around an axis in the plane of the disk.
EVALUATE (a) Along the axis shown for the first case, I1  121 Ma2. Along a second, perpendicular, axis in the
plane of the plate, I 2  121 Mb2. The sum of these two is I  121 M (a2  b2 ), which is the equation given for the plate
around an axis perpendicular to the plate.
(b) The rotational inertia of a disk through the center, I  12 MR2, is equal to the sum of the rotational inertias
around two perpendicular diameters. Since the rotational inertia around any diameter has the same value,
Icenter  2Idiameter  Idiameter  12 Icenter  12 ( 12 MR2 )  14 MR2.
ASSESS We have shown that Table 10.2 is consistent, and we have used this theorem to find a new rotational
inertia value not shown on the table.
73.
INTERPRET
DEVELOP
We find the rotational inertia of a right cylindrical cone, using the integral form for I.
We divide the cone into circular slices, and by integrating over all these slices we find the total
rotational inertia of the cone. The height of the cone is h and the base radius is R, so the radius of each slice
is r 
x, where x is the distance from the apex. The volume of the cone is V  13 Ah, where A is the area of the
2
base, A   R 2 , so V   R3 h. The volume of each disk-shaped slice is dV   r 2 dx. The cone has uniform mass
density MV , so each disk has mass dm  MV dV  3RM2 h ( r 2 dx). The rotational inertia of each disk is dI  12 r 2 dm.
R
h
EVALUATE
2
h
h R x 
3M
3M
I   dI  1  r 2 dm  1  
 r 2 dx  3
0
2 0
2 0  h   R 2 h
2h
2
2
3MR h 4
3MR  1 5  3
I
x dx 
h   MR 2
5 0

5
2h
2 h  5  10
h

h
0
2
 Rx 
x2 
 dx
 h 
ASSESS The units are correct. The value of I is less than that of a cylinder, since more of the mass is concentrated
along the axis of the cone.
10.18
Chapter 10
74.
INTERPRET We are to show that the rotational inertia of a uniform solid spheroid, about the axis of revolution, is
2
MR2 regardless whether the spheroid is prolate, oblate, or spherical. R in this case is the semi-axis perpendicular
5
to the axis of revolution.
DEVELOP We imagine slicing a sphere into a large number of thin slices, cut perpendicularly to the rotation axis.
Now if we stretch the sphere into a prolate spheroid, it’s the same as increasing the thickness of each slice, without
changing the mass or mass distribution of that slice. We have not moved any mass toward or away from the axis of
rotation, though, so the rotational inertia of each disk stays the same. Since the rotational inertia of the spheroid is
just the sum of the rotational inertias of the disks, which don’t change, the rotational inertia of the spheroid is the
same as that of the sphere.
EVALUATE We’re actually done already. For the oblate spheroid, we do the same thing but compress each disk.
Again, the rotational inertia of each disk does not change, so the rotational inertia of the spheroid does not change.
ASSESS Following this same reasoning, the rotational inertia of a rectangular plate, through an axis centered in
the plane of the plate, should be the same as for a thin rod through the center. We can check this with Table 10.2,
and we find that it is the case.
75.
INTERPRET We are designing a space station, and need to know the thrust of the rockets that spin the station up
to speed. We know the mass and geometry of the station, so we can find the rotational inertia. We know the desired
centripetal acceleration, so we can find the angular speed necessary. We know the time it takes to get up to speed,
so we can use the equations for constant angular acceleration to find the force necessary.
DEVELOP
use ac 
v2
R
The rotational inertia is I  MR 2 , the mass is M  6  105 kg, and the radius is R  50 m. We can
 R 2 to find the desired angular speed. The rotational equivalent of Newton’s second law
is   I  , which we can use to find the angular acceleration  . We also use the angular acceleration in Equation
10.7,   0  t. The torque is provided by 4 rockets at the rim of the station, each with thrust F, so   4RF . Put
all these pieces together to find F necessary to get to speed  in time t  5 hours  18 103 s.
EVALUATE
First, find the angular speed needed: R 2  g   
needed:    0  t   

t

g
Rt 2
g
R
. Next, find the angular acceleration
. Newton’s second law tells us that   I     I 
4F

these two expressions for , and solve for F: MR
ASSESS The units in our final equation are N  kg
g
Rt 2
F 
m m/s2
s2
Rg
M
4
 kg
t2
m2
s4
4 RF
MR2

4F
MR
. Equate
 0.23 N.
 kg m/s2, which is correct. This seems like a
small force, but it is acting at a great distance, the time needed is long, and the final speed is not great.
76.
INTERPRET What angular acceleration is required to change a wheel’s speed in a given time? We can use the
equations for constant angular acceleration.
DEVELOP The initial angular speed is 7.48 rev/s, and the final is 9.39 rev/s. We’ll need to change these to
radians/s. The time needed is 140 s. We solve Equation 10.7,   0  t , for  .
EVALUATE
The initial angular speed is 0  7.48 rev
 21revrad  47.0 rad/s. The final angular speed
s
 
 21revrad  59.0 rad/s.   0  t    t 0  0.0857 rad/s2.
is   9.39 rev
s
ASSESS A fairly small angular acceleration, for a small change in angular speed in a large time.
77.
INTERPRET Find a force from a torque and a distance. Use the definition of torque, and since no angular
information is given, assume that the force is applied at 90°.
DEVELOP   RF. We solve this for F.
EVALUATE
F

R
kN m
 10.1
 10.6 kN.
0.95 m
ASSESS Since the distance is approximately 1 meter, the torque and the force have nearly the same numeric value.
78.
INTERPRET We are checking whether marketing figures about flywheel energy storage and power availability are
correct. We will use angular kinetic energy, and the definition of power.
DEVELOP The flywheel is ring-shaped, so we’ll use I  MR 2 , with M  48 kg and R  0.411 m. Rotational
rev
1 min
kinetic energy is K  12 I 2, and the initial angular velocity is   30,000 min
 3600
 21revrad  52.4 rad/s.
s
Rotational Motion
10.19
EVALUATE The energy stored in this flywheel is K  12 I 2  12 (MR2 ) 2  11.1 kJ. This is almost a thousand
times less than the claimed energy storage. The power P  tK tells us that if power were taken from this flywheel
at a rate of 25 kW, the time until it stopped would be t  PK  0.44 s, or if we needed power for t  400 s, the
power it could supply for that time would be only P  tK  27.8 W.
ASSESS This flywheel might be able to run the taillight of the car for 400 seconds, but other than that it’s nearly
useless. The specs are wildly incorrect!