CHAPTER 8
ROTATIONAL KINEMATICS
CONCEPTUAL QUESTIONS
4.
REASONING AND SOLUTION The tangential speed, vT, of a point on the earth's
surface is related to the earth's angular speed  according to vT  r , Equation 8.9,
where r is the perpendicular distance from the point to the earth's rotation axis. At
the equator, r is equal to the earth's radius. As one moves away from the equator
toward the north or south geographic pole, the distance r becomes smaller. Since the
earth's rotation axis passes through the geographic poles, r is effectively zero at those
locations. Therefore, your tangential speed would be a minimum if you stood as
close as possible to either the north or south geographic pole.
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7.
REASONING AND SOLUTION
The wheel
and the two points under consideration are shown
in the figure at the right.
a. Each point must undergo the same angular
displacement in the same time interval.
Therefore, at any given instant, the angular
velocity of both points is the same.
1
2
C
b. Each point must increase its angular velocity
at the same rate; therefore, each point has the
same angular acceleration.
c. The tangential speed is given by vT  r . Since both points have the same
angular speed , but point 1 is further from the center than point 2, point 1 has the
larger tangential speed.
d. The tangential acceleration is given by aT  r . Since both points have the same
angular acceleration , but point 1 is further from the center than point 2, point 1 has
the larger tangential acceleration.
e. The centripetal acceleration is given by ac  r 2 . Since both points have the
same angular speed , but point 1 is further from the center than point 2, point 1 has
the larger centripetal acceleration.
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9.
REASONING AND SOLUTION The centripetal acceleration of any point in the
space station is given by ac  r 2 (Equation 8.11),where r is the distance from the
point to the axis of rotation of the space station. If the centripetal acceleration is
adjusted to be g at a given value of r (such as the astronaut's feet), the centripetal
acceleration will be different at other values of r. Therefore, if the adjustment is
made so that the centripetal acceleration at the astronaut's feet equals g, then the
centripetal acceleration will not equal g at the astronaut's head.
12. REASONING AND SOLUTION The bicycle wheel has an angular acceleration.
The arrows are perpendicular to the radius of the wheel. The magnitude of the
arrows increases with increasing distance from the center in accordance with
vT  r , or aT  r . The arrows in the picture could represent either the tangential
velocity or the tangential acceleration. The arrows are not directed radially inward;
therefore, they cannot represent the centripetal acceleration.
PROBLEMS
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1.
SSM REASONING AND SOLUTION Since there are 2 radians per
revolution, and it is stated in the problem that there are 100 grads in one-quarter of a
circle, we find that the number of grads in one radian is
 1 rev 100 grad 
(1.00 rad)
 63.7 grad
2 rad 0.250 rev 
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2.
REASONING The average angular velocity is equal to the angular displacement
divided by the elapsed time (Equation 8.2). Thus, the angular displacement of the engine
is equal to the product of the average angular velocity and the elapsed time.
SOLUTION
a.
Since 2 rad = 1 rev and 1 min = 60 s, the angular speed (in rad/s) of the
engine is
 2500 rev   2  rad  1 min 
 min   1 rev  60 s  




260 rad/s
b.
The average angular velocity  of the engine is equal to the angular
displacement  divided by the elapsed time t. The angular displacement is
 60 s 
4
 =  t   260 rad/s  5.1 min  
  8.0  10 rad
 1 min 
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(8.2)
4.
REASONING AND SOLUTION In one revolution the pulsar turns through
2 radians . The average angular speed of the pulsar is, from Equation 8.2,

2 rad


 1.9  102 rad/s
t 0.033 s
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5.
REASONING AND SOLUTION Using Equation 8.4 and the appropriate
conversion factors, the average angular acceleration of the CD in rad/s 2 is
2
  210 rev/ min  480 rev/ min  2 rad  1 min 
3
2




  – 6.4  10 rad/s
t 
74 min
  1 rev  60 s 
The magnitude of the average angular acceleration is 6.4 10 –3 rad/s 2 .
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9.
REASONING AND SOLUTION The angular displacements of the astronauts
are equal.
For A
 = sA/rA
For B
 = sB/rB
Equating these two equations for  and solving for sB gives
sB = (rB/rA)sA = [(1.10  103 m)/(3.20  102 m)](2.40  102 m) = 825 m
11.
SSM REASONING The time required for the bullet to travel the distance d is
equal to the time required for the discs to undergo an angular displacement of 0.240 rad.
The time can be found from Equation 8.2; once the time is known, the speed of the bullet
can be found using Equation 2.2.
SOLUTION From the definition of average angular velocity:

the required time is
t 




t
0.240rad
2.53  103 s
95.0rad/s
(8.1)
Note that    because the angular speed is constant. The (constant) speed of
the bullet can then be determined from the definition of average speed:
v
17.
x
d
0.850m



t
t
2.53  103 s
336 m/s
REASONING AND SOLUTION
a. From Equation 8.7 we obtain
SSM
   0 t  12  t 2  (5.00 rad/s)(4.00 s)  12 (2.50 rad/s 2 )(4.00 s) 2  4.00 101 rad
b. From Equation 8.4, we obtain
   0   t  5.00 rad/s + (2.50 rad/s 2 )(4.00 s) = 15.0 rad/s
21.
REASONING AND SOLUTION
a.
 = 12 ( + 0 ) t =
1 (1420 rad/s + 420 rad/s)(5.00
2
s) =
3
4.60  10 rad
b.
=
  0
t
=
1420 rad/s  420 rad/s
= 2.00  102 rad / s2
5.00 s
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24.
REASONING The angular displacement for the interval from 0 to 25.0 s can be
obtained by breaking the interval into two segments. During the first segment, from 0 to
15.0 s, the wheel has an constant angular acceleration. During the second segment, from
15.0 to 25.0 s, the angular velocity remains constant so the angular acceleration is zero.
We will find the angular displacement for each segment by using the equations of
kinematics for rotational motion. The angular displacement for the entire interval is the
sum of the two displacements.
SOLUTION The angular acceleration 1 during the first segment can be found,
since we know that the angular velocity changes from 0 to +7.50 rad/s during a time of
5.00 s:
1 
  0
t  t0

7.50 rad /s  0 rad /s
2
  1.50 rad /s
5.00 s  0 s
The angular displacement 1 for the first segment (0 to 15.0 s) is
(8.4)
1  0,1 t1  12 1 t12   0 rad /s 15.0 s  
1
2
 1.50 rad /s  15.0 s 
2
2
  169 rad
(8.7)
The angular velocity 1 at the end of the first segment is


1  0,1  1 t1  0 rad /s + +1.50 rad /s 2 15.0 s    22.5 rad /s
(8.4)
The angular displacement 2 for the second segment (15.0 to 25.0 s) is
 2  0,2 t  12 2 t 2   22.5 rad /s 10.0 s  
1
2
0 rad /s  10.0 s 
2
2
  225 rad
where we have used the fact that the final angular velocity 1 of the first segment
becomes the initial angular velocity 0,2 for the second segment.
The total displacement is 1 + 2 = 169 rad + 225 rad = 394 rad .
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28.
REASONING AND SOLUTION
a. According to Equation 8.9, the tangential speed of the sun is
vT  r  (2.2 10 20 m)(1.2 10 –15 rad/s) = 2.6 105 m/s
b. According to Equation 8.2,    / t . Since the angular speed of the sun is
constant,    . Solving for t , we have
t 

2 rad
1y

  1 h   1 day  
8


  1.7  10 y






15
  1.2  10 rad/s   3600 s   24 h   365.25 day 

33.
REASONING AND SOLUTION The average tangential speed v T of points on
the rim of the reel is given by Equation 8.9,
vT  r   (2.5  102 m)(1.9rad/s)  0.048m/s
where  is the average angular speed. The average tangential speed must equal
the average linear speed v of the tape. The length L of tape that passes around the reel in
a time t can be found from the definition of average velocity:
x
t
Solving for x and setting x = L gives
v
(8.7)
L  v t  (0.048 m/s)(7.1 s)  0.34 m
40.
REASONING
a.
According to Equation 8.2, the average angular speed is equal to the
magnitude of the angular displacement divided by the elapsed time. The magnitude of the
angular displacement is one revolution, or 2 rad. The elapsed time is one year,
expressed in seconds.
b.
The tangential speed of the earth in its orbit is equal to the product of its
orbital radius and its orbital angular speed (Equation 8.9).
c.
Since the earth is moving on a nearly circular orbit, it has a centripetal
acceleration that is directed toward the center of the orbit. The magnitude ac of the
centripetal acceleration is given by Equation 8.11 as ac = r2.
SOLUTION
a.
The average angular speed is
 =
b.
2 rad


 1.99  107 rad /s
7
t 3.16  10 s
The tangential speed of the earth in its orbit is



vT  r   1.50  1011 m 1.99  107 rad/s  2.98  104 m/s
c.
the sun is
(8.2)
(8.9)
The centripetal acceleration of the earth due to its circular motion around


a c  r  2  1.50  1011 m 1.99  107 rad /s

2
 5.94  103 m /s2
The acceleration is directed toward the center of the orbit.
42.
REASONING AND SOLUTION The linear speed of all points on either
sprocket, as well as the linear speed of all points on the chain, must be equal (otherwise
the chain would bunch up or break).
a. Therefore, according to Equation 8.9, the linear speed of the chain as it moves
between the sprockets is
v T  r  (9.00 cm)(9.40 rad/s) = 84.6 cm/s
(8.11)
b. The centripetal acceleration of the chain as it passes around the rear sprocket is,
according to Equation 8.11, a c  rrear  2 , where   vT / rrear . Therefore,
2
 v T 
v T 2 (84.6 cm/s) 2
a c  rrear 

 1.40 10 3 cm/s 2
 
5.10 cm
rrear  rrear
47.
SSM WWW REASONING AND SOLUTION From Equation 2.4, the linear
acceleration of the motorcycle is
v  v0 22.0m/s0 m/s
a

 2.44m/s 2
t
9.00s
Since the tire rolls without slipping, the linear acceleration equals the tangential
acceleration of a point on the outer edge of the tire: a  aT . Solving Equation 8.13 for 
gives

aT 2.44 m/s 2

 8.71 rad/s2
r
0.280 m