CHAPTER 8
ROTATIONAL KINEMATICS
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The figures below show two axes in the plane of the
paper and located so that the points B and C move in circular paths having the same radii
(radius = r).
C
r
A
r
Axis
B
C
r
Axis
A
B
r
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2.
REASONING AND SOLUTION When a pair of scissors is used to cut a string, each blade
of the scissors does not have the same angular velocity at a given instant during the cut. The
angular speed of each blade is the same; however, each blade rotates in the opposite
direction. Therefore, it is correct to conclude that the blades have opposite angular velocities
at any instant during the cut.
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384 ROTATIONAL KINEMATICS
3.
REASONING AND SOLUTION Just before the battery is removed, the second hand is
rotating so that its angular velocity is clockwise. The second hand moves with a constant
angular velocity, so the angular acceleration is zero. When the battery is removed, the
second hand will continue to rotate with its clockwise angular velocity, but it will slow
down. Therefore, the angular acceleration must be opposite to the angular velocity, or
counterclockwise.
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4.
REASONING AND SOLUTION The tangential speed, vT, of a point on the earth's surface
is related to the earth's angular speed ω according to vT = rω , 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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5.
SSM REASONING AND SOLUTION
a. A thin rod rotates at a constant angular speed about an axis of rotation that is
perpendicular to the rod at its center. As the rod rotates, each point at a distance r from the
center on one half of the rod has the same tangential speed as the point at a distance r from
the center on the other half of the rod. This is true for all values of r for 0 < r ≤ (L / 2) where
L is the length of the rod.
b. If the rod rotates about an axis that is perpendicular to the rod at one end, no two points
are the same distance from the axis of rotation. Therefore, no two points on the rod have the
same tangential speed.
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6.
REASONING AND SOLUTION The wheels are rotating with a constant angular velocity.
a. Since the angular velocity is constant, each wheel has zero angular acceleration,
α = 0 rad/s. Since the tangential acceleration aT is related to the angular acceleration
through Equation 8.10, aT = rα , every point on the rim has zero tangential acceleration.
b. Since the particles on the rim of the wheels are moving along a circular path, they must
have a centripetal acceleration. This can be supported by Equation 8.11, ac = rω 2 , where
ac is the magnitude of the centripetal acceleration. Since ω is nonzero, ac is nonzero.
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Chapter 8 Conceptual Questions
7.
385
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 ω . Since both points have the same angular
speed ω, 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 = rα . Since both points have the same
angular acceleration α, but point 1 is further from the center than point 2, point 1 has the
larger tangential acceleration.
2
e. The centripetal acceleration is given by ac = rω . Since both points have the same
angular speed ω, but point 1 is further from the center than point 2, point 1 has the larger
centripetal acceleration.
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8.
REASONING AND SOLUTION For a building that is located on the earth's equator, the
points in the building rotate about the center of the earth. The tangential speed of any point
is given by vT = r ω , where, in this case, r is measured from the center of the earth. The
top floor of the building has a larger value of r than the other floors; therefore, the top floor
has the greater tangential speed.
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9.
SSM REASONING AND SOLUTION The centripetal acceleration of any point in the
2
space station is given by ac = rω (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.
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10. REASONING AND SOLUTION The tangential speed for any point on the tip of a clock
hand is given by vT = r ω . The angular speeds of the second hand, the minute hand, and the
hour hand differ greatly, with the second hand having the largest angular speed and the hour
hand having the smallest. If one desired to create a clock in which the tips of the second
386 ROTATIONAL KINEMATICS
hand, the minute hand, and the hour hand moved with the same tangential speed, the lengths
of the arms would also have to differ greatly, with the second hand having the smallest arm
and the hour hand having the largest. Such a clock would not be very practical.
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11. REASONING AND SOLUTION Any point on a rotating object possesses a centripetal
acceleration that is directed radially toward the axis of rotation. This also applies to a tire on
a moving car and is true regardless of whether the car has a constant linear velocity or
whether it is accelerating.
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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 ω , or aT = rα . 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.
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13. SSM REASONING AND SOLUTION The speedometer of a truck uses a device that
measures the angular speed of the tires. The angular speed is related to the linear speed of
the truck by v = rω .
Suppose two trucks are traveling side-by-side along a highway at the same linear speed v,
and one truck has larger wheels than the other. The angular speed of the larger-diameter
wheels is less than that of the smaller-diameter wheels. Since the speedometer uses a device
that measures the angular speed of the tires, the speedometer reading on the truck with the
larger wheels will indicate a smaller linear speed than that on the truck with the smaller
wheels.
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14. REASONING AND SOLUTION When a fan is shut off, the fan blades gradually slow
down until they eventually stop. The angular acceleration is never really constant because it
gradually decreases, becoming zero at the instant the blades stop rotating. In solving
problems, it is sometimes convenient to approximate such motion as having constant
angular acceleration. This is a fairly good approximation for a short time interval just after
the switch is turned off.
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15. REASONING AND SOLUTION Three examples that involve rotation about an axis that
is not fixed:
(1) The motion of a Frisbee. The axis of the rotating Frisbee moves through the air
with the Frisbee.
(2) The motion of the earth in its orbit. The rotation axis of the earth changes
position as the earth revolves around the sun.
(3) The motion of a twirling baton that has been thrown into the air. The rotation axis
constantly changes location as the baton rises and falls.
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Chapter 8 Problems
387
CHAPTER 8 ROTATIONAL KINEMATICS
PROBLEMS
1.
REASONING AND SOLUTION Since there are 2π 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
SSM
⎛ 1 rev ⎞ ⎛ 100 grad ⎞
= 63.7 grad
(1.00 rad)⎝
2π rad ⎠ ⎝ 0.250 rev ⎠
2.
REASONING The average angular velocity ω is defined as the angular displacement
∆θ divided by the elapsed time ∆t during which the displacement occurs: ω = ∆θ / ∆t
(Equation 8.2). This relation can be used to find the average angular velocity of the earth as
it spins on its axis and as it orbits the sun.
SOLUTION
a. As the earth spins on its axis, it makes 1 revolution (2π rad) in a day. Assuming that the
positive direction for the angular displacement is the same as the direction of the earth’s
rotation, the angular displacement of the earth in one day is ( ∆θ )spin = +2π rad . The
average angular velocity is (converting 1 day to seconds):
ω=
( ∆θ )spin
( ∆t )spin
=
+2π rad
= +7.3 × 10−5 rad/s
⎛ 24 h ⎞ ⎛ 3600 s ⎞
1 day ⎜
⎜ 1 day ⎟⎟ ⎜⎝ 1 h ⎟⎠
⎝
⎠
(
)
b. As the earth orbits the sun, the earth makes 1 revolution (2π rad) in one year. Taking the
positive direction for the angular displacement to be the direction of the earth’s orbital
motion, the angular displacement in one year is ( ∆θ )orbit = +2π rad . The average angular
velocity is (converting 365¼ days to seconds):
ω=
( ∆θ )orbit
=
( ∆t )orbit
(
+2π rad
⎛ 24 h
365 14 days ⎜
⎜ 1 day
⎝
)
⎞ ⎛ 3600 s ⎞
⎟⎟ ⎜
⎟
⎠⎝ 1 h ⎠
= +2.0 ×10−7 rad/s
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3.
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 baseball
is equal to the product of the average angular velocity and the elapsed time. However, the
SSM
388 ROTATIONAL KINEMATICS
problem gives the travel time in seconds and asks for the displacement in radians, while the
angular velocity is given in revolutions per minute. Thus, we will begin by converting the
angular velocity into radians per second.
SOLUTION Since 2π rad = 1 rev and 1 min = 60 s, the average angular velocity ω (in
rad/s) of the baseball is
⎛ 330 rev ⎞ ⎛ 2 π rad ⎞ ⎛ 1 min ⎞
ω =⎜
⎟⎜
⎟ = 35 rad/s
⎟⎜
⎝ min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠
Since the average angular velocity of the baseball is equal to the angular displacement ∆θ
divided by the elapsed time ∆t, the angular displacement is
∆θ = ω ∆t = ( 35 rad/s )( 0.60 s ) = 21 rad
4.
REASONING AND SOLUTION In one revolution the pulsar turns through 2π radians .
The average angular speed of the pulsar is, from Equation 8.2,
ω=
5.
(8.2)
2π rad
∆θ
=
= 1.9 × 102 rad/s
∆t 0.033 s
REASONING AND SOLUTION Equation 8.4 gives the desired result. Assuming
t0 = 0 s, the final angular velocity is
SSM
ω = ω 0 + α t = 0 rad/s + (328 rad/s 2 )(1.50 s) = 492 rad/s
6.
Equation 8.4 ⎡⎣α = (ω − ω 0 ) / t ⎤⎦ indicates that the average angular
acceleration is equal to the change in the angular velocity divided by the elapsed time. Since
the wheel starts from rest, its initial angular velocity is ω0 = 0 rad/s. Its final angular
velocity is given as ω = 0.24 rad/s. Since the average angular acceleration is given as
α = 0.030 rad/s 2 , Equation 8.4 can be solved to determine the elapsed time t.
REASONING
SOLUTION Solving Equation 8.4 for the elapsed time gives
t=
7.
ω − ω 0 0.24 rad/s − 0 rad/s
=
= 8.0 s
α
0.030 rad/s 2
REASONING AND SOLUTION Using Equation 8.4 and the appropriate conversion
factors, the average angular acceleration of the CD in rad/s2 is
Chapter 8 Problems
389
2
∆ω ⎛ 210 rev/ min − 480 rev/ min ⎞ ⎛ 2π rad ⎞⎛ 1 min ⎞
−3
2
=⎜
α=
⎟⎜
⎟⎜
⎟ = – 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/s2 .
8.
REASONING The average angular velocity ω is defined as the angular displacement
∆θ divided by the elapsed time ∆t during which the displacement occurs: ω = ∆θ / ∆t
(Equation 8.2). Solving for the elapsed time gives ∆t = ∆θ / ω . We are given ∆θ and can
calculate ω from the fact that the earth rotates on its axis once every 24.0 hours.
SOLUTION The sun itself subtends an angle of 9.28 × 10−3 rad. When the sun moves a
distance equal to its own diameter, the angle through which it moves is also 9.28 × 10−3 rad;
thus, ∆θ = 9.28 × 10−3 rad. The average angular velocity ω at which the sun appears to
move across the sky is the same as that of the earth rotating on its axis, ωearth , so
ω = ωearth . Since the earth makes one revolution (2π rad) every 24.0 h, its average angular
velocity is
ωearth =
∆θearth
∆tearth
=
2π rad
=
24.0 h
2π rad
= 7.27 × 10−5 rad/s
( 24.0 h ) ⎛⎜ 3600 s ⎞⎟
⎝ 1 h ⎠
The time it takes for the sun to move a distance equal to its diameter is
∆t =
∆θ
ωearth
=
9.28 ×10−3 rad
= 128 s (a little over 2 minutes)
7.27 ×10−5 rad/s
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9.
REASONING AND SOLUTION The angular displacements of the astronauts are equal.
For A
θ = sA/rA
For B
θ = sB/rB
Equating these two equations for θ and solving for sB gives
sB = (rB/rA)sA = [(1.10 × 103 m)/(3.20 × 102 m)](2.40 × 102 m) = 825 m
(8.1)
390 ROTATIONAL KINEMATICS
10. REASONING AND SOLUTION The people meet at time t. At this time the magnitudes of
their angular displacements must total 2π rad.
θ1 + θ2 = 2π rad
Then
ω1t + ω2t = 2π rad
t=
2π rad
2π rad
=
= 1200 s
−
3
ω1 + ω 2 1.7 × 10 rad/s + 3.4 × 10−3 rad/s
11. REASONING AND SOLUTION The baton will make four revolutions in a time t given by
t=
θ
ω
Half of this time is required for the baton to reach its highest point. The magnitude of the
initial vertical velocity of the baton is then
v0 = g
( 12 t ) = g ⎛⎜⎝ 2θω ⎞⎟⎠
With this initial velocity the baton can reach a height of
2
h=
v0
2g
=
gθ
2
8ω
2
(9.80 m/s ) (8π rad )
2
=
2
⎡⎛
rev ⎞ ⎛ 2π rad ⎞ ⎤
8 ⎢⎜1.80
⎟⎥
⎟⎜
s ⎠ ⎝ 1 rev ⎠ ⎦
⎣⎝
2
= 6.05 m
12. REASONING AND SOLUTION
a. If the propeller is to appear stationary, each blade must move through an angle of 120° or
2π / 3 rad between flashes. The time required is
t=
θ
=
ω
(2π / 3) rad
= 2.00 ×10 –2 s
2
π
rad
⎛
⎞
(16.7 rev/s) ⎜
⎟
⎝ 1 rev ⎠
b. The next shortest time occurs when each blade moves through an angle of 240°, or
4π / 3 rad, between successive flashes. This time is twice that found in part a, or
4.00 × 10
−2
s.
Chapter 8 Problems
13.
391
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.
SSM
SOLUTION From the definition of average angular velocity:
ω=
the required time is
∆t =
∆θ
ω
=
∆θ
∆t
0.240 rad
= 2.53 × 10−3 s
95.0 rad/s
Note that ω = ω because the angular speed is constant. The (constant) speed of the bullet
can then be determined from the definition of average speed:
v=
d
∆x
0.850 m
=
=
=
∆t
∆t
2.53 × 10−3 s
336 m/s
14. REASONING The golf ball must travel a distance equal to its diameter in a maximum
time equal to the time required for one blade to move into the position of the previous blade.
SOLUTION The time required for the golf ball to pass through the opening between two
blades is given by ∆t = ∆θ / ω , with ω = 1.25 rad/s and ∆θ = (2π rad)/16 = 0.393 rad .
Therefore, the ball must pass between two blades in a maximum time of
∆t =
0.393 rad
= 0.314 s
1.25 rad/s
The minimum speed of the ball is
v=
∆x 4.50 × 10 –2 m
–1
=
= 1.43 × 10 m/s
∆t
0.314 s
15. REASONING AND SOLUTION The ball has a time of flight given by the kinematics
equation 0 m = v0 t − 12 a y t 2 . Solving for the time gives
t=
2v0 2 (19 m/s )( sin 55° )
=
= 3.2 s
2
g
9.80 m/s
The ball then makes (7.7 rev/s)(3.2 s) = 25 rev
392 ROTATIONAL KINEMATICS
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1
2
16. REASONING AND SOLUTION From Equation 8.6, θ = (ω 0 + ω ) t .
Solving for t gives
t=
17.
2θ
2(85.1 rad)
=
= 5.22 s
ω 0 + ω 18.5 rad/s + 14.1 rad/s
SSM REASONING AND SOLUTION
a. From Equation 8.7 we obtain
θ = ω 0 t + 12 α 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
ω = ω 0 + α t = 5.00 rad/s + (2.50 rad/s 2 )(4.00 s) = 15.0 rad/s
18. REASONING AND SOLUTION
19.
a.
ω = ω0 + α t = 0 rad/s + (3.00 rad/s2)(18.0 s) = 54.0 rad/s
b.
θ=
1 (ω +
0
2
ω)t =
1 (0
2
rad/s + 54.0 rad/s)(18.0 s) = 486 rad
REASONING AND SOLUTION
a. Since the flywheel comes to rest, its final angular velocity is zero. Furthermore, if the
initial angular velocity ω 0 is assumed to be a positive number and the flywheel decelerates,
the angular acceleration must be a negative number. Solving Equation 8.8 for θ , we obtain
SSM
WWW
ω 2 − ω 02 ( 0 rad/s ) − (220 rad/s) 2
=
= 1.2 × 104 rad
2
2α
2(–2.0 rad/s )
2
θ=
b. The time required for the flywheel to come to rest can be found from Equation 8.4.
Solving for t, we obtain
t=
ω − ω 0 0 rad/s − 220 rad/s
2
=
= 1.1 × 10 s
2
α
–2.0 rad/s
Chapter 8 Problems
393
20. REASONING
a. The time t for the wheels to come to a halt depends on the initial and final velocities, ω0
and ω, and the angular displacement θ : θ =
time yields
t=
1
2
(ω0 + ω ) t
(see Equation 8.6). Solving for the
2θ
ω0 + ω
b. The angular acceleration α is defined as the change in the angular velocity, ω − ω0,
divided by the time t:
ω − ω0
α=
(8.4)
t
SOLUTION
a. Since the wheel comes to a rest, ω = 0 rad/s. Converting 15.92 revolutions to radians
(1 rev = 2π rad), the time for the wheel to come to rest is
⎛ 2π rad ⎞
2 ( +15.92 rev ) ⎜
⎟
2θ
⎝ 1 rev ⎠ = 10.0 s
t=
=
+20.0 rad/s + 0 rad/s
ω0 + ω
b. The angular acceleration is
ω − ω0
0 rad/s − 20.0 rad/s
= −2.00 rad/s 2
t
10.0 s
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α=
=
21. REASONING Equation 8.8 (ω 2 = ω 02 + 2αθ ) from the equations of rotational kinematics
can be employed to find the final angular velocity ω. The initial angular velocity is
ω0 = 0 rad/s since the top is initially at rest, and the angular acceleration is given as
α = 12 rad/s2. The angle θ (in radians) through which the pulley rotates is not given, but it
can be obtained from Equation 8.1 (θ = s/r), where the arc length s is the 64-cm length of
the string and r is the 2.0-cm radius of the top.
SOLUTION Solving Equation 8.8 for the final angular velocity gives
ω = ± ω 02 + 2αθ
We choose the positive root, because the angular acceleration is given as positive and the
top is at rest initially. Substituting θ = s/r from Equation 8.1 gives
⎛s⎞
⎝ ⎠
ω = + ω 02 + 2α ⎜ ⎟ = +
r
( 0 rad/s )
2
⎛ 64 cm ⎞
+ 2 (12 rad/s 2 ) ⎜
⎟ = 28 rad/s
⎝ 2.0 cm ⎠
394 ROTATIONAL KINEMATICS
22. REASONING The equations of kinematics for rotational motion cannot be used directly to
find the angular displacement, because the final angular velocity (not the initial angular
velocity), the acceleration, and the time are known. We will combine two of the equations,
Equations 8.4 and 8.6 to obtain an expression for the angular displacement that contains the
three known variables.
SOLUTION The angular displacement of each wheel is equal to the average angular
velocity multiplied by the time
θ=
1
2
(ω 0 + ω ) t
(8.6)
ω
The initial angular velocity ω0 is not known, but it can be found in terms of the angular
acceleration and time, which are known. The angular acceleration is defined as (with t0 = 0 s)
α=
ω − ω0
t
or
ω0 = ω − α t
(8.4)
Substituting this expression for ω0 into Equation 8.6 gives
θ = 12 ⎡(ω − α t ) + ω ⎤ t = ω t − 12 α t 2
⎢
⎣⎢
ω0
⎥
⎦⎥
= ( +74.5 rad /s )( 4.50 s ) −
1
2
( +6.70 rad /s ) ( 4.50 s )
2
2
= +267 rad
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23. REASONING There are three segments to the propeller’s angular motion, and we will
calculate the angular displacement for each separately. In these calculations we will
remember that the final angular velocity for one segment is the initial velocity for the next
segment. Then, we will add the separate displacements to obtain the total.
SOLUTION For the first segment the initial angular velocity is ω0 = 0 rad/s, since the
propeller starts from rest. Its acceleration is α = 2.90 × 10-3 rad/s2 for a time
t = 2.10 × 103 s. Therefore, we can obtain the angular displacement θ1 from Equation 8.7 of
the equations of rotational kinematics as follows:
[First segment]
θ1 = ω 0t + 12 α t 2 = ( 0 rad/s ) ( 2.10 ×103 s ) + 12 ( 2.90 ×10−3 rad/s 2 )( 2.10 ×103 s )
= 6.39 × 103 rad
2
Chapter 8 Problems
395
The initial angular velocity for the second segment is the final velocity for the first segment,
and according to Equation 8.4, we have
ω = ω 0 + α t = 0 rad/s + ( 2.90 ×10−3 rad/s 2 )( 2.10 × 103 s ) = 6.09 rad/s
Thus, during the second segment, the initial angular velocity is ω0 = 6.09 rad/s and remains
constant at this value for a time of t = 1.40 × 103 s. Since the velocity is constant, the
angular acceleration is zero, and Equation 8.7 gives the angular displacement θ2 as
[Second segment]
θ 2 = ω 0t + 12 α t 2 = ( 6.09 rad/s ) (1.40 ×103 s ) + 12 ( 0 rad/s 2 )(1.40 ×103 s ) = 8.53 ×103 rad
2
During the third segment, the initial angular velocity is ω0 = 6.09 rad/s, the final velocity is
ω = 4.00 rad/s, and the angular acceleration is α = -2.30 × 10-3 rad/s2. When the propeller
picked up speed in segment one, we assigned positive values to the acceleration and
subsequent velocity. Therefore, the deceleration or loss in speed here in segment three
means that the acceleration has a negative value. Equation 8.8 (ω 2 = ω 02 + 2αθ 3 ) can be
used to find the angular displacement θ3. Solving this equation for θ3 gives
[Third segment]
ω 2 − ω 02 ( 4.00 rad/s ) − ( 6.09 rad/s )
θ3 =
=
= 4.58 ×103 rad
−3
2
2α
2 ( −2.30 × 10 rad/s )
2
2
The total angular displacement, then, is
θ Total = θ1 + θ 2 + θ 3 = 6.39 ×103 rad + 8.53 ×103 rad + 4.58 ×103 rad = 1.95 ×104 rad
24. REASONING Since the time t and angular acceleration α are known, we will begin by
using Equation 8.7 from the equations of kinematics to determine the angular
displacement θ :
θ = ω0t + 12 α t 2
However, the initial angular velocity ω0 is not given. We can determine it by resorting to
another equation of kinematics, ω = ω0 + α t (Equation 8.4), which relates ω0 to the final
angular velocity ω, the angular acceleration, and the time, all of which are known.
SOLUTION Solving Equation 8.4 for ω0 gives ω0 = ω − α t . Substituting this result into
θ = ω0t + 12 α t 2 gives
396 ROTATIONAL KINEMATICS
θ = ω0t + 12 α t 2 = (ω − α t ) t + 12 α t 2 = ω t − 12 α t 2
= ( +1.88 rad/s )(10.0 s ) − 12 ( −5.04 rad/s 2 ) (10.0 s ) = +2.71×102 rad
2
____________________________________________________________________________
25.
REASONING The time required for the change in the angular velocity to occur can
be found by solving Equation 8.4 for t. In order to use Equation 8.4, however, we must
know the initial angular velocity ω 0 . Equation 8.6 can be used to find the initial angular
velocity.
SSM
SOLUTION From Equation 8.6 we have
1
2
θ = (ω 0 + ω )t
Solving for ω0 gives
ω0 =
2θ
−ω
t
Since the angular displacement θ is zero, ω0 = –ω. Solving Equation 8.4 for t, and using
the fact that ω0 = –ω, gives
t=
2ω
α
=
2(− 25.0 rad/s)
=
− 4.00 rad/s2
12.5 s
____________________________________________________________________________________________
26. REASONING According to Equation 3.5b, the time required for the diver to reach the
water, assuming free-fall conditions, is t = 2 y / a y . If we assume that the "ball" formed
by the diver is rotating at the instant that she begins falling vertically, we can use Equation
8.2 to calculate the number of revolutions made on the way down.
SOLUTION Taking upward as the positive direction, the time required for the diver to
reach the water is
2(–8.3 m)
t=
= 1.3 s
–9.80 m/s 2
Solving Equation 8.2 for ∆ θ , we find
∆θ = ω ∆t = (1.6 rev/s)(1.3 s)= 2.1 rev
27.
REASONING The angular displacement of the child when he catches the
horse is, from Equation 8.2, θ c = ω c t . In the same time, the angular displacement of the
SSM
WWW
Chapter 8 Problems
397
horse is, from Equation 8.7 with ω 0 = 0 rad/s, θ h = 12 α t 2 . If the child is to catch the horse
θ c = θ h + (π / 2).
SOLUTION Using the above conditions yields
1 αt2
2
or
1 (0.0100
2
− ωct + 12 π = 0
rad/s 2 )t 2 − ( 0.250 rad/s ) t +
1
2
(π
rad ) = 0
The quadratic formula yields t = 7.37 s and 42.6 s; therefore, the shortest time needed to
catch the horse is t = 7.37 s .
28. REASONING AND SOLUTION
a.
ωA = v/r = (0.381 m/s)/(0.0508 m) = 7.50 rad/s
b.
ωB = v/r = (0.381 m/s)(0.114 m) = 3.34 rad/s
α=
ωB −ωA
t
=
3.34 rad/s − 7.50 rad/s
= − 1.73 × 10 −3 rad/s 2
3
2.40 × 10 s
The angular velocity is decreasing .
29.
REASONING The angular speed ω and tangential speed vT are related by
Equation 8.9 (vT = rω), and this equation can be used to determine the radius r. However,
we must remember that this relationship is only valid if we use radian measure. Therefore, it
will be necessary to convert the given angular speed in rev/s into rad/s.
SSM
SOLUTION Solving Equation 8.9 for the radius gives
r=
vT
ω
=
54 m/s
= 0.18 m
⎛ 2π rad ⎞
( 47 rev/s ) ⎜
⎟
⎝ 1 rev ⎠
Conversion from rev/s into rad/s
where we have used the fact that 1 rev corresponds to 2π rad to convert the given angular
speed from rev/s into rad/s.
398 ROTATIONAL KINEMATICS
30. REASONING AND SOLUTION
a. In one lap, the car undergoes an angular displacement of 2π radians. Therefore, from the
definition of average angular speed
ω=
∆θ 2π rad
=
=
∆t
18.9 s
0.332 rad/s
b. Equation 8.9 relates the average angular speed ω of an object moving in a circle to its
average linear speed v T tangent to the path of motion:
v T = rω
Solving for r gives
r=
vT
ω
=
42.6 m/s
=
0.332 rad/s
128 m
Notice that the unit "rad," being dimensionless, does not appear in the final answer.
31. REASONING The length of tape that passes around the reel is just the average tangential
speed of the tape times the time t. The average tangential speed vT is given by Equation 8.9
( vT = rω ) as the radius r times the average angular speed ω
in rad/s.
SOLUTION The length L of tape that passes around the reel in t = 13 s is L = vT t . Using
Equation 8.9 to express the tangential speed, we find
L = vT t = rω t = ( 0.014 m )( 3.4 rad/s )(13 s ) = 0.62 m
32. REASONING The angular speed ω of the reel is related to the tangential speed vT of the
fishing line by vT = rω (Equation 8.9), where r is the radius of the reel. Solving this
equation for ω gives ω = vT / r . The tangential speed of the fishing line is just the distance x
it travels divided by the time t it takes to travel that distance, or vT = x/t.
SOLUTION Substituting vT = x/t into ω = vT / r and noting that 3.0 cm = 3.0 × 10−2 m, we
find that
x
2.6 m
vT t
9.5 s
ω=
= =
= 9.1 rad/s
r
r 3.0 × 10−2 m
______________________________________________________________________________
Chapter 8 Problems
399
33. REASONING AND SOLUTION
a. A person living in Ecuador makes one revolution (2π rad) every 23.9 hr (8.60 × 104 s).
The angular speed of this person is ω = (2π rad)/(8.60 × 104 s) = 7.31 × 10−5 rad/s.
According to Equation 8.9, the tangential speed of the person is, therefore,
( 6.38 ×106 m )( 7.31 × 10−5 rad/s ) =
vT = rω =
4.66 ×102 m/s
b. The relevant geometry is shown in the
drawing at the right. Since the tangential
speed is one-third of that of a person
living in Ecuador, we have,
r
θ
θ
r
vT
= rθ ω
3
θ
or
rθ =
The angle θ is, therefore,
vT
4.66 ×102 m/s
=
= 2.12 × 106 m
5
−
3ω 3 7.31 × 10 rad/s
(
)
⎛ 2.12 × 10 6 m ⎞
⎟ = 70.6°
⎝ 6.38 × 10 6 m ⎠
θ = cos −1 ⎜
34. REASONING The tangential speed vT of a point on the “equator” of the baseball is given
by Equation 8.9 as vT = rω, where r is the radius of the baseball and ω is its angular speed.
The radius is given in the statement of the problem. The (constant) angular speed is related
to that angle θ through which the ball rotates by Equation 8.2 as ω = θ /t, where we have
assumed for convenience that θ0 = 0 rad when t0 = 0 s. Thus, the tangential speed of the ball
is
⎛θ ⎞
vT = r ω = r ⎜ ⎟
⎝t⎠
The time t that the ball is in the air is equal to the distance x it travels divided by its linear
speed v, t = x/v, so the tangential speed can be written as
⎛θ
vT = r ⎜
⎝t
⎞
⎛ θ ⎞ rθ v
⎟ = r⎜ x ⎟ =
x
⎠
⎜ ⎟
⎝v⎠
400 ROTATIONAL KINEMATICS
SOLUTION The tangential speed of a point on the equator of the baseball is
vT =
35.
−2
r θ v ( 3.67 × 10 m ) ( 49.0 rad )( 42.5 m/s )
=
= 4.63 m/s
x
16.5 m
REASONING AND SOLUTION According to Equation 8.9, the crank handle has
an angular speed of
v
1.20 m/s
ω= T = 1
= 6.00 rad/s
r
(0.400 m)
2
SSM
The crank barrel must have the same angular speed as the handle; therefore, the tangential
speed of a point on the barrel is, again using Equation 8.9,
vT = r ω = (5.00 × 10–2 m)(6.00 rad/s) = 0.300 m/s
If we assume that the rope does not slip, then all points on the rope must move with the
same speed as the tangential speed of any point on the barrel. Therefore the linear speed
with which the bucket moves down the well is v = 0.300 m/s .
36. REASONING AND SOLUTION The figure below shows the initial and final states of the
system.
m
L
L
v
INITIAL CONFIGURATION
FINAL CONFIGURATION
a. From the principle of conservation of mechanical energy:
E0 = Ef
Initially the system has only gravitational potential energy. If the level of the hinge is
chosen as the zero level for measuring heights, then: E0 = mgh0 = mgL. Just before the
object hits the floor, the system has only kinetic energy.
Therefore
Chapter 8 Problems
1
2
mgL = mv
401
2
Solving for v gives
v=
2 gL
From Equation 8.9, vT = rω. Solving for ω gives ω = vT/r. As the object rotates downward, it
travels in a circle of radius L. Its speed just before it strikes the floor is its tangential speed.
Therefore,
vT
2 gL
v
ω=
= =
r
L
L
=
2g
=
L
2(9.80 m/s 2 )
= 3.61 rad/s
1.50 m
b. From Equation 8.10:
aT = rα
Solving for α gives α = aT/r. Just before the object hits the floor, its tangential acceleration
is the acceleration due to gravity. Thus,
aT
2
g 9.80 m/s
α=
= =
=
r
L
1.50 m
6.53 rad/s
2
37. REASONING AND SOLUTION The stone leaves the circular path with a horizontal speed
v0 = vT = rω
so ω = v0/r. We are given that r = x/30 so ω = 30v0/x. Kinematics gives x = v0t and
h=
1 gt2.
2
Using the above yields
2
9.80 m/s
g
ω = 30
= 30
= 14.8 rad/s
2h
2 ( 20.0 m )
38. REASONING The magnitude ω of each car’s angular speed can be evaluated from ac = rω2
(Equation 8.11), where r is the radius of the turn and ac is the magnitude of the centripetal
acceleration. We are given that the centripetal acceleration of each car is the same. In
addition, the radius of each car’s turn is known. These facts will enable us to determine the
ratio of the angular speeds.
SOLUTION Solving Equation 8.11 for the angular speed gives ω = ac / r . Applying this
relation to each car yields:
402 ROTATIONAL KINEMATICS
Car A:
ωA = ac, A / rA
Car B:
ωB = ac, B / rB
Taking the ratio of these two angular speeds, and noting that ac, A = ac, B, gives
ac, A
ωA
=
ωB
rA
ac, B
=
ac, A rB
ac, B rA
=
36 m
= 0.87
48 m
rB
______________________________________________________________________________
39.
REASONING Since the car is traveling with a constant speed, its tangential
acceleration must be zero. The radial or centripetal acceleration of the car can be found
from Equation 5.2. Since the tangential acceleration is zero, the total acceleration of the car
is equal to its radial acceleration.
SSM
SOLUTION
a. Using Equation 5.2, we find that the car’s radial acceleration, and therefore its total
acceleration, is
vT2 (75.0 m/s) 2
a = aR =
=
= 9.00 m/s 2
r
625 m
b. The direction of the car’s total acceleration is the same as the direction of its radial
acceleration. That is, the direction is radially inward .
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 2π 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 = rω2.
Chapter 8 Problems
403
SOLUTION
a. The average angular speed is
ω= ω=
2π rad
∆θ
=
= 1.99 × 10−7 rad /s
7
∆t 3.16 × 10 s
(8.2)
b. The tangential speed of the earth in its orbit is
(
11
vT = r ω = 1.50 × 10
)(
m 1.99 × 10
−7
)
4
rad/s = 2.98 × 10 m/s
(8.9)
c. The centripetal acceleration of the earth due to its circular motion around the sun is
(
)(
a c = r ω 2 = 1.50 × 1011 m 1.99 × 10−7 rad /s
)
2
= 5.94 × 10−3 m /s2
(8.11)
The acceleration is directed toward the center of the orbit.
41.
REASONING The top of the racket has both tangential and centripetal acceleration
components given by Equations 8.10 and 8.11, respectively: aT = rα and a c = rω 2 . The
total acceleration of the top of the racket is the resultant of these two components. Since
these acceleration components are mutually perpendicular, their resultant can be found by
using the Pythagorean theorem.
SSM
SOLUTION Employing the Pythagorean theorem, we obtain
a=
aT2 + a 2c = ( rα ) 2 + ( rω 2 )2 = r α 2 + ω 4
Therefore,
a = (1.5 m) (160 rad/s 2 )2 + (14 rad/s)4 = 380 m/s2
42. REASONING AND SOLUTION
a. 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). Therefore,
according to Equation 8.9, the linear speed of the chain as it moves between the sprockets is
v T = rω = (9.00 cm)(9.40 rad/s) = 84.6 cm/s
b. The centripetal acceleration of the chain as it passes around the rear sprocket is,
according to Equation 8.11, a c = rrear ω 2 , where ω = vT / rrear . Therefore,
404 ROTATIONAL KINEMATICS
⎛ v T ⎞ 2 v T 2 (84.6 cm/s) 2
a c = rrear ⎜
=
= 1.40 ×10 3 cm/s2
⎟ =
rrear
5.10 cm
⎝ rrear ⎠
43. REASONING AND SOLUTION The tangential acceleration of point A is aTA = rAα and
of point B is aTB = rBα. Eliminating α gives
rA/rB = aTA/aTB
We are given that aTA = 2aTB,
so rA/rB = 2.
From the figure in the text it is seen that
2
2
rA = L1 + L2
and
rB = L1
which give
(L2/L1)2 = (rA/rB)2 – 1
Then
L1
=
L2
1
3
44. REASONING
a. The tangential speed vT of the sun as it orbits about the center of the Milky Way is
related to the orbital radius r and angular speed ω by Equation 8.9, vT = rω. Before we use
this relation, however, we must first convert r to meters from light-years.
b. The centripetal force is the net force required to keep an object, such as the sun, moving
on a circular path. According to Newton’s second law of motion, the magnitude Fc of the
centripetal force is equal to the product of the object’s mass m and the magnitude ac of its
centripetal acceleration (see Section 5.3): Fc = mac. The magnitude of the centripetal
acceleration is expressed by Equation 8.11 as ac = rω2, where r is the radius of the circular
path and ω is the angular speed of the object.
SOLUTION
a. The radius of the sun’s orbit about the center of the Milky Way is
⎛ 9.5 × 1015 m ⎞
20
r = 2.3 × 104 light-years ⎜
⎟ = 2.2 × 10 m
⎜ 1 light-year ⎟
⎝
⎠
(
)
Chapter 8 Problems
405
The tangential speed of the sun is
vT = rω = ( 2.2 × 1020 m )(1.1 × 10−15 rad/s ) = 2.4 × 105 m/s
(8.9)
b. The magnitude of the centripetal force that acts on the sun is
= mac = m r ω 2
Fc
Centripetal
force
(
)(
)(
= 1.99 × 1030 kg 2.2 × 1020 m 1.1 × 10−15 rad /s
45.
)
2
= 5.3 × 1020 N
REASONING The tangential acceleration and the centripetal acceleration of a
point at a distance r from the rotation axis are given by Equations 8.10 and 8.11,
respectively: aT = rα and a c = rω 2 . After the drill has rotated through the angle in
question, a c = 2a T , or
SSM
rω 2 = 2rα
This expression can be used to find the angular acceleration α . Once the angular
acceleration is known, Equation 8.8 can be used to find the desired angle.
SOLUTION Solving the expression obtained above for α gives
α=
ω2
2
Solving Equation 8.8 for θ (with ω0 = 0 rad/s since the drill starts from rest), and using the
expression above for the angular acceleration α gives
θ=
⎛ω2 ⎞⎛ 2 ⎞
ω2
ω2
⎜
⎟
=
=
= 1.00 rad
2α 2(ω 2 / 2) ⎝ 2 ⎠ ⎝ ω 2 ⎠
Note that since both Equations 8.10 and 8.11 require that the angles be expressed in radians,
the final result for θ is in radians.
46. REASONING AND SOLUTION The bike would travel with the same speed as a point on
the wheel v = rω . It would then travel a distance
⎛ 60 s ⎞
x = v t = r ω t = ( 0.45 m )( 9.1 rad/s )( 35 min ) ⎜
⎟=
⎝ 1 min ⎠
8.6 ×103 m
406 ROTATIONAL KINEMATICS
____________________________________________________________________________________________
47.
REASONING AND SOLUTION
acceleration of the motorcycle is
SSM
WWW
a=
v − v0
t
=
From Equation 2.4, the linear
22.0 m/s − 0 m/s
= 2.44 m/s 2
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 α gives
α=
aT 2.44 m/s 2
2
=
= 8.71 rad/s
r
0.280 m
48. REASONING AND SOLUTION
a. If the wheel does not slip, a point on the rim rotates about the axle with a speed
vT = v = 15.0 m/s
For a point on the rim
ω = vT/r = (15.0 m/s)/(0.330 m) = 45.5 rad/s
b.
vT = rω = (0.175 m)(45.5 rad/s) = 7.96 m/s
49. REASONING The angular displacement θ of each wheel is given by Equation 8.7
(θ = ω0t + 12 α t 2 ) , which is one of the equations of rotational kinematics. In this expression
ω0 is the initial angular velocity, and α is the angular acceleration, neither of which is given
directly. Instead the initial linear velocity v0 and the linear acceleration a are given.
However, we can relate these linear quantities to their analogous angular counterparts by
means of the assumption that the wheels are rolling and not slipping. Then, according to
Equation 8.12 (v0 = rω0), we know that ω0 = v0/r, where r is the radius of the wheels.
Likewise, according to Equation 8.13 (a = rα), we know that α = a/r. Both Equations 8.12
and 8.13 are only valid if used with radian measure. Therefore, when we substitute the
expressions for ω0 and α into Equation 8.7, the resulting value for the angular displacement
θ will be in radians.
SOLUTION Substituting ω0 from Equation 8.12 and α from Equation 8.13 into Equation
8.7, we find that
Chapter 8 Problems
⎛v ⎞
407
⎛a⎞
θ = ω 0t + 12 α t 2 = ⎜ 0 ⎟ t + 12 ⎜ ⎟ t 2
⎝r⎠
⎝r ⎠
2
2
⎛ 20.0 m/s ⎞
1 ⎛ 1.50 m/s ⎞
=⎜
+
8.00
s
(
)
⎜
⎟ ( 8.00 s ) = 693 rad
⎟
2
⎝ 0.300 m ⎠
⎝ 0.300 m ⎠
50. REASONING
a. The constant angular acceleration α of the wheel is defined by Equation 8.4 as the change
in the angular velocity, ω −ω0, divided by the elapsed time t, or α = (ω − ω0 ) / t . The time is
known. Since the wheel rotates in the positive direction, its angular velocity is the same as
its angular speed. However, the angular speed is related to the linear speed v of a wheel and
its radius r by v = rω (Equation 8.12). Thus, ω = v / r , and we can write for the angular
acceleration that
v v0
ω − ω0 r − r v − v0
=
=
α=
t
t
rt
b. The angular displacement θ of each wheel can be obtained from Equation 8.7 of the
equations of kinematics: θ = ω0t + 12 α t 2 , where ω0 = v0/r and α can be obtained as
discussed in part (a).
SOLUTION
a. The angular acceleration of each wheel is
α=
v − v0
rt
=
2.1 m/s − 6.6 m/s
= −1.4 rad/s 2
( 0.65 m )( 5.0 s )
b. The angular displacement of each wheel is
⎛ v0 ⎞ 1 2
⎟t + α t
⎝ r ⎠ 2
θ = ω0t + 12 α t 2 = ⎜
2
⎛ 6.6 m/s ⎞ (
2
1
=⎜
⎟ 5.0 s ) + 2 ( −1.4 rad/s ) ( 5.0 s ) = +33 rad
⎝ 0.65 m ⎠
______________________________________________________________________________
51.
REASONING The angle through which the tire rotates is equal to its
average angular velocity ω multiplied by the elapsed time t, θ = ω t. According to Equation
8.6, this angle is related to the initial and final angular velocities of the tire by
SSM
WWW
θ =ωt =
1
2
(ω 0 + ω ) t
408 ROTATIONAL KINEMATICS
The tire is assumed to roll at a constant angular velocity, so that ω0 = ω and θ = ω t. Since
the tire is rolling, its angular speed is related to its linear speed v by Equation 8.12, v = rω,
where r is the radius of the tire. The angle of rotation then becomes
⎛v⎞
⎝r⎠
θ =ωt =⎜ ⎟t
The time t that it takes for the tire to travel a distance x is equal to t = x/v, according to
Equation 2.1. Thus, the angle that the tire rotates through is
⎛v⎞
⎝r⎠
⎛v⎞⎛ x⎞
⎝r⎠⎝v⎠
θ =⎜ ⎟t =⎜ ⎟⎜ ⎟=
x
r
SOLUTION Since 1 rev = 2π rad, the angle (in revolutions) is
θ=
3
x 96 000 × 10 m
8
=
= 3.1 × 10 rad
r
0.31 m
) ⎛⎝ 21πrevrad ⎞⎟⎠ =
(
θ = 3.1 × 108 rad ⎜
7
4.9 × 10 rev
52. REASONING AND SOLUTION
a. If the rope is not slipping on the cylinder, then the tangential speed of the teeth on the
larger gear (gear 1) is 2.50 m/s. The angular speed of gear 1 is then
ω1 = v/r1 = (2.50 m/s)/(0.300 m) = 8.33 rad/s
The direction of the larger gear is counterclockwise .
b. The gears are in contact and do not slip. This requires that the teeth on both gears move
with the same tangential speed.
vT1 = vT2
or
ω1r1 = ω2r2
So
⎛r ⎞
⎛ 0.300 m ⎞
ω 2 = ⎜ 1 ⎟ ω1 = ⎜
⎟ ( 8.33 rad/s ) = 14.7 rad/s
⎝ 0.170 m ⎠
⎝ r2 ⎠
Chapter 8 Problems
409
The direction of the smaller gear is clockwise .
53. REASONING AND SOLUTION The angular speed of the ball can be found from
Equation 8.12, v = rω. Solving for ω gives
ω=
v 3.60 m/s
=
= 18.0 rad/s
r 0.200 m
The time it takes for the ball to fall through the vertical distance y can be found from
Equation 3.5b:
1
2
y = v0 y t + a y t
2
Solving for t (with v0y = 0 m/s, and taking "up" to be positive) gives
t=
2y
=
ay
2(−2.10 m)
2
(−9.80 m/s )
= 0.655 s
Since the angular speed of the ball remains constant while it is falling,
θ = ω t = (18.0 rad/s)(0.655 s) = 11.8 rad
54. REASONING As a penny-farthing moves, both of its wheels roll without slipping. This
means that the axle for each wheel moves through a linear distance (the distance through
which the bicycle moves) that equals the circular arc length measured along the outer edge
of the wheel. Since both axles move through the same linear distance, the circular arc length
measured along the outer edge of the large front wheel must equal the circular arc length
measured along the outer edge of the small rear wheel. In each case the arc length s is equal
to the number n of revolutions times the circumference 2π r of the wheel (r = radius).
SOLUTION Since the circular arc length measured along the outer edge of the large front
wheel must equal the circular arc length measured along the outer edge of the small rear
wheel, we have
nRear 2π rRear = nFront 2π rFront
Arc length for rear
wheel
Solving for nRear gives
nRear =
nFront rFront
rRear
=
Arc length for front
wheel
276 (1.20 m )
= 974 rev
0.340 m
410 ROTATIONAL KINEMATICS
55.
REASONING AND SOLUTION By inspection, the distance traveled by the "axle"
or the center of the moving quarter is
SSM
d = 2 π (2 r ) = 4π r
where r is the radius of the quarter. The distance d traveled by the "axle" of the moving
quarter must be equal to the circular arc length s along the outer edge of the quarter. This
arc length is s = rθ , where θ is the angle through which the quarter rotates. Thus,
so that θ = 4π rad . This is equivalent to
4 π r = rθ
⎛ 1 rev ⎞
⎟ = 2 revolutions
(4π rad)⎜
⎝ 2π rad ⎠
56. REASONING AND SOLUTION
a. According to Equation 8.9, the tangential speed of the sun is
vT = r ω = (2.2 × 10 20 m)(1.2 × 10 –15 rad/s) = 2.6 × 105 m/s
b. According to Equation 8.2, ω = ∆θ / ∆t . Since the angular speed of the sun is constant,
ω = ω . Solving for ∆ t , we have
∆t =
57.
⎞
2π rad
1y
⎛
⎞ ⎛ 1 h ⎞ ⎛ 1 day ⎞ ⎛
8
=⎜
⎜
⎟ = 1.7 × 10 y
⎜
⎟
⎜
⎟
⎟
−
15
ω ⎝ 1.2 × 10 rad/s ⎠ ⎝ 3600 s ⎠ ⎝ 24 h ⎠ ⎝ 365.25 day ⎠
∆θ
REASONING The tangential acceleration aT of the speedboat can be found by
using Newton's second law, FT = ma T , where FT is the net tangential force. Once the
tangential acceleration of the boat is known, Equation 2.4 can be used to find the tangential
speed of the boat 2.0 s into the turn. With the tangential speed and the radius of the turn
known, Equation 5.2 can then be used to find the centripetal acceleration of the boat.
SSM
SOLUTION
a. From Newton's second law, we obtain
F
550 N
aT = T =
= 2.5 m/s2
m 220 kg
b. The tangential speed of the boat 2.0 s into the turn is, according to Equation 2.4,
v T = v 0T + aT t = 5.0 m/s + (2.5 m/s 2 )(2.0 s) = 1.0 × 101 m/s
Chapter 8 Problems
411
The centripetal acceleration of the boat is then
ac =
v 2T
r
=
(1.0 × 10 1 m/s) 2
= 3.1 m/s 2
32 m
58. REASONING The angular displacement is given as θ = 0.500 rev, while the initial angular
velocity is given as ω0 = 3.00 rev/s and the final angular velocity as ω = 5.00 rev/s. Since
we seek the time t, we can use Equation 8.6 ⎡⎣θ =
rotational kinematics to obtain it.
1
2
(ω
0
+ ω ) t ⎤⎦ from the equations of
SOLUTION Solving Equation 8.6 for the time t, we find that
t=
59.
2 ( 0.500 rev )
2θ
=
= 0.125 s
ω 0 + ω 3.00 rev/s + 5.00 rev/s
REASONING AND SOLUTION Since the angular speed of the fan decreases, the
sign of the angular acceleration must be opposite to the sign for the angular velocity.
Taking the angular velocity to be positive, the angular acceleration, therefore, must be a
negative quantity. Using Equation 8.4 we obtain
SSM
ω 0 = ω − α t = 83.8 rad/s – (–42.0 rad/s 2 )(1.75 s) = 157.3 rad/s
60. REASONING AND SOLUTION
a. The tangential acceleration of the train is given by Equation 8.10 as
aT = r α = (2.00 × 102 m)(1.50 × 10−3 rad/s 2 ) = 0.300 m/s 2
The centripetal acceleration of the train is given by Equation 8.11 as
ac = r ω 2 = (2.00 × 102 m)(0.0500 rad/s)2 = 0.500 m/s 2
The magnitude of the total acceleration is found from the Pythagorean theorem to be
2
2
a = aT + ac = 0.583 m/s
2
b. The total acceleration vector makes an angle relative to the radial acceleration of
412 ROTATIONAL KINEMATICS
θ=
⎛a
tan −1 ⎜⎜ T
⎜a
⎝ c
⎞
⎟ =
⎟⎟
⎠
⎛
−1 ⎜ 0.300
tan ⎜
⎜ 0.500
⎝
m/s 2 ⎞⎟
⎟ =
m/s 2 ⎟⎠
31.0°
61. REASONING The angular speed ω of the sprocket can be calculated from the tangential
speed vT and the radius r using Equation 8.9 (vT = rω). The radius is given as
r = 4.0 × 10-2 m. The tangential speed is identical to the linear speed given for a chain link
at point A, so that vT = 5.6 m/s. We need to remember, however, that Equation 8.9 is only
valid if radian measure is used. Thus, the value calculated for ω will be in rad/s, and we
will have to convert to rev/s using the fact that 2π rad equals 1 rev.
SOLUTION Solving Equation 8.9 for the angular speed ω gives
ω=
vT
r
=
5.6 m/s
= 140 rad/s
4.0 × 10−2 m
Using the fact that 2π rad equals 1 rev, we can convert this result as follows:
⎛ 1 rev ⎞
⎟ = 22 rev/s
⎝ 2π rad ⎠
ω = (140 rad/s ) ⎜
62. REASONING
a. Since the angular velocity of the fan blade is
changing, there are simultaneously a tangential
acceleration aT and a centripetal acceleration ac that are
oriented at right angles to each other. The drawing
shows these two accelerations for a point on the tip of
one of the blades (for clarity, the blade itself is not
shown). The blade is rotating in the counterclockwise
(positive) direction.
a
aT
φ
ac
The magnitude of the total acceleration is
a = ac2 + aT2 , according to the Pythagorean theorem.
The magnitude ac of the centripetal acceleration can be
evaluated from ac = rω 2 (Equation 8.11), where ω is the final angular velocity. The final
angular velocity can be determined from Equation 8.4 as ω = ω0 + α t . The magnitude aT of
the tangential acceleration follows from aT = rα (Equation 8.10).
b. From the drawing we see that the angle φ can be obtained by using trigonometry,
ϕ = tan −1 ( aT / ac ) .
Chapter 8 Problems
413
SOLUTION
a. Substituting ac = rω 2 (Equation 8.11) and aT = rα (Equation 8.10) into a = ac2 + aT2
gives
a = ac2 + aT2 =
( rω 2 )2 + ( rα )2
= r ω4 +α 2
The final angular velocity ω is related to the initial angular velocity ω0 by ω = ω0 + α t (see
Equation 8.4). Thus, the magnitude of the total acceleration is
a = r ω4 + α 2 = r
(ω0 + α t )4 + α 2
= ( 0.380 m ) ⎡⎣1.50 rad/s + ( 2.00 rad/s 2 ) ( 0.500 s ) ⎤⎦ + ( 2.00 rad/s 2 ) = 2.49 m/s 2
4
2
b. The angle φ between the total acceleration a and the centripetal acceleration ac is (see the
drawing above)
⎛ aT
⎝ ac
ϕ = tan −1 ⎜
⎞
α
⎡
⎤
−1 ⎛ r α ⎞
= tan −1 ⎢
⎟ = tan ⎜
2⎟
2⎥
⎝ rω ⎠
⎠
⎢⎣ (ω0 + α t ) ⎥⎦
⎧
⎫⎪
2.00 rad/s 2
= tan −1 ⎪⎨
= 17.7°
2⎬
2 )(
⎡
⎤
(
)
⎩⎪ ⎣1.50 rad/s + 2.00 rad/s 0.500 s ⎦ ⎭⎪
where we have used the same substitutions for aT, ac, and ω as in part (a).
______________________________________________________________________________
63.
REASONING AND SOLUTION
a. From Equation 8.9, and the fact that 1 revolution = 2π radians, we obtain
SSM
rev ⎞ ⎛ 2π rad ⎞
⎛
= 1.25 m/s
vT = r ω = (0.0568 m)⎝ 3.50
s ⎠ ⎝ 1 rev ⎠
b. Since the disk rotates at constant tangential speed,
v T1 = v T2
or
ω 1 r1 = ω 2 r2
Solving for ω 2 , we obtain
ω2 =
ω 1 r1
r2
=
(3.50 rev/s)(0.0568 m)
= 7.98 rev/s
0.0249 m
414 ROTATIONAL KINEMATICS
64. REASONING The average angular velocity is defined as the angular displacement divided
by the elapsed time (Equation 8.2). Therefore, the angular displacement is equal to the
product of the average angular velocity and the elapsed time The elapsed time is given, so
we need to determine the average angular velocity. We can do this by using the graph of
angular velocity versus time that accompanies the problem.
+15 rad/ s
Angular velocity
SOLUTION The angular displacement ∆θ is
related to the average angular velocity ω and
the elapsed time ∆t by Equation 8.2, ∆θ = ω ∆t .
The elapsed time is given as 8.0 s. To obtain the
average angular velocity, we need to extend the
graph that accompanies this problem from a
time of 5.0 s to 8.0 s. It can be seen from the
graph that the angular velocity increases by
+3.0 rad/s during each second. Therefore, when
the time increases from 5.0 to 8.0 s, the angular
velocity increases from +6.0 rad/s to
6 rad/s + 3×(3.0 rad/s) = +15 rad/s. A graph of
the angular velocity from 0 to 8.0 s is shown at
the right. The average angular velocity during
this time is equal to one half the sum of the
initial and final angular velocities:
+3.0 rad/s
ω
0
Time (s)
3.0 s
8.0 s
–9.0 rad/ s
ω = 12 (ω 0 + ω ) = 12 ( −9.0 rad/s + 15 rad/s ) = + 3.0 rad/s
The angular displacement of the wheel from 0 to 8.0 s is
∆θ = ω ∆t = ( +3.0 rad/s )( 8.0 s ) = +24 rad
65. REASONING AND SOLUTION
circumstance.
The angular acceleration is found for the first
ω 2 − ω 02 ( 3.14 × 104 rad/s ) − (1.05 × 104 rad/s )
=
= 2.33 × 104 rad/s 2
α=
2θ
2 (1.88 × 104 rad )
2
2
For the second circumstance
t=
ω − ω 0 7.85 × 104 rad/s − 0 rad/s
=
= 3.37 s
α
2.33 × 104 rad/s 2
____________________________________________________________________________________________
Chapter 8 Problems
66. REASONING The drawing shows a top view of
the race car as it travels around the circular turn.
Its acceleration a has two perpendicular
components: a centripetal acceleration ac that
arises because the car is moving on a circular path
and a tangential acceleration aT due to the fact that
the car has an angular acceleration and its angular
velocity is increasing. We can determine the
magnitude of the centripetal acceleration from
Equation
8.11
as
2
ac = rω , since both r and ω are given in the
statement of the problem. As the drawing shows,
we can use trigonometry to determine the
magnitude a of the total acceleration, since the
angle (35.0°) between a and ac is given.
415
aT
a
35.0°
ac
Race car
SOLUTION Since the vectors ac and a are one side and the hypotenuse of a right triangle,
we have that
a=
ac
cos 35.0°
The magnitude of the centripetal acceleration is given by Equation 8.11 as ac = rω2, so the
magnitude of the total acceleration is
( 23.5 m )( 0.571 rad /s )
rω2
=
=
= 9.35 m /s 2
a=
cos 35.0° cos 35.0°
cos 35.0°
2
ac
67. REASONING AND SOLUTION The distance d traveled by the axle of the wheel is given
by Equation 8.1 as d = r θ , where r is the distance from the center of the circular hill to the
axle (r = 9.00 m − 0.400 m = 8.60 m) and θ = 0.960 rad. Thus,
d = (8.60 m)(0.960 rad) = 8.26 m
According to the discussion in Section 8.6, the distance d traveled by the axle equals the
circular arc length s along the outer edge of the rotating wheel; d = s. But s = rw θw , where
rw is the radius of the wheel and θw is the angle through which the wheel rotates; so, d = s =
rw θw. Solving for θw,
θw =
d
8.26 m
=
= 20.6 rad
rw 0.400 m
416 ROTATIONAL KINEMATICS
68. REASONING AND SOLUTION
The figure at the right shows the
relevant angles and dimensions for
either one of the celestial bodies
under consideration.
r
θ
s
person on earth
celestial body
a. Using the figure above
θ moon =
θ sun =
smoon 3.48 × 10 6 m
–3
=
= 9.04 × 10 rad
8
rmoon 3.85 × 10 m
s sun 1.39 × 109 m
–3
=
= 9.27 × 10 rad
rsun 1.50 × 1011 m
b. Since the sun subtends a slightly larger angle than the moon, as measured by a person
standing on the earth, the sun cannot be completely blocked by the moon. Therefore,
a "total" eclipse of the sun is not really total .
c. The relevant geometry is shown below.
r sun
R sun
s sun
r moon
R
b
s
b
θ sun
moon
θ moon
person on earth
The apparent circular area of the sun as measured by a person standing on the earth is given
2
by: Asun = π Rsun
, where Rsun is the radius of the sun. The apparent circular area of the sun
that is blocked by the moon is Ablocked = π Rb2 , where Rb is shown in the figure above. Also
from the figure above, it follows that
Rsun = (1/2) ssun
and
Rb = (1/2) sb
Therefore, the fraction of the apparent circular area of the sun that is blocked by the moon is
Chapter 8 Problems
Ablocked
Asun
=
π Rb2
2
π Rsun
=
⎛θ
= ⎜⎜ moon
⎝ θ sun
⎛ s
=⎜ b
2
⎜s
/ 2)
⎝ sun
π ( sb / 2) 2
π ( ssun
2
⎞
⎛ θ moon rsun
⎟⎟ = ⎜⎜
⎠
⎝ θ sun rsun
⎞
⎟⎟
⎠
417
2
2
2
⎛ 9.04 × 10−3 rad ⎞
⎞
=
⎜
⎟ = 0.951
⎟⎟
⎜ 9.27 × 10−3 rad ⎟
⎠
⎝
⎠
The moon blocks out 95.1 percent of the apparent circular area of the sun.
69. CONCEPT QUESTIONS
a. The average angular velocity ω has the same direction as θ − θ 0 , because ω =
θ − θ0
t − t0
according to Equation 8.2. If θ is greater than θ0, then ω is positive. If θ is less than θ0,
then ω is negative.
b. The direction (positive or negative) of the average angular velocity for each entry in the
table is:
(a) ω is positive, because 0.75 rad is greater than 0.45 rad.
(b) ω is negative, because 0.54 rad is less than 0.94 rad.
(c) ω is negative, because 4.2 rad is less than 5.4 rad.
(d) ω is positive, because 3.8 rad is greater than 3.0 rad.
SOLUTION The average angular velocity is given by Equation 8.2 as ω =
t – t0 = 2.0 s is the elapsed time:
(a )
ω=
(b)
ω=
(c)
ω=
(d)
ω=
θ − θ0
t − t0
θ − θ0
t − t0
θ − θ0
t − t0
θ − θ0
t − t0
=
0.75 rad − 0.45 rad
= +0.15 rad /s
2.0 s
=
0.54 rad − 0.94 rad
= −0.20 rad /s
2.0 s
=
4.2 rad − 5.4 rad
= −0.60 rad /s
2.0 s
=
3.8 rad − 3.0 rad
= +0.4 rad /s
2.0 s
θ − θ0
t − t0
, where
418 ROTATIONAL KINEMATICS
70. CONCEPT QUESTIONS
a. The average angular acceleration has the same direction as ω − ω0, because
ω − ω0
α=
, according to Equation 8.4. If ω is greater than ω0, α is positive. If ω is less
t − t0
than ω0, α is negative.
b. The direction (positive or negative) of the average angular acceleration for each entry in
the table is:
(a) α is positive, because 5.0 rad/s is greater than 2.0 rad/s.
(b) α is negative, because 2.0 rad/s is less than 5.0 rad/s.
(c) α is positive, because −3.0 rad/s is greater than −7.0 rad/s.
(d) α is negative, because −4.0 rad/s is less than +4.0 rad/s.
SOLUTION The average angular acceleration is given by Equation 8.4 as α =
ω − ω0
t − t0
,
where t – t0 = 4.0 s is the elapsed time.
(a )
α=
(b)
α=
(c)
α=
(d)
α=
ω − ω0
t − t0
ω − ω0
t − t0
ω − ω0
t − t0
ω − ω0
t − t0
=
+5.0 rad /s − 2.0 rad /s
2
= +0.75 rad /s
4.0 s
=
+2.0 rad /s − 5.0 rad /s
2
= −0.75 rad /s
4.0 s
=
=
−3.0 rad /s − ( −7.0 rad /s )
4.0 s
−4.0 rad /s − ( +4.0 rad /s )
4.0 s
= +1.0 rad /s
2
= −2.0 rad /s
2
71. CONCEPT QUESTION The relation between the final angular velocity ω, the initial
angular velocity ω0, and the angular acceleration α is given by Equation 8.4 (with t0 = 0 s)
as
ω = ω0 + α t
If α has the same sign as ω0, then the angular speed, which is the magnitude of the angular
velocity ω, is increasing. On the other hand, If α and ω0 have opposite signs, then the
angular speed is decreasing.
Chapter 8 Problems
419
(a) ω0 and α have the same sign, so the angular speed is increasing.
(b) ω0 and α have opposite signs, so the angular speed is decreasing.
(c) ω0 and α have opposite signs, so the angular speed is decreasing.
(d) ω0 and α have the same sign, so the angular speed is increasing.
SOLUTION According to Equation 8.4, we know that ω = ω0 + α t. Therefore, we find:
(
) ( 2.0 s ) = + 18 rad /s . The angular speed is 18 rad/s .
(
)
(
)
(
) ( 2.0 s ) = − 18 rad /s. The angular speed is 18 rad/s .
(a) ω = + 12 rad /s + +3.0 rad /s 2
(b) ω = + 12 rad /s + −3.0 rad /s2 ( 2.0 s ) = + 6.0 rad /s. The angular speed is 6.0 rad/s .
(c) ω = − 12 rad /s + +3.0 rad /s2 ( 2.0 s ) = − 6.0 rad /s. The angular speed is 6.0 rad /s .
(d) ω = − 12 rad /s + −3.0 rad /s 2
72. CONCEPT QUESTIONS
a. No. The angular velocity does not change, so the angular acceleration α is zero. The
tangential acceleration aT is given by Equation 8.10 as aT = rα, so it is also zero.
b. Yes, because the nonzero angular acceleration α gives rise to a nonzero tangential
acceleration aT according to Equation 8.10, aT = rα.
SOLUTION Let r be the radial distance of the point from the axis of rotation. Then,
according to Equation 8.10, we have
g = rα
aT
Thus,
r=
g
α
=
9.80 m /s 2
12.0 rad /s 2
= 0.817 m
73. CONCEPT QUESTIONS
a. The angular displacement is greater than ω0t. When the angular displacement θ is given
by the expression θ = ω0t, it is assumed that the angular velocity remains constant at its
420 ROTATIONAL KINEMATICS
initial (and smallest) value of ω0 for the entire time. This expression does not account for
the additional angular displacement that occurs because the angular velocity is increasing.
b. The angular displacement is less than ω t. When the angular displacement is given by the
expression θ = ωt, it is assumed that the angular velocity remains constant at its final (and
largest) value of ω for the entire time. This expression does not account for the fact that the
wheel was rotating at a smaller angular velocity during the time interval. Thus, this
expression “overestimates” the angular displacement.
c. The angular displacement is given as the product of the average angular velocity and the
time
θ = ω t = 12 (ω0 + ω ) t
Average angular
velocity
SOLUTION
a. If the angular velocity is constant and equals the initial angular velocity ω0, then ω = ω0
and the angular displacement is
θ = ω 0 t = ( +220 rad /s )(10.0 s ) = +2200 rad
b. If the angular velocity is constant and equals the final angular velocity ω, then ω = ω
and the angular displacement is
θ = ω t = ( +280 rad /s )(10.0 s ) = +2800 rad
c. Using the definition of average angular velocity, we have
θ=
1
2
(ω0 + ω ) t = 12 ( +220 rad /s + 280 rad /s )(10.0 s ) =
+2500 rad
(8.6)
74. CONCEPT QUESTIONS
a. The wheels on the winning dragster roll without slipping. The wheels on the losing
dragster slip during part of the time. During the slippage, the wheels rotate, but the speed of
the car does not increase. Thus, its speed at any given instant is less than that of the winning
car.
b. For the wheels that roll without slipping, the relationship between their linear speed v
and the angular speed ω is given by Equation 8.12 as v = rω, where r is the radius of a
wheel.
c. For the wheels that roll without slipping, the relationship between the magnitude a of
their linear acceleration and the magnitude α of the angular acceleration is given by
Equation 8.13 as a = rα, where r is the radius of a wheel.
Chapter 8 Problems
421
SOLUTION
a. From Equation 8.12 we have that
v = r ω = ( 0.320 m )( 288 rad /s ) = 92.2 m /s
b. The magnitude of the angular acceleration is given by Equation 8.13 as α = a/r. The
linear acceleration a is related to the initial and final linear speeds and the displacement x by
v 2 − v02
Equation 2.9 from the equations of kinematics for linear motion; a =
. Thus, the
2x
magnitude of the angular acceleration is
v 2 − v02
a
α = = 2x
r
r
=
v 2 − v02
2x r
( 92.2 m /s )2 − ( 0 m /s )2
=
2 ( 384 m )( 0.320 m )
= 34.6 rad /s 2
75. CONCEPT QUESTIONS
a. It does not matter whether the arrow is aimed closer to or farther away from the axis. The
blade edge sweeps through the open angular space as a rigid unit. This means that a point
closer to the axis has a smaller distance to travel along the circular arc in order to bridge the
angular opening and correspondingly has a smaller tangential speed. A point farther from
the axis has a greater distance to travel along the circular arc but correspondingly has a
greater tangential speed. These speeds have just the right values so that all points on the
blade edge bridge the angular opening in the same time interval.
b. The rotational speed of the blades must not be so fast that one blade rotates into the open
angular space while part of the arrow is still there. A faster arrow speed means that the
arrow spends less time in the open space. Thus, the blades can rotate more quickly into the
open space without hitting the arrow, so the maximum value of the angular speed ω
increases with increasing arrow speed v.
c. A longer arrow traveling at a given speed means that some part of the arrow is in the
open space for a longer time. To avoid hitting the arrow, then, the blades must rotate more
slowly. Thus, the maximum value of the angular speed ω decreases with increasing arrow
length L.
SOLUTION The time during which some part of the arrow remains in the open angular
space is the time it takes the arrow to travel at a speed v through a distance equal to its own
length L. This time is tArrow = L/v. The time it takes for the edge to rotate at an angular
speed ω through the angle θ between the blades is tBlade = θ/ω. The maximum angular speed
is the angular speed such that these two times are equal. Therefore, we have
422 ROTATIONAL KINEMATICS
L
θ
=
v
ω
Arrow
Blade
In this expression we note that the value of the angular opening is θ = 60.0º, which is
θ = 16 ( 2π ) rad = 13 π rad . Solving the expression for ω gives
ω=
θv
L
=
πv
3L
Substituting the given values for v and L into this result, we find that
a.
ω=
b.
ω=
c.
ω=
πv
3L
πv
3L
πv
3L
=
=
=
π ( 75.0 m/s )
3 ( 0.71 m )
π ( 91.0 m/s )
3 ( 0.71 m )
π ( 91.0 m/s )
3 ( 0.81 m )
= 111 rad/s
= 134 rad/s
= 118 rad/s
These answers are consistent with the answers to the Concept Questions.
76. CONCEPT QUESTIONS
a. In addition to knowing the initial angular velocity ω0 and the acceleration α, we know
that the final angular velocity ω is 0 rev/s, because the wheel comes to a halt. With values
available for these three variables, the unknown angular displacement θ can be calculated
from Equation 8.8 (ω 2 = ω 02 + 2αθ ) .
b. When using any of the equations of rotational kinematics, it is not necessary to use
radian measure. Any self-consistent set of units may be used to measure the angular
quantities, such as revolutions for θ, rev/s for ω0 and ω, and rev/s2 for α.
c. A greater initial angular velocity does not necessarily mean that the wheel will come to a
halt on an angular section labeled with a greater number. It is certainly true that greater
initial angular velocities lead to greater angular displacements for a given deceleration.
However, remember that the angular displacement of the wheel in coming to a halt may
consist of a number of complete revolutions plus a fraction of a revolution. In deciding on
which number the wheel comes to a halt, the number of complete revolutions must be
Chapter 8 Problems
423
subtracted from the angular displacement, leaving only the fraction of a revolution
remaining.
SOLUTION Solving Equation 8.8 for the angular displacement gives θ =
ω 2 − ω 02
.
2α
a. We know that ω0 = +1.20 rev/s, ω = 0 rev/s, and α = -0.200 rev/s2, where ω0 is positive
since the rotation is counterclockwise and, therefore, α is negative because the wheel
decelerates. The value obtained for the displacement is
ω 2 − ω02 ( 0 rev/s ) − ( +1.20 rev/s )
θ=
=
= +3.60 rev
2α
2 ( −0.200 rev/s 2 )
2
2
To decide where the wheel comes to a halt, we subtract the three complete revolutions from
this result, leaving 0.60 rev. Converting this value into degrees and noting that each angular
section is 30.0º, we find the following number n for the section where the wheel comes to a
halt:
⎛ 360° ⎞ ⎛ 1 angular section ⎞
n = ( 0.60 rev ) ⎜
⎟⎜
⎟ = 7.2
30.0°
⎝ 1 rev ⎠ ⎝
⎠
A value of n = 7.2 means that the wheel comes to a halt in the section following number 7.
Thus, it comes to a halt in section 8 .
b. Following the same procedure as in part a, we find that
ω 2 − ω02 ( 0 rev/s ) − ( +1.47 rev/s )
θ=
=
= +5.40 rev
2α
2 ( −0.200 rev/s 2 )
2
2
Subtracting the five complete revolutions from this result leaves 0.40 rev. Converting this
value into degrees and noting that each angular section is 30.0º, we find the following
number n for the section where the wheel comes to a halt:
⎛ 360° ⎞ ⎛ 1 angular section ⎞
n = ( 0.40 rev ) ⎜
⎟⎜
⎟ = 4.8
30.0°
⎝ 1 rev ⎠ ⎝
⎠
A value of n = 4.8 means that the wheel comes to a halt in the section following number 4.
Thus, it comes to a halt in section 5 . We see, then, that a greater initial velocity does not
mean that the wheel comes to a halt in a section labeled with a greater number.