CHAPTER 8
ROTATIONAL KINEMATICS
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) Using Equation 8.1 (θ = Arc length / Radius) to calculate the angle (in radians) that each
object subtends at your eye shows that θMoon = 9.0 × 10−3 rad, θPea = 7.0 × 10−3 rad, and
θDime = 25 × 10−3 rad. Since θPea is less than θMoon, the pea does not completely cover your
view of the moon. However, since θDime is greater than θMoon, the dime does completely
cover your view of the moon.
2. 2.20 cm
3. 38.2 s
4. (a) An angular acceleration of zero means that the angular velocity has the same value at all
times, as in statements A or B. However, statement C is also consistent with a zero angular
acceleration, because if the angular displacement does not change as time passes, then the
angular velocity remains constant at a value of 0 rad/s.
5. (c) A non-zero angular acceleration means that the angular velocity is either increasing or
decreasing. The angular velocity is not constant.
6. (b) Since values are given for the initial angular velocity ω0, the final angular velocity ω, and
the time t, Equation 8.6 [ θ = 12 (ω0 + ω ) t ] can be used to calculate the angular displacement
θ.
7. 32 rad/s
8. 88 rad
9. (c) According to Equation 8.9 (vT = rω), the tangential speed is proportional to the radius r
when the angular speed ω is constant, as it is for the earth. As the elevator rises, the radius,
which is your distance from the center of the earth, increases, and so does your tangential
speed.
10. (b) According to Equation 8.9 (vT = rω), the tangential speed is proportional to the radius r
when the angular speed ω is constant, as it is for the merry-go-round. Thus, the angular
⎛ 2.1 m ⎞
speed of the second child is vT = ( 2.2 m/s ) ⎜
⎟.
⎝ 1.4 m ⎠
392 ROTATIONAL KINEMATICS
11. 367 rad/s2
12. (e) According to Newton’s second law, the centripetal force is given by Fc = mac , where m
is the mass of the ball and ac is the centripetal acceleration. The centripetal acceleration is
given by Equation 8.11 as ac = rω2, where r is the radius and ω is the angular speed.
Therefore, Fc = mrω 2 , and the centripetal force is proportional to the radius when the mass
and the angular speed are fixed, as they are in this problem. As a result,
⎛ 33 cm ⎞
Fc = (1.7 N ) ⎜
⎟.
⎝ 12 cm ⎠
13. (d) Since the angular speed ω is constant, the angular acceleration α is zero, according to
Equation 8.4. Since α = 0 rad/s2, the tangential acceleration aT is zero, according to
Equation 8.10. The centripetal acceleration ac, however, is not zero, since it is proportional
to the square of the angular speed, according to Equation 8.11, and the angular speed is not
zero.
14. 17.8 m/s2
15. (a) The number N of revolutions is the distance s traveled divided by the circumference 2πr
of a wheel: N = s/(2πr).
16. 27.0 m/s
Chapter 8 Problems
393
CHAPTER 8 ROTATIONAL KINEMATICS
PROBLEMS
1.
SSM 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
(1.00
2.
⎛ 1 rev ⎞ ⎛ 100 grad
rad ⎜
⎟⎜
⎝ 2π rad ⎠ ⎝ 0.250 rev
)
⎞
⎟ = 63.7 grad
⎠
REASONING The average angular velocity ω has the same direction as θ − θ 0 , because
θ − θ0
ω=
according to Equation 8.2. If θ is greater than θ0, then ω is positive. If θ is
t − t0
less than θ0, then ω is negative.
SOLUTION The average angular velocity is given by Equation 8.2 as ω =
t – t0 = 2.0 s is the elapsed time:
3.
(a )
ω=
θ − θ0
(b)
ω=
θ − θ0
(c)
ω=
θ − θ0
(d)
ω=
θ − θ0
t − t0
t − t0
t − t0
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
REASONING The average angular velocity ω is defined as the angular displacement
Δθ divided by the elapsed time Δt during which the displacement occurs: ω = Δθ / Δ 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.
394 ROTATIONAL KINEMATICS
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 ( Δθ )spin = +2π rad . The
average angular velocity is (converting 1 day to seconds):
ω=
( Δθ )spin
=
( Δt )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 ( Δθ )orbit = +2π rad . The average angular
velocity is (converting 365¼ days to seconds):
ω=
( Δθ )orbit
=
( Δt )orbit
(
+2π rad
⎛ 24 h
365 14 days ⎜
⎜ 1 day
⎝
)
⎞ ⎛ 3600 s ⎞
⎟⎟ ⎜
⎟
⎠⎝ 1 h ⎠
= +2.0 ×10−7 rad/s
____________________________________________________________________________________________
4.
REASONING The average angular velocity of either mandible is given by ω = Δθ Δt
(Equation 8.2), where Δθ is the angular displacement of the mandible and Δt is the elapsed
time. In order to calculate the average angular velocity in radians per second, we will first
convert the angular displacement Δθ from degrees to radians.
SOLUTION Converting an angular displacement of 90° into radians, we find that the
angular displacement of the mandible is
⎛
1 rev
Δθ = 90 degrees ⎜
⎜ 360 degrees
⎝
(
)
⎞ ⎛ 2π rad ⎞ π
= rad
⎟⎜
⎟ ⎝ 1 rev ⎟⎠ 2
⎠
The average angular velocity of the mandible is
π
rad
Δθ
ω=
= 2 −4 = 1.2 × 104 rad/s
Δt 1.3 × 10 s
(8.2)
Chapter 8 Problems
5.
395
SSM 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
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 Δθ
divided by the elapsed time Δt, the angular displacement is
Δθ = ω Δt = ( 35 rad/s )( 0.60 s ) = 21 rad
6.
(8.2)
REASONING The jet is maintaining a distance of r = 18.0 km from the air traffic control
tower by flying in a circle. The angle that the jet’s path subtends while its nose crosses over
the moon is the same as the angular width θ of the moon. The corresponding distance the jet
travels is the length of arc s subtended by the moon’s diameter. We will use the relation
s = rθ (Equation 8.1) to determine the distance s.
SOLUTION In order to use the relation s = rθ (Equation 8.1), the angle θ must be
expressed in radians, as it is. The result will have the same units as r. Because s is required
in meters, we first convert r to meters:
⎛ 1000 m ⎞
4
r = 18.0 km ⎜
⎟ = 1.8 × 10 m
⎝ 1 km ⎠
(
)
Therefore, the distance that the jet travels while crossing in front of the moon is
(
)(
)
s = rθ = 1.80 ×104 m 9.04 ×10−3 rad = 163 m
7.
REASONING 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.
396 ROTATIONAL KINEMATICS
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.
8.
(a )
α=
ω − ω0
(b)
α=
ω − ω0
(c)
α=
ω − ω0
(d)
α=
ω − ω0
t − t0
t − t0
t − t0
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
REASONING 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.
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 /s 2 ( 2.0 s ) = + 6.0 rad /s. The angular speed is 6.0 rad/s .
(c) ω = − 12 rad /s + +3.0 rad /s 2 ( 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
Chapter 8 Problems
9.
397
SSM REASONING 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.
SOLUTION Solving Equation 8.4 for the elapsed time gives
t=
ω − ω 0 0.24 rad/s − 0 rad/s
=
= 8.0 s
0.030 rad/s 2
α
10. REASONING AND SOLUTION Using Equation 8.4 and the appropriate conversion
factors, the average angular acceleration of the CD in rad/s2 is
α=
2
Δω ⎛ 210 rev/ min − 480 rev/ min ⎞ ⎛ 2π rad ⎞ ⎛ 1 min ⎞
2
−3
=⎜
⎟⎜
⎟⎜
⎟ = – 6.4 × 10 rad/s
Δt ⎝
74 min
⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠
The magnitude of the average angular acceleration is 6.4 × 10−3 rad/s2 .
11. REASONING The average angular velocity ω is defined as the angular displacement
Δθ divided by the elapsed time Δt during which the displacement occurs: ω = Δθ / Δ t
(Equation 8.2). Solving for the elapsed time gives Δ t = Δθ / ω . We are given Δθ 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 diameter, it moves through an angle that is also 9.28 × 10−3 rad; thus,
Δθ = 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 =
Δθearth
Δtearth
=
2π rad
=
24.0 h
2π rad
= 7.27 × 10−5 rad/s
3600 s ⎞
⎛
( 24.0 h ) ⎜
⎟
⎝ 1 h ⎠
The time it takes for the sun to move a distance equal to its diameter is
Δt =
Δθ
ωearth
=
9.28 × 10−3 rad
= 128 s (a little over 2 minutes)
7.27 × 10−5 rad/s
398 ROTATIONAL KINEMATICS
____________________________________________________________________________________________
12. REASONING AND SOLUTION The angular displacements of the astronauts are equal.
For A
θ = sA/rA
For B
θ = sB/rB
(8.1)
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
13. 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
ω1 + ω 2 1.7 × 10−3 rad/s + 3.4 × 10−3 rad/s
14. REASONING 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.
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.
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.
Chapter 8 Problems
399
The time during which some part of the arrow remains in the open angular space is tArrow.
The time it takes for the edge of a propeller blade to rotate through the open angular space
between the blades is tBlade. The maximum angular speed is the angular speed such that these
two times are equal.
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
L
θ
=
ω
v
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
15. 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
400 ROTATIONAL KINEMATICS
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
16. 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:
ω=
the required time is
Δt =
Δθ
ω
=
Δθ
Δt
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=
Δx
d
0.850 m
=
=
=
−3
Δt
Δt
2.53 × 10 s
336 m/s
17. SSM 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
= 2.00 × 10 s
⎛ 2π 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 the value that we found in part a,
or 4.00 × 10
−2
s.
Chapter 8 Problems
18. REASONING AND SOLUTION
The figure at the right shows the
relevant angles and dimensions for
either one of the celestial bodies
under consideration.
401
r
s
θ
person on earth
celestial body
a. Using the figure above
θ moon =
θ sun =
smoon 3.48 × 10 6 m
=
= 9.04 × 10 –3 rad
rmoon 3.85 × 10 8 m
ssun 1.39 × 109 m
–3
=
= 9.27 × 10 rad
11
rsun 1.50 × 10 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
b
r moon
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
, where Rsun is the radius of the sun. The apparent circular area of the sun
by: Asun = π Rsun
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
402 ROTATIONAL KINEMATICS
Ablocked
Asun
=
π Rb2
2
π Rsun
=
⎛θ
= ⎜⎜ moon
⎝ θ sun
π ( sb / 2) 2
π ( ssun
⎛ sb
=
⎜⎜
2
/ 2)
⎝ ssun
2
⎞
⎛ θ moon rsun
⎟⎟ = ⎜⎜
⎠
⎝ θ sun rsun
⎞
⎟⎟
⎠
2
2
2
⎛ 9.04 × 10−3 rad ⎞
⎞
⎟ = 0.951
⎟⎟ = ⎜⎜
−3
⎟
⎠
⎝ 9.27 × 10 rad ⎠
The moon blocks out 95.1 percent of the apparent circular area of the sun.
19. REASONING AND SOLUTION Since the ball spins at 7.7 rev/s, it makes (7.7 rev/s)t
revolutions while in flight, where t is the time of flight and must be determined. The ball’s
vertical displacement is y = 0 m since the ball returns to its launch point. The vertical
component of the ball’s initial velocity is v0y = (19 m/s) sin 55°, assuming upward to be the
positive direction. The acceleration due to gravity is ay = −9.80 m/s2. With these three
1
2
pieces of information at hand, we use y = v0 y t + a y t 2 (Equation 3.5b) to determine the
time of flight. Noting that y = 0 m, we can solve this expression for t and find that
t=−
2v0 y
ay
=−
2 (19 m/s ) sin 55°
−9.80 m/s
2
=3.2 s
and
Number of = 7.7 rev/s 3.2 s = 25 rev
)(
)
revolutions (
____________________________________________________________________________________________
20. 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=
2 ( 0.500 rev )
2θ
=
= 0.125 s
ω 0 + ω 3.00 rev/s + 5.00 rev/s
21. 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
Chapter 8 Problems
403
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
22. REASONING We are given the turbine’s angular acceleration α, final angular velocity ω,
and angular displacement θ. Therefore, we will employ ω 2 = ω02 + 2αθ (Equation 8.8) in
order to determine the turbine’s initial angular velocity ω0 for part a. However, in order to
make the units consistent, we will convert the angular displacement θ from revolutions to
radians before substituting its value into Equation 8.8. In part b, the elapsed time t is the
only unknown quantity. We can, therefore, choose from among ω = ω0 + α t (Equation 8.4),
θ = 12 (ω0 + ω ) t (Equation 8.6), or θ = ω0t + 12 α t 2 (Equation 8.7) to find the elapsed time.
Of the three, Equation 8.4 offers the least algebraic complication, so we will solve it for the
elapsed time t.
SOLUTION
a. One revolution is equivalent to 2π radians, so the angular displacement of the turbine is
⎛ 2π rad ⎞
4
⎟ = 1.80 ×10 rad
⎝ 1 rev ⎠
θ = ( 2870 rev ) ⎜
Solving ω 2 = ω02 + 2αθ (Equation 8.8) for the square of the initial angular velocity, we
obtain ω02 = ω 2 − 2αθ , or
ω0 = ω 2 − 2αθ =
(137 rad/s )2 − 2 ( 0.140 rad/s2 )(1.80 ×104 rad ) = 117 rad/s
b. Solving ω = ω0 + α t (Equation 8.4) for the elapsed time, we find that
t=
ω − ω0 137 rad/s − 117 rad/s
=
= 140 s
α
0.140 rad/s 2
23. SSM WWW REASONING AND SOLUTION
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
404 ROTATIONAL KINEMATICS
24. REASONING The angular displacement is given by Equation 8.6 as the product of the
average angular velocity and the time
θ =ωt =
1
2
(ω0 + ω )
t
Average angular
velocity
This value for 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 initial (and smallest) value of ω0 for the entire time, which does not, however, account
for the additional angular displacement that occurs because the angular velocity is
increasing.
The angular displacement is also less than ω t. When the angular displacement is given by
the expression θ = ω t , it is assumed that the angular velocity remains constant at its final
(and largest) value of ω for the entire time, which does not account for the fact that the
wheel was rotating at a smaller angular velocity during the time interval.
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)
25. 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
2θ
ω0 + ω
(see Equation 8.6). Solving for the
Chapter 8 Problems
405
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
10.0 s
t
______________________________________________________________________________
α=
=
26. 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 ⎠
27. SSM 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.
406 ROTATIONAL KINEMATICS
SOLUTION The angular displacement of each wheel is equal to the average angular
velocity multiplied by the time
θ=
1
2
(ω0 + 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
or
t
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
______________________________________________________________________________
28. 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 )
4
2
α=
=
= 2.33 × 10 rad/s
4
2θ
2 (1.88 × 10 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
____________________________________________________________________________________________
29. 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.
Chapter 8 Problems
407
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 )
2
= 6.39 × 103 rad
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
2
−3
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
30. 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 :
408 ROTATIONAL KINEMATICS
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
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
31. 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/s2
Solving Equation 8.2 for Δ θ , we find
Δθ = ω Δt = (1.6 rev/s)(1.3 s)= 2.1 rev
32. REASONING 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 ) .
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.
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
Chapter 8 Problems
409
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
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 .
410 ROTATIONAL KINEMATICS
33. SSM WWW 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
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 .
The angular acceleration α gives rise to a tangential acceleration aT
according to aT = rα (Equation 8.10). Moreover, it is given that aT = g, where g is the
magnitude of the acceleration due to gravity.
34. REASONING
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
35. 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
3
2.40 × 10 s
The angular velocity is decreasing .
= −1.73 × 10 −3 rad/s 2
Chapter 8 Problems
411
36. REASONING The tangential speed vT of a point on a rigid body rotating at an angular
speed ω is given by vT = rω (Equation 8.9), where r is the radius of the circle described by
the moving point. (In this equation ω must be expressed in rad/s.) Therefore, the angular
speed of the bacterial motor sought in part a is ω = vT r . Since we are considering a point
on the rim, r is the radius of the motor itself. In part b, we seek the elapsed time t for an
angular displacement of one revolution at the constant angular velocity ω found in part a.
We will use θ = ω0t + 12 α t 2 (Equation 8.7) to calculate the elapsed time.
SOLUTION
a. The angular speed of the bacterial motor is, from ω = vT r (Equation 8.9),
ω=
vT 2.3 × 10 −5 m/s
=
= 1500 rad/s
r
1.5 × 10−8 m
b. The bacterial motor is spinning at a constant angular velocity, so it has no angular
acceleration. Substituting α = 0 rad/s2 into θ = ω0t + 12 α t 2 (Equation 8.7), and solving for
the elapsed time yields
(
)
θ = ω0t + 12 0 rad/s 2 t 2 = ω0t
or
t=
θ
ω0
The fact that the motor has a constant angular velocity means that its initial and final angular
velocities are equal: ω0 = ω = 1500 rad/s , the value calculated in part a, assuming a
counterclockwise or positive rotation. The angular displacement θ is one revolution, or 2π
radians, so the elapsed time is
t=
37. SSM
REASONING
θ
2π rad
=
= 4.2 × 10−3 s
ω0 1500 rad/s
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.
412 ROTATIONAL KINEMATICS
SOLUTION Solving Equation 8.9 for the radius gives
r=
vT
ω
54 m/s
2π rad ⎞
( 47 rev/s ) ⎛⎜
⎟
⎝ 1 rev ⎠
=
= 0.18 m
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.
38. 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
______________________________________________________________________________
39. 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 ) ⎜
Chapter 8 Problems
413
40. 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,
or
r
θ
θ
r
vT
= rθ ω
3
θ
rθ =
vT
4.66 × 102 m/s
=
= 2.12 × 106 m
5
−
3ω 3 7.31 × 10 rad/s
(
)
The angle θ is, therefore,
6
⎛ rθ ⎞
−1 ⎛ 2.12 × 10 m ⎞
⎟⎟ = 70.6°
⎟ = cos ⎜⎜
6
⎝ r ⎠
⎝ 6.38 × 10 m ⎠
θ = cos −1 ⎜
41. SSM REASONING AND SOLUTION
a. From Equation 8.9, and the fact that 1 revolution = 2π radians, we obtain
rev ⎞ ⎛ 2π rad ⎞
⎛
vT = rω = (0.0568 m) 3.50
= 1.25 m/s
⎝
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 ω2 , we obtain
ω2 =
ω 1 r1
r2
=
(3.50 rev/s)(0.0568 m)
= 7.98 rev/s
0.0249 m
414 ROTATIONAL KINEMATICS
42. REASONING The linear speed v1 with which the bucket moves down the well is the same
as the linear speed of the rope holding the bucket. The rope, in turn, is wrapped around the
barrel of the hand crank, and unwinds without slipping. This ensures that the rope’s linear
speed is the same as the tangential speed vT = r1ω (Equation 8.9) of a point on the surface of
the barrel, where ω and r1 are the angular speed and radius of the barrel, respectively.
Therefore, we have v1 = r1ω . When applied to the linear speed v2 of the crank handle and the
radius r2 of the circle the handle traverses, Equation 8.9 yields v2 = r2ω . We are justified in
using the same symbol ω to represent the angular speed of the barrel and the angular speed
of the hand crank, since both make the same number of revolutions in any given amount of
time. Lastly, we note that the radii r1 of the crank barrel and r2 of the hand crank’s circular
motion are half of the respective diameters d1 = 0.100 m and d2 = 0.400 m shown in the
drawing provided in the text.
SOLUTION Solving the relations v1 = r1ω and v2 = r2ω for the angular speed ω and the
linear speed v1 of the bucket, we obtain
v v
ω= 1= 2
r1 r2
or
( )
1
v d
v2 r1 v2 2 d1
=
v1 =
= 2 1
1d
d2
r2
2 2
The linear speed of the bucket, therefore, is
v1 =
(1.20 m/s) ( 0.100 m )
= 0.300 m/s
( 0.400 m )
43. REASONING AND SOLUTION The figure below shows the initial and final states of the
system.
m
L
L
v
INITIAL CONFIGURATION
FINAL CONFIGURATION
Chapter 8 Problems
415
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
1
2
mgL = mv 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
b. From Equation 8.10:
=
2g
=
L
2
2(9.80 m/s )
= 3.61 rad/s
1.50 m
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,
α=
aT
r
=
g 9.80 m/s 2
=
=
L
1.50 m
6.53 rad/s 2
44. 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. With this
substitution for x the expression for ω becomes ω = 30/t. Kinematics also gives for the
1
2
vertical displacement y that y = v0 y t + a y t 2 (Equation 3.5b). In Equation 3.5b we know
1
2
that v0y = 0 m/s since the stone is launched horizontally, so that y = a y t 2 or t = 2 y / a y .
Using this result for t in the expression for ω and assuming that upward is positive, we find
416 ROTATIONAL KINEMATICS
ω = 30
ay
2y
−9.80 m/s 2
= 14.8 rad/s
2 ( −20.0 m )
= 30
45. SSM 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:
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
46. 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.
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 .
Chapter 8 Problems
417
47. SSM REASONING AND SOLUTION
a. The tangential acceleration of the train is given by Equation 8.10 as
2
aT = r α = (2.00 × 10 m)(1.50 × 10
−3
2
rad/s ) = 0.300 m/s
2
The centripetal acceleration of the train is given by Equation 8.11 as
2
2
2
ac = r ω = (2.00 × 10 m)(0.0500 rad/s) = 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
⎛a
T
⎜a
⎝ c
θ = tan −1 ⎜⎜
⎞
⎟ =
⎟⎟
⎠
tan
⎛
−1 ⎜ 0.300
⎜
⎜ 0.500
⎝
m/s 2 ⎞⎟
⎟ =
m/s 2 ⎟⎠
31.0°
48. 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.
SOLUTION
a. The average angular speed is
ω= ω=
2π rad
Δθ
=
= 1.99 × 10−7 rad /s
Δt 3.16 × 107 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)
418 ROTATIONAL KINEMATICS
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.
49. REASONING The centripetal acceleration ac at either corner is related to the angular speed
ω of the plate by ac = rω 2 (Equation 8.11), where r is the radial distance of the corner from
the rotation axis of the plate. The angular speed ω is the same for all points on the plate,
including both corners. But the radial distance rA of corner A from the rotation axis of the
plate is different from the radial distance rB of corner B. The fact that the centripetal
acceleration at corner A is n times as great as the centripetal acceleration at corner B yields
the relationship between the radial distances:
(
rA ω 2 = n rB ω 2
Centripetal
acceleration
at corner A
)
rA = n rB
or
(1)
Centripetal
acceleration
at corner B
The radial distance rB at corner B is the length of the short side of the rectangular plate:
rB = L1. The radial distance rA at corner A is the length of a straight line from the rotation
axis to corner A. This line is the diagonal of the plate, so we obtain rA from the Pythagorean
theorem (Equation 1.7): rA = L12 + L22 .
SOLUTION Making the substitutions rA = L12 + L22 and rB = L1 in Equation (1) gives
L12 + L22 = n L1
rA
(2)
rB
Squaring both sides of Equation (2) and solving for the ratio L1/L2 yields
n 2 L12 = L12 + L22
or
( n2 − 1) L12 = L22
or
L1
1
=
L2
n2 − 1
Thus, when n = 2.00, the ratio of the lengths of the sides of the rectangle is
Chapter 8 Problems
L1
=
L2
1
22 − 1
419
1
= 0.577
3
=
The centripetal acceleration of the trainee is given by ac = rω 2
50. REASONING
(Equation 8.11), where ω is the angular speed (in rad/s) of the centrifuge, and r is the radius
of the circular path the trainee follows. Because this radius is the length of the centrifuge
arm, we will solve Equation 8.11 for r. In the second exercise, the trainee’s total
acceleration gains a tangential component aT = rα (Equation 8.10) due to the angular
acceleration α (in rad/s2) of the centrifuge. The angular speed ω and the length r of the
centrifuge arm are both the same as in the first exercise, so the centripetal component of
acceleration ac = rω 2 is unchanged in the second exercise. The two components of the
trainee’s acceleration are perpendicular and, thus, are related to the trainee’s total
acceleration a by the Pythagorean theorem: a 2 = ac2 + aT2 (Equation 1.7). We will solve the
Pythagorean theorem for the trainee’s tangential acceleration aT, and then use aT = rα to
determine the angular acceleration α of the centrifuge.
SOLUTION
a. Solving ac = rω 2 (Equation 8.11) yields the length r of the centrifuge arm:
r=
ac
ω
2
=
(
3.2 9.80 m/s 2
( 2.5 rad/s )
2
) = 5.0 m
b. Solving a 2 = ac2 + aT2 for the tangential component of the trainee’s total acceleration, we
obtain aT = a 2 − ac2 . Then, using aT = rα (Equation 8.10), we find that the angular
acceleration of the centrifuge is
α=
aT
r
=
a
2
− ac2
r
=
(
)
2
(
)
2
⎡( 4.8 ) 9.80 m/s 2 ⎤ − ⎡( 3.2 ) 9.80 m/s 2 ⎤
⎣
⎦ ⎣
⎦
= 7.0 rad/s 2
5.0 m
51. SSM 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.
420 ROTATIONAL KINEMATICS
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 ⎟
⎝
⎠
(
)
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
Fc
= mac = m r ω 2
Centripetal
force
(
)(
)(
= 1.99 × 1030 kg 2.2 × 1020 m 1.1 × 10−15 rad /s
)
2
= 5.3 × 1020 N
52. 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 = 2 a T ,
or
r ω 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
Chapter 8 Problems
421
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
⎛ω2 ⎞⎛ 2 ⎞
ω2
ω2
⎜
⎟
θ=
=
=
= 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.
53. SSM WWW REASONING AND SOLUTION
acceleration of the motorcycle is
a=
v − v0
t
=
From Equation 2.4, the linear
22.0 m/s − 0 m/s
2
= 2.44 m/s
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
aT 2.44 m/s 2
2
α=
=
= 8.71 rad/s
r
0.280 m
54. 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 ⎠
3
8.6 ×10 m
____________________________________________________________________________________________
55. 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.
422 ROTATIONAL KINEMATICS
SOLUTION Substituting ω0 from Equation 8.12 and α from Equation 8.13 into Equation
8.7, we find that
⎛v ⎞
⎛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 ⎠
56. 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
vT = rω = (0.175 m)(45.5 rad/s) = 7.96 m/s
b.
57. 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 )
Chapter 8 Problems
423
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 ⎠
______________________________________________________________________________
58. REASONING For a wheel that rolls without slipping, the relationship between its linear
speed v and its angular speed ω is given by Equation 8.12 as v = rω, where r is the radius of
a wheel.
For a wheel that rolls without slipping, the relationship between the magnitude a of its 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. The linear acceleration can be obtained using the
equations of kinematics for linear motion, in particular, Equation 2.9.
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 / ( 2 x )
a
α= =
r
r
=
v 2 − v02
2x r
2
2
92.2 m /s ) − ( 0 m /s )
(
=
2 ( 384 m )( 0.320 m )
= 34.6 rad /s 2
59. 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
424 ROTATIONAL KINEMATICS
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
r
0.170
m
⎝
⎠
⎝ 2⎠
The direction of the smaller gear is clockwise .
60. REASONING The distance d traveled by the axle of a rolling wheel (radius = r) during one
complete revolution is equal to the circumference (2π r) of the wheel: d = 2π r. Therefore,
when the bicycle travels a total distance D in a race, and the wheel makes N revolutions, the
total distance D is N times the circumference of the wheel:
D = Nd = N ( 2π r )
(1)
We will apply Equation (1) first to the smaller bicycle wheel to determine its radius r1.
Equation (1) will then also determine the number of revolutions made by the larger bicycle
wheel, which has a radius of r2 = r1 + 0.012 m.
SOLUTION Because 1 km = 1000 m, the total distance traveled during the race is
D = (4520 km)[(1000 m)/(1 km)] = 4520×103 m. From Equation (1), then, the radius r1 of
the smaller bicycle wheel is
r1 =
D
4520 × 103 m
=
= 0.330 m
2π N1 2π 2.18 × 106
(
)
The larger wheel, then, has a radius r2 = 0.330 m + 0.012 m = 0.342 m. Over the same
distance D, this wheel would make N2 revolutions, where, by Equation (1),
N2 =
D
4520 × 103 m
=
= 2.10 × 106
2π r2 2π ( 0.342 m )
Chapter 8 Problems
425
61. 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
62. REASONING While the ball is in the air, its angular speed ω is constant, and thus its
angular displacement is given by θ = ω t (Equation 8.2). The angular speed of the ball is
found by considering its rolling motion on the table top, because its angular speed does not
change after it leaves the table. For rolling motion, the angular speed ω is related to the
linear speed v by ω = v r (Equation 8.12), where r is the radius of the ball. In order to
determine the time t the ball spends in the air, we treat it as a projectile launched
horizontally. The vertical displacement of the ball is then given by y = v0 yt + 12 ayt 2
(Equation 3.5b), which we will use to determine the time t that elapses while the ball is in
the air.
SOLUTION Since the ball is launched horizontally in the projectile motion, its initial
velocity v0 has no vertical component: v0y = 0 m/s. Solving y = v0 yt + 12 ayt 2 (Equation 3.5b)
for the elapsed time t, we obtain
y = ( 0 m/s ) t + 12 a y t 2 = 12 a y t 2
or
t=
2y
ay
(1)
426 ROTATIONAL KINEMATICS
Substituting Equation (1) and ω =
v
(Equation 8.12) into θ = ω t (Equation 8.2) yields
r
⎛ v ⎞ 2y
θ =⎜ ⎟
⎝ r ⎠ ay
ω t
(2)
We choose the upward direction to be positive. Once the ball leaves the table, it is in free
fall, so the vertical acceleration of the ball is that due to gravity: ay = −9.80 m/s2. Further,
the ball’s vertical displacement is negative as the ball falls to the floor: y = −2.10 m. The
ball’s angular displacement while it is in the air is, from Equation (2),
θ=
3.60 m/s 2 ( −2.10 m )
v 2y
=
= 11.8 rad
r a y 0.200 m/s −9.80 m/s 2
63. SSM REASONING AND SOLUTION
"axle" or the center of the moving quarter is
By inspection, the distance traveled by the
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,
4 π r = rθ
so that θ = 4π rad . This is equivalent to
⎛ 1 rev ⎞
⎟ = 2 revolutions
(4π rad)⎜
⎝ 2π rad ⎠
64. REASONING AND SOLUTION
According to Equation 8.2, ω = Δθ / Δt . Since the angular speed of the sun is constant,
ω = ω . Solving for Δ t , we have
Δt =
Δθ
⎞
2π rad
1y
⎛
⎞ ⎛ 1 h ⎞ ⎛ 1 day ⎞ ⎛
8
=⎜
⎟
⎜
⎟ = 1.8 × 10 y
⎜
⎟
⎜
⎟
−
15
ω ⎝ 1.1×10 rad/s ⎠ ⎝ 3600 s ⎠ ⎝ 24 h ⎠ ⎝ 365.25 day ⎠
Chapter 8 Problems
427
65. SSM REASONING The tangential acceleration aT of the speedboat can be found by
using Newton's second law, FT = maT , 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.
SOLUTION
a. From Newton's second law, we obtain
aT =
FT 550 N
=
= 2.5 m/s 2
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
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
1
2
66. REASONING AND SOLUTION From Equation 8.6, θ = (ω 0 + ω ) t . Solving for t gives
t=
2θ
2(85.1 rad)
=
= 5.22 s
ω 0 + ω 18.5 rad/s +14.1 rad/s
67. SSM 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
ω 0 = ω − α t = 83.8 rad/s – (–42.0 rad/s2 )(1.75 s) = 157.3 rad/s
68. 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 = r ω 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.
428 ROTATIONAL KINEMATICS
SOLUTION Employing the Pythagorean theorem, we obtain
a=
aT2 + a 2c = (rα ) 2 + (rω 2 )2 = r α 2 + ω 4
Therefore,
2 2
4
2
a = (1.5 m) (160 rad/s ) + (14 rad/s) = 380 m/s
69. 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
70. 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.
The
magnitude
of
the
total
acceleration
a
aT
φ
ac
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
429
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
4
= ( 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
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 ⎜
= tan
−1 ⎧
⎪
⎞
α
⎡
⎤
−1 ⎛ r α ⎞
= tan −1 ⎢
⎟ = tan ⎜
2⎟
2⎥
⎝ rω ⎠
⎠
⎣⎢ (ω0 + α t ) ⎦⎥
⎫⎪
= 17.7°
⎨
2⎬
⎪⎩ ⎡⎣1.50 rad/s + ( 2.00 rad/s 2 ) ( 0.500 s ) ⎤⎦ ⎪⎭
2.00 rad/s 2
where we have used the same substitutions for aT, ac, and ω as in part (a).
______________________________________________________________________________
71. SSM 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
430 ROTATIONAL KINEMATICS
⎛θ
vT = r ⎜
⎝t
⎞
⎛ θ ⎞ rθ v
⎟ = r⎜ x ⎟ =
x
⎠
⎜ ⎟
⎝v⎠
SOLUTION The tangential speed of a point on the equator of the baseball is
−2
r θ v ( 3.67 × 10 m ) ( 49.0 rad )( 42.5 m/s )
vT =
=
= 4.63 m/s
x
16.5 m
72. 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.
Angular velocity
SOLUTION The angular displacement Δθ is
+15 rad/ s
related to the average angular velocity ω and
the elapsed time Δt by Equation 8.2, Δθ = ω Δt .
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
ω
time of 5.0 s to 8.0 s. It can be seen from the
+3.0 rad/s
graph that the angular velocity increases by
0
+3.0 rad/s during each second. Therefore, when
3.0 s
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
–9.0 rad/ s
the right. The average angular velocity during
this time is equal to one half the sum of the initial and final angular velocities:
Time (s)
8.0 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
Δθ = ω Δt = ( +3.0 rad/s )( 8.0 s ) = +24 rad
73. 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.
431
Chapter 8 Problems
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 ω = ω0 + α t (Equation 8.4) for
t and using the fact that ω0 = –ω give
t=
2ω
α
=
2(−25.0 rad/s)
−4.00 rad/s2
=
12.5 s
____________________________________________________________________________________________
74. 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 ac = rω2, 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.
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
ac
a=
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
2
( 23.5 m )( 0.571 rad /s )
rω
2
a=
=
=
= 9.35 m /s
cos 35.0° cos 35.0°
cos 35.0°
ac
2
432 ROTATIONAL KINEMATICS
75. 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 Δt = Δθ / ω , with ω = 1.25 rad/s and Δθ = (2π rad)/16 = 0.393 rad .
Therefore, the ball must pass between two blades in a maximum time of
Δt =
0.393 rad
= 0.314 s
1.25 rad/s
The minimum speed of the ball is
v=
Δx 4.50 × 10 –2 m
–1
=
= 1.43 × 10 m/s
Δt
0.314 s
76. REASONING The wheels on both sides of the car have the same radius r = 0.350 m and
undergo rolling motion, so we will use v = rω (Equation 8.12) to calculate their individual
angular speeds:
ωleft =
vleft
r
and
ωright =
vright
(1)
r
In Equations (1) the linear speeds vleft and vright at which the wheels on opposite sides of the
car travel around the track differ. This is because the wheels on one side of the car are closer
to the center of the track than are the wheels on the other side. As the car makes one
complete lap of the track, therefore, both sets of wheels follow circular paths of different
radii Rleft and Rright. The linear speed of each wheel is the circumference of its circular path
divided by the elapsed time t, which is the same for both sets of wheels:
vleft =
2π Rleft
t
and
vright =
2π Rright
(2)
t
Substituting Equations (2) into Equations (1), we obtain
ωleft
2π Rleft
2π Rleft
t
=
=
r
rt
2π Rright
and
ωright =
t
r
=
2π Rright
rt
(3)
SOLUTION We will assume that the wheels on the left side of the car are closer to the
center of the track than the wheels on the right side. We do not know the radii of the circular
paths of either set of wheels, but the difference between them is Rright − Rleft = 1.60 m . We
Chapter 8 Problems
433
can now calculate the difference between the angular speeds of the wheels on the left and
right sides of the car by subtracting ωleft from ωright [see Equations (3)]:
ωright − ωleft =
2π Rright
rt
−
2π Rleft
rt
=
(
2π Rright − Rleft
rt
)=
2π (1.60 m )
= 1.47 rad/s
( 0.350 m )(19.5 s )