CHAPTER 8 ROTATIONAL KINEMATICS
PROBLEMS
1.
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 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
21.
SSM 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
23. 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
θ = ( 2870 rev ) ⎜
⎟ = 1.80 ×10 rad
1
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
36. REASONING In one lap, the car undergoes an angular displacement of 2π radians. The
average angular speed is the magnitude of the average angular velocity ω , which is the
angular displacement Δθ divided by the elapsed time Δ t . Therefore, since the time for
Δθ
(Equation
Δt
8.2). In addition, the given average tangential speed vT and the average angular speed are
one lap is given, we can apply the definition of average angular velocity ω =
related by vT = rω (Equation 8.9), which we can use to obtain the radius r of the track.
SOLUTION
a. Using the facts that Δθ = 2π rad and Δ t = 18.9 s in Equation 8.2, we find that the
average angular speed is
Δθ 2π rad
=
=
Δ t 18.9 s
ω=
0.332 rad/s
b. Solving Equation 8.9 for the radius r gives
vT
r=
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.
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,
(
)(
)
vT = rω = 6.38 ×106 m 7.31 × 10−5 rad/s = 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,
vT
= rθ ω
3
or
rθ =
vT
3ω
=
4.66 ×102 m/s
(
3 7.31 × 10
−5
rad/s
)
= 2.12 ×106 m
The angle θ is, therefore,
⎛ rθ
⎝ r
θ = cos−1 ⎜
6
⎞
−1 ⎛ 2.12 ×10 m ⎞
=
cos
⎜
⎟
⎜ 6.38 ×106 m ⎟⎟ = 70.6°
⎠
⎝
⎠
46. 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
47.
SSM 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
−7
=
= 1.99 × 10 rad /s
7
Δt 3.16 × 10 s
(8.2)
b. The tangential speed of the earth in its orbit is
(
)(
)
vT = r ω = 1.50 × 1011 m 1.99 × 10−7 rad/s = 2.98 × 104 m/s
c. The centripetal acceleration of the earth due to its circular motion around the sun is
(8.9)
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.
53.
SSM REASONING AND SOLUTION From Equation 2.4, the linear acceleration of the
motorcycle is
a=
v − v0
t
=
22.0 m/s − 0 m/s
= 2.44 m/s2
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
=
= 8.71 rad/s2
r
0.280 m
56. 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 ⎞
3
x = v t = r ω t = ( 0.45 m )(9.1 rad/s )(35 min ) ⎜
= 8.6 ×10 m
⎟
⎝ 1 min ⎠
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