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2θ
=
= 0.125 s
ω 0 + ω 3.00 rev/s + 5.00 rev/s SSM REASONING AND SOLUTION
a. From Equation 8.7 we obtain θ = ω 0 t + 1 α t 2 = (5.00 rad/s)(4.00 s) + 1 (2.50 rad/s 2 )(4.00 s) 2 = 4.00 × 101 rad
2
2 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,
2
and angular displacement θ. Therefore, we will employ ω 2 = ω0 + 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 = 1 (ω0 + ω ) t (Equation 8.6), or θ = ω0t + 1 α t 2 (Equation 8.7) to find the elapsed time.
2
2
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 ) 2
Solving ω 2 = ω0 + 2αθ (Equation 8.8) for the square of the initial angular velocity, we 2
obtain ω0 = 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= 23. ω − ω0 137 rad/s − 117 rad/s
=
= 140 s
α
0.140 rad/s 2 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 ω +ω
2
140 3
24 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 = 1 ( +220 rad /s + 280 rad /s ) (10.0 s ) =
2 +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
t
10.0 s
______________________________________________________________________________ α= = 2
26. REASONING Equation 8.8 (ω 2 = ω 0 + 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 64cm length of
the string and r is the 2.0cm radius of the top.
SOLUTION Solving Equation 8.8 for the final angular velocity gives
2
ω = ± ω0 + 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 2
ω = + ω 0 + 2α = +
r 27. ( 0 rad/s ) 2 64 cm + 2 (12 rad/s 2 ) = 28 rad/s 2.0 cm SSM REASONING The equations of kinematics for rotational motion cannot be used
dire...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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