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Unformatted text preview: ctly 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
14243 (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 = 1 ( ω − α t ) + ω t = ω t − 1 α t 2
2
2
43 1 24 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 2
ω 2 − ω 0 ( 3.14 × 104 rad/s ) − (1.05 × 104 rad/s )
4
2
=
= 2.33 × 10 rad/s
α=
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
4
2
α
2.33 × 10 rad/s ____________________________________________________________________________________________ 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 + 1 α t 2 = ( 0 rad/s ) ( 2.10 ×103 s ) + 1 ( 2.90 × 10−3 rad/s 2 ) ( 2.10 ×103 s )
2
2 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 + 1 α t 2 = ( 6.09 rad/s ) (1.40 ×103 s ) + 1 ( 0 rad/s 2 ) (1.40 ×103 s ) = 8.53 ×103 rad
2
2
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
2
means that the acceleration has a negative value. Equation 8.8 (ω 2 = ω 0 + 2αθ 3 ) can be
used to find the angular displacement θ3. Solving this equation for θ3 gives [Third segment]
2
ω 2 − ω 0 ( 4.00 rad/s ) − ( 6.09 rad/s )
=
= 4.58 ×103 rad
θ3 =
−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
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 = ω0 t + 1 α t 2
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 + 1 α t 2 gives
2 θ = ω0t + 1 α t 2 = (ω − α t ) t + 1 α t 2 = ω t − 1 α t 2
2
2
2
= ( +1.88 rad/s )(10.0 s ) − 1 ( −5.04 rad/s 2 ) (10.0 s ) = +2.71× 102 rad
2
2 31. REASONING According to Equation 3.5b, the time required for the diver to reach the
water, assuming freefall 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=
2 = 1.3 s
–9.80 m/s Solving Equation 8.2 for ∆ θ , we find
∆θ = ω ∆t = (1.6 rev/s)(1.3 s)= 2.1 rev 32....
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 Spring '13
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 Physics, The Lottery

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