Unformatted text preview: 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 ) 144 2444 4 1 rev3 = = 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
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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 onethird of that of a person
living in Ecuador, we have, r
θ θ
r vT
= rθ ω
3 θ 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, 2.12 × 106 m rθ = cos −1 6.38 × 106 m = 70.6° r 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 = r ω = ( 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 ω= v1
r1 = v2
r2 or v1 = v2 r1
r2 = v2 ( d )=v d
1
21 1d
22 21 d2 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 po...
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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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