A 235-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.800 rev/s in 2.00 s? (State the magnitude of the force.)

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Asked Nov 9, 2019
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A 235-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.800 rev/s in 2.00 s? (State the magnitude of the force.)

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Expert Answer

Step 1

Given,

Mass m 235 kg
Radius r = 1.5 m
Angular speed o=0.8 rev/s = 0.8 x 27 rad/s = 5.02 rad/s
Time t 2.0 s
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Mass m 235 kg Radius r = 1.5 m Angular speed o=0.8 rev/s = 0.8 x 27 rad/s = 5.02 rad/s Time t 2.0 s

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Step 2

By the equation of motion,

+at
=
5.02 0ax 2
5.02
2
a 2.51 rad/s
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+at = 5.02 0ax 2 5.02 2 a 2.51 rad/s

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Step 3

Now,

...
Torque Ia
1
Fxr mra
2
1
F =-mra
2
1
F x 235x 1.5 x 2.51
2
F = 442.38 Newton
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Torque Ia 1 Fxr mra 2 1 F =-mra 2 1 F x 235x 1.5 x 2.51 2 F = 442.38 Newton

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Angular Motion

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