Question

A 150-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 150-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
must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s
in 2.00 s?

Answer

mass, m = 150 kg 

radius, r = 1.5 m 

initial angular velocity, wo = 0  

final angular velocity, w = 0.5 rps 

time, t =  2 s 

mass, m = 150 kg 

radius, r = 1.5 m 

initial angular velocity, wo = 0  

final angular velocity, w = 0.5 rps 

time, t =  2 s 

Moment of inertia of the disk, I = 0.5 mr2 = 0.5 x 150 x 1.5 x 1.5 = 168.75 kgm2

The angular acceleration is given by

w = wo + α t 2 x 3.14 x 0.5 = 0 + α x 2 α = 1.57 rad/s2

According to the equation of torque, let the force is F. 

τ= I α = r F

168.75 x 1.57 = 1.5 x F 

F = 176.625 N

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