Question

A 170-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 0.600 rev/s in 2.00 s? (State the magnitude of the force.)

Answer 1

Given

mass of the horizontal disc (merry-go-round) m = 170 kg

radius R = 1.50 m

angular speed is w = 0.6 rev/s = 0.6*2pi rad/s = 3.76991 rad/s

time t =2 s  

the angular speed is w = 3.76991 rad/s

so the angular acceleration alpha = W/t = 3.76991/2 rad/s2 = 1.884955 rad/s2

due to the force applied by the rope on the rim of the disc the torque will act on it

by definition torque T = R X F = R*F sin theta

here theta = 90 degrees

and from rotational motion the torque T = I*alpha

I is moment of inertia of the disc I = 0.5*M*R^2

I = 0.5*170*1.5^2 kg m2 = 191.25 kg m2

equating the torques  

R*F sin theta = I*alpha

1.5*F sin90 = 191.25*1.884955

solving for F

F = 240.332 N

constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s is 240.332 N