A 346 kg merry-go-round in the shape of a

horizontal disk with a radius of 1.7 m is set in
motion by wrapping a rope about the rim of
the disk and pulling on the rope.
How large a torque would have to be exerted to bring the merry-go-round from rest
to an angular speed of 4.7 rad/s in 3.4 s?

1 answer

angular acceleration = a = 4.7/3.4 = 1.38 radians/s^2

I = (1/2)mr^2 = (1/2)346(1.7^2) = 500kg m^2

Torque = I a = 500 (1.38) = 690 Nm