A 1.80 m radius playground merry-go-round has a mass of 120 kg and is rotating with an angular velocity of 0.350 rev/s. What is its angular velocity after a 28.0 kg child gets onto it by grabbing its outer edge? The child is initially at rest.

Answer 1

The moment of inertia of the empty merry-go-round, assuming it to be a uniform circular platform, is
I = (1/2)MR^2 = 194.4 kg*m^2

Angular momentum is conserved when the child gets on, but the new moment of inertia becomes (with the boy's contribution added),
I' = I + mR^2 = 194.4 + 90.7
= 285.1 kg*m^2

To conserve angular momentum, the rotation speed must drop by a factor 194.4/285.1 = 0.6819

That makes the new rotation rate 0.239 rev/s