A playground merry-go-round of radius R = 1.92m has a moment of inertia of I = 221kgm^2 and is rotating at 14.0rev/min about a frictionless vertical axle. Facing the axle, a 29.8kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed (in revolutions per minute) of the merry-go-round?

Question

A playground merry-go-round of radius R = 1.92m has a moment of inertia of I = 221kgm^2 and is rotating at 14.0rev/min about a frictionless vertical axle. Facing the axle, a 29.8kg child hops onto the merry-go-round and manages to sit down on its edge. What is the new angular speed (in revolutions per minute) of the merry-go-round?
How much work is done by the child?

I got 9.35 rev/min or 0.979 rad/s but I have no clue how to approach the child's work

Damon · Answer

find ke before child jumps
(1/2)I omega^2

find new I (which you did already)
find new omega (which you did already)
Find new ke = (1/2) Inew (omega new)^2
now
work done = new ke - old ke

I went up a little but omega went down so I suspect the work will be negative
In other words the ride did work on the child to bring the kid up to speed.