physics quize?

Question

A playground merry go round of radisu 2.15 m has a moment of inertia 245 kg. m^2 and is rotatings at 5.6 rev/min about a friction-less vertical axle. Facing the axle, a 21.3 kg child hops onto the merry go round, and sits down on the edge.
What is the new angular speed of the merry go round? Answer in units of rev/min
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Answer 1

Ke=0.5 I w^2

Ke (before) =Ke (after) assuming that child did not add any more energy when hops on to marry-go-around

I(new)= I old + mR^2 where mR^2 is child's moment of inertia

0.5 I (old)w1^2=0.5( I(old) + mR^2) w2^2
I (old)w1^2=( I(old) + mR^2) w2^2

w2= w1sqrt(I(old)/[ I(old) + mR^2])


w2= 5.6 sqrt(245/[245+ 21.3x(2.15 )^2])
w2=4.7 rev/min