Help pleaseeee?

Question

A 70 kg child is standing at the rim of a merry-go-round that is rotating at an angular speed of 1.8 rad/s. The moment of inertia of the merry-go-round and the child is 380 kg*m^2. The radius of the merry-go-round is 2.0 m. The child moves a certain distance toward the center of the merry-go-round and the angular speed of the merry-go-round and the child is changed to 2.2 rad/s.

(a) What is the new moment of inertia of the child and the merry-go-round? I figured this one out. The answer is 311 kg*m^2.

I need help with (b).

(b) Find the change of kinetic energy of the merry-go-round and the child when the child moved from the rim to his new position.

Thanks!

Answer 1

no... idea... wow! i stopped after the first sentence

Answer 2

Rotational kinetic energy is given by KE = ½Iω², where I is the moment of inertia, and ω is the angular velocity. Start of by plugging the initial and final values into the formula. The initial kinetic energy uses the values of 380 kg*m^2 and 1.8 rad/s. The final kinetic energy uses the values of 311 kg*m^2 and 2.2 rad/s. The change in kinetic energy is this:

∆KE = KE (final) - KE (initial)

I did the calculations and I got 137 J for the change in kinetic energy.