Question

A playground merry-go-round of radius 1.96 m has a moment of inertia 335 kg m2 and is rotating at 12.3 rev/min. A child with mass 43.1 kg jumps on the edge of the merry-go-round. What is the new moment of inertia of the merry-go-round and child, together? Answer in units of kg m2.

Answer 1

add them: Inew=335 kg m^2+43.5*1.96^2 kg m^2

Answer 2

Thank you, but what equation do you use? I cant seem to get the correct answer.What happens to the rev/min? Is it: inertia+child mass*radius^2

Answer 3

The rev/min is not needed.

here is the equation:

momentofINertiatotal=momentinertiaMerryGoRound + momentinertiaKid.

Then you add them.

That is it. Now, if the problem goes on, you may need to find what the new angular speed is, but that is a different problem, you didn't ask that.

Answer 4

if you want to find the final angular speed, wf, you take:

momentinertiaMerryGoRound(initial angular speed) = (momentinertiaMerryGoRound + momentinertiaKid)(wf)

then solve for wf

Answer 5

Correct

Answer 6

What is the moment of inertia of the kid?

Answer 7

The moment of inertia of a point mass (i.e. the kid in this case) is MR^2 (we use radius R of the merry-go-round because the kid's on the edge). So the complete expression of the final angular momentum is: L_f = (335 + 43.1 * 1.96^2) (w_f)