Posted by **HELP ME!!** on Monday, April 19, 2010 at 11:08pm.

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.

- Physics Inertia : Merry-go-round -
**bobpursley**, Monday, April 19, 2010 at 11:12pm
add them: Inew=335 kg m^2+43.5*1.96^2 kg m^2

- Physics Inertia : Merry-go-round -
**HELP ME!!**, Monday, April 19, 2010 at 11:22pm
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

- Physics Inertia : Merry-go-round -
**bobpursley**, Monday, April 19, 2010 at 11:27pm
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.

- Physics Inertia : Merry-go-round -
**nora**, Friday, April 1, 2011 at 1:59pm
if you want to find the final angular speed, wf, you take:

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

then solve for wf

- Physics Inertia : Merry-go-round -
**John**, Thursday, May 7, 2015 at 7:25pm
Correct

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