# Angular momentum of merry go round

This question was asked a while ago by another user, but I do not understand how he got the answer:

"A runner of mass m=36 kg and running at 2.9 m/s runs and jumps on the rim of a playground merry-go-round which has a Moment Of Inertia of 404 kgm^2 and a radius of 2 m. Assuming the merry-go-round is initially at rest, what is its final angular velocity to three decimal places?

"According to the back of the book, the answer is 0.381 rad/s; however, I can never come up with that answer."

He says he used the formula mvr = (Imgr + Ir)w where Imgr is the inertia of the merry-go-round and Ir is the inertia of runner, then solved for w and was able to get the correct answer. When I plug in the numbers, I don't get the right answer at all. Is he using a different "I" value for the runner, if so how does he calculate it? Please help!

angular momentum is given by $L=I \omega = mvr$
So, the moment of inertia of the merry-go-round alone is $404 kgm^2$, what is the moment of inertia of the runner at a distance of 2m from the centre of rotation of the merry-go-round? Then what would be the total moment of inertia of the entire system?