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Angular momentum of merry go round

  1. May 13, 2008 #1
    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!
  2. jcsd
  3. May 13, 2008 #2


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    angular momentum is given by [itex]L=I \omega = mvr[/itex]

    Angular momentum is always conserved. So that the angular momentum as the runner jumps on the merry go round should be the same as the angular momentum as merry go round and the runner.

    So, the moment of inertia of the merry-go-round alone is [itex]404 kgm^2[/itex], 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?
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