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Angular momentum problem

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A merry-go-round rotates at an angular velocity of 0.2 rev/s with an 80 kg man standing at a point 2 m form the axis of rotation. what is the new angular velocity when the man walks to a point 1 m from the center? Assume the merry-go-round is a solid cylinder of mass 25 kg and radius 2 m.


    2. Relevant equations
    I= MR^2
    L = Iω



    3. The attempt at a solution

    First I found the moment of inertia of the platform
    25 x 4 = 100
    Then I found the original moment of inertia of the man
    80(4) = 320

    Then I found the total angular momentum
    L = (100 + 320)0.2/2pi
    L = 42/pi

    Then I found the moment of inertia after the man moved
    I = 80(1)
    I = 80

    Then I compared the old momentum to the new momentum
    42/pi = 180ω
    ω = .074 rad/sec
    .074 x 2pi = .467 rev/sec

    The correct answer is .569 rev/sec
     
  2. jcsd
  3. Jul 12, 2014 #2

    Nathanael

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    Homework Helper

    The problem said the merry-go-round was a solid cylinder, this is the incorrect moment of inertia.

    Edit:
    All your steps are good though; I walked through your own steps with the correct moment of inertia and got the correct answer.
     
    Last edited: Jul 12, 2014
  4. Jul 30, 2014 #3
    I got it right. Thanks!
     
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