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Angular momentum [Help Needed]

  1. Apr 21, 2007 #1
    1. The problem statement, all variables and given/known data

    A skater extends her arms horizontally, holding a 5kg mass in each hand. She is rotating about a vertical axis with an angular velocity of 1rev/s. If she drops her hands to her sides, what will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5 kg.m^2, and the distance of the masses from the axis changes from 1 m to 0.1m?


    2. Relevant equations

    I tried using the conservation of momentum. I1w1=I2w2. But since the qtn states that moment of inertia is constant, isn't w1=w2?


    3. The attempt at a solution

    The answer is 3 rev/s but I can't seem to get it.

    Hope to find help
    Thank you
     
  2. jcsd
  3. Apr 21, 2007 #2

    Hootenanny

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    The moment of inertia of the skater's body remains constant, but don't neglect the overall change in inertia, since she changes the position of the two masses.
     
  4. Apr 21, 2007 #3
    oh ok..

    but I still can't seem to solve it..

    I1w1 + inertia of skater's body*w1 = inertia of skater's body*w2 + I2w2

    and I = mr^2

    so:

    5 (1)^2 * 1 + 5*1 = 5*w2 + 5 (0.1)^2 * w2

    5.05w2= 10
    w2 = 1.98 rev/s

    have I made any mistakes again?
     
  5. Apr 21, 2007 #4

    Hootenanny

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    Your nearly there, but don't forget she's holding one mass in each hand ... :wink:
     
  6. Apr 21, 2007 #5
    oh ya..

    Thanks for your help! =)
     
  7. Apr 21, 2007 #6

    Hootenanny

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    My pleasure :smile:
     
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