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Moment of inertia and angular speed

  1. Jul 22, 2007 #1
    1. The problem statement, all variables and given/known data
    A 60kg skater begins a spin with an angular speed of 6 rad/s. By changing the position of her arms, the skater decreases her moment of intertia by 50%. What is the skater's final angular speed?


    2. Relevant equations
    I understand that I=m*(r^2), so if the radius decreases when she puts her arns in, then the initertia decreases.


    3. The attempt at a solution

    My attempt at this is that since it decreases by 50%, then the angular speed must increase by two to compensate. Is this correct and that her final speed is 9 rad/s?
     
  2. jcsd
  3. Jul 22, 2007 #2

    Dick

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    I=m*r^2 describes how to decrease moment of inertia - but doesn't tell you much about angular speed. You'll want to be thinking about angular momentum, which is conserved.
     
  4. Jul 22, 2007 #3
    "My attempt at this is that since it decreases by 50%, then the angular speed must increase by two to compensate"

    you are correct but 6 x 2 does not equal 9
     
  5. Jul 22, 2007 #4
    Ah.....

    Lf = Lo
    [I(final) * w(final)] = [I(initial) * w(initial)]
    w(final) = [I(initial) * w(initial)] / I(final)

    *If I(initial) = 2, and, I(initial) * 50%= 1, then her final speed is 12 rad/s.
     
  6. Jul 22, 2007 #5
    Why I typed 9 I don't know... :D, I meant 12....
     
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