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Rotational Kinetic Energy

  1. Apr 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A skater spins with an angular speed of 17.4 rad/s with his arms outstretched. He lowers his arms, decreasing his moment of inertia from 43 kg/m^2 to 37 kg/m^2. Calculate his initial and final rotational kinetic energy.



    2. Relevant equations
    L=I[tex]\omega[/tex]
    KE=1/2I[tex]\omega[/tex]^2



    3. The attempt at a solution
    Not sure if im on the right track here, for initial kinetic energy I came up with 3.53 J. I manipulated L=I[tex]\omega[/tex] to get [tex]\omega[/tex]=L/I to find my angular velocity. Then plugged that in the KE=(1/2)(43 kg/m^2)(.405^2) to get 3.53 J. Did I do this correctly?
     
  2. jcsd
  3. Apr 28, 2009 #2

    rl.bhat

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    A skater spins with an angular speed of 17.4 rad/s
    This is not the angular momentum L, but is is the angular velocity w.
     
  4. Apr 28, 2009 #3
    Okay, so the speed would be [tex]\omega[/tex]?
     
  5. Apr 28, 2009 #4

    rl.bhat

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    Yes. Angular speed is w.
     
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