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Homework Help: Angular Velocity of Ice Skater Around a Pole

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data
    An Ice skater of mass m = 70kg is initially traveling at a speed v = 4 m/s along a straight path that brings his center of mass to within a distance b = .6m of a vertical pole fixed in the ice. He reaches out with his hand as he passes the pole and hangs on so that he pulls himself into a circular path of radius r = .8m (measured from the axis of rotation to his center of mass) around the pole. If the skater's moment of inertia about his center of mass is I = 1.40 kgm2 , what is his final angular velocity ω about the pole, assuming that no torques act during this maneuver?

    2. Relevant equations
    Li = Lf
    Iiωi = Ifωf
    or maybe...
    L = mvr = Iω
    or possibly...
    mviri = mvfrf

    or if energy is conserved...
    .5mv2 = .5Iω2

    I'm mostly confused about which formula to use.

    3. The attempt at a solution

    mvri = Iω
    (70)(4)(.6) = (1.4)ω
    ω = 120 rad/s
    But I thought that this was a relatively unreasonable speed for the skater to be going, so I tried this...

    .5(70)(4)2 = .5(1.4)ω2
    ω = 28.3 rad/s

    But I don't think energy is conserved, so I also tried this...

    mvr = mvr
    (70)(4)(.6) = (70)(v)(.8)
    v = 2.14 m/s
    3 = ω(.8)
    ω = 3.75 rad/s

    This seemed reasonable to me, but I didn't use the moment of inertia, and I have nothing to check my work against. How do I know which formula is the correct one to use in this situation?

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  3. Nov 22, 2013 #2


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    That I is the MoI about the skater's centre of mass, but the rotation is about the pole. How do you correct for that?
  4. Nov 22, 2013 #3
    I = Icm + Md2
    I = (1.4) + (70)(.8)2
    I = 46.2 kgm2

    mvr = Iω

    (70)(4)(.6) = (46.2)ω
    3.64 rad/s = ω

    Would this be the correct answer?
  5. Nov 22, 2013 #4


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    Looks right to me.
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