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Dynamics - spinning skater

  1. May 18, 2008 #1
    Practice question for a general physics exam - I'm a 4th year undergraduate but have mostly taken astrophysics courses this year so am a bit stuck going back to general physics...

    1. The problem statement, all variables and given/known data

    A skater spins with angular velocity [tex]\omega = 6 rad/s[/tex] with his arms extended. How fast will he spin with his arms by his sides? Treat the skater's body as a uniform cylinder of radius R = 20cm, approximate his arms as thin uniform rods of length L = 70 cm and mass m = 4.5 kg. His total mass excluding arms is M = 70 kg.

    After ceasing to spin the skater now steps onto the outer edge of a large uniform disk of radius [tex]R_{disk} = 2.5 m[/tex] and mass [tex]M_{disk} = 500 kg[/tex]. Assume that the disk is mounted on a frictionless bearing with a vertical axis of rotation and is initially at rest. If the skater begins skaing around the edge of the disk at a speed of 2 m/s how fast does the disk turn and what is its angular momentum?

    2. Relevant equations

    For a body of mass M with a moment of inertia I about an axis through its ventre of mass the
    moment of inertia about a parallel axis a distance d from the first is [tex]I + Md^{2}[/tex]

    The moment of inertia of a uniform cylinder of mass M, radius R and length L rotating about its axis is [tex]\frac{1}{2}ML^{2}[/tex]

    3. The attempt at a solution

    I think I've done the first part. I called the situation with the arms outstretched 1 and arms by his side 2 so

    [tex]I_{1} = \frac{1}{2}MR^{2} = \frac{1}{2}(79)(0.2)^{2} = 1.58[/tex]
    [tex]I_{2} = \frac{1}{2}MR^{2} + 2(\frac{1}{12}ML^{2} +Md^{2})= \frac{1}{2}(70)(0.2)^{2} + 2(\frac{1}{12}(45)(0.7)^{2} + (4.5)(0.55)^{2})= 7.798[/tex]

    [tex]E = \frac{1}{2}I_{1}\omega_{1}^{2} = \frac{1}{2}I_{2}\omega_{1}^{2}[/tex]

    [tex]\omega_{2} = 13.33 rad/s[/tex]

    Its the second part I'm struggling with. I've said

    [tex]v = r\omega[/tex] and using conservation of momentum

    [tex]v_{disk} = \frac{m_{s}v_{s}}{m_{disk}} = 0.316 m/s [/tex]

    [tex]\omega_{s} = \frac{v_{s}}{r} = 0.83 rad/s [/tex]

    Is this right because I'm not convinced...
     
  2. jcsd
  3. May 18, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks like you reversed your notation. I_1 looks like an attempt at I for arms by his side.

    What's the I for the torso? That should not change.

    What's the I for the arms in each case?

    Energy is not conserved, but something else is.
     
  4. May 18, 2008 #3
    Oops yeah that notation is the wrong way round. I can't seem to edit my original post though.

    So I do the moment of inertia for body+arms for each case and then assume momentum is conserved? That's the only other thing I can think of that is conserved! I'll have a go at that now then. I feel like I'm being really stupid not being able to do this but I was taught this stuff 3 and a half years ago and I don't think I ever fully grasped it!
     
  5. May 18, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes. And it's angular momentum that's conserved. Assuming no friction, then there's no external torque and thus angular momentum is conserved.
     
  6. May 18, 2008 #5
    I've hit a snag when trying to calcuate the moment of inertia for when his arms are by his side because I don't know the radius of his arms. I've said

    I_arm = I + Md^2 where I = MR^2 and d is the distance from the axis i.e. R_body+R_arm
     
  7. May 18, 2008 #6

    Doc Al

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    Staff: Mentor

    The arms are modeled as thin rods, not cylinders. Radius is small and irrelevant.
     
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