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Homework Help: Spinning ice skater (angular momentum)

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    A 45 kg figure skater is spinning on the toes of her skates at 1.0 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 66 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 kg, 20 cm diameter, 200 cm-tall cylinder.


    2. Relevant equations
    Conservation of angular momentum
    Moment of Inertia

    3. The attempt at a solution

    Li = (Ib + 2(Ia))wi
    wi = 6.2831 rad/s
    Ib = (1/2)(40)(.1)(.1)
    Ia = (1/3)(2.5)(.66)2

    Lf = (If)(wf)
    If = (1/2)(45)(.1)2

    Li = Lf

    Solve for wf

    wf = (.926)(6.2831)/(.225)

    Any ideas where I went wrong?
     
  2. jcsd
  3. Apr 25, 2010 #2
    The arms are attached at the outside of the torso, so [itex] I_a [/itex] is wrong.
     
  4. Apr 26, 2010 #3
    Okay, I get that I should use the parallel-axis theorem, but I'm a little confused about how to apply it.

    Would it be the moment of inertia as I calculated it + the mass of the rod times the radius of the skater's torso?

    The definition says that you should take the moment of inertia about the center of mass and then add the distance of the new axis (times the mass), so I'm a little confused with the terminology.
     
  5. Apr 26, 2010 #4
    No. For the parallel-axis theorem you need the moment of inertia around the center of mass.
    (and you add that to the mass of the rod times the square of the distance from the center of mass to the axis of rotation
     
  6. Apr 26, 2010 #5
    Ugh.

    Ia = (1/12)(2.5)(.12) + 2.5(.432)

    Plugging in the new values gives me a final rotational speed of .98 rev/s, which is slower.

    I hate rotational motion.
     
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