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Conservation of Angular Momentum-Ice Skater

  1. Nov 5, 2013 #1
    Conservation of Angular Momentum--Ice Skater

    1. The problem statement, all variables and given/known data
    Real-life situations are generally complicated, but some can be approximately analyzed by using simple models. Such a model for a skater’s spin is shown, with a cylinder and rods representing the skater. In (a), the skater goes into the spin with the “arms” out, and in (b) the “arms” are over the head to achieve a faster spin by conservation of angular momentum. If the initial spin rate is 1 revolution per 1.5 s, what is the angular speed when the arms are tucked in?

    The cylinder represents the skater's torso and has radius r = 20 cm, mass M = 75 kg. The two rods represent the skater's arms and have length L = 80 cm, mass m = 5 kg.


    2. Relevant equations
    Conservation of angular momentum (L) = I(1)ω(1) = I(2)ω(2)
    Moment of Inertia of cylinder = (1/2)MR^2
    Angular velocity = 2∏/T
    (where T = period)
    Parallel axis theorem = I = Icm + Md^2


    3. The attempt at a solution
    Where i indicates before her arms have been pulled in and f indicates after her arms have been pulled in:

    ωi = 2∏/T = 2∏/1.5 = 4.189 rad/s.
    Ii(cylinder) = .5Mr^2 = .5(75)(.2)^2 = 1.5 kg m^2
    Ii(rod) = I(cylinder) + md^2 = 1.5 + 5 ((.8/2)+.2)^2 = 3.3 kg m^2
    Ii(system) = 1.5 + 2(3.3) = 8.1 kg m ^2
    La = Iiωi = (4.189)(8.1) = 33.931 kg m^2 / s

    If(rod) = I(cylinder) + md^2 = 1.5 + 5(.2)^2 = 1.7 kg m^2
    If(system) = 1.5 + 2(1.7) = 4.9 kg m^2

    Iiωi = Ifωf => 33.931 = 4.9 (ωf)
    ωf = 6.925 rad/s

    I feel like I'm missing something huge and obvious, but for the life of me I can't see it. Please help!
    According to my professor, the correct answer is somewhere between 12.5 and 13 radians/s (accounting for rounding)
     

    Attached Files:

    Last edited: Nov 5, 2013
  2. jcsd
  3. Nov 5, 2013 #2
    Explain why you compute MoI of the rods that way, both cases.
     
  4. Nov 5, 2013 #3
    I managed to get the right answer, actually. I had been calculating the moment of inertia incorrectly. Here's the right solution:

    Ii = I(cylinder) + 2( (1/12) mL^2 + md^2) where d = (L/2) + r
    = 1.5 + 2 ( (1/12)(5)(.8)^2 + (5)(.6)^2)
    = 5.633 kg m^2
    La = Ii ωi = (5.633)(4.189) = 23.598 rad/s

    If = I(cylinder) + 2(md^2) where d = r
    = 1.5 + 2((5)(.2)^2)
    = 1.9 kg m^2

    Ii ωi = If ωf => 23.598 = 1.9 ωf
    ωf = 12.42 rad/s

    You use the parallel axis theorem to determine the moment of inertia of the arms in relation to the center of mass of the cylinder, or the body. When the arms are above the head, the distance d in the parallel axis thm equation is the same as the radius of the cylinder because the arms can be taken as point masses on the edge of the cylinder.
    I apologize if this explanation makes no sense...
     
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