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Rotational kinetic energy problem

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    An ice skater executes a spin about a vertical axis with her feet on a frictionless ice surface. In each hand she holds a small 5kg mass of which are both 1m from the rotation axis and the angular velocity of the skater is 10rad/s. The skater then moves her arms so that both masses are 0.5m from the rotation axis. The skaters own moment of intertia can be taken as being 50kgm^2, independent of her arm position.

    Find the total kinetic energy of the skater and the masses both before and after the arm movement. Explain any difference.





    2. Relevant equations
    KErotational = 1/2 Iω2

    3. The attempt at a solution
    KErotational initial = 1/2 Iω2 = 1/2 ((10⋅(1)2)+50)⋅(10)2
    = 3000J
    Then using the same method to find the final rotational KE once the skater has moved her arms in
    KErotational initial = 1/2 Iω2 = 1/2 ((10⋅(0.5)2)+50)⋅(10)2
    = 2625J
    which doesn't make intuitive sense to me, since moving her arms in will increase her speed and should in turn increase her kinetic rotational energy?? on the worked answers for the exam they use the equation
    KE= L2/2I
    where L is the angular momentum

    can someone explain to me how the got that equation and why you cant use the other one, thanks.

     
  2. jcsd
  3. Oct 30, 2016 #2
    You are right that moving her arms in will increase her angular velocity and her kinetic energy. But her angular momentum should remain the same.

    So you wrote 2 equations for KE, both with a subscript of "rotational initial". I'm pretty sure that you intended the second equation to be "rotational final".
     
  4. Oct 30, 2016 #3
    oh whoops yeah i meant the second equation to be sub final, i just copied the equation for the initial ke so i didnt have to rewrite it haha.
    So how did they come to that second equation involving the angular momentum ?
    thanks.
     
  5. Oct 30, 2016 #4
    I'm not familiar with that second equation, but I can substitute Iω for L into it and see that it is a valid equation. So maybe they calculated the final momentum and substituted that into that second equation to get the final kinetic energy.

    L = Iω is easier for me to remember. For these kinds of problems, unless there is something that is dissipating energy, angular momentum will be conserved. Once you calculate the final angular velocity, you can calculate the final kinetic energy.
     
  6. Oct 30, 2016 #5
    The other equation is useful as it contains quantities that are constant(angular Momentum)
     
  7. Oct 30, 2016 #6

    rcgldr

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    The relevant equations should have included L = I ω. As already answered, KE = L = 1/2 I ω2 = (I ω)2 / 2.

    How technical should this part be? The skater performs internal work to move the weights in. For more detail, as the weights are pulled in, their path is not circular, somewhat spiral like, and during the transition, a component of the tension in the skaters arms is in the direction of the inwards spiral like path of the weights, increasing their speed. The skater arms also oppose some of this tendency for the weights to speed up, otherwise the skater's arms would wrap around the skaters body, so the skater's body also experiences an increase in angular velocity.
     
  8. Oct 30, 2016 #7

    haruspex

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    Quite so, so why did you plug in 10rad/s in both energy calculations?
     
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