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Angular Momentum of an ice skater

  1. Nov 10, 2003 #1
    An ice skater doing a toe spin with outstretched arms has an angular velocity of 4 rad/s. She then tucks in her arms, decreasing her moment of inertia by 7.5%

    a. What is the resulting angular velocity?
    b. By what factor does the skater's kinetic energy change?

    For a, I use IW = I'W' >> 1(4rad/s) = (1-.075)W', then solve for W'
    I got 4.32 rad/s for W'. I don't know if I'm doing it correctly. Also, I need help on part b.

    For part b I got .925. Is that correct?

  2. jcsd
  3. Nov 10, 2003 #2
    For part a. you got the correct answer, but I'm not sure whether you are "shortcutting" the calculation or whether you just got lucky.

    I would say:
    the initial angular momentum
    L = Ii* ωi
    L = 4*Ii
    then the moment of inertia decreases by 7.5% so If = .925*Ii and thus
    L = .925*Ii * ωf
    4*Ii = .925*Ii * ωf
    4 = .925 * ωf
    ωf = 4 / .925
    ωf = 4.32 rad/s

    So if that is what you intended, you were doing it correctly.

    For part b, your answer is not correct, but you probably have the right idea. You're just not answering the right question. You don't really mean that the energy changed by 92.5%, do you?
  4. Nov 10, 2003 #3


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    You did part a correctly.

    You didn't say how you did part b, but it goes something like this:

    Ko = .5*I*ω^2

    you're looking for K/Ko

    K = .5*I'*(ω')^2 = .5*(.925*I)*(I*ω/(.925*I))^2

    (velocity calculation from conservation of angular momentum (part a))

    From this, K/Ko = 1/.925
  5. Nov 10, 2003 #4


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    I see I'm too slow with the response again; maybe I need to increase my momentum.
  6. Nov 10, 2003 #5
    Yeah, well, yesterday I was real quick with some dumb answers, so I'll be taking my time from now on...
  7. Nov 10, 2003 #6
    what I don't understand is why 1/.925?
  8. Nov 10, 2003 #7
    I guess the question is a little vague. I would have said the energy decreases by 7.5%. At first I took your original answer to be that it decreased by 92.5%, but I guess you meant that it was multiplied by 92.5%, so maybe we're saying the same thing & the only difference is semantics.

    But I agree with you -- I wouldn't say 1/.925
  9. Nov 10, 2003 #8


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    At the risk of being slow and stupid, wouldn't the kinetic energy increase? I can't see what I did wrong in my previous post. Let me try again:

    we all agree that ω = 4.32 rad/s

    initial kinetic energy Ko = .5*I*ωi^2 = .5*I*4^2 = 8*I

    final kinetic energy K = .5*(.925I)*(4.32)^2 = 8.65*I

    There may be a semantics issue here, but I'm calling the factor by which the kinetic energy changes k, so that K = k*Ko

    In that case k = K/Ko = 8.65/8 = 1.08 (=1/.925)
  10. Nov 10, 2003 #9
    You're right James. I guess I'm still going too fast. I did everything exactly the same as you until the very end, when somehow that .925 slipped from the denominator up to the numerator.

    The final kinetic energy is 8I/.925 so it increases by 1/.925 or 8.1%
  11. Nov 10, 2003 #10


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    No worries. I do the same kind of thing all the time; the avatar to my left is actually a scale drawing of the size of my brain relative to my cranium :wink:.
  12. Nov 10, 2003 #11
    Thank y'all very much for taking time to help me. I appreciate it.
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