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Figure skater spinning (rate of rotation)

  1. Nov 10, 2006 #1
    A figure skater is spinning at a rate of 1.40 rev/s with her arms outstretched. She then draws her arms in to her chest, reducing her rotational inertia to 68.0% of its original value. What is her new rate of rotation?



    I am having trouble starting this problem.
    I know that
    (I)(alpha) = (I)(omega)(t)

    however, the problem doesn't mention time..
     
  2. jcsd
  3. Nov 10, 2006 #2

    Doc Al

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    Staff: Mentor

    What physical quantity is conserved in this problem?
     
  4. Nov 10, 2006 #3
    Angular momentum is conserved.
    So...
    (I1)(omega1) = (I2)(omega2)
    however i am confused as to how to find omega when i do not have time
     
  5. Nov 10, 2006 #4

    Doc Al

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    Staff: Mentor

    Where do you see time mentioned in that equation? What makes you think you need it? (It doesn't matter how fast or how slowly the skater pulls her arms in, since all you care about is the end result.)
     
  6. Nov 10, 2006 #5
    sorry i am struggling in physics so i don't really understand.

    so
    if she reduces her inertia by 68% of its original value....
    it would be 0.68*omega

    I don't know where to start
     
  7. Nov 10, 2006 #6

    Doc Al

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    Staff: Mentor

    Express that mathematically. If her original rotational inertia is I1, what's I2? Note that the precise instructions were:
    reducing her rotational inertia to 68.0% of its original value​

    That's "to", not "by".
     
  8. Nov 10, 2006 #7
    ok thanks. i got it

    1.40 rev/s = (0.68)I2
    = 2.06 rev/s
     
  9. Nov 11, 2006 #8

    Doc Al

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    Staff: Mentor

    Good! I'll rewrite it more systematically.
    Conservation of angular momentum says:
    (I1)(omega1) = (I2)(omega2)

    Given:
    omega1 = 1.4 rev/s
    I2 = 0.68 I1

    So:
    (I1)(1.40 rev/s) = (0.68 I1)(omega2)
    (1.40 rev/s) = (0.68)(omega2)
    omega2 = (1.40 rev/s)/(0.68)
     
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