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Angular acceleration of a skater

  1. Nov 2, 2005 #1
    Hi,
    I need help with the following problem.

    A skater does a pirouette at the rate of 2.0 rev / second and then stop within 3/4 rev. Assume that the angular decceleration is constant and calculate its magnitude.

    I tried to compute it but I found -18 rad/s2.
    BUt the textbook gives -17 rad/s2.
    Please can someone help me with this problem?

    Thank you
    B
     
  2. jcsd
  3. Nov 2, 2005 #2

    andrevdh

    User Avatar
    Homework Helper

    brad sue,
    use
    [tex]\omega^2 \ = \ {\omega^2}_o \ + \ 2\alpha \ \Delta\theta[/tex]
    The rotational equivalent of
    [tex]v^2 \ = \ {v^2}_o \ + \ 2as[/tex]
     
    Last edited: Nov 2, 2005
  4. Nov 2, 2005 #3

    Astronuc

    User Avatar

    Staff: Mentor

    What formula did you use to compute 18 rad/s2?

    What is the angular analogy to v2 = vo2 + 2ax?

    The 17 rad/s would be rounded.
     
  5. Nov 2, 2005 #4
    Yes the -17 has been rounded.
    Thank you very much you all..

    B
     
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