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Moment of inertia and Kinetic Energy (Rotational Motion)

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A skater spins with an angular speed of 8.3 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 8.9. Ignoring the friction on the skates, determine the percent of change in her kinetic energy. Answer in percent.


    2. Relevant equations
    KE = (1/2)Iw^2
    Li = Lf

    3. The attempt at a solution[/b
    wi = omega initial

    wf = omega final

    If Li = Lf,

    Iwi = (I/8.9)wf

    8.9wi = wf

    So KEi = (1/2)I(wi^2)

    KEf = (1/2)(I/8.9)(wf^2)

    So KEf = (1/2)(I/8.9)(8.9wi^2)

    So I(wi^2)/2 = I(8.9wi^2)/17.8

    Does that mean there is no change in kinetic energy? Also, if there is a change in kinetic energy how do I convert that to a percent?
     
  2. jcsd
  3. Dec 6, 2009 #2
    So I just realized I could simplify the last part to:

    Code (Text):
    I(wi^2)       I(wi^2)
    -------   =  -------
        2          4.45
    So is the percent change 4.45 /2 = 2.225 = 222.5 %?
     
  4. Dec 6, 2009 #3
    Anybody?
     
  5. Dec 6, 2009 #4

    nrqed

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    You forgot to square the factor of 8.9 that comes with omega_f.
    It should be (8.9 omega_i)^2 , not 8.9 omega_i^2
     
  6. Dec 6, 2009 #5
    I didn't correct it in my second post?
     
  7. Dec 7, 2009 #6
    wait I am so lost in this problem, can somebody help me please?
     
  8. Dec 7, 2009 #7
    Oh wait I just figured it out. If wf = 8.9wi, then (8.9wi - wi)/wi = 7.9 *100, 790%
     
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