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Work and Rotational Kinetic Energy

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A 38.0 kg wheel, essentially a thin hoop with radius 1.13 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
    (a) How much work must be done to stop it?


    2. Relevant equations
    change in KE = (1/2)Iwi^2 - (1/2)Iwf^2
    I = (1/2)mr^2


    3. The attempt at a solution
    (1/2)[(1/2)(38)(1.13)^2](28pi/3) = 355.686 kg(m)^2/s

    I don't see where the time comes into play though..can anyone help?
     
    Last edited: Oct 20, 2009
  2. jcsd
  3. Oct 20, 2009 #2

    rl.bhat

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    (1/2)[(1/2)(38)(1.13)^2](28pi/3)
    It should be
    (1/2)[(1/2)(38)(1.13)^2](28pi/3)^2

    To find the work done, time is not necessary.
     
  4. Oct 20, 2009 #3
    Oh ok, I get it now!
    Thank you so much!
     
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