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Work and rotational Kinetic Energy

  1. May 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A 32.0 kg wheel, essentially a thin hoop with r=1.20 m is rotating at 280 rev/min. It must be brought to stop in 15.0 s. How much work must be done to stop it?


    2. Relevant equations/3. The attempt at a solution
    I=mr^2=46.08 kg*m^2
    W=(1/2)I[tex]\omega[/tex]f^2 - (1/2)I[tex]\omega[/tex]i^2
    W=[tex]\tau[/tex]([tex]\Delta[/tex][tex]\theta[/tex])
    [tex]\tau[/tex]=I[tex]\alpha[/tex]
    Rev in 15 sec = 70 rev => 140[tex]\Pi[/tex] radians in 15 sec.

    I'm not sure which equation(s) to use. I've tried plugging numbers into all of them and getting stuck or wrong answers. Thanks!!
     
  2. jcsd
  3. May 1, 2008 #2

    Nabeshin

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    Remember, work is a change in kinetic energy. Do you know the formula for kinetic energy of a rotating object?
     
  4. May 2, 2008 #3
    i thought it was
    W = delta K = 1/2(I)(omega final)^2 - 1/2(I)(omega initial)^2...?
    or do i use the formula
    W= (integral from theta initial to theta final) torque d-theta?
     
  5. May 3, 2008 #4

    tiny-tim

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    … one step at a time …

    Hi frig0018! :smile:

    First step: what angular acceleration is needed to stop it in 15 s? :smile:
     
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