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Too easy and wrong: consv of angular momentum

  1. Nov 3, 2004 #1
    Here is my problem:
    A man stands at the center of a platform that rotates without friction with an angular speed of 1.60rev/s. His arms are outstretched, and he holds a heavy weight in each hand. The moment of inertia of the man, the extended weights, and the platform is 6.30 kg m2. When the man pulls the weights inward toward his body, the moment of inertia decreases to 2.30 kg m2. What is the resulting angular speed of the platform? ( Give answer in rad/s)
    I got this first part using conservation of Angular momentum or:
    Ii *wi = If *wf
    and then I converted of course.
    It then asks :
    What is the change in kinetic energy of the system?
    I thought finding this answer would be easy but I cant seem to get it.
    I tried using KE = .5 I w^2
    and finding the KE for before and after the arm extension and then subtracting. I guess that was too easy because it was wrong!








    Thanks for the help guys (and gals hopefully!)
     
  2. jcsd
  3. Nov 4, 2004 #2

    Galileo

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    The method is correct though.

    What'd you get? I got a difference of 553.67 J.
     
  4. Nov 4, 2004 #3
    I swear I must have tried that problem a milllion different times last night but I just couldnt get it. When I tried again I got the right answer (553.7). I wasnt converting the angular speed from the first answer back into rev/s. Thanks so much for your help!
     
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