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KE of a rotational system

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 5 kg weight is hanging on the string. The system of the weight and disk is released from rest.

    a) When the 5 kg weight is moving with a speed of 1.7 m/s, what is the kinetic energy of the entire system?

    2. Relevant equations




    KEtotal = KEwheel + KEweight

    3. The attempt at a solution

    KE(wheel) = (.5)((15)(.25)2)((1.7)/(.25))2 = 21.675


    KE(weight) = (.5)(5)(1.7)2 = 7.225

    = 28.9 [This answer is incorrect though, I'm unsure as to what I'm doing incorrectly.]

    Any help is much appreciated, thanks! :)
  2. jcsd
  3. Oct 25, 2008 #2


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    Homework Helper

    Hi CaptainSFS,

    This line is incorrect. It is saying that the rotational inertia I for this uniform disk is mr2, which is not true. There should be a table in your book that gives the formula for I for different shapes.
  4. Oct 27, 2008 #3
    Thanks, :P. It needed that constant for a disk (1/2). Thanks for your help. :)
  5. Oct 28, 2008 #4


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    Homework Helper

    Glad to help!
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