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Rotational kinetic energy of falling mass

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]
    Engineers are designing a system by which a falling mass {m} imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum ( View Figure ). There is no appreciable friction in the axle of the drum, and everything starts from rest.

    That is top part of the problem


    2. Relevant equations
    mgh = rotational + translation kinetic energy


    3. The attempt at a solution

    I've tried various methods mainly using energy conservation laws, but I can't get it right. the problem didn't give I or R so ive been using .200J as the total rotational energy.

    Any hits on how I should be going on this problem?

    thanks
     
  2. jcsd
  3. Apr 20, 2008 #2

    Dick

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    The presentation of your problem is headache inducing. Can you maybe just include the diagram, and the problem statement and leave off the irrelevent blank graphic parts. Just type out what the question is. You also left out any description of how you attempted to use those conservation laws. There is a relation between the velocity of the descending weight and the rotational velocity of the disk. v=omega*r. Did you use that? No way to know, is there?
     
    Last edited: Apr 20, 2008
  4. Apr 20, 2008 #3
    that was all that's given in the problem. The question is under Part A.
     
  5. Apr 20, 2008 #4

    Dick

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    You don't need I or R. The question is to compare the operation of the thing on earth and on Mars. The only difference is 'g'. If you don't know something leave it as a letter. Now, write down one of these conservation laws.
     
  6. Apr 20, 2008 #5
    In this situation there are no non conservative forces so

    Kinetic energy = potential energy.

    So in our situation we have
    (Point 1 to be taken at before the system is released and point 2 to be taken just before it hits the ground)

    mgh= .200 (which is the rotational kinetic energy that was given to us)

    so in mars, just replace 9.8 with the given 3.71 (mass is the same) and solve for h. But when I tried that, they told me it's wrong.
    I got h to be .003594m
     
  7. Apr 20, 2008 #6

    Dick

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    Absolutely right. But how did you get that silly answer? If m*(9.8m/s^2)*h_earth=m*(3.7m/s^2)*h_mars, then h_mars must be LARGER than h_earth. Can you post the details of how you did it?
     
  8. Apr 21, 2008 #7
    ah, I got it. Forgot to take account linear kinetic energy also

    so the system is actually

    Potential energy = Kinetic rotational + kinetic linear

    thanks.
     
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