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Two rotating discs

  1. Mar 13, 2009 #1

    TG3

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    1. The problem statement, all variables and given/known data
    A disk of mass M1 = 350 g and radius R1 = 10 cm rotates about its symmetry axis at f initial = 151 rpm. A second disk of mass M2 = 251 g and radius R2 = 8 cm, initially not rotating, is dropped on top of the first. Frictional forces act to bring the two disks to a common rotational speed f final.
    In the process, how much kinetic energy is lost due to friction?

    2. Relevant equations
    K = 1/2 I W^2 + 1/2 mv^2
    I =1/2 mr^2
    W = V/R
    F Final = 103 rpm. (I calculated this in an earlier part and got it right.)

    3. The attempt at a solution
    Given that I have an equation and data to punch in, this should be really simple. However, to reconcile the fact that there are two radius' in the final solution, I decided to calculate KE for each disc separately. That is correct, right?
    Anyways, here goes:
    Initial K:
    1/2 I W^2 + 1/2 mv^2
    1/2 1/2 mr^2 151^2 + 1/2 m (Wr)^2
    1/4 35,000 22801 + 1/2 350 2280100
    199508750 + 399017500
    Initial K = 598526250

    Final K: (Done in two parts)

    First, the same disc as above:
    1/4 35,000 10609 + 1/2 350 1060900
    92828750 + 185657500
    Final K for first disc= 278486250

    Next, for the other disc:
    1/4 251 64 103^2 + 1/2 251 (103 8)^2
    42605744 + 85211488
    Final K for second disc = 127817232

    Total Final K= 278486250+127817232
    406303482

    Change in K = 598526250 - 406303482
    Change in K =192222768
    However, this is wrong. Where did I slip up?
     
  2. jcsd
  3. Mar 13, 2009 #2

    tiny-tim

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    Hi TG3! :smile:

    (I haven't checked the rest of what you've done, but …)

    it would help if you stated the units at all times …

    i think you need to change rpm to radians per second :wink:
     
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