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Homework Help: Rotating disks

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data

    NOTE: Use g=9.809 m/s^2
    A solid disk of radius R and mass 2.74 kg is spinning at angular velocity 61.1 rpm about a vertical axis. A solid disk of radius 2R drops concentrically on the first (sort of like dropping an old-fashioned 33 rpm record on a 45 rpm one), so that eventually the two cease to slip relative to each other and end up spinning at 24.7 rpm. What is the mass of the second disk?

    2. Relevant equations

    KE=1/2 mv^2 + 1/2 Iw^2 = 1/2 mv^2 (1+(I/(mr^2)))

    3. The attempt at a solution

    Is there another equation that is supposed to be used to solve this problem? Because I thought that this would be the only equation (where I have initial equal to final according to the conservation of energy), but, at the beginning, they hint that I need to use a value of g.
    Then I started thinking about the conservation of angular momentum, which had the equation alpha=mgr / (I+mr^2), where I would have to use the value of g and I could set it up as intial equal to final. However, this setup did not work either.
    Another possibility could be that I am just plugging in the wrong numbers. For the inital mass I am plugging in the first mass that is mentioned and the final mass is the two masses added together, but solving for the second mass. Initial velocity is 61.1 m/s and final velocity is 49.4 m/s (v=rw). (I=1/2 mr^2). inital radius is 1 and final radius is 2.....can you please help me figure out where I went wrong???
  2. jcsd
  3. Oct 29, 2007 #2


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    In this case angular momentum is conserved, not rotational kinetic energy. Find the angular momentum of the first disk and equate it to the combined angular momentum of the two disks combined.
  4. Oct 30, 2007 #3
    okay....so where would I incorporate g if angular momentum = Iw^2 ?
  5. Oct 30, 2007 #4
    You may not need "g". Does it bother you very much if you get the correct solution without using "g"?
    angular momentum = Iw^2 ? -- It is wrong, dimensionally or otherwise!
    Last edited: Oct 30, 2007
  6. Oct 30, 2007 #5
    oh sorry! I meant L=Iw
  7. Oct 30, 2007 #6
    okay... so I set it up as Iw=Iw....number-wise it's 1/2 mr^2*w= 1/2 mr^2*w....or 1/2(2.74)(1)^2(61.1)=1/2(2.74+x)(2)^2(24.7)....but the answer that I am getting for 2.74+x is not larger than 2.74...so if I subtract it, I will get a negative number....what am I doing wrong?
  8. Oct 30, 2007 #7
    Use I1*w1 = (I1 + I2)*wf.
    Note that for I1 radius is R; for I2 radius is 2R.
  9. Oct 30, 2007 #8
    oh...okay...thank you so much!
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