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Atwood Machine, Rotational Inertia, and Energy

  1. Dec 12, 2008 #1
    In the figure below, the pulley is a solid disk of mass M and radius R, with rotational inertia (MR2)/2. Two blocks, one of mass m1, and one of mass m2, hang from either side of the pulley by a light cord. Initially, the system is at rest, with Block 1 on the floor and Block 2 held at height h above the floor. Block 2 is then released and allowed to fall.
    a. What is the speed of Block 2 just before it strikes the ground?
    b. What is the angular speed of the pulley at this moment?
    c. What's the angular displacement of the pulley?
    d. How long does it take for Block 2 to fall to the floor?




    (MR2)/2 = I
    mgh=GPE
    (mv2)/2 =KEtrans
    Iw(omega)2=KErot




    I can't figure out how to do it, but the answers are
    a. v = sqrt((2gh(m2-m1))/(m1+m2+(M/2)))
    b. w(omega) = (1/R) * sqrt((2gh(m2-m1))/(m1+m2+(M/2)))
    c. ???????
    d. t = h * sqrt((2m1+2m2+M)/(gh(m2-m1)))
     
  2. jcsd
  3. Dec 13, 2008 #2
    Think about it in terms of energy conservation. Do a sketch of the system at the start and at the end of the period you're interested in.

    Then play spot the difference. Where has energy been transferred?
     
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