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Rotational Inertia of an Atwood Machine

  1. Mar 1, 2010 #1
    1. In the figure below, the pulley is a solid disk of mass M and radius R with rotational inertia .5MR. 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 at height h above the floor. Block 2 is then released and allowed to fall.

    a) What is the acceleration of block 1 and block 2?
    b) What is the tension on the cord?
    c) What is the speed of block 2 just before it strikes the ground?
    d) What is the angular speed of the pulley at this point?
    e) What is the angular displacement of the pulley?
    f) How long does it take for block 2 to fall to the floor?

    2. Equations:
    r=radius
    s=arc length
    theta= angular displacement
    omega=(theta2-theta1)/time
    alpha=(omega2-omega1)/time
    theta=omega1xtime+.5alphaxtime^2
    omega2^2=omega1^2+2alphaxtheta
    omega2=omega1+alphaxtime
    theta=.5(omega1+omega2)time
    v^2/r=omega^2/r
    i=moment of inertia
    torque=radiusxforce
    a=g(m1-m2)/(.5mp+m1+m2) --> mp is mass of pulley

    3. Solution attempt:
    a) a=g(m1-m2)/(.5mp+m1+m2)
    Since only variables were given in the problem, then I am assuming that this is the answer.
    b) t1=m1g-m1a or t2=m2g+m2a
    I am not sure, which one or both or neither is correct.
    c) omega2=omega1+alphaxtime
    I do not know if I need to go any further on this part of the problem.
    d) omega=(theta2-theta1)/time
    Once again, I believe this is all I need.
    e) theta
    Since theta=angular displacement.
    f) theta=.5(omega1+omega2)time
    Then, solve for time to find how long it takes for block 2 to fall to the floor.

    I am just want to know if I am on the right track, I believe this is correct (or mostly correct), but I was wondering if I could have some verification? Thanks.
     
  2. jcsd
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