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Torque and Velocity

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    An Atwood Machine consists of two masses m1 and m2 attached to each other by a single massless string passing over a cylindrical pulley of mass M without slipping. The masses are released from rest, and allowed to accelerate until each has moved a distance h = 1.8 m. Assuming that upwards is positive, rank the velocity of m2 (from most negative to most positive) at this moment for the following cases:
    Case A: m1 = 21 kg, m2 = 16 kg, M = 28 kg

    Case B: m1 = 13 kg, m2 = 16 kg, M = 31 kg

    Case C: m1 = 25 kg, m2 = 16 kg, M = 26 kg

    Case D: m1 = 18 kg, m2 = 16 kg, M = 34 kg

    Case E: m1 = 17 kg, m2 = 16 kg, M = 33 kg


    prob15a.gif
    2. Relevant equations
    sum of the torques = I*alpha
    vf^2 = vi^2 + 2a (yf - yi)


    3. The attempt at a solution
    I did a free body diagram of the cylinder. I did sum of the Torques = I(alpha). I found that the acceleration was equal to
    a = [2g(m1+m2)]/M

    Then I used a constant acceleration equation Vf^2 = Vi^2 + 2a(yf-yi). I said the initial velocity is 0, I substituted [2g(m1+m2)]/M in for a, I said that yf is the height change (1.8 m) and yi is 0.
    I simplified this to say Vf = the square root of [4gh(m1+m2)]/M

    I plugged in the values for the examples A through E. I made B a negative velocity because mass 2 is heavier so it would go down.

    My ranking was B<D=E<A<C
    Thanks for helping :)
     
  2. jcsd
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