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Atwood's machine with angular momentum

  1. Jan 2, 2015 #1
    1. The problem statement, all variables and given/known data

    An Atwood's machine consists of two masses, mA and mB, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R0 and moment of inertia I about its axle, determine the acceleration of the masses mA and mB.

    2. Relevant equations

    L = mvr
    T = dL/ dt

    3. The attempt at a solution

    This is an example from my textbook. The solution involves:

    angular momentum about the axle = (mA +mB)vR0 + Iv/R0
    torque about the axle = mBgR0 - mAgR0

    Then it uses the equation T = dL/dt and finds accelerations of the masses.

    What confuses me is the way it calculates the torque. When we were solving this exact same problem in rotational motion, the equations were like these:

    mB*g - FTB = mB*a
    FTA - mA*g = mA*a
    (FTB - FTA )*r = I*a/r

    How can angular momentum solution get rid of tensions in the torque equation?

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    Last edited: Jan 2, 2015
  2. jcsd
  3. Jan 2, 2015 #2


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    The torque equation mBgR0 - mAgR0 is not correct.
    At the pulley you have net torque = TA - TB

    At mA you have TA - mA g R = mA aA
    At mB you have TB - mB g R = mB aB
    since aB = - aA the differerence has an extra term ( mA + mB ) aA

    Yuo can also check that this gives the right relationship for a massless pulley, where the torque difference at the pulley must be zero.
  4. Jan 2, 2015 #3
    I've uploaded the picture.
  5. Jan 2, 2015 #4


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    Can't argue with that, can we :) ?
    Book calls that the external torque. I'm using a simpleton approach looking at the upper half and the lower half separately. So my net torque is on the wheel only and it gives the wheel an angular acceleration Iα. Book's external torque gives an angular acceleration to the whole lot: ( (mA + mB) R02 + I ) α

    So fortunately we end up doing the same thing to calculate the same α. As we should ;)
  6. Jan 2, 2015 #5


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    So, for the record: I was a bit hasty stating "The torque equation is not correct" . Depends on what torque is meant.

    And the answer to your original question is now clear, I hope: the book solution can avoid the tensions by considering them internal. In exchange the angular momentum then has to be for the whole system.
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