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Torque on Atwoods

  1. Nov 27, 2007 #1
    A pulley with radius R is free to rotate on a horizontal fixed axis through its center. A string passes over the pulley. Mass m1 is attached to one end and mass m2 is attached to the other. The portion of the string attached to m1 has tension T1 and the portion attached to m2 has tension T2. The magnitude of the total external torque, about the pulley center, acting on the masses and pulley, considered as a system, is given by:

    (m1 – m2)gR

    Can someone explain how to derive this relation? Why is the distance just R?
     
  2. jcsd
  3. Nov 27, 2007 #2

    Doc Al

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    Staff: Mentor

    The force (mg) acts vertically and R is the perpendicular distance to the axis. So each torque is just:

    [tex]\tau = \vec{r} \times \vec{F} = RF = mgR[/tex]
     
  4. Nov 27, 2007 #3
    I thought the force a acting perpendicular to the center of pulley is tension, not gravity.
     
  5. Nov 27, 2007 #4

    Doc Al

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    The force on the pulley is the tension, not the weight. But I thought you were asking about the net torque on the entire system, not just on the pulley. (A perfectly valid way to analyze the problem, but I would not recommend it.)
     
  6. Nov 27, 2007 #5
    Hmm,

    The problem asks for the net torque on the entire system about the center of the pulley. I believe the gravitational forces, m1g and m2g are acting at the center of masses of the two masses, which are some distance from the center. I am confused how to calculate the net torque because I thought you had to add all the individual torques, which would mean I would need to know the distance at any time between the center of mass of each mass and the center of the pulley.

    I'm sorry, but I do not see the reasoning behind just adding up all the forces on the system and multiplying by R.
     
  7. Nov 27, 2007 #6

    Doc Al

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    You can certainly do it that way also. You'll get the same answer.
    All you need is the perpendicular distance. Consider this:
    [tex]\tau = \vec{r} \times \vec{F} = rF\sin\theta[/tex]

    While the distance r is constantly changing, what counts is [itex]r\sin\theta[/itex]--which is constant and equal to R.
    You need to firm up your understanding of torque and how to calculate it.
     
  8. Nov 27, 2007 #7
    Yeah, you're right. I forgot it was the moment arm, since it's the cross product. Wow.
     
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