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Torque question

  1. Nov 27, 2006 #1
    If you have two blocks connected over a pulley, and are using the torque on the pulley to calculate the acceleration, then you should disregard the sign of the torque, always treating it as positive, no matter which direction it rotates?

    Is this right?

    Thanks,
    Dorothy
     
  2. jcsd
  3. Nov 27, 2006 #2

    Doc Al

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    I'd say no. Pick an overall direction for the acceleration. (For example: block one goes up, block two goes down, pulley goes clockwise.) Choose the sign of your torque to match.

    The blocks and pulley are connected, so the acceleration of all three must be aligned.
     
  4. Nov 27, 2006 #3
    Thanks, Doc Al. Well, here is my problem, then.

    A m1 is connected over a pulley to a m2 on an incline. Let's say the incline is to the right of the pulley, so that the acceleration is to the right.

    T1 - friction1 = m1a
    m2g(sin theta) - friction2 - T2 = m2a
    (m1 + m2)R = -I(alpha) = -Ma/2 (where M is mass of pulley).

    If I rotate the diagram 180 degrees, so the incline is on the left, acceleration is to the left, keeping positive in the direction of the acceleration, the only equation that would change is the torque:

    (m1 + m2)R = Ma/2

    Which is obviously going to lead to a different acceleration by just changing the horizontal orientation, which is a physical impossibility, of course.

    Or I'm confused (probably)... Clear this up for me, please?

    Thanks, as always,
    Dorothy
     
  5. Nov 27, 2006 #4
    Oh, I'm thinking m1 is on a horizontal surface, if that makes any difference.
     
  6. Nov 27, 2006 #5

    Doc Al

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    These two make sense.
    But this one doesn't. Why are m1 and m2 mentioned in the equation for the pulley?

    What forces act on the pulley? What torques do they exert? Set the net torque equal to I(alpha).

    This is an artifact of this strange equation. Fix it. :wink:
     
  7. Nov 27, 2006 #6
    I'm an idiot. I meant to write:

    (T2 - T1)R = Ma/2

    And really, that's what I was thinking. My fingers typed something else. :yuck:

    But that still leaves me with the same question, doesn't it?

    With the incline on the right, (T2 - T1)R = -Ma/2

    With it on the left, (T2 - T1)R = Ma/2

    Since the only that has changed is the sense of the torque?
     
  8. Nov 27, 2006 #7

    Doc Al

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    OK.
    Nope. The signs of T2 and T1 will change also.

    If the sense of the torques change, so must their signs.
     
  9. Nov 27, 2006 #8
    But how is that possible? Don't the signs of T2 and T1 (the tensions on the cord) depend on the sign of the acceleration? T2 will always be positive, since it is pulling on the pulley in the direction of a, and T1 will always be negative.

    Gosh, I'm sorry to be difficult, but I am just not getting this.
     
  10. Nov 27, 2006 #9

    Doc Al

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    But why did you change the sign of the acceleration but not the torques?

    If all you did was flip the diagram, so that the acceleration now goes left instead of right, then you can:
    (1) Use a convention that right is + and left is -, in which case the signs of T1, T2, and "a" all get reversed;
    (2) Stick to the convention that "a" is always positive, in which case all the signs stay the same.

    Pick one!

    Either way, it's all or nothing. :smile:
     
  11. Nov 27, 2006 #10
    Well, I picked (2). But the sign of the torque has to change, right? If the acceleration is to the right, won't the pulley be rotating clockwise?

    Now, if I flip the diagram, keeping convention 2, then the pulley is rotating counterclockwise. Doesn't that change the sign of the torque?

    Which is why I thought I should ignore the sense of the torque in a problem like this, and always treat it as a positive number. Ok, I should give it the same sign as the acceleration, in this case, positive. Well, in spite of your good efforts, I am still thinking this :eek:
     
  12. Nov 27, 2006 #11

    Doc Al

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    The sign of the torque depends on your convention. Using convention 2, if it accelerates clockwise then clockwise is positive. Everything flips, so the signs all stay the same.

    Counting clockwise as + and counterclockwise as - (or vice versa) is the same as using convention 1, not convention 2.

    Giving torque the same sign as the acceleration is not the same as "ignoring the sense of the torque"--it's following convention 2.
     
    Last edited: Nov 27, 2006
  13. Nov 27, 2006 #12
    Ah.. Penetration. Thank you. Well, I feel dumb, but enlightened. I wish textbooks would be clearer about this.

    Thank you for your patience and help, as always.

    Dorothy
     
  14. Nov 27, 2006 #13

    Doc Al

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    You are always welcome, Dorothy! :smile:
     
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