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Forces and Pulleys

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data
    A pulley has a moment of inertia of 5.0 kg*m^2 and a radius of 0.5m. The cord supporting the masses m_1 and m_2 does not slip, and the axle is frictionless. Find the acceleration of each mass when m_1 = 2.0 kg and m_2 = 5.0 kg.


    2. Relevant equations
    F=ma, Fr=I(alpha)


    3. The attempt at a solution
    My work is attached in the image. The thing that appears or be wrong is that T_2 - T_1 = Ia/r^2, not -Ia/r^2. However, I can't figure out why it shouldn't be negative, as the tangential acceleration for the pulley should be negative, as it is the same as the acceleration downwards, which is negative.
     

    Attached Files:

  2. jcsd
  3. Jul 31, 2011 #2
    Torque=I*aangular so (T2+(-T1))=I*alinear/r^2

    Since It's T2+(-T1), not the other way around, you're taking a tension that goes in the counterclockwise direction and adding a tension that goes in the clockwise direction. When adding these tensions/torques to find the net torque, you have defined the counterclockwise direction as positive. It follows logically that if T2 is larger, you'll get a positive aangular and that the pulley will spin in a counterclockwise direction (The net torque will be in the counterclockwise direction). Otherwise, if T1 is greater, you'll get a negative aangular that will rotate in the clockwise direction. By choosing T2 as the positive tension, you made a counterclockwise rotation positive because T2 acts on the pulley in the counterclockwise direction.

    While a counterclockwise aangular may rotation "down" on the left, it rotates "up" on the right, so you can't call it negative or positive based on the linear positive and negative directions. Hope that helps
     
    Last edited: Aug 1, 2011
  4. Aug 1, 2011 #3
    Thanks for answering.
    Well, the equation you gave (T1-T2 = Ia/r^2) is the same as the one I got, just multiplied by -1 on both sides (T2-T1 = -Ia/r^2). In your case, you are designating T2 as negative and thus counterclockwise. Shouldn't the acceleration still be negative then?

    As for the tensions, shouldn't T2 be the one that's pulling counterclockwise, as it is pulling m2 up, and T1 be clockwise, as it is pulling m1 up?

    Thanks.
     
  5. Aug 1, 2011 #4
    Oh, my bad, I was a little inconsistent with our definitions of T1 and T2. For mine, I mean T1 to be the counterclockwise tension above the mass that is going to fall. In other words, I would have written:

    T1-m1g=-m1a, instead of T2-m2g=-m2g

    To make our posts consistent, I'll switch T1 with T2 and vice versa, then you can re-read it and hopefully it'll make more sense, sorry!

    With the definition changes you are right, T2 pulls counterclockwise and T1 pulls clockwise, if the falling mass and T2 are on the left.
     
  6. Aug 1, 2011 #5
    No, you were right. Let me rephrase that.

    Let m1 be the mass on the left, and m2 the mass on the right. Let T1 be the tension holding up m1 and T2 be the tension holding up m2. So would T1 also be the force that applies the counterclockwise torque, and T2 the clockwise torque?

    I always thought that the T2 applied the counterclockwise torque because it is the result of m1g.
     
  7. Aug 1, 2011 #6
    T1 is on the left, T2 on the right.

    T1 applies the counterclockwise torque to the pulley, and T2 applies the clockwise torque to the pulley. You can think of it like this, the rope transmits a particular force (What we call tension) from the mass to the pulley and vice versa. Because of Newton's Third Law, the rope is pulling up on the mass just as much as the mass is pulling down on the rope (The rope is also going down with the mass, so you can't say the mass is pulling down on the rope with force of mg). The rope transmits this tension onto the pulley and pushes the pulley down by this tension, and the pulley pulls up on the rope. In the end, the rope is pushing DOWN on the pulley with the tension. So, T2 is facing downward, and thus applies a clockwise torque.

    In other words, you are thinking about the tension only as an "upward" force. Think of it also pushing down on the pulley. Hopefully the diagram with help from wikipedia: http://upload.wikimedia.org/wikipedia/commons/8/8f/Tension_figure.svg

    Just pretend the scaffold is the pulley. You'll notice a downward force/tension "on scaffold by rope". This is the torque.
     
    Last edited: Aug 1, 2011
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