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Homework Help: Torque on a pulley with a mass

  1. Sep 25, 2012 #1
    If there is a load of say, mass =m Kg which is being raised on the end of a light inextensible rope by means of a winch.

    see attachment.

    the pulley of the winch has a radium x meters and is lets say that it is frictionless.
    the acceleration of the load is a ms^-2

    can someone tell me how would i calculate the Torque / moment on the pulley ?
    i know that torque = Force * Displacement

    here is what i think shud be the correct equation :

    net force =N = T - mg

    N[x]= Torque

    But the equation in my notes says otherwise :

    let K = the torque applied by the winch (the torque driving the pulley)

    Torque on the pulley = K - (x)(T)

    which one is correct and why ?

    Attached Files:

  2. jcsd
  3. Sep 25, 2012 #2
    The Force here is the force applied on the pulley. From the figure you shared, what is the force on the pulley?
  4. Sep 25, 2012 #3

    i stated the Torque instead, and u have the radius (where the force is acting tangential)

    the force on the pulley which causes it to rotate CLOCKwise = K/x
  5. Sep 25, 2012 #4
    Where's K in your drawing?

    my guess is that you mean K is some other given torque.
    It should be in your drawing.

    Torque depends on 2 things. The force and the lever arm.
    so you need to keep track of what you're torque-ing
    about. in this case the natural origin is the center of
    the pulley.

    Torques add up - just keep track of the sign. say we pick clockwise
    as positive. and calculate torque (on the pulley) about the pivot

    then the net torque is the sum of
    torque due to K - which is just K
    torque due to m_pulley g - which is 0 (since the lever arm is 0)
    torque due to T - which is -xT (x is lever arm , and minus sign is down)
    so net torque is τ=K - xT

    also the word is should not shud, it hurts my eyes to see the latter.
  6. Sep 25, 2012 #5
    I dont understand this point .

    Would you please explain it in more detail ? :confused:
  7. Sep 25, 2012 #6
    Firstly, as qbert points out, we don't see 'K' in the image, so I cannot comment on it.

    Secondly, my question exactly was "From the figure you shared, what is the force on the pulley?". The figure serves as a free body diagram of the pulley, so the force on the pulley is: T, the tension acting on the rim, downwards.

    Now you can use your equation: torque = Force * Displacement

    To calculate the tension force: Since you know the acceleration of the block and its mass, you can calculate the tension force. This tension force is the same as the one acting at the rim of the pulley.
  8. Sep 26, 2012 #7

    Why is that ?

    i dont understand the free body diagram.

    Why is TENSION of the string (applied on the pulley cuz the string is in contact with the pulley) causeing the Torque, why is weight not causing the torque ?
  9. Sep 26, 2012 #8
    I see no pulley in your picture. It looks to me like a simple hoist drum lifting a load.

    If you add a frictionless Lilly to the rigging then the torque on it is zero. That is the whole point of putting good bearings in a pulley.
  10. Sep 26, 2012 #9
    Because the weight itself is not directly attached or connected to the pulley.

    The string tension is transferred to the pulley due to friction. If there were no friction, the string would slide over the pulley and produce no torque.
  11. Sep 26, 2012 #10
    Because we are calculating torque about the center of mass of the pulley and weight of the pulley is acting at the center of mass.

    There is no torque at a point due to a force at the same point. Do you know why?

    EDIT: Above post answers why weight of the block is not causing torque, this post answers why weight of pulley isn't causing torque.
  12. Sep 26, 2012 #11
    Because the weight of the pulley acts downwards passing through the axis of rotation, thus the Torque = W*r

    W = weight of pulley
    r= zero
  13. Sep 26, 2012 #12
    Right, so weight is not causing the torque. Tension is applied at the rim of the pulley, it's not passing through the axis of rotation, so it does have a torque.
  14. Sep 27, 2012 #13
    Another related problem for me [here]
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