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Torque on shaft of pulley question

  1. Jul 23, 2013 #1
    Hi

    Here is my question:
    A solid disc pulley with a 5kg mass, radius of 0.5m has one layer of thin rope coiled around it. On the other end of the rope is a metal block with a mass of 10kg.
    Calculate the following (without take rope friction into account. also the rope does not slip):
    1. The torque on the pulleys shaft.

    To solve this I did the following:
    Torque = Force x Radius
    Torque = 10kg x 0.5m
    Torque on pulleys shaft is = 5Nm

    Would you please let me know if this is correct?

    Thanks.
     
  2. jcsd
  3. Jul 23, 2013 #2

    Doc Al

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    (1) 10 km is a mass, not a force.
    (2) What is the tension in the rope?
     
  4. Jul 23, 2013 #3
    The tension in the rope is mass x gravity = 10 x 9.8 = 98N

    So my answer should be:
    Torque = Force x Radius
    Torque = (10kg x 9.8) x 0.5m
    Torque on pulleys shaft is = 49Nm

    Is this correct?

    Thanks
     
  5. Jul 23, 2013 #4

    Doc Al

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    That's the weight of the hanging mass, not necessarily the tension in the rope.

    Assuming the pulley is free to rotate, there will be acceleration. You'll have to figure out the tension by applying Newton's 2nd law to both the pulley and the hanging mass.
     
  6. Jul 23, 2013 #5
    To calculate acceleration 1st calculate pulley inertia:

    pulley inertia(I) = 1/2mr2 (2 represents squared)
    I = 1/2 x 5 x 0.25
    I = 0.625

    acceleration(a) = hanging mass weight divided by mass of hanging mass plus pulleys inertia divided by radius squared.
    a = 98 / (10 + (0.625 / 0.25))
    a = 98 / (10 + 2.5)
    a = 7.84 m/s

    I think the above is the correct acceleration. I don't know how to calculate the tension on the rope, please would you give me another hint?

    thanks
     
  7. Jul 24, 2013 #6

    haruspex

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    The standard way to solve a problem like this is to introduce an unknown, T, for the tension, another, a, for the linear acceleration, then analyse each of the masses wrt forces/torques/accelerations separately. That would give you a pair of equations from which you could deduce a and T.
    You've finessed that by treating the rotational inertia / radius-squared as equivalent to a mass. That works. Having done that, it's easy to get the tension by considering just the hanging mass. There are two forces on it, you know one of them, you know the mass, and you know the acceleration.
     
  8. Jul 24, 2013 #7
    Thanks for your reply haruspex

    So f = ma
    f = 10 x 7.84
    f = 78.4N

    So the Torque on the pulley shaft - T = fr
    T = 78.4N x 0.5m
    T = 39.2Nm

    Is this correct?

    Thanks.
     
  9. Jul 24, 2013 #8

    haruspex

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    What exactly does f mean here?
     
  10. Jul 24, 2013 #9
    f = force

    Is that not correct?
     
  11. Jul 25, 2013 #10

    haruspex

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    Yes, but which force?
     
  12. Jul 25, 2013 #11
    I was thinking the for I was calculating was the tension force on the rope.
     
  13. Jul 25, 2013 #12

    Doc Al

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    When you apply Newton's 2nd law, ƩF = ma, you must decide what system or object you are analyzing, in this case the hanging mass. You then must identify all the forces acting on that mass. The rope tension is just one of the forces acting on the mass.

    To really learn how to solve this and similar problems, do not rely on "canned" formulas, like the one you used for acceleration in post #5. Instead, derive your own formulas by applying Newton's 2nd law to both the pulley and the hanging mass. (Just like haruspex has advised.)

    Also, do not create multiple threads on the same problem! https://www.physicsforums.com/showthread.php?t=702575
     
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