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Homework Help: Torque for rotating a shaft of unbalanced weight

  1. Oct 28, 2011 #1

    sekar_507

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    1. Hi all. Can anyone please help me find the torque required for rotating a shaft which has two masses 1000 Kg each at 120 Degree angles. Please refer to the attachment for clarity.



    2. Relevant equations



    3. The attempt at a solution
     

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    sekar_507, Oct 28, 2011
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  3. Oct 28, 2011 #2

    rude man

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    I think you need the lengths of the two bars.
     
    rude man, Oct 28, 2011
  4. Oct 30, 2011 #3

    sekar_507

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    The bars are 500mm long each. How do I calculate the torque?
     
    sekar_507, Oct 30, 2011
  5. Oct 30, 2011 #4

    rude man

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    Torque = force times length.
    So, what do you think you should use for length and force?

    You sure the bars are of equal length? Looks like bar 1 is longer than bar 2.
     
    Last edited: Oct 30, 2011
    rude man, Oct 30, 2011
  6. Oct 30, 2011 #5

    sekar_507

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    should i use formula torque = I * Angular velocity for calculating the torque for rotating the shaft? Please clarify
     
    sekar_507, Oct 30, 2011
  7. Oct 30, 2011 #6

    rude man

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    No, since that's the formula for angular momentum, not torque.
     
    rude man, Oct 30, 2011
  8. Oct 30, 2011 #7

    sekar_507

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    ok then. since both the masses are at 120 deg angle. what is the total mass i should i consider. should i add 1000+1000 = 2000 Kg or should i calculate the resultant mass? can u help me in finding the resultant mass?
     
    sekar_507, Oct 30, 2011
  9. Oct 30, 2011 #8

    rude man

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    You never answered my question: are you sure the masses are of equal length?
     
    rude man, Oct 30, 2011
  10. Oct 30, 2011 #9

    sekar_507

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    Yes. Both the masses are of equal length.
     
    sekar_507, Oct 30, 2011
  11. Oct 31, 2011 #10

    rude man

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    OK, (though I'm still wondering why the drawing shows them as clearly of different lengths).

    How about writing an expression for the torque about the axis of rotation as a function of angular position? You can call zero angle at any position you like.

    Take one bar at a time. What's the torque gravity applies to one of the bars as a function of its angular position?
     
    Last edited: Oct 31, 2011
    rude man, Oct 31, 2011
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