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Does Torque act on a disc rotating with constant angular velocity

  1. Oct 16, 2009 #1
    Hey Folks
    I have rather a silly doubt i guess...
    It goes like this

    Consider a disc rotating about a fixed axis with constant angular velocity. Which means it has no angular acceleration ie it must be 0. Now since Torque=(moment of inertia)*(angular acceleration). hence the torque acting on the disc must be zero.
    But i have come across situations where torque comes into the picture while calculting the power produced by say a Turbine(rotating with a constant angular velocity) . where Power=(torque)*(angular velocity)...?? Why is it so? isn't the torque 0 on circular bodies rotating with constant angular velocity?
     
  2. jcsd
  3. Oct 16, 2009 #2

    FredGarvin

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    Science Advisor

    The torque in a real system will never be 0, especially in a turbine engine. The aerodynamic loads alone load the rotating components highly. The only time you would have the torque = 0 is in a perfect vacuum with no losses due to things like friction and windage.
     
  4. Oct 16, 2009 #3
    Ok under those conditions(vaccum,no friction losses etc) will it be appopriate to say the power produced by a turbine rotating at constant angular velocity is 0 since torque is 0???
     
  5. Oct 16, 2009 #4

    Doc Al

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    Staff: Mentor

    If the angular velocity is constant, then the net torque is zero. But as FredGarvin says, you may need to apply a torque to overcome friction, etc.
     
  6. Oct 16, 2009 #5
    If the angular acceleration of the disk is zero that means the net torque is zero.

    In the case of a turbine rotating at constant angular velocity, you have equal and opposite torques acting on both sides of the central shaft. The first torque is the output torque produced by the turbine engine, and the second torque is the resistive torque of the prime mover (which might be a generator). Since they are equal and opposite, the turbine rotates at constant angular velocity.

    The power equation you gave is correct.


     
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