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Conservation of angular momentum and change in rotational kinetic energy

  1. May 3, 2012 #1
    I have a question regarding angular momentum and rot kinetic energy. For example, if angular momentum is conserved, and the radius is cut in half, then moment of inertia is reduced by a fourth, which will result in increase in angular velocity by factor of 4. My question is why is the rotational kinetic energy increase by a factor of 4 also. Since the equation of rot kinetic energy is (1/2)Iw^2, the new w increased by factor of 4 but the new I also decreased by factor of 1/4. So wouldn't KE stay the same???
     
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  3. May 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi chamddol! welcome to pf! :smile:
    as you know, angular momentum is always conserved (if there's no external torque, of course)

    in a "collision" situation, (mechanical) energy usually isn't conserved, but if the changes are gradual (as here), yes we would usually expect it to be conserved

    however, that's forgetting the work energy theorem … work done = change in mechanical energy

    imagine that you're rotating on ice, and you're holding onto a heavy mass on a rope

    if you pull the rope in, the total energy increases because you are doing work (force "dot" distance) by pulling the mass in :wink:

    when you reduce the moment of inertia of any rotating mass, you have to do work! :smile:
     
  4. May 4, 2012 #3

    haruspex

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    Er - no. Think that through again. You've reduced I by a factor of 4, and increased w by a factor of 4, so you've increased w^2 by a factor of ...?
     
  5. May 4, 2012 #4

    Cleonis

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    attachment.php?attachmentid=46972&stc=1&d=1336163618.png

    Expanding on the answer given by tiny-tim:
    The curved line in the diagram represents the trajectory of an object that is pulled closer to the center. The dark grey arrow represents the centripetal force.

    Now, in the case of perfectly circular motion the centripetal force is at all times perpendicular to the instantaneous velocity, and hence there is no change of kinetic energy. But here, with the object being pulled closer to the center, the exerted force is not perpendicular to the instantaneous velocity.

    You can think of the force as decomposed, one component perpendicular to the instantaneous velocity and one paralllel to the instantaneous velocity. The perpendicular component causes change of direction, the parallel component causes acceleration.

    So in the process of pulling closer to the center you are doing work upon the object. This is big: by the time you've managed to reduce the radial distance by half you have quadrupled the kinetic energy.


    In general: when you have an object in circumnavigating motion, and you pull that object closer to the center, then the acceleration occurs because you are doing work upon that object.
     

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  6. May 5, 2012 #5

    Rap

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    haruspex is right. Its a simple mathematical error.
     
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