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Total kinetic energy

  1. Feb 16, 2015 #1
    K = Ktranslational + Krotational

    My book says that in order for this equation to be valid the moment of inertia must be taken about an axis through the center of mass. Why is this true?
  2. jcsd
  3. Feb 16, 2015 #2


    Staff: Mentor

    The rotational KE is defined so as to make that equation true.
  4. Feb 17, 2015 #3
    i may be wrong but you can think of that because the rotational kinetic energy have nothing different respect the translational kinetic energy.
    If you mathematically divide an object rotating through an axis in "particles" the KE of one of these is dKE=0.5 dm (w r)^2
    so to find the entire KE of the rotating object you will have to integrate that stuff. w (the derivative of the angle) is a constant so you are just going to find the moment of inertia of the body with respect to the axis of rotation.
  5. Feb 17, 2015 #4
    If the body is free it's a common physics proof that the axis will pass through CM, if th ebody is constrained tha axis may be different. So to answer your question. It's because of the Newton's laws.
    If you have a strange translating mechanical device with a rotating part, not rotaing through the CM the inertia is going to be taken with respect to another axis.
  6. Feb 17, 2015 #5


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    Science Advisor

    Isn't this because Ktranslational is computed based on the velocity of the center of mass? Could you not pick some other point, as long you stay consistent?
  7. Feb 17, 2015 #6
    If you pick any other point the energy will contain cross terms.
    Only in the center of mass system the KE can be decomposed in a pure translation part and a pure rotation part.
  8. Feb 17, 2015 #7


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    But will you still get the same total kinetic energy K?

    What do you mean by "pure"? Isn't rotational KE just the sum of tangential translational KE, of all point masses forming the object? And what counts as "tangential" depends on the center point you pick.
    Last edited: Feb 17, 2015
  9. Feb 17, 2015 #8
    In respect to the lab system should be the same.

    You will have terms containing the product between the velocity of this other point (not the COM, let say P) and the individual velocities of the points in the system.
    If the velocity of the COM is used, the mixed term will have the product between the velocity of P in the lab system and the velocity of P in the COM system.
    I don't know if you will call this "rotation".

    See here, for example:
    https://books.google.ca/books?id=Ya...#v=onepage&q=konig's theorem mechanics&f=true
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