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Rotational energy conservation

  1. Jun 18, 2013 #1
    I am a little confused about the energy conservation of a pin ended free falling rod.
    When i try to derive energy conservation equation i am not sure including angular and linear velocity at the same time. I try to visualize the problem in the attached picture and put my derivation also.
    Any explanation will be appretiated.

    Attached Files:

  2. jcsd
  3. Jun 18, 2013 #2
    This is the kinetic energy in general:

    [itex]KE = \frac{1}{2}m v_{cm}^2 + \frac{1}{2}I_{cm}\omega^2[/itex]

    for this example the velocity of the center of mass
    [itex]v_{cm} = \frac{L}{2}\omega[/itex]

    [itex]KE = \frac{1}{2}m \frac{L^2}{4}\omega^2 + \frac{1}{2}I_{cm}\omega^2[/itex]

    Notice if the center of rotation is at the end of the rod, the angular velocity [itex]\omega[/itex] is the same as for the center of rotation being the center of mass.

    The moment of inertia changes by the parallel axis theorem
    [itex]I_0 = I_{cm} + m\left(\frac{L}{2}\right)^2 = I_{cm} + m\frac{L^2}{4}[/itex]

    So we find that the rotational kinetic energy about the end of the rod is equal to the total kinetic energy from before.
    [itex]\frac{1}{2}I_0\omega^2 = \frac{1}{2}\left(I_{cm} + m\frac{L^2}{4}\right)\omega^2 = KE[/itex]
  4. Jun 19, 2013 #3
    MisterX thanks for the answer.
    Could you also suggest me an alternative way of determining angular velocity?
  5. Jun 19, 2013 #4
    No, he's saying that your solution is incorrect: either you use the kinetic energy of rotation with respect to the centre of mass (and then the moment of inertia is not the one you wrote) *and* the centre of mass' kinetic energy, or the kinetic energy of rotation with respect to the pin, *only* (with the moment of inertia you wrote).
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