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Total Kinetic Energy of a Body

  1. Jul 7, 2004 #1
    [SOLVED] Total Kinetic Energy of a Body

    I was reading my physics text and it came up with the standard formula for the kinetic energy of a rigid body in motion: the sum of the translational and rotational kinetic energy. That got me wondering whether we could include the internal energy by simply taking the sum of the kinetic energy of each particle when viewed in a frame where the translational and rotational velocity appear to be zero. In other words the total kinetic energy of the body should be (where v_Hi is the velocity in addition to trans and rot):

    [tex] K = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + \sum\frac{1}{2}\, m_i v_\text{Hi}^{\;\; 2} = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + H [/tex]

    So I took a hand at the calculations, but the best I could seem to come up with was the following:

    [tex] K = \frac{1}{2}\,Mv_\text{cm}^{\;\; 2} + \frac{1}{2}\,I_\text{cm}\,\omega^2 + H + \omega \, \cdot \, \sum m_i \, (r_i \times v_\text{Hi}) [/tex]

    In the last term the quantities are vectors. This would seem to imply that the total kinetic energy of a body is more or less than the sum of its translation, rotational, and internal energy (assuming the internal energy is as i have defined it). Have I made some sort of mistake or is there some better way of looking at it?
  2. jcsd
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