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How to prove this

  1. Aug 4, 2007 #1
    how does one prove this
    if a system is rotating about three different perpendicular axes then
    total kinetic energy of the sytem is
    [tex]T=\frac{I_{x}\omega_{x}^{2}}{2} + \frac{I_{y}\omega_{y}^{2}}{2}+\frac{I_{z}\omega_{z}^{2}}{2}[/tex]
     
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  3. Aug 4, 2007 #2

    AlephZero

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    Consider the body as a set of particles.

    [tex]T = \frac{1}{2}\iiint \rho (\dot x^2 + \dot y^2 + \dot z^2) dx\,dy\,dz[/tex]

    Then write the translational velocity at (x,y,z) in terms of the rotation velocities and the position (the 3-D version of [tex]v = r\omega[/tex]).

    Rearrange the result and the integrals that define [tex]I_x[/tex] etc will appear.
     
  4. Aug 4, 2007 #3

    olgranpappy

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    well, in general you'll find:
    [tex]
    T=\frac{1}{2}\sum_{i,j}I_{ij}\omega_i\omega_j
    [/tex]
    then you can change coordinates to diagonalize the I tensor... or, if you made a good choice to begin with the tensor will already be diagonal giving you the expression you want.
     
  5. Aug 4, 2007 #4

    olgranpappy

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    I.e.
    [tex]
    \vec v = \vec \omega \times \vec r
    [/tex]
    or
    [tex]
    v_i = \epsilon_{ijk}\omega_j r_k
    [/tex]

    And use the definition
    [tex]
    I_{ij}=\int\rho\left(
    r^2\delta_{ij}-r_ir_j
    \right)
    [/tex]
     
  6. Aug 4, 2007 #5

    Andrew Mason

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    I think you can start with the given that:

    [tex]KE = \frac{1}{2}I|\vec\omega|^2[/tex]

    What is the square of the length of the vector [itex]\omega[/itex] in terms of the x, y and z components?

    AM
     
  7. Aug 5, 2007 #6

    olgranpappy

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    no. he can't start with that. Only in the special case of
    [tex]
    I_x=I_y=I_z=I
    [/tex]
     
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