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For a torque-free symmetric top, the Inertia tensor has an inverse $I^{-1}$, and $L=I\omega$. Which implies that $\omega=I^{-1}L$. But since $I, L$ are constants, $\vec\omega$ is a constant. However, $\vec\omega$ precesses. Why is there this paradox in argument?

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Does your "paradox" also stand when you use the standard $L=I\omega$? $\omega$ changes (precession) but $I$ is constant, so $L$ should vary. Yet, being torque-free implies $L$ should be constant. –  BMS Nov 29 '13 at 20:47
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up vote 4 down vote accepted

The moment of inertia tensor is not constant in the external reference frame (http://en.wikipedia.org/wiki/Precession#Torque-free )

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+1: That was fast. –  joshphysics Nov 29 '13 at 20:56
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The point being, that as the top moves, it's distribution of mass changes, so $I$ changes. –  lionelbrits Nov 29 '13 at 22:37
    
But the position of the masses with respect to the principal axes and the center of mass remains the same, doesn't it? So, as long as the 'rigid' body remains rigid, how can $I$ change? –  Artemisia Nov 30 '13 at 1:27
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@Artemisia: The laws of dynamics of a rigid body can be written either in an inertial frame of reference, or in a rotating frame of reference (en.wikipedia.org/wiki/… ). In the inertial frame of reference, one uses the moment of inertia tensor in the inertial frame of reference, which is not constant, but angular momentum is preserved. In the rotating frame of reference, one uses the moment of inertia tensor in the rotating frame of reference, which is constant, but angular momentum is not preserved. –  akhmeteli Nov 30 '13 at 3:10
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@Artemisia To convince you further that akhmeteli is right see physics.stackexchange.com/a/89304/19976 and lionelbrits's answer to the same question. –  joshphysics Dec 9 '13 at 6:09
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