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

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