newtonian mechanics - Relation between torque and moment of inertia

I have seen 2 formulas: $\tau = I \alpha$ and $\tau = I \dot\omega + \omega \times I \omega$. Which one shall I use in which case please? Also, am I right to think that $\alpha$ from the first formula is $\dot\omega$ from the second one?

Answer

Yes, $\alpha = \dot{\omega}$, being the angular acceleration. The first equation is special case of the second equation.

For a general object the moment of inertia is not just a scalar (a single value) but a tensor, in that case you have to use your second equation. $\tau$ and $\omega$ are then vectors and $I$ is a 3x3 matrix.

But when you spin an object around one of its high symmetry axes (one of the eigenvectors of the inertia matrix $I$), the equation simplifies to your first equation. Proof:

If $\vec{\omega}$ is an eigenvector of $\hat{I}$ it holds that: $\hat{I} \vec{\omega} = \lambda \space \vec{\omega} = \vec{\omega} \space \lambda$ Therefore your second equation becomes: $\vec{\tau} = \hat{I}\dot{\omega} + \vec{\omega} \times \vec{\omega} \space \lambda$ and $\vec{\tau} = \hat{I}\dot{\omega} + 0$ since the crossproduct of a vector with itself is 0.