classical mechanics - Cases in which angular velocity and angular momentum point into same direction

I know that angular momentum $\vec{L}$ and angular velocity $\vec{\omega}$ of a rigid body doesn't point into the same direction in general. However if your body spins around a principal axis, $\vec{L}$ and $\vec{\omega}$ point into the same direction.

In which other situations is this is exactly or at least approximately true? Please give reasons or a mathematical derivations why it is true in a given situation.

Answer

There are no "other" examples. The condition that $\vec \omega$ and $$ \vec L = I_{\rm tensor} \cdot \vec \omega $$ point to the same direction i.e. $$ (\vec L=) I_{\rm tensor} \cdot \vec \omega = k \vec \omega $$ where $k$ is a real number (and no longer a tensor) is a definition of an eigenvector of $I_{\rm tensor}$: both $\vec \omega$ and $\vec L$ are eigenvectors of the moment of inertia in such a case.

Eigenvectors of the moment of inertia tensor are called the principal axes. They may always be chosen to be orthogonal to each other and the tensor has the form $$ I_{\rm tensor} = {\rm diag} (I_1,I_2,I_3) $$ in the coordinate system given by these principal axes. In a generic case, the three principal axes are uniquely determined by the tensor (and the axes may be shown to be orthogonal to each other, as eigenvectors of any Hermitian operator).

The only subtlety appears if some of the entries $I_1,I_2,I_3$ coincide. If two of them are equal, any combination of the two vectors is an eigenvector and there is a freedom in choosing these two principal axes (the third axis corresponding to a different eigenvalue is still unambiguous).

If $I_1=I_2=I_3$, then the tensor is proportional to the unit matrix and any vector is an eigenvector of it with the same eigenvalue $I_1=I_2=I_3$. In that case, the choice of principal axes is totally arbitrary. If 2 or 3 eigenvalues coincide, the orthogonality of the basis isn't guaranteed but we still usually impose the extra conditions that the "principal axes" should be orthogonal to each other.

One could discuss various examples – particular shapes with particular tensors of the moment of inertia. But the important fact here is that for the purposes of the spin and angular momentum, only the tensor of the moment of inertia matters. It may be calculated from any matter distribution but everything else besides the directions of the three axes and the three eigenvalues $I_1,I_2,I_3$ is irrelevant for discussions on the angular momentum.