Suppose that at a certain instant the angular momentum with respect to the center of mass is not parallel to the angular velocity. Does this necessarily imply that the angular momentum is rotating around the axis of rotation? If so, then an isolated body must necessarily have angular momentum with respect to the center of mass parallel angular velocity (otherwise the angular momentum varies and the system is not isolated). However, this gives me a perplexity: an irregular rigid body has only 3 axes such that $ \vec{L}_{cm} $ and $ \vec{\omega} $ are parallel. Does this mean that the irregular rigid body can only rotate in three directions? It seems completely absurd to me, but where is the mistake?
Does an irregular rigid body can only rotate in three directions
angular momentumangular velocityrigid-body-dynamicsrotational-dynamics
Related Solutions
Does this mean we can somehow add the 3 rotations (which are about different axes) and get an equivalent rotation about some other (single) axis?
Yes. It's one of many consequences of Euler's rotation theorem. This neat little trick of finding the Euler rotation axis works in three dimensional space, and in three dimensional space only. The Lie group SO(3) is a very special space.
A more generic way of looking at rotation is that rotation in any Euclidean space comprises a combinations of rotations parallel to a two dimensional plane rather than about an axis. There's only one plane in two dimensional space, so rotation in 2D space requires but one parameter. There are three pairs of planes of rotation in three dimensional space (the YZ, ZX, and XY planes), so rotation in 3D space requires three parameters. You can think of those planar rotations as being about an axis (with an arbitrary decision as to what qualifies as positive or negative rotation). There are six planes of rotation in four dimensional space, so rotations in 4D space requires six parameters. This can result in something very weird, something you don't see in 2D or 3D space, a Clifford rotation.
As you mentioned, $\vec {L_o}_⊥$ is proportional to $\omega$. So the middle term means $\vec {L_o}_⊥$ varies by varying angular velocity.
You can also derive this. formula like this:
$$\frac{d \vec {L_o}_⊥ }{dt}= \frac {d\vec{\omega}}{dt}A \hat{L_o}_⊥+ \vec{\omega}A \frac{d\hat{L_o}_⊥}{dt}= \frac{1}{\omega}
\frac{d \omega }{dt} \vec{L_o}_⊥ + \vec{ \omega } \times \vec{L_o}_⊥ $$.
Best Answer
A general rigid body has 3 principal axis. Suppose $I_1<I_2<I_3$. If it rotates around $I_1$ or $I_3$ without external torques, the angular velocity doesn't change and is parallel to the angular momentum. Theoretically the same happens with rotation around $I_2$, but in a unstable mode. Any deviation and it starts to wobble (tennis racket effect).
The angular momentum is conserved if there is no applied torques, but the angular velocity can varies chaotically. https://www.youtube.com/watch?v=WvrbejgDL3A