Angular momentum in non-inertial frame of reference (König’s theorem)

Question

König's theorem states that:
$ \vec{L} =\vec {r}_{CoM}\times \sum \limits _{i}m_{i}{\vec {v}}_{CoM}+{\vec {L}}'=\vec {L}_{CoM}+\vec {L}' $
where $ \vec{L}'$ is the angular momentum with respect to the center of mass in a non-inertial frame of reference integral with the CoM.
For instance, we could take a rigid body rotating around its center of mass with velocity $\vec{\omega}$ and we could use as non-inertial frame of reference to calculate $\vec{L}'$ a frame of reference centered in the center of mass and rotating with the same angular velocity $\vec{\omega}$. In this case, we can notice that each particle of the rigid body will have zero velocity in our non-inertial frame of reference; therefore, $\vec{L}'$ will be 0. Hence, $ \vec{L} =\vec {L}_{CoM} $. If we consider, instead, as non-inertial frame of reference a frame of reference integral with the CoM, but without rotation, is easy to demonstrate that $\vec{L}'$ is not zero and $ \vec{L} =\vec {L}_{CoM}+\vec {L}' $.
It is clear that something is wrong with my considerations. May you explain me where I am wrong?

Answer 1

In König's theorem, you aren't free to choose any non-inertial frame for $\vec{L}'$.

$\vec{L}'$ is defined in what's called the center of mass frame: its origin is the center of mass of the system (as you already stated), but its axes must be identical to those of an inertial frame. In other words, this frame is in a pure translation with respect to the inertial frame.

So you can't compute $\vec{L}'$ in a frame where all points of the system are at rest, unless of course the system isn't rotating at all in the inertial frame.

Edit: that last paragraph is true for a rigid body. For a generic system of moving points, $\vec{L}'$ could be zero without every point being at rest, but it doesn't change the fact that $\vec{L}'$ is defined in the center-of-mass frame defined above.