# Parallel axis theorem and Koenig theorem for angular momentum

Are the parallel axis theorem and the Koenig theorem for angular momentum linked with each other in rigid body dynamics?

The parallel axis theorem states that $$I_{z}=I_{cm}+ma^2$$

Koenig theorem for angular momentum states that $$\vec{L}=\vec{L_{cm}}+\vec{L'}$$ Where $\vec{L'}$ is the angular momentum measured in cm frame.

They are different of course but in which way are they related in rigid body description?

Is there a general proof of the fact that these two are related?

Let the body rotate about the $z$-axis, then by the definition of angular momentum

$$\vec{L}=\vec{\omega} I_z.$$

where $\omega$ is the angular velocity about the $z$-axis. So we could take the parallel axis theorem and multiply it by $\omega$:

$$\vec{\omega}I_{z}=\vec{\omega}I_{cm}+\vec{\omega}ma^2$$

Now ponder the terms in it. If I understand the notation in the König theorem correctly, we have that $\vec{L}_{cm}$ is the angular momentum of the centre of mass about the rotation axis (i.e. as if the mass was concentrated at the COM). This is indeed the last term, so:

$$\vec{L}_{cm}=\vec{\omega}ma^2$$

The term $\vec{\omega}I_{cm}$ can then be defined as $\vec{L}'$, which gives the König relation, as the OP required. A further trivial step would be giving $\vec{L}'$ further physical interpretation (e.g. it is the angular momentum about the COM).

• The crucial point is: "rotate about the z-axis". The two are equivalent only for pure rotation. If there is circular translation (e.g. revolution of a planet) they are not. Commented Apr 19, 2016 at 14:11
• @L.Levrel Thank you so much for the really useful comment to this great answer, and also for your answer to my other post! The fact that $v_{cm}=\omega a$ (i.e. the motion is a pure rotation about the axis with respect to which we want to calculate the moment of inertia with parallel axis theorem, called $z$) is crucial in order say that $\vec{\omega}ma^2=\vec{L_{cm}}$. And a similar thing holds for the Koenig theorem about kinetic energy. So if the motion, described taking $z$ as axis of rotation, is also traslational in part, Koenig and parallel axis are not equivalent. Commented Apr 21, 2016 at 14:58
• I'd like to make an example if I can. Consider a disk rolling and slipping : the contact point of the disk with the floor $O$ is not istantaneously steady, if we call $z$ the axis through this point, the motion about $z$ is not a pure rotational motion and this implies that the parallel axis theorem cannot be used in that case (instead of Koenig theorems) in order to find the kinetic energy and the angular momentum taking as pivot $O$. While Koenig theorems still give the correct answers. Is this possibly correct? Commented Apr 21, 2016 at 15:00
• Yes, this is correct. Commented Apr 21, 2016 at 19:11