Conservation of Angular Momentum w.r.t a Reference point

Question

When angular momentum is conserved, does it mean that it does not matter what the reference point is at? Say for example with this image below, the observer stands at two possible points P1 and P2. Since L = m(r x v), In the case where there is P1's observation point, there is a smaller r, but larger v magnitude, and in P2 observation point, it is a large r, but seemingly smaller v magnitude?

A generalisation of the question would then be: does the point of reference make a difference for the calculation of angular momentum?

Answer 1

If the center of mass isn’t moving, then the observation point doesn’t matter.

To prove this, first find the angular momentum as a sum over parts:

$ L = \Sigma m_i \vec{v_i} \times \vec{r_i}$

Now to find the angular momentum around some other point offset by $\vec{R}$:

$ L(\vec{R}) = \Sigma m_i \vec{v_i} \times (\vec{r_i}+\vec{R})$ $ = L + (\Sigma m_i \vec{v_i}) \times \vec{R}$

But the part in parentheses is just the c.m. motion so that term is zero, and the angular momentum doesn’t depend on $\vec{R}$.

If the c.m. is moving, you get the Parallel Axis theorem.

Answer 2

Angular momentum does depend on your choice of an origin. But angular momentum is conserved regardless of which origin you choose! So choose whichever origin makes the computation convenient.