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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?

enter image description here

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

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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.

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  • $\begingroup$ Thanks for the response! I think the two L's are different in that second equation right? On the second line, right most side, how does ( Sum of mi * vi ) cross R become zero? $\endgroup$ – Sam Low May 26 at 7:01
  • $\begingroup$ (Sum of m v) is the motion of the center of mass. We’re assiming the simple case that it not moving. $\endgroup$ – Bob Jacobsen May 26 at 8:06
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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.

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  • $\begingroup$ Thanks for the response! When you say it depends, do you mean it depends in the vectorial sense, but regardless of the origin, the scalar magnitude would be conserved nonetheless? Thank you. $\endgroup$ – Sam Low May 26 at 5:08
  • $\begingroup$ No. The scalar magnitude can depend on the origin. For example, consider a mass moving in a straight line. It has nonzero angular momentum around a point off that line, but zero angular momentum around a point on that line. $\endgroup$ – G. Smith May 26 at 5:11
  • $\begingroup$ This is such a simple but effective counterproof to my statement above. Thank you G. Smith :) $\endgroup$ – Sam Low May 26 at 5:15
  • $\begingroup$ Glad to be of help! $\endgroup$ – G. Smith May 26 at 5:20

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