Angular momentum is composed of intrinsic angular momentum and center of mass angular momentum

up:: Angular Momentum

Given an origin $O$, a system of particles has total angular momentum

$$\vec{L}=\sum _i{\vec{r}}_{i;CM}\times {\vec{p}}_{i;CM}+\overrightarrow{R_O}\times \vec{P}$$

where variables ${\vec{x}}_{i;CM}$ are centered on the center of mass and uppercase ones are referring directly to the CoM’s properties (with respect to the origin $O$). This assumes that all particles’ masses are constant!

Proof

Given an origin $O$, each particle has position¹

$${\vec{r}}_{i;O}=\vec{R}+{\vec{r}}_{i;CM}$$

with $i;O(CM)$ denoting the $i$-th particle’s position with respect to $O$ (CM); $\vec{R}$ is measured with respect to $O$.

The total angular momentum is the sum of all angular momenta

$$\vec{L}=\sum _i{\vec{r}}_i\times {\vec{p}}_i=\sum _i{m}_i{\vec{r}}_i\times {\dot{\vec{r}}}_i$$

Opening with above equations yields, and using $\vec{R}={\sum }_i{m}_i{\vec{r}}_i/M$,

$$\vec{L}=\vec{R}\times \vec{P}+\sum _i{m}_i{\vec{r}}_{i;CM}\times \dot{\vec{R}}+\vec{R}\times \sum _i{m}_i{\dot{\vec{r}}}_{i;CM}+\sum _i{m}_i{\vec{r}}_{i;CM}\times {\dot{\vec{r}}}_{i;CM}$$

The second term equals $0$, since

$$\sum _i{m}_i({\vec{r}}_i-\vec{R})=M\vec{R}-M\vec{R}=0$$

The third term equals $0$ provided that all masses are constant in time.

Corollaries

• A system of particles always has a Spin Angular Momentum which is independent of the Reference Frame (i.e. choice of origin). Thus, the “most natural” frame is the one centered on the Center of Mass, in which it is the only term of $\vec{L}$

Footnotes

1. ${\vec{r}}_{i;CM}$ points from the center of mass to particle $i$. ↩