Question

A certain system of particles possesses a total momentum $\overrightarrow{p}$ and an angular momentum $\overrightarrow{M}$ relative to a point $O$. Find its angular momentum ${\overrightarrow{M}}^{\prime }$ relative to a point $O^{\prime }$ whose position with respect to the point $O$ is determined by the radius vector $r_0$. Find out when the angular momentum of the system of particles does not depend on the choice of the point $O$.

Answer 1

Let position vectors of the particles of the system be $\overrightarrow{r_i}$ and ${\overrightarrow{r_i}}^{\prime }$ with respect to the points $O$ and $O^{\prime }$ respectively. Then we have,
$\overrightarrow{r_i}={\overrightarrow{r_i}}^{\prime }+\overrightarrow{r_0}$ (1)
where $\overrightarrow{r_0}$ is the radius vector of $O^{\prime }$ with respect to $O$.
Now, the angular momentum of the system relative to the point $O$ can be written as follows,
$\overrightarrow{M}=\sum (\overrightarrow{r_i}\times \overrightarrow{p_i})=\sum ({\overrightarrow{r_i}}^{\prime }\times \overrightarrow{p_i})+\sum (\overrightarrow{r_0}\times \overrightarrow{p_i})$ [using (1)]
or, $\overrightarrow{M}={\overrightarrow{M}}^{\prime }+(\overrightarrow{r_0}\times \overrightarrow{p})$, where, $\overrightarrow{p}=\sum \overrightarrow{p_i}$ (2)
From (2), if the total linear momentum of the system $\overrightarrow{p}=0$, then its angular momentum does not depend on the choice of the point $O$.
Note that in the C.M. frame, the system of particles, as a whole is at rest.