Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass: Show $\text{L}=\text{L}'+\text{R}\times\text{MV}$ where $\text{L}'=\sum\text{r}'_\text{i}\times\text{p}'_\text{i}$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $\text{r}'_\text{i}=\text{r}_\text{i}-\text{R},$ rest of the notation is the standard notation used in the chapter. Note: $\text{L}'$ and $\text{MR}\times\text{V}$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
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Position vector of the $i^{th}$ particle with respect to origin = $r_i$ Position vector of the $i^{th}$ particle with respect to the centre of mass = $r'_i$_ Position vector of the centre of mass with respect to the origin = R It is given that, $\text{r}'_\text{i}=\text{r}_\text{i}-\text{R}$
$\text{r}_\text{i}=\text{r}'_\text{i}+\text{R}$ We have from part (a), $\text{P}_\text{i}=\text{}\text{p}'_\text{i}+\text{m}_\text{i}\text{V}$ Taking the cross product of this relation by $\text{r}_\text{i},$ we get $\sum\limits_\text{i}\text{r}_\text{i}\times\text{p}_\text{i}=\sum\limits_\text{i}\text{r}_\text{i}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}_\text{i}\times\text{m}_{\text{i}}\text{V}$
$\text{L}=\sum\limits_\text{i}(\text{r}'_\text{i}+\text{R})\times\text{p}'_1+\sum\limits_\text{i}(\text{r}'_\text{i}+\text{R})\times\text{m}_\text{i}\text{V}$
$=\sum\limits_\text{i}\text{r}'_\text{i}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{R}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}'_\text{i}\times\text{m}_\text{i}\text{V}+\sum\limits_\text{i}\text{R}\times\text{m}_\text{i}\text{V}$
$=\text{L}'+\sum\limits_\text{i}\text{R}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}'_\text{i}\times\text{m}_\text{i}\text{V}+\sum\limits_\text{i}\text{R}\times\text{m}_\text{i}\text{V}$ Where, $\text{R}\times\sum\limits_\text{i}\text{p}'_\text{i}=0$ and $\Big(\sum\limits_\text{i}\text{r}'_\text{i}\Big)\times\text{MV}=0$
$\sum\limits_\text{i}\text{m}_\text{i}=\text{M}$
$\therefore\ \text{L}=\text{L}'+\text{R}\times\text{MV}$
$\text{r}_\text{i}=\text{r}'_\text{i}+\text{R}$ We have from part (a), $\text{P}_\text{i}=\text{}\text{p}'_\text{i}+\text{m}_\text{i}\text{V}$ Taking the cross product of this relation by $\text{r}_\text{i},$ we get $\sum\limits_\text{i}\text{r}_\text{i}\times\text{p}_\text{i}=\sum\limits_\text{i}\text{r}_\text{i}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}_\text{i}\times\text{m}_{\text{i}}\text{V}$
$\text{L}=\sum\limits_\text{i}(\text{r}'_\text{i}+\text{R})\times\text{p}'_1+\sum\limits_\text{i}(\text{r}'_\text{i}+\text{R})\times\text{m}_\text{i}\text{V}$
$=\sum\limits_\text{i}\text{r}'_\text{i}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{R}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}'_\text{i}\times\text{m}_\text{i}\text{V}+\sum\limits_\text{i}\text{R}\times\text{m}_\text{i}\text{V}$
$=\text{L}'+\sum\limits_\text{i}\text{R}\times\text{p}'_\text{i}+\sum\limits_\text{i}\text{r}'_\text{i}\times\text{m}_\text{i}\text{V}+\sum\limits_\text{i}\text{R}\times\text{m}_\text{i}\text{V}$ Where, $\text{R}\times\sum\limits_\text{i}\text{p}'_\text{i}=0$ and $\Big(\sum\limits_\text{i}\text{r}'_\text{i}\Big)\times\text{MV}=0$
$\sum\limits_\text{i}\text{m}_\text{i}=\text{M}$
$\therefore\ \text{L}=\text{L}'+\text{R}\times\text{MV}$
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