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\begin{document}

\title{Rigid Grouping}

\maketitle

Suppose we have a system of particles with masses, positions, and
velocities $m_{i},\mathbf{x}_{i},\mathbf{v}_{i}$. This system has
center of mass $\mathbf{x}=\frac{\sum m_{i}\mathbf{x}_{i}}{\sum m_{i}}$,
and linear and angular momenta:

\begin{itemize}
\item Linear momentum: $\mathbf{p}=\sum m_{i}\mathbf{v}_{i}$
\item Angular momentum about $\mathbf{x}$: $\mathbf{L}=\sum\mathbf{r}_{i}\times m_{i}\mathbf{v}_{i}$
\end{itemize}
Let $\mathbf{r}_{i}=\mathbf{x}_{i}-\mathbf{x}$. Note that $\sum m_{i}\mathbf{r}_{i}=0$
(easy to show using definition of $\mathbf{x}$). If these particles
were moving as a rigid group with linear velocity $\mathbf{v}$ and
angular velocity $\omega$ then $\mathbf{v}_{i}=\mathbf{v}+\omega\times\mathbf{r}_{i}$
and we would have

\begin{itemize}
\item Linear momentum: $\mathbf{p}'=\sum m_{i}\left(\mathbf{v+\omega\times}\mathbf{r}_{i}\right)=\sum m_{i}\mathbf{v}+\underbrace{\omega\times\sum m_{i}\mathbf{r}_{i}}_{=0\, since\,\sum m_{i}\mathbf{r}_{i}=0}=\sum m_{i}\mathbf{v}$
\item Angular momentum: $\mathbf{L}'=\sum\mathbf{r}_{i}\times m_{i}\left(\mathbf{v}+\omega\times\mathbf{r}_{i}\right)=\underbrace{\sum m_{i}\mathbf{r}_{i}\times\mathbf{v}}_{=0\, since\,\sum m_{i}\mathbf{r}_{i}=0}+\sum m_{i}\mathbf{r}_{i}\times\left(\omega\times\mathbf{r}_{i}\right)$

\begin{itemize}
\item Note $\mathbf{r}_{i}\times\left(\omega\times\mathbf{r}_{i}\right)=\left(\mathbf{r}_{i}\cdot\mathbf{r}_{i}\right)\omega-\left(\mathbf{r}_{i}\cdot\omega\right)\mathbf{r}_{i}=\left(\mathbf{r}_{i}^{T}\mathbf{r}_{i}-\mathbf{r}_{i}\mathbf{r}_{i}^{T}\right)\omega$
\item Letting $I=\sum m_{i}\left(\mathbf{r}_{i}^{T}\mathbf{r}_{i}-\mathbf{r}_{i}\mathbf{r}_{i}^{T}\right)$,
we get $\mathbf{L}'=I\omega$
\end{itemize}
\end{itemize}
We want the rigid group to conserve momentum, hence we can solve for
$\mathbf{v}$ and $\omega$:

$\mathbf{p}=\mathbf{p}'\Rightarrow\sum m_{i}\mathbf{v}_{i}\mathbf{=\sum}m_{i}\mathbf{v}\Rightarrow\mathbf{v}=\frac{\sum m_{i}\mathbf{v}_{i}}{\sum m_{i}}$

$\mathbf{L=L}'\Rightarrow\sum\mathbf{r}_{i}\times m_{i}\mathbf{v}_{i}=I\omega\Rightarrow\omega=I^{-1}\left(\sum\mathbf{r}_{i}\times m_{i}\mathbf{v}_{i}\right)$
\end{document}
