4.3: Center of Mass

If an inertial frame of reference \(\begin{equation}
K^{\prime}
\end{equation}\) is moving at constant velocity \(\begin{equation}
\vec{V}
\end{equation}\) relative to inertial frame K, the velocities of individual particles in the frames are related by \(\begin{equation}
\vec{v}_{i}=\vec{v}_{i}^{\prime}+\vec{V}
\end{equation}\), so the total momenta are related by

\begin{equation}
\vec{P}=\sum_{i} m_{i} \vec{v}_{i}=\sum_{i} m_{i} \vec{v}_{i}^{\prime}+\vec{V} \sum_{i} m_{i}=\vec{P}^{\prime}+M \vec{V}, \quad M=\sum_{i} m_{i}
\end{equation}

If we choose \(\begin{equation}
\vec{V}=\vec{P} / M, \text { then } \vec{P}^{\prime}=\sum_{i} m_{i} \vec{v}_{i}^{\prime}=0
\end{equation}\), the system is “at rest” in the frame \(\begin{equation}
K^{\prime}
\end{equation}\). Of course, the individual particles might be moving, what is at rest in \(\begin{equation}
\overline{K^{\prime}}
\end{equation}\) is the center of mass defined by

\(\begin{equation}
M \vec{R}_{\mathrm{cm}}=\sum_{i} m_{i} \vec{r}_{i}
\end{equation}\)

(Check this by differentiating both sides with respect to time.)

The energy of a mechanical system in its rest frame is often called its internal energy, we’ll denote it by \(\begin{equation}
E_{\text {int }}
\end{equation}\) (This includes kinetic and potential energies.) The total energy of a moving system is then

\begin{equation}
E=\frac{1}{2} M \vec{V}^{2}+E_{\text {int }}
\end{equation}

(Exercise: verify this.)