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4.12: Center of Mass

  • Page ID
    29998
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    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=\dfrac{1}{2} M \vec{V}^{2}+E_{\text {int }}
    \end{equation}

    (Exercise: verify this.)


    This page titled 4.12: Center of Mass is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler.

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