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10.5: Center of Mass

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    Consider two point-like particles with masses \(m_{1}\) and \(m_{2}\). Choose a coordinate system with a choice of origin such that body 1 has position \(\overrightarrow{\mathbf{r}}_{1}\) and body 2 has position \(\overrightarrow{\mathbf{r}}_{2}\) (Figure 10.5).

    clipboard_e14b0effc3d2bc97e29c0aa6bd2fdf04e.png
    Figure 10.5 Center of mass coordinate system.

    The center of mass vector, \(\overrightarrow{\mathbf{R}}_{\mathrm{cm}}\), of the two-body system is defined as

    \[\overrightarrow{\mathbf{R}}_{\mathrm{cm}}=\frac{m_{1} \overrightarrow{\mathbf{r}}_{1}+m_{2} \overrightarrow{\mathbf{r}}_{2}}{m_{1}+m_{2}} \nonumber \]

    We shall now extend the concept of the center of mass to more general systems. Suppose we have a system of N particles labeled by the index \(i=1,2,3, \cdots, N\). Choose a coordinate system and denote the position of the \(i^{t h}\) particle as \(\overrightarrow{\mathbf{r}}_{i}\). The mass of the system is given by the sum

    \[m_{\mathrm{sys}}=\sum_{i=1}^{i=N} m_{i} \nonumber \]

    and the position of the center of mass of the system of particles is given by

    \[\overrightarrow{\mathbf{R}}_{\mathrm{cm}}=\frac{1}{m_{\mathrm{sys}}} \sum_{i=1}^{i=N} m_{i} \overrightarrow{\mathbf{r}}_{i} \nonumber \]

    (For a continuous rigid body, each point-like particle has mass dm and is located at the position \(\overrightarrow{\mathbf{r}}^{\prime}\) The center of mass is then defined as an integral over the body,

    \[\overrightarrow{\mathbf{R}}_{\mathrm{cm}}=\frac{\int_{\mathrm{bdy}} d m \overrightarrow{\mathbf{r}}^{\prime}}{\int_{\mathrm{body}} d m} \nonumber \]

    Example 10.2 Center of Mass of the Earth-Moon System

    The mean distance from the center of the earth to the center of the moon is \(r_{e m}=3.84 \times 10^{8} \mathrm{m}\). The mass of the earth is \(m_{e}=5.98 \times 10^{24} \mathrm{kg}\) and the mass of the moon is \(m_{m}=7.34 \times 10^{22} \mathrm{kg}\). The mean radius of the earth is \(r_{e}=6.37 \times 10^{6} \mathrm{m}\). The mean radius of the moon is \(r_{m}=1.74 \times 10^{6} \mathrm{m}\). Where is the location of the center of mass of the earthmoon system? Is it inside the earth’s radius or outside?

    Solution: The center of mass of the earth-moon system is defined to be

    \[\overrightarrow{\mathbf{R}}_{c m}=\frac{1}{m_{\mathrm{sys}}} \sum_{i=1}^{i=N} m_{i} \overrightarrow{\mathbf{r}}_{i}=\frac{1}{m_{e}+m_{m}}\left(m_{e} \overrightarrow{\mathbf{r}}_{e}+m_{m} \overrightarrow{\mathbf{r}}_{m}\right) \nonumber \]

    Choose an origin at the center of the earth and a unit vector \(\hat{\mathbf{i}}\) pointing towards the moon, then \(\overrightarrow{\mathbf{r}}_{e}=\overrightarrow{\mathbf{0}}\). The center of mass of the earth-moon system is then

    \[\overrightarrow{\mathbf{R}}_{c m}=\frac{1}{m_{e}+m_{m}}\left(m_{e} \overrightarrow{\mathbf{r}}_{e}+m_{m} \overrightarrow{\mathbf{r}}_{m}\right)=\frac{m_{m} \overrightarrow{\mathbf{r}}_{e m}}{m_{e}+m_{m}}=\frac{m_{m} r_{e m}}{m_{e}+m_{m}} \hat{\mathbf{i}} \nonumber \]

    \[\overrightarrow{\mathbf{R}}_{c m}=\frac{\left(7.34 \times 10^{22} \mathrm{kg}\right)\left(3.84 \times 10^{8} \mathrm{m}\right)}{\left(5.98 \times 10^{24} \mathrm{kg}+7.34 \times 10^{22} \mathrm{kg}\right)} \hat{\mathbf{i}}=4.66 \times 10^{6} \mathrm{m} \hat{\mathbf{i}} \nonumber \]

    The earth’s mean radius is \(r_{e}=6.37 \times 10^{6} \mathrm{m}\) so the center of mass of the earth-moon system lies within the earth.

    Example 10.3 Center of Mass of a Rod

    A thin rod has length L and mass M.

    clipboard_ef4ad610018fe118ecb13c9c7f0b0b5dd.png
    Figure 10.6 a) Uniform rod and b) non-uniform rod

    (a) Suppose the rod is uniform (Figure 10.6a). Find the position of the center of mass with respect to the left end of the rod. (b) Now suppose the rod is not uniform (Figure 10.6b) with a linear mass density that varies with the distance x from the left end according to

    \[\lambda(x)=\frac{\lambda_{0}}{L^{2}} x^{2} \nonumber \]

    where \(\lambda_{0}\) is a constant and has SI units \(\left[\mathrm{kg} \cdot \mathrm{m}^{-1}\right]\). Find \(\lambda_{0}\) and the position of the center of mass with respect to the left end of the rod.

    Solution: (a) Choose a coordinate system with the rod aligned along the x -axis and the origin located at the left end of the rod. The center of mass of the rod can be found using the definition given in Equation (10.5.4). In that expression dm is an infinitesimal mass element and \(\overrightarrow{\mathbf{r}}\) is the vector from the origin to the mass element dm (Figure 10.6c).

    clipboard_e62e2c6dcdbd48d7b623b15080b63f86b.png
    Figure 10.6c Infinitesimal mass element for rod

    Choose an infinitesimal mass element dm located a distance \(x^{\prime}\) from the origin. In this problem \(x^{\prime}\) will be the integration variable. Let the length of the mass element be \(d x^{\prime}\). Then

    \[d m=\lambda d x^{\prime} \nonumber \]

    The vector \(\overrightarrow{\mathbf{r}}=x^{\prime} \hat{\mathbf{i}}\). The center of mass is found by integration

    \[\overrightarrow{\mathbf{R}}_{c m}=\frac{1}{M} \int_{\text {body }} \overrightarrow{\mathbf{r}} d m=\frac{1}{L} \int_{x^{\prime}=0}^{x} x^{\prime} d x^{\prime} \hat{\mathbf{i}}=\left.\frac{1}{2 L} x^{\prime 2}\right|_{x^{\prime}=0} ^{x^{\prime}=L} \hat{\mathbf{i}}=\frac{1}{2 L}\left(L^{2}-0\right) \hat{\mathbf{i}}=\frac{L}{2} \hat{\mathbf{i}} \nonumber \]

    (b) For a non-uniform rod (Figure 10.6d),

    clipboard_e87192e5713915b6dfa71de557245086a.png
    Figure 10.6d Non-uniform rod

    the mass element is found using Equation (10.5.8)

    \[d m=\lambda\left(x^{\prime}\right) d x^{\prime}=\lambda=\frac{\lambda_{0}}{L^{2}} x^{\prime 2} d x^{\prime} \nonumber \]

    The vector \(\overrightarrow{\mathbf{r}}=x^{\prime} \hat{\mathbf{i}}\). The mass is found by integrating the mass element over the length of the rod

    \[M=\int_{\text {body }} d m=\int_{x^{\prime}=0}^{x=L} \lambda\left(x^{\prime}\right) d x^{\prime}=\frac{\lambda_{0}}{L^{2}} \int_{x^{\prime}=0}^{x=L} x^{\prime 2} d x^{\prime}=\left.\frac{\lambda_{0}}{3 L^{2}} x^{3}\right|_{x^{\prime}=0} ^{x^{\prime}=L}=\frac{\lambda_{0}}{3 L^{2}}\left(L^{3}-0\right)=\frac{\lambda_{0}}{3} L \nonumber \]

    Therefore

    \[\lambda_{0}=\frac{3 M}{L} \nonumber \]

    The center of mass is again found by integration

    \[\begin{array}{l}
    \overrightarrow{\mathbf{R}}_{c m}=\frac{1}{M} \int_{\mathrm{body}} \overrightarrow{\mathbf{r}} d m=\frac{3}{\lambda_{0} L} \int_{x^{\prime}=0}^{x} \lambda\left(x^{\prime}\right) x^{\prime} d x^{\prime} \hat{\mathbf{i}}=\frac{3}{L^{3}} \int_{x^{\prime}=0}^{x} x^{\prime 3} d x^{\prime} \mathbf{\hat { \mathbf { i } }} \\
    \overrightarrow{\mathbf{R}}_{c m}=\left.\frac{3}{4 L^{3}} x^{\prime 4}\right|_{x^{\prime}=0} \hat{\mathbf{i}}=\frac{3}{4 L^{3}}\left(L^{4}-0\right) \hat{\mathbf{i}}=\frac{3}{4} L \hat{\mathbf{i}}
    \end{array} \nonumber \]


    This page titled 10.5: Center of Mass is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.