# PlanetPhysics/Center of Mass Examples

The center of mass of a system of equal particles is their average position; in other words, it is that point whose distance from any fixed plane is the average of the distances of all the particles of the system.

Let ${\displaystyle x_{1},x_{2},x_{3},...x_{n}}$ denote the distances of the particles of a system from the yz-plane; then, by the above definition, the distance of the center of mass from the same plane is

${\displaystyle x_{cm}={\frac {x_{1}+x_{2}+x_{3}+...+x_{n}}{n}}={\frac {1}{n}}\sum x}$

When the particles have different masses their distances must be weighted, that is, the distance of each particle must by multiplied by the masss of the particle before taking the average. In this case the distance of the center of mass from the yz-plane is defined by the following equation:

${\displaystyle (m_{1}+m_{2}+...+m_{n})x_{cm}=m_{1}x_{1}+m_{2}x_{2}+...+m_{n}x_{n}}$

or

${\displaystyle x_{cm}={\frac {\sum mx}{\sum m}}}$

${\displaystyle y_{cm}={\frac {\sum my}{\sum m}}}$

${\displaystyle z_{cm}={\frac {\sum mz}{\sum m}}}$

Evidently ${\displaystyle x_{cm},y_{cm},z_{cm}}$ are the coordinates of the center of mass.

### Illustrative Examples

{\mathbf 1.} Find the center of mass of two particles of masses ${\displaystyle m}$  and ${\displaystyle nm}$ , which are separated by a distance ${\displaystyle a}$ . Taking the origin of the axes at the particle which has the mass ${\displaystyle m}$ , figure 72, and taking as the z-axis the line which joins the two particles we get

${\displaystyle x_{cm}={\frac {0+nma}{m+nm}}={\frac {n}{n+1}}a}$  ${\displaystyle y_{cm}=z_{cm}=0}$

\begin{figure} \includegraphics[scale=.85]{Figure72.eps} \vspace{20 pt} \end{figure}

{\mathbf 2.} Find the center of mass of three particles of masses ${\displaystyle m,2m,3m}$ , which are at the vertices of an equilateral triangle of sides ${\displaystyle a}$ . Choosing the axes as shown if Fig. 73 we have

${\displaystyle x_{cm}={\frac {0+2ma+3ma\cos 60^{o}}{m+2m+3m}}={\frac {7}{12}}a}$  ${\displaystyle y_{cm}={\frac {0+0+3ma\sin 60^{o}}{6m}}={\frac {1}{4}}{\sqrt {3}}a}$  ${\displaystyle z_{cm}=0}$

\begin{figure} \includegraphics[scale=.85]{figure73.eps} \vspace{20 pt} \end{figure}

### Center of Mass of Continuous Bodies

When the particles form a continuous body we can replace the summation signs of equation (1) by integration signs and obtain the following expressions for the coordinates of the center of mass:

${\displaystyle x_{cm}={\frac {\int _{0}^{m}xdm}{\int _{0}^{m}dm}}}$

${\displaystyle y_{cm}={\frac {\int _{0}^{m}ydm}{\int _{0}^{m}dm}}}$

${\displaystyle z_{cm}={\frac {\int _{0}^{m}zdm}{\int _{0}^{m}dm}}}$

where ${\displaystyle m}$  is the mass of the body.

### Illustrative Examples

{\mathbf 1.} Find the center of mass of the parabolic lamina bounded by the curves ${\displaystyle y^{2}=2px}$  and ${\displaystyle x=a}$ , Fig. 74.

\begin{figure} \includegraphics[scale=.85]{figure74.eps} \vspace{20 pt} \end{figure}

Obviously the center of mass lies on the x-axis. Therefore we need to determine ${\displaystyle x_{cm}}$  only. Taking a strip of width ${\displaystyle dx}$  for the element of mass we have

${\displaystyle dm=\sigma 2ydx=2\sigma {\sqrt {2px}}dx}$

where ${\displaystyle \sigma }$  is the mass per unit area. Therefore substituting this expression of ${\displaystyle dm}$  in equation (2) nd changing the limits of integration we obtain

${\displaystyle x_{cm}={\frac {2\sigma \int _{0}^{a}x{\sqrt {2px}}dx}{2\sigma \int _{0}^{a}{\sqrt {2px}}dx}}}$  ${\displaystyle x_{cm}={\frac {\int _{0}^{a}x^{3/2}dx}{\int _{0}^{a}x^{1/2}dx}}}$  ${\displaystyle x_{cm}={\frac {3a}{5}}}$

{\mathbf 2.} Find the center of mass of the lamina bounded by the curves ${\displaystyle y^{2}=4ax}$  and ${\displaystyle y=bx}$ , Fig. 75. Let ${\displaystyle dx\,dy}$  be the area of the element of mass, then

${\displaystyle dm=\sigma dxdy}$

\begin{figure} \includegraphics[scale=.85]{figure75.eps} \vspace{20 pt} \end{figure}

Therefore substituting in equation (2) and introducing the proper limits of integration we obtain

${\displaystyle x_{cm}={\frac {\int _{0}^{\frac {4a}{b^{2}}}\int _{bx}^{2{\sqrt {ax}}}xdydx}{\int _{0}^{\frac {4a}{b^{2}}}\int _{bx}^{2{\sqrt {ax}}}dydx}}}$

${\displaystyle x_{cm}={\frac {\int _{0}^{\frac {4a}{b^{2}}}\left(2{\sqrt {ax}}-bx\right)xdx}{\int _{0}^{\frac {4a}{b^{2}}}\left(2{\sqrt {ax}}-bx\right)dx}}}$

${\displaystyle x_{cm}={\frac {8a}{5b^{2}}}}$

${\displaystyle y_{cm}={\frac {\int _{0}^{\frac {4a}{b^{2}}}\int _{bx}^{2{\sqrt {ax}}}ydydx}{\int _{0}^{\frac {4a}{b^{2}}}\int _{bx}^{2{\sqrt {ax}}}dydx}}}$

${\displaystyle y_{cm}={\frac {\int _{0}^{\frac {4a}{b^{2}}}\left(2ax-{\frac {b^{2}}{2}}x^{2}\right)dx}{\int _{0}^{\frac {4a}{b^{2}}}\left(2{\sqrt {ax}}-bx\right)dx}}}$

${\displaystyle y_{cm}={\frac {2a}{b}}}$

{\mathbf 3.} Find the center of mass of a semicircular lamina. Selecting the coordinates and the element of mass as shown in Fig. 76 we have

${\displaystyle dm=\sigma \cdot \rho d\theta \cdot d\rho }$  ${\displaystyle y_{cm}={\frac {\int _{0}^{\pi }\int _{0}^{a}y\cdot \sigma \rho d\rho d\theta }{\int _{0}^{\pi }\int _{0}^{a}\sigma \rho d\rho d\theta }}}$  ${\displaystyle y_{cm}={\frac {\int _{0}^{\pi }\int _{0}^{a}\rho ^{2}\sin \theta d\rho d\theta }{\int _{0}^{\pi }\int _{0}^{a}\rho d\rho d\theta }}}$  ${\displaystyle y_{cm}={\frac {4a}{3\pi }}}$  ${\displaystyle x_{cm}=0}$

\begin{figure} \includegraphics[scale=.85]{figure76.eps} \vspace{20 pt} \end{figure}