# 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 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

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:

or

Evidently are the coordinates of the center of mass.

### Illustrative Examples

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

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

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

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

### Center of Mass of Continuous Bodies

editWhen 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:

where is the mass of the body.

### Illustrative Examples

edit{\mathbf 1.} Find the center of mass of the parabolic lamina bounded by the curves and , 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 only. Taking a strip of width for the element of mass we have

where is the mass per unit area. Therefore substituting this expression of in equation (2) nd changing the limits of integration we obtain

{\mathbf 2.} Find the center of mass of the lamina bounded by the curves and , Fig. 75. Let be the area of the element of mass, then

\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

{\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

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

### References

editThis article is a derivative of the public domain work, "Analytical mechanics" by Haroutune M. Dadourian, 1913. Made available by the internet archive