Talk:PlanetPhysics/Center of Mass Examples

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The \htmladdnormallink{center of mass}{http://planetphysics.us/encyclopedia/CenterOfGravity.html} of a \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html} of equal \htmladdnormallink{particles}{http://planetphysics.us/encyclopedia/Particle.html} is their average \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html}; 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 $x_1, x_2, x_3,. . . x_n$ 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

$$ x_{cm} = \frac{ x_1 + x_2 + x_3 +. . . + x_n}{n} = \frac{1}{n} \sum x$$

When the particles have different \htmladdnormallink{masses}{http://planetphysics.us/encyclopedia/Mass.html} 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:

$$ (m_1 + m_2 + ... + m_n)x_{cm} = m_1 x_1 + m_2 x_2 + ... + m_n x_n $$

or

$$x_{cm} = \frac{ \sum mx}{\sum m}$$ \begin{equation} y_{cm} = \frac{ \sum my}{\sum m} \end{equation} $$z_{cm} = \frac{ \sum mz}{\sum m}$$

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

\subsection{Illustrative Examples}

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

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

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

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

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

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

\subsection{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:

$$x_{cm} = \frac{ \int_0^m x dm}{ \int_0^m dm}$$ \begin{equation} y_{cm} = \frac{ \int_0^m y dm}{ \int_0^m dm} \end{equation} $$z_{cm} = \frac{ \int_0^m z dm}{ \int_0^m dm}$$

where $m$ is the mass of the body.

\subsection{Illustrative Examples}

{\bf 1.} Find the center of mass of the parabolic lamina bounded by the curves $y^2 = 2px$ and $x = a$, 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 $x_{cm}$ only. Taking a strip of width $dx$ for the element of mass we have

$$ dm = \sigma 2 y dx = 2 \sigma \sqrt{2 px} dx $$

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

$$ x_{cm} = \frac{ 2 \sigma \int_0^a x\sqrt{2px} dx}{2 \sigma \int_0^a \sqrt{2px} dx} $$ $$ x_{cm} = \frac{ \int_0^a x^{3/2} dx}{\int_0^a x^{1/2} dx} $$ $$ x_{cm} = \frac{3a}{5} $$

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

$$ dm = \sigma dx dy $$

\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

$$x_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \int_{bx}^{2\sqrt{ax}} x dy dx}{\int_0^{\frac{4 a}{b^2}} \int_{bx}^{2\sqrt{ax}} dy dx} $$

$$x_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \left ( 2\sqrt{ax} - bx \right )x dx}{\int_0^{\frac{4 a}{b^2}} \left ( 2\sqrt{ax} - bx \right ) dx} $$

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

$$y_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \int_{bx}^{2\sqrt{ax}} y dy dx}{\int_0^{\frac{4 a}{b^2}} \int_{bx}^{2\sqrt{ax}} dy dx} $$

$$y_{cm} = \frac{\int_0^{\frac{4a}{b^2}} \left ( 2ax - \frac{b^2}{2}x^2 \right ) dx}{\int_0^{\frac{4 a}{b^2}} \left ( 2\sqrt{ax} - bx \right ) dx} $$

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

{\bf 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

$$ dm = \sigma \cdot \rho d\theta \cdot d\rho $$ $$ y_{cm} = \frac{\int_0^{\pi} \int_0^a y \cdot \sigma \rho d\rho d\theta}{\int_0^{\pi} \int_0^a \sigma \rho d\rho d\theta} $$ $$ y_{cm} = \frac{\int_0^{\pi} \int_0^a \rho^2 \sin \theta d\rho d\theta}{\int_0^{\pi} \int_0^a \rho d\rho d\theta} $$ $$ y_{cm} = \frac{4a}{3\pi} $$ $$ x_{cm} = 0 $$

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

\subsection{References}

This article is a derivative of the public \htmladdnormallink{domain}{http://planetphysics.us/encyclopedia/Bijective.html} \htmladdnormallink{work}{http://planetphysics.us/encyclopedia/Work.html}, "Analytical \htmladdnormallink{mechanics}{http://planetphysics.us/encyclopedia/Mechanics.html}" by Haroutune M. Dadourian, 1913. Made available by the \htmladdnormallink{internet archive}{http://www.archive.org/index.php}

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