OBJECTIVES

When you finish this section, you should be able to:

1. Find the center of mass of a finite system of objects (p. 458)
2. Find the centroid of a homogeneous lamina (p. 460)
3. Find the volume of a solid of revolution using the Pappus Theorem (p. 464)

Find the center of mass of the system when a mass of $90\textrm{ kg}$ is placed at $6$ and a mass of $40\textrm{ kg}$ is placed at $2$.

Solution The system is shown in Figure 67, where the two masses are placed on a weightless seesaw. The center of mass $\bar{x}$ will be at some number where a fulcrum balances the two masses. Then for equilibrium, $$\begin{equation*} \bar{x}=\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}=\dfrac{40(2)+90(6)}{40+90}=\dfrac{620}{130}\approx 4.769 \end{equation*}$$

The center of mass is at $\bar{x}\approx 4.769$.

Figure 67 At the center of mass $\bar{x}$ the system is in equilibrium.

NOW WORK

Problem 7.

Center of Mass of a System of $n$ Objects on a Line

If $n$ objects with masses $m_{1}, m_{2}, \ldots , m_{n}$ are placed on a line at $x_{1}, x_{2}, \ldots , x_{n},$ respectively, then for equilibrium $$\begin{equation*} (m_{1}+m_{2}+\cdots +m_{n})\bar{x}=m_{1}x_{1}+m_{2}x_{2}+\cdots +m_{n}x_{n} \end{equation*}$$

and the center of mass is $\bar{x}$, where $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{m_{1}x_{1}+m_{2}x_{2}+\cdots +m_{n}x_{n}}{ m_{1}+m_{2}+\cdots +m_{n}}=\dfrac{\sum\limits_{i=1}^{n}\left( m_{i}x_{i}\right) }{\sum\limits_{i=1}^{n}m_{i}}=\dfrac{\sum\limits_{i=1}^{n} ( m_{i}x_{i}) }{M}} \end{equation*}$$

where $M=\sum\limits_{i=1}^{n}m_{i}$ is the mass of the system.

Center of Mass of a System of $n$ Objects in a Plane

If $n$ objects $m_{1}, m_{2}, \ldots , m_{n}$ are located at the points $(x_{1},y_{1})$, $(x_{2},y_{2}), \ldots , (x_{n},y_{n})$ in a plane, then the center of mass of the system is located at the point $(\bar{x},\bar{y})$, where $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{\sum\limits_{i=1}^{n}( m_{i}x_{i}) }{M} \qquad \bar{y}=\dfrac{ \sum\limits_{i=1}^{n}( m_{i} y_{i}) }{M} } \end{equation*}$$

and $M=\sum\limits_{i=1}^{n}m_{i}$ is the mass of the system.

Find the center of mass of the system of objects having masses $4$, $6$, and $9$ kg, located at the points $(-2,1), (3,-2)$, and $(4,3),$ respectively.

Solution See Figure 68. Where is a good estimate for the center of mass? Certainly, it will lie within the rectangle $-2\leq x\leq 4$; $-2\leq y\leq 3$.

To find the exact position of the center of mass, we first find the moment of the system about the $y$-axis, $M_{y},$ and the moment of the system about the $x$-axis, $M_{x}$. $$\begin{align*} M_{y}=\sum\limits_{i=1}^{3} m_{i}x_{i} =4(-2)+6(3)+9(4)=46\\ M_{x}=\sum\limits_{i=1}^{3} m_{i} y_{i} =4(1)+6(-2)+9(3)=19 \end{align*}$$

Since the mass $M$ of the system is $M=4+6+9=19,$ we have $$\begin{equation*} \bar{x}=\frac{M_{y}}{M}= \frac{46}{19},\qquad \bar{y}=\frac{M_{x}}{M}=\frac{19}{19}=1 \end{equation*}$$

The center of mass of these objects is at the point $\left( \dfrac{46}{19},1\right)$. Notice that the center of mass lies in the rectangle $-2\leq x\leq 4$; $-2\leq y\leq 3$.

NOW WORK

Problem 13.

The Centroid of a Homogeneous Lamina

Let $R$ be a lamina with a constant mass density $\rho .$ If $R$ is bounded by the graph of a function $f,$ the $x$-axis, and the lines $x=a$ and $x=b,$ where $f$ is continuous and nonnegative on the closed interval $[a,b]$, then the centroid $\left( \bar{x},\bar{y}\right)$ of $R$ is $$\begin{align*} \bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{1}{A}\displaystyle \int_{a}^{b} x f(x) ~{\it dx} \qquad \bar{y}=\dfrac{1}{2A}\displaystyle \int_{a}^{b}[f(x)] ^{2}\,{\it dx}}\tag{2} \end{align*}$$

where $$\begin{align*} \bbox[5px, border:1px solid black, #F9F7ED]{ A=\displaystyle \int_{a}^{b} f(x)\,{\it dx}} \end{align*}$$

is the area under the graph of $f$ from $a$ to $b.$

Find the centroid of the homogeneous lamina bounded by the graph of $f(x)=x^{2},$ the $x$-axis, and the line $x=1$.

Solution The lamina is shown in Figure 71.

Figure 71 $f(x)=x^{2}$, $0\leq x\leq 1$.

The area $A$ of the lamina is $$\begin{equation*} A=\int_{0}^{1}x\,^{2}{\it dx}=\left[ \dfrac{x^{3}}{3}\right] _{0}^{1}=\dfrac{1}{3} \end{equation*}$$

Using formulas (2), the centroid of the lamina is $$\begin{equation*} \bar{x} = \dfrac{1}{A}\int_{0}^{1}xf(x)~{\it dx}=\dfrac{1}{\dfrac{1}{3}} \displaystyle \int_{0}^{1}~x\cdot x^{2}\,{\it dx}=3\displaystyle \int_{0}^{1}x^{3}\,{\it dx}=3\left[ \dfrac{x^{4}}{4}\right] _{0}^{1}=\dfrac{3}{4} \\[4pt] \bar{y} = \dfrac{1}{2A}\displaystyle \int_{0}^{1}[f(x)] ^{2}\,{\it dx}=\dfrac{1}{\dfrac{2}{3}}\displaystyle \int_{0}^{1}( x^{2}) ^{2}\,{\it dx}=\dfrac{3}{2} \displaystyle \int_{0}^{1}x^{4}\,{\it dx}=\dfrac{3}{2}\left[ \dfrac{x^{5}}{5}\right] _{0}^{1}=\dfrac{3}{10} \end{equation*}$$

The centroid of the lamina is $\left( \dfrac{3}{4},\dfrac{3}{10}\right) .$

Find the centroid of one-quarter of a circular plate of radius $R$.

Solution We place the quarter-circle in the first quadrant, as shown in Figure 72(a). The equation of the quarter circle can be expressed as $f(x)= \sqrt{R^{2}-x^{2}}$, where $0\leq x\leq R$.

Figure 72

If you guessed that because the quarter of the circular plate is symmetric with respect to the line $y=x$, the centroid will lie on this line, you are correct. See Figure 72(b). So, $\bar{x}=\bar{y}$. The area $A$ of the quarter circular region is $A=\dfrac{\pi R^{2}}{4}$. $$\begin{equation*} \begin{array}{lll} \bar{x}=\dfrac{1}{A}\displaystyle \int_{a}^{b}\left[ xf(x)\right] {\it dx}=\dfrac{1}{A} \int_{0}^{R}x\sqrt{R^{2}-x^{2}}{\it dx}\\ \bar{y}=\dfrac{1}{2A}\displaystyle \int_{a}^{b}\left[ f(x)\right] ^{2}{\it dx}=\dfrac{1}{2A} \int_{0}^{R}\left( R^{2}-x^{2}\right) {\it dx} \end{array} \end{equation*}$$

Since $\bar{x}=\bar{y},$ and $\bar{y}$ is easier to find, we evaluate $\bar{y }.$ $$\begin{equation*} \bar{x}=\bar{y}=\dfrac{1}{2A}\int_{0}^{R}( R^{2}-x^{2})\, {\it dx}=\dfrac{ 1}{2\left( \dfrac{\pi R^{2}}{4}\right) }\left[ R^{2}x-\dfrac{x^{3}}{3}\right] _{0}^{R}=\dfrac{2}{\pi R^{2}}\left( \dfrac{2}{3}R^{3}\right) =\dfrac{4}{3\pi }R \end{equation*}$$

The centroid of the lamina, as shown in Figure 72(c), is $\left( \bar{x},\bar{y}\right) =\left( \dfrac{4% }{3\pi }R,\dfrac{4}{3\pi }R\right)$.

Find the centroid of the lamina bounded by the graph of $y=f(x)=x^{2}+1$, the $x$-axis, and the lines $x=-2$ and $x=2$.

Solution Figure 73(a) shows the graph of the lamina.

Figure 73

We notice two properties of $f$.

• The graph of $f$ is symmetric about the $y$-axis, so by the symmetry principle, $\bar{x}=0.$
• $f$ is an even function, so $\displaystyle \int_{-2}^{2}f(x) ~{\it dx}=2\displaystyle \int_{0}^{2}f(x) ~{\it dx}$.

The area $A$ of the region is $$\begin{equation*} A=2\displaystyle \int_{0}^{2}( x^{2}+1) ~{\it dx}=2\left[ \dfrac{x^{3}}{3} +x \right] _{0}^{2}=2\left( \dfrac{8}{3}+2\right) =\dfrac{28}{3}\ \end{equation*}$$ Using (2), we get $$\begin{eqnarray*} \bar{y} &=&\dfrac{1}{2A}\displaystyle \int_{a}^{b}[ f(x) ] ^{2}\,{\it dx}=\dfrac{1}{2A}\displaystyle \int_{-2}^{2}[f(x)] ^{2}\,{\it dx}=\dfrac{1}{2A} \cdot 2\displaystyle \int_{0}^{2}[f(x)] ^{2}\,{\it dx}\\[5pt] &=&\dfrac{3}{28}\displaystyle \int_{0}^{2}( x^{2}+1) ^{2}\,{\it dx}=\dfrac{3}{28}\displaystyle \int_{0}^{2}( x^{4}+2x^{2}+1) ~{\it dx} \\[5pt] &=&\dfrac{3}{28}\left[ \dfrac{x^{5}}{5}+\dfrac{2x^{3}}{3}+x\right] _{0}^{2}= \dfrac{3}{28}\left( \dfrac{32}{5}+\dfrac{16}{3}+2\right) =\dfrac{103}{70} \approx 1.471 \end{eqnarray*}$$

The centroid of the lamina, as shown in Figure 73(b), is $( \bar{x},\bar{y}) =\left( 0, \dfrac{103}{70}\right) .$

NOW WORK

Problem 19.

IN WORDS

The volume of a solid formed by revolving a plane region around an axis equals the product of area of the region and the distance its centroid travels around the axis.

THEOREM The Pappus Theorem for Volume

Let $R$ be a plane region of area $A$ and let $V$ be the volume of the solid of revolution obtained by revolving $R$ about a line that does not intersect $R$. Then the volume $V$ of the solid of revolution is $$\begin{align*} \bbox[5px, border:1px solid black, #F9F7ED]{ V=2\pi Ad } \end{align*}$$

where $d$ is the distance from the centroid of $R$ to the line.

Proof

We give a proof for the special case where the region $R$ is bounded by the graph of a function $f$ that is continuous and nonnegative on the interval $[a,b]$, the $x$-axis, and the lines $x=a$ and $x=b,$ and where $R$ is revolved about the $y$-axis, as shown in Figure 74. Using the shell method to find the volume of the solid of revolution and the centroid formula (2) for $\bar{x}$, we find $$\begin{eqnarray*} V&\underset{\underset{\color{#0066A7}{\hbox{Shell Method}}}{\color{#0066A7}{\uparrow }}} {=}&2\pi\int_{a}^{b}xf(x){\it dx}\underset{\underset{\color{#0066A7}{\hbox{(2)}}}{\color{#0066A7}{\uparrow }}} {=} 2\pi (A \bar{x}) =2\pi Ad\\[-9pt] \end{eqnarray*}$$

where $\bar{x}=d$ is the distance of the centroid $( \bar{x},\bar{y})$ of $R$ from the $y$-axis.

ORIGINS

The Greek mathematician Pappus of Alexandria (c. 300 AD) produced a mathematical collection containing a record of much of classical Greek mathematics. In it he shows a relationship between volume and centroids.

Use the Pappus Theorem to find the volume of the solid formed by revolving the region enclosed by the circle $(x-3)^{2}+y^{2}=1$ about the $y$-axis.

Solution By symmetry, the centroid of a circular region is the center of the circle. Here, the centroid is the point $(3,0)$. See Figure 75(a).

Figure 75

NOTE

A torus is a doughnut-shaped surface.

IN WORDS

A disk moving in a circle generates a solid torus. The volume of the torus equals the product of the circumference of the circle traveled by the centroid of the generating disk and the area of the disk.

NOW WORK

Problem 23.