6.8 Center of Mass; Centroid; the Pappus Theorem
458
OBJECTIVES
When you finish this section, you should be able to:
- Find the center of mass of a finite system of objects (p. 458)
- Find the centroid of a homogeneous lamina (p. 460)
- Find the volume of a solid of revolution using the Pappus Theorem (p. 464)
Anytime you have been able to balance an object on your fingertip, you have located its center of mass. See Figure 64. The center of mass of an object or a system of objects is the point that acts as if all the mass is concentrated at that point.
1 Find the Center of Mass of a Finite System of Objects
We begin with a system of two objects with masses \(m_{1}\) and \(m_{2}\) that are placed on the ends of a nearly weightless rod of length \(d\) that is hung from a wire. When the wire is placed at the center of mass of the objects, the rod will be horizontal. Mathematically, the rod is balanced, or in equilibrium, when \begin{equation*} m_{1}d_{1}=m_{2}d_{2} \end{equation*}
where \(d_{1}\) and \(d_{2}\) are the distances of the objects from the vertical wire, as shown in Figure 65. The quantities \(m_{1}d_{1}\) and \(m_{2}d_{2},\) called moments, represent the tendency of the objects to rotate about the balancing point. When \(m_{1}d_{1}=m_{2}d_{2},\) the tendency of the objects to rotate is equal, so no rotation occurs and equilibrium is attained.
If the rod were placed on a positive \(x\)-axis, as in Figure 66, with mass \( m_{1}\) at \(x_{1},\) mass \(m_{2}\) at \(x_{2},\) and the center of mass at \( \bar{x},\) then the rod is balanced when \begin{eqnarray*} m_{1}( \bar{x}-x_{1}) &=&m_{2} ( x_{2}-\bar{x}) \\[4pt] m_{1}\bar{x}-m_{1} x_{1} &=&m_{2} x_{2}-m_{2} \bar{x} \\[4pt] ( m_{1}+m_{2}) \bar{x} &=&m_{1} x_{1}+m_{2} x_{2} \end{eqnarray*}
The center of mass \(\bar{x}\) of the two objects satisfies the equation \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{m_{1}\,x_{1}+m_{2}\,x_{2}}{m_{1}+m_{2}} } \end{equation*}
Finding the Center of Mass of a System of Objects on a Line
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 \(\bar{x}\approx 4.769\).
At the center of mass \(\bar{x}\) the system is in equilibrium.NOW WORK
Problem 7.
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The formula for the center of mass of two objects can be extended to any number \(n\) of objects.
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.
The numbers \(m_{1}x_{1}\), \(m_{2}x_{2},\ldots, m_{n}x_{n}\) are called the moments about the origin of the masses \(m_{1},\) \(m_{2}, \ldots , m_{n}\). So, the center of mass \(\bar{x}\) is found by adding the moments about the origin and dividing by the total mass \(M\) of all the objects.
Now suppose the objects are not in a line, but are scattered in a plane.
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.
The sum \(M_{y}=\sum\limits_{i=1}^{n}( m_{i} x_{i}) \) is called the moment of the system about the \(y\)-axis and the sum \(M_{x}=\sum\limits_{i=1}^{n}(m_{i} y_{i})\) is called the moment of the system about the \(x\)-axis. The center of mass formulas then can be written as \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{M_{y}}{M}\qquad \bar{y}=\dfrac{M_{x}}{M} }\tag{1} \end{equation*}
Physically, \(M_{y}\) measures the tendency of the system to rotate about the \(y\)-axis; \(M_{x}\) measures the tendency of the system to rotate about the \(x\)-axis.
Finding the Center of Mass of a System of Objects in a Plane
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 \(-2\leq x\leq 4\); \(-2\leq y\leq 3\).
460
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.
2 Find the Centroid of a Homogeneous Lamina
With the formulas (1), we can approximate the center of mass of a thin, flat sheet of material, called a lamina, and express its center of mass in terms of definite integrals. We will assume that matter is continuously distributed throughout the lamina; that is, the mass density function \(\rho \) is continuous on the domain of the lamina. In the special case where the mass density function \(\rho \) is a constant function, the lamina is called homogeneous.
The mass of a homogeneous lamina is \(\rho A,\) where \(A\) is the area of the lamina and \(\rho \) is its constant mass density. The center of mass of a homogeneous lamina is located at the centroid, the geometric center of the lamina.
Suppose a homogeneous lamina is determined by a region \(R\) bounded by the graph of a function \(f\), the \(x\)-axis, and the lines \(x=a\) and \(x=b\). Also suppose that \(f\) is continuous and nonnegative on the closed interval \([a,b]\). See Figure 69.
As before, we partition the interval \([a,b]\) into \(n\) subintervals: \begin{equation*} \lbrack a,x_{1}], [x_{1},x_{2}], \ldots , [x_{i-1},x_{i}], \ldots , [x_{n-1},b]\quad a=x_{0}\quad b=x_{n}\quad i=1,2,\ldots ,n \end{equation*} each of width \(\Delta x=\dfrac{b-a}{n}\), and select a number \(u_{i}\) in each subinterval. Here, we choose \(u_{i}\) to be the midpoint of the \(i\)th subinterval. That is, \begin{equation*} u_{i}=\dfrac{x_{i-1}+x_{i}}{2} \end{equation*}
The lamina is now partitioned into \(n\) nonoverlapping rectangular regions, \(R_{i}\), where \(i=1,2,\ldots ,n\), each of which is a homogeneous lamina. From the symmetry of a rectangle, the centroid of \(R_{i}\) is \(\left(u_{i},\dfrac{1}{2}f(u_{i})\right).\) See Figure 70.
The mass \(m_{i}\) of \(R_{i}\) is \begin{equation*} m_{i}=\rho A_{i}=\rho f(u_{i})\Delta x \end{equation*}
The moment of \(R_{i}\) about the \(y\)-axis, \(M_{y}(R_{i})\), is the product of the mass of \(R_{i}\) and the distance from \(\left(u_{i},\dfrac{1}{2} f(u_{i})\right)\) to the \(y\)-axis, which is \(u_{i}\). Then \begin{equation*} M_{y}(R_{i})=m_{i}u_{i}=[ \rho f(u_{i}) \Delta x] u_{i}=\rho u_{i} f(u_{i}) \Delta x \end{equation*}
Summing the \(n\) moments gives an approximation of the moment \(M_{y}\) of the lamina about the \(y\)-axis. That is, \begin{equation*} M_{y}\approx \rho \sum\limits_{i=1}^{n}u_{i }f(u_{i})\Delta x \end{equation*}
461
As the number \(n\) of subintervals increases, the sums \(\rho \sum\limits_{i=1}^{n}u_{i}f(u_{i})\Delta x\) become better approximations of \(M_{y}\). These sums are Riemann sums, and since \(f\) is continuous on \( [a,b] ,\) the limit of the sums is a definite integral. That is, \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ M_{y}=\rho \displaystyle \int_{a}^{b}x f(x) ~{\it dx} } \end{equation*}
Similarly, the moment of \(R_{i}\) about the \(x\)-axis, \(M_{x}(R_{i})\), is the product of its mass and the distance from the point \(\left( u_{i},\dfrac{1}{2 }f(u_{i})\right) \) to the \(x\)-axis, which is \(\dfrac{1}{2}f(u_{i})\). \begin{equation*} M_{x}(R_{i})=m_{i}\left[ \dfrac{1}{2}f( u_{i}) \right] =\rho f(u_{i})\Delta x\left[ \dfrac{1}{2}f(u_{i})\right] =\dfrac{1}{2}\rho [ f(u_{i})] ^{2} \Delta x \end{equation*}
Again, adding these moments gives an approximation of the moment \(M_{x}\) of the lamina about the \(x\)-axis, and as the number of subintervals increases, the sums \(\dfrac{1}{2}\rho \sum\limits_{i=1}^{n}[ f(u_{i})] ^{2} \Delta x\) become better approximations of \(M_{x}\). Since the sums are Riemann sums and \(f\) is continuous on \([a,b], \) we have \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ M_{x}=\dfrac{1}{2}\rho \displaystyle \int_{a}^{b}[ f(x) ] ^{2}~{\it dx} } \end{equation*}
For the region \(R\), the mass of a homogeneous lamina is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ M=\rho \displaystyle \int_{a}^{b}f(x) ~{\it dx} } \end{equation*}
Then the centroid \(( \bar{x},\bar{y}) \) is given by \begin{equation*} \bar{x}=\dfrac{M_{y}}{M}=\dfrac{\rho \displaystyle \int_{a}^{b} xf(x) ~{\it dx}}{ \rho \displaystyle \int_{a}^{b}f(x) ~{\it dx}}=\dfrac{\displaystyle \int_{a}^{b}xf(x) ~{\it dx}}{\displaystyle \int_{a}^{b}f(x) ~{\it dx}}=\dfrac{\displaystyle \int_{a}^{b}x\,f (x) ~{\it dx}}{A} \end{equation*}
and \[ \bar{y}=\dfrac{M_{x}}{M}=\dfrac{\dfrac{1}{2}\rho \displaystyle \int_{a}^{b}[ f(x) ] ^{2}~{\it dx}}{\rho \displaystyle \int_{a}^{b}f(x) ~{\it dx}}= \dfrac{\dfrac{1}{2}\displaystyle \int_{a}^{b}[ f(x) ] ^{2}~{\it dx}}{ \displaystyle \int_{a}^{b}f(x) ~{\it dx}}=\dfrac{\displaystyle \int_{a}^{b}[ f(x) ] ^{2}~{\it dx}}{2A} \]
where \(A=\int^b_a f(x)\,{\it dx}\) is the area under the graph of \(f\) from \(a\) to \(b\).
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 \(b.\)
462
Finding the Centroid of a Homogeneous Lamina
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.
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 \(\left( \dfrac{3}{4},\dfrac{3}{10}\right) .\)
Finding the Centroid of a Homogeneous Lamina
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 \(0\leq x\leq R\).
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*}
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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 \(\left( \bar{x},\bar{y}\right) =\left( \dfrac{4% }{3\pi }R,\dfrac{4}{3\pi }R\right) \).
Example 4 illustrates the symmetry principle: If a homogeneous lamina is symmetric about a line \(L\) or a point \(P\), then the centroid of the lamina lies on \(L\) or at the point \(P\).
Finding the Centroid of a Homogeneous Lamina
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.
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 \(( \bar{x},\bar{y}) =\left( 0, \dfrac{103}{70}\right) .\)
Notice that the centroid in Example 5 does not lie within the lamina.
NOW WORK
Problem 19.
464
In general, laminas are not homogeneous. The Challenge Problems at the end of the section investigate the center of mass of a lamina for which the density of the material varies with respect to \(x.\) The more general case, where the density of a lamina varies with both \(x\) and \(y,\) requires double integration and is treated in Chapter 14.
3 Find the Volume of a Solid of Revolution Using the Pappus Theorem
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.
Using the Pappus Theorem for Volume
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).
NOTE
A torus is a doughnut-shaped surface.
The distance \(d\) from the centroid to the axis of revolution, which is the \(y \)-axis, is \(d=3\). The area of the circle is \(A=\pi R^{2}=\pi \cdot 1^{2}=\pi\). It follows from the Pappus Theorem that the volume \(V\) of the solid of revolution [Figure 75(b)] is \[ V=2\pi \cdot \hbox{(Area of the circle)}\cdot d=2\pi \cdot \pi \cdot 3=6\pi ^{2}\approx 59.218\hbox{ cubic units.} \]
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.