OBJECTIVES

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

1. Find the mass and the center of mass of a lamina (p. 929)
2. Find moments of inertia (p. 933)

NEED TO REVIEW?

The center of mass of a homogeneous lamina is discussed in Section 6.8, pp. 460-464.

RECALL

The mass density of a two-dimensional material is defined as the mass per unit area of the material. In SI units, mass density is measured in kilograms per meter squared; in U.S. customary units, it is measured in slugs per foot squared.

NOTE

Observe that if the mass density \(\rho \) is constant, then the formula for the mass of the lamina reduces to \(M=\iint\limits_{\kern-3ptR}\rho \,{\it dA}=\rho \iint\limits_{\kern-3ptR} \,{\it dA}=\rho A\) where \(A\) is the area of the region \(R.\)

NEED TO REVIEW?

Moments of mass were discussed in Section 6.8, p. 459.

NOTE

Sometimes \(M_{a}\) is referred to as the first moment of the particle about the axis \(a\).

THEOREM Mass of the Lamina

Suppose a lamina of variable mass density \(\rho =\rho (x,y)\) is represented by a closed, bounded region \(R\) of the \(xy\)-plane. If \(\rho \) is continuous on \(R\), then the mass of the lamina is \[\bbox[5px, border:1px solid black, #F9F7ED]{ M=\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA} } \]

The moment of mass about the the \(x\)-axis is \[\bbox[5px, border:1px solid black, #F9F7ED]{ M_{x}=\displaystyle\iint\limits_{\kern-3ptR}y\rho (x,y)\,{\it dA} } \]

The moment of mass about the \(y\)-axis is \[\bbox[5px, border:1px solid black, #F9F7ED]{ M_{y}=\displaystyle\iint\limits_{\kern-3ptR}x \rho (x,y)\,{\it dA} } \]

The center of mass of a lamina is the point \((\bar{x},\bar{ y})\) whose coordinates satisfy the equations \[\bbox[5px, border:1px solid black, #F9F7ED]{ \bar{x}=\dfrac{M_{y}}{M}=\dfrac{\displaystyle\iint\limits_{\kern-3ptR}x\rho (x,y)\,{\it dA}}{\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}}\qquad \qquad \bar{y}= \dfrac{M_{x}}{M}=\dfrac{\displaystyle\iint\limits_{\kern-3ptR}y\rho (x,y)\,{\it dA}}{ \displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}} } \]

Finding the Mass and the Center of Mass of a Lamina

Find the mass and center of mass of a lamina in the shape of an isosceles right triangle. The mass density function \(\rho =\rho (x,y) \) is directly proportional to the square of the distance from the vertex at the right angle.

Solution We position the lamina in the \(xy\)-plane so that the vertex at the right angle is at the origin and the two equal sides lie along the positive coordinate axes. See Figure 33. Suppose the equal sides of the triangle each measure \(a.\)

Figure 33 \(\rho(x,y)=k(x^2+y^2)\)

Since the distance of a point \((x,y) \) from the origin is \(\sqrt{ x^{2}+y^{2}},\) the mass density of the lamina is \[ \rho (x,y)=k(x^{2}+y^{2}) \]

where \(k\) is the constant of proportionality. Since \(\rho =\rho (x,y)\) is continuous on \(R,\) the mass \(M\) of this lamina is given by a double integral. \[ \begin{eqnarray*} M& =&\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}k(x^{2}+y^{2})\,d\!A=k\int_{0}^{a} \int_{0}^{a-x}(x^{2}+y^{2})\,{\it dy}\,{\it dx}\\[4pt] &=&k\int_{0}^{a}\left[ \left( x^{2}y+\dfrac{ y^{3}}{3}\right) \right] _{0}^{a-x}{\it dx} = k\int^a_0 \left[x^2a -x^3+\dfrac{(a-x)^3}{3}\right] {\it dx}\\[4pt] & =& k\int_{0}^{a}\dfrac{1}{3}(a^{3}-3a^{2}x+6ax^{2}-4x^{3})\,{\it dx} =\dfrac{k}{3} \left[a^3x-3a^2 \dfrac{x^2}{2} +2ax^3 -x^4\right]^a_0\\[4pt] &=&\dfrac{k}{3}\left( a^{4}-\dfrac{3a^{4}}{2}+2a^{4}-a^{4}\right) =\dfrac{ka^{4}}{6} \end{eqnarray*} \]

Due to the symmetry of both the region \(R\) and the mass density, the center of mass \((\bar{x},\bar{y})\) lies on the line \(y=x\). So, once we find \(\bar{ x}\), we also know \(\bar{y}\). The moment of mass with respect to the \(y\)-axis, \(M_{y}\), is \[ \begin{eqnarray*} M_{y}& =& \displaystyle\iint\limits_{\kern-3ptR}x\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}xk(x^{2}+y^{2})\,d\!A=k\int_{0}^{a} \int_{0}^{a-x}(x^{3}+xy^{2})\,{\it dy}\,{\it dx} \\[4pt] & =&\dfrac{k}{3}\int_{0}^{a}(a^{3}x-3a^{2}x^{2}+6ax^{3}-4x^{4})\,{\it dx}=\dfrac{ka^{5}}{15} \end{eqnarray*} \]

The center of mass \((\bar{x},\bar{y})\) is: \[ \bar{x}=\dfrac{M_{y}}{M}=\dfrac{\dfrac{ka^{5}}{15}}{\dfrac{ka^{4}}{6}}=\dfrac{2}{5}a=\overline{y} \]

The center of mass of the lamina is the point \(\left( \dfrac{2}{5}a,\dfrac{2}{5}a\right) \).

NOW WORK

Problem 11.

Finding the Mass and the Center of Mass of a Lamina

Find the mass \(M\) and the center of mass \((\bar{x},\bar{y})\) of a lamina in the shape of a region \(R\) in the \(xy\)-plane that lies outside the circle \( x^{2}+y^{2}=a^{2}\) and inside the circle \(x^{2}+y^{2}=2ax.\) The mass density \(\rho \) is inversely proportional to the distance from the origin.

Figure 34 \(a\leq r\leq 2a\cos\theta\), \(-\dfrac{\pi}{3}\leq \theta\leq \dfrac{\pi}{3}\)

Solution Figure 34 illustrates the region \(R\). Since \(R\) involves two circles, we use polar coordinates. Then the two circles are given by \( r=a \) and by \(r=2a\cos \theta \). The circles intersect when \(a=2a\cos \theta \) or \(\cos \theta =\dfrac{1}{2},\) and the points of intersection are \(\left( a,\dfrac{\pi }{3}\right) \) and \(\left( a,-\dfrac{\pi }{3}\right) .\) The region \(R\) is given by \(a\leq r\leq 2a\cos \theta ,\) \(-\dfrac{\pi }{3}\leq \theta \leq \dfrac{\pi }{3}.\)

Since the distance of a point \((x,y)\) from the origin is \(\sqrt{x^{2}+y^{2}}\) , the mass density of the lamina at any point \((x,y)\) in \(R\) is \[ \rho (x,y)=\dfrac{k}{\sqrt{x^{2}+y^{2}}}=\dfrac{k}{r} \qquad {\color{#0066A7}{\hbox{\(r^{2} =x^{2} + y^{2}\)}}} \]

where \(k\) is the constant of proportionality. Since \(R\) does not contain the origin, the mass density \(\rho =\rho (x,y)\) is continuous on \(R.\) The mass \(M\) of the lamina is \[ \begin{eqnarray*} M &=&\displaystyle\iint\limits_{\kern-3ptR}\rho (x,y)\,{\it dA}=k\displaystyle\iint\limits_{\kern-3ptR}\dfrac{rdr\,d\theta }{r} =k\int_{-\pi /3}^{\pi /3}\int_{a}^{2a\cos \theta }\,dr\,d\theta \\[4pt] &=&k \int_{-\pi /3}^{\pi /3}(2a\cos \theta -a)\,d\theta =k\big[2a\sin \theta -a\theta \big] _{-\pi /3}^{\pi /3}= k\!\left( 2a\sqrt{3}-a\dfrac{2\pi }{3}\right)\\[4pt] &=&\dfrac{2ka}{3}(3\sqrt{3}-\pi ) \end{eqnarray*} \]

Both the mass density function and the region \(R\) are symmetric about the \(x\)-axis. So, the center of mass lies on the \(x\)-axis; that is, \(\bar{y}=0\). To find \(\bar{x}\), we first find \(M_{y}\). \[ \begin{eqnarray*} M_{y}& =&\displaystyle\iint\limits_{\kern-3ptR}x\rho (x,y)\,{\it dA}=\displaystyle\iint\limits_{\kern-3ptR}( r\cos \theta ) \left( \dfrac{k}{r}\right) r\,dr\,d\theta =k\int_{-\pi /3}^{\pi /3}\int_{a}^{2a\cos \theta }r\cos \theta \,dr\,d\theta\\[4pt] &=&\dfrac{k}{2}\int_{-\pi /3}^{\pi /3}(4a^{2}\cos ^{2}\theta -a^{2})\cos \theta \,d\theta = a^{2}\dfrac{k}{2}\int_{-\pi /3}^{\pi /3}(3-4\sin ^{2}\theta )\cos \theta \,d\theta\\[4pt] &=&\dfrac{a^{2}k}{2}\left[ 3\sin \theta -\dfrac{4\sin ^{3}\theta }{3} \right] _{-\pi /3}^{\pi /3}=ka^{2}\sqrt{3} \end{eqnarray*} \]

Then \[ \bar{x}=\dfrac{M_{y}}{M}=\dfrac{ka^{2}\sqrt{3}}{\dfrac{2ka}{3}(3\sqrt{3}-\pi )} =\dfrac{3a\sqrt{3}}{2(3\sqrt{3}-\pi )}\approx 1.265a \]

The center of mass is approximately \((1.265a,0)\).

NOW WORK

Problem 17.

NOTE

The moment of inertia \(I_{a}\) is often referred to as the second moment of the particle about the axis \(a\). It is called the second moment because the distance is squared.

Finding Moments of Inertia

1. Find the polar moment of inertia \(I_{O}\) of a homogeneous lamina of mass density \(\rho \) in the shape of a region \(R\) in the \(xy\)-plane enclosed by the circle \(x^{2}+y^{2}=a^{2}\), \(a \gt 0\).
2. Use the polar moment to find the moments of inertia \(I_{x}\) and \(I_{y}\) of the lamina.

Solution (a) The region \(R\) is circular so we use polar coordinates \((r, \theta).\) Then \(0\leq r\leq a,\) \(0\leq \theta \leq 2\pi \), and \({\it dA}=r\,dr\, d\theta \). The polar moment of inertia is \[ \begin{eqnarray*} I_{O}&=& \displaystyle\iint\limits_{\kern-3ptR}(x^{2}+y^{2}) \rho (x,y) \,{\it dA} \underset{\underset{\color{#0066A7}{\hbox{\(x^{2}+y^{2}=r^{2}\)}}}{\color{#0066A7}{\uparrow }}}{=}\iint\limits_{\kern-3ptR}r^{2}\rho \,r\,{\it dr}\,d\theta \underset{\underset{\color{#0066A7}{\hbox{\(\rho\) is constant}}}{\color{#0066A7}{\uparrow }}}{=}\rho \int_{0}^{2\pi }\!\!\!\int_{0}^{a}r^{3}\,dr\,d\theta\\ &=&\rho \int_{0}^{2\pi }\dfrac{a^{4}}{4} \,d\theta =\dfrac{\pi a^{4}\rho }{2} \end{eqnarray*} \]

(b) By symmetry, the moment of inertia about the \(x\)-axis, \(I_{x}\), equals the moment of inertia about the \(y\)-axis, \(I_{y}\). Then \[ I_{x}=I_{y}=\dfrac{1}{2}I_{O}=\dfrac{1}{2}\!\left( \dfrac{\pi a^{4}\rho }{2} \right) =\dfrac{\pi a^{4}\rho }{4} \]

NOW WORK

Problem 25.