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Q.62

Expert-verifiedFound in: Page 1040

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Use the results of Exercises 59 and 60 to find the centers of masses of the laminæ in Exercises 61–67.

Use the lamina from Exercise 61, but assume that the density is proportional to the distance from the x-axis.

The center of mass of lumina is at $\left(\overline{)x},\overline{)y}\right)=\left(\frac{5}{12},\frac{5}{9}\right).$

Given lamina is a composition of rectangles.

Density is proportional to the distance from the *x*-axis.

Density is proportional to the distance from the *x*-axis so substitute $\rho \left(x,y\right)=ky$in the formula of the *x* coordinate of the center of mass.

$\overline{x}=\frac{{\iint}_{\Omega}x\rho (x,y)dA}{{\iint}_{\Omega}\rho (x,y)dA}\phantom{\rule{0ex}{0ex}}\overline{x}=\frac{{\int}_{0}^{b}{\int}_{0}^{h}xkydydx}{{\int}_{0}^{b}{\int}_{0}^{h}kydydx}\phantom{\rule{0ex}{0ex}}\overline{x}=\frac{\frac{k{h}^{2}}{2}{\int}_{0}^{b}xdx}{\frac{k{h}^{2}}{2}{\int}_{0}^{b}dx}\phantom{\rule{0ex}{0ex}}\overline{x}=\frac{b}{2}$

Similarly, substitute $\rho (x,y)=ky$in the formula of the *y* coordinate of the center of mass.

$\begin{array}{rl}\overline{y}& =\frac{{\iint}_{\mathrm{\Omega}}y\rho (x,y)dA}{{\iint}_{\mathrm{\Omega}}\rho (x,y)dA}\\ \overline{y}& =\frac{{\int}_{0}^{b}{\int}_{0}^{h}k{y}^{2}dydx}{{\int}_{0}^{b}{\int}_{0}^{h}kydydx}\\ \overline{y}& =\frac{k\frac{{h}^{3}}{3}{\int}_{0}^{b}dx}{k\frac{{h}^{2}}{2}{\int}_{0}^{h}dx}\\ \overline{y}& =\frac{2h}{3}\end{array}$

So the center of mass of rectangular lamina whose Density is proportional to the distance from the *x*-axis is at $\left(\overline{)x},\overline{)y}\right)=\left(\frac{b}{2},\frac{2h}{3}\right).$

Consider lumina ${\Omega}_{1},{\Omega}_{2},\&{\Omega}_{3}.$

As the center of mass of each rectangle is at $\left(\overline{)x},\overline{)y}\right)=\left(\frac{b}{2},\frac{2h}{3}\right).$

The graph state that the center of mass of ${\Omega}_{1}$is role="math" localid="1650339550815" $\left(\overline{){x}_{1}},\overline{){y}_{1}}\right)=\left(\frac{1}{4},1\right).$

Similarly center of mass of ${\Omega}_{2}$is role="math" localid="1650339565700" $\left(\overline{){x}_{2}},\overline{){y}_{2}}\right)=\left(\frac{1}{4},\frac{1}{3}\right).$

center of mass of ${\Omega}_{3}$is role="math" localid="1650339579105" $\left(\overline{){x}_{3}},\overline{){y}_{3}}\right)=\left(\frac{3}{4},\frac{1}{3}\right).$

center of mass $\overline{)x}$is the ratio of the sum of all linear moments of the mass about the *y*-axis to the sum of all masses.

$\overline{)x}=\frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}+{m}_{3}{x}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\phantom{\rule{0ex}{0ex}}\overline{)x}=\frac{m\frac{1}{4}+m\frac{1}{4}+m\frac{3}{4}}{m+m+m}\phantom{\rule{0ex}{0ex}}\overline{)x}=\frac{m\frac{5}{4}}{3m}=\frac{5}{12}$

center of mass $\overline{)y}$is the ratio of the sum of all linear moments of the mass about the *x*-axis to the sum of all masses.

$\overline{)y}=\frac{{m}_{1}{y}_{1}+{m}_{2}{y}_{2}+{m}_{3}{y}_{3}}{{m}_{1}+{m}_{2}+{m}_{3}}\phantom{\rule{0ex}{0ex}}\overline{)y}=\frac{m1+m\frac{1}{3}+m\frac{1}{3}}{m+m+m}\phantom{\rule{0ex}{0ex}}\overline{)y}=\frac{m\frac{5}{3}}{3m}=\frac{5}{9}$

So the center of mass of total lamina is at role="math" localid="1650339903228" $\left(\overline{)x},\overline{)y}\right)=\left(\frac{5}{12},\frac{5}{9}\right).$

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