• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q.62

Expert-verified Found in: Page 1040 ### Calculus

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 $\left(\overline{)x},\overline{)y}\right)=\left(\frac{5}{12},\frac{5}{9}\right).$

See the step by step solution

## Step 1. Given information.

Given lamina is a composition of rectangles.

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

## Step 2. The formula for the center of mass

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.

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

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

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

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

## Step 3. center of mass of individual lumina.

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" $\left(\overline{){x}_{3}},\overline{){y}_{3}}\right)=\left(\frac{3}{4},\frac{1}{3}\right).$

## Step 4. Center of mass of composition of lumina.

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" $\left(\overline{)x},\overline{)y}\right)=\left(\frac{5}{12},\frac{5}{9}\right).$ ### Want to see more solutions like these? 