Q. 63 - Calculus

Question

In the following lamina, all angles are right angles and the density is constant:

Step-by-Step Solution

Answer

The center of mass of lumina is at  $(\overline{X},\overline{Y})=\left(\frac{b_1^2{h}_2+b_2^2{h}_1-b_1^2{h}_1}{2(b_1{h}_2+b_2{h}_1-b_1{h}_1)},\frac{h_2^2{b}_1+h_1^2{b}_2-h_1^2{b}_1}{2(b_1{h}_2+b_2{h}_1-b_1{h}_1)}\right)$

Step 1. Given information.

Given lamina is a composition of rectangles.

density is constant. 

Step 2. The formula for the center of mass

Density is constant so substitute $\rho (x,y)=k$in the formula of the x coordinate of the center of mass $\overline{x}.$

$\begin{array}{rl}\overline{x}& =\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\\ \overline{x}& =\frac{{\int }_{h_a}^{h_b}{\int }_{b_a}^{b_b}xkdxdy}{{\int }_{h_a}^{h_b}{\int }_{b_a}^{b_b}kdxdy}\\ \overline{x}& =\frac{k\frac{\left({b_b}^2-{b_a}^2\right)}{2}{\int }_{h_a}^{h_b}dy}{k\left(b_b-b_a\right){\int }_{h_a}^{h_b}dy}\\ \overline{x}& =\frac{k\frac{\left({b_b}^2-{b_a}^2\right)}{2}\left(h_b-h_a\right)}{k\left(b_b-b_a\right)\left(h_b-h_a\right)}\\ \overline{x}& =\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)}\end{array}$

Similarly, substitute $\rho (x,y)=k$formula of the y coordinate of the center of mass $\overline{y}$

$\begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\\ \overline{y}& =\frac{{\int }_{h_a}^{h_b}{\int }_{b_a}^{b_b}ykdxdy}{{\int }_{h_a}^{h_b}{\int }_{b_a}^{b_b}kdxdy}\\ \overline{y}& =\frac{k\frac{\left({h_b}^2-{h_a}^2\right)}{2}{\int }_{b_a}^{b_b}dx}{k\left(h_b-h_a\right){\int }_{b_a}^{b_b}dx}\\ \overline{y}& =\frac{k\frac{\left({h_b}^2-{h_a}^2\right)}{2}\left(b_b-b_a\right)}{k\left(h_b-h_a\right)\left(b_b-b_a\right)}\\ \overline{y}& =\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\end{array}$

So the center of mass of rectangular lumina of constant density is  $\left(\overline{x},\overline{y}\right)=\left(\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)},\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\right).$

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 $\left(\overline{x},\overline{y}\right)=\left(\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)},\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\right).$

The graph state that the center of mass of ${\Omega }_1$is at  $\left(\overline{x_1},{\overline{y}}_1\right)=\left(\frac{{b_1}^2\left(h_2-h_1\right)}{2b_1\left(h_2-h_1\right)},\frac{\left({h_2}^2-{h_1}^2\right)b_1}{\left(h_2-h_1\right)b_1}\right).$

Similarly center of mass of ${\Omega }_2$is at  $\left(\overline{x_2},{\overline{y}}_2\right)=\left(\frac{{b_1}^2{h}_1}{2b_1{h}_1},\frac{{h_1}^2{b}_1}{h_1{b}_1}\right).$

center of mass of ${\Omega }_3$is at  $\left(\overline{x_3},{\overline{y}}_3\right)=\left(\frac{\left({b_2}^2-{b_1}^2\right)h_1}{2\left(b_2-b_1\right)h_1},\frac{{h_1}^2\left(b_2-b_1\right)}{h_1\left(b_2-b_1\right)}\right).$

Step 4. Center of mass of composition of lumina.

The Center of mass is at the sum of all centers of mass.

$$\begin{gathered} \left(\overline{X},\overline{Y}\right)=\left(\overline{x_1},{\overline{y}}_1\right)+\left(\overline{x_2},{\overline{y}}_2\right)+\left(\overline{x_3},{\overline{y}}_3\right) \\ =\left(\frac{{b_1}^2\left(h_2-h_1\right)}{2b_1\left(h_2-h_1\right)},\frac{\left({h_2}^2-{h_1}^2\right)b_1}{\left(h_2-h_1\right)b_1}\right)+\left(\frac{{b_1}^2{h}_1}{2b_1{h}_1},\frac{{h_1}^2{b}_1}{h_1{b}_1}\right)+\left(\frac{\left({b_2}^2-{b_1}^2\right)h_1}{2\left(b_2-b_1\right)h_1},\frac{{h_1}^2\left(b_2-b_1\right)}{h_1\left(b_2-b_1\right)}\right) \\ =\left(\frac{{b_1}^2\left(h_2-h_1\right)+{b_1}^2{h}_1+\left({b_2}^2-{b_1}^2\right)h_1}{2b_1\left(h_2-h_1\right)+2b_1{h}_1+2\left(b_2-b_1\right)h_1},\frac{\left({h_2}^2-{h_1}^2\right)b_1+{h_1}^2{b}_1+{h_1}^2\left(b_2-b_1\right)}{\left(h_2-h_1\right)b_1+h_1{b}_1+h_1\left(b_2-b_1\right)}\right) \\ (\overline{X},\overline{Y})=\left(\frac{b_1^2{h}_2+b_2^2{h}_1-b_1^2{h}_1}{2(b_1{h}_2+b_2{h}_1-b_1{h}_1)},\frac{h_2^2{b}_1+h_1^2{b}_2-h_1^2{b}_1}{2(b_1{h}_2+b_2{h}_1-b_1{h}_1)}\right) \end{gathered}$$