Q.67 - Calculus

Question

Step-by-Step Solution

Answer

The center of mass of the lamina is at $\left(\overline{x},\overline{y}\right)=\left(0,\frac{15a}{14}\right).$

Step 1. Given information.

The given lamina is the following.

The density of the given lamina is constant.

Step 2. x coordinate Center of mass of the left lamina

substituting $\rho (x,y)=ky$in the formula of the center of mass $\overline{x}$for left lamina.

$$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ {\overline{x}}_1=\frac{{\int }_{-a}^0{\int }_0^{x+2a}xkydydx}{{\int }_{-a}^0{\int }_0^{x+2a}kydydx} \\ {\overline{x}}_1=\frac{{\displaystyle k{\int }_{-a}^0{\left[\frac{{\displaystyle y^2}}{{\displaystyle 2}}\right]}_0^{x+2a}}}{{\displaystyle k{\int }_{-a}^0{\left[\frac{{\displaystyle y^2}}{{\displaystyle 2}}\right]}_{-a}^{x+2a}}}dx \\ {\overline{x}}_1=\frac{{\displaystyle {\int }_{-a}^0(x^3+4a{x}^2+4a^{2}x)dx}}{{\displaystyle {\int }_{-a}^0(x^2+4ax+4a^2)dx}} \\ {\overline{x}}_1=\frac{{\displaystyle {[\frac{{\displaystyle x^4}}{{\displaystyle 4}}+\frac{{\displaystyle 4a{x}^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^2{x}^2}}{{\displaystyle 2}}]}_{-a}^0}}{{\displaystyle {[\frac{{\displaystyle x^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a{x}^2}}{{\displaystyle 2}}+4a^{2}x]}_{-a}^0}} \\ {\overline{x}}_1=\frac{{\displaystyle -\left(\frac{{\displaystyle a^4}}{{\displaystyle 4}}-\frac{{\displaystyle 4a^4}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^4}}{{\displaystyle 2}}\right)}}{{\displaystyle -\left(-\frac{{\displaystyle a^3}}{{\displaystyle 3}}+\frac{{\displaystyle 4a^3}}{{\displaystyle 2}}-4a^3\right)}} \\ {\overline{x}}_1=\frac{{\displaystyle -11a}}{{\displaystyle 28}} \end{gathered}$$

Step 3. y coordinate the Center of mass of the left lamina

substituting $\rho (x,y)=ky$in the formula of the center of mass $\overline{y}$for left lamina.

$$\begin{gathered} \begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\end{array} \\ \overline{y_1}=\frac{{\int }_{-a}^0{\int }_0^{x+2a}y ky dydx}{{\int }_{-a}^0{\int }_0^{x+2a}ky dydx} \\ \overline{y_1}=\frac{k{\int }_{-a}^0{\left[\frac{y^3}{3}\right]}_0^{x+2a}dx}{k{\int }_{-a}^0{\left[\frac{y^2}{2}\right]}_0^{x+2a}dx} \\ \overline{y_1}=\frac{{\int }_{-a}^0\left[\frac{x^3+6a{x}^2+12a^{2}x+8a^3}{3}\right]dx}{{\int }_{-a}^0\left[\frac{x^2+4ax+4a^2}{2}\right]dx} \\ \overline{y_1}=\frac{{\left[\frac{\frac{x^4}{4}+2a{x}^3+6a^2{x}^2+8a^{3}x}{3}\right]}_{-a}^0}{{\left[\frac{\frac{x^3}{3}+2a{x}^2+4a^{2}x}{2}\right]}_{-a}^0} \\ \overline{y_1}=\frac{-\left(\frac{\frac{a^4}{4}-2a^4+6a^4-8a^4}{3}\right)}{-{\displaystyle \left(\frac{-\frac{a^3}{3}+2a^3-4a^3}{2}\right)}} \\ \overline{y_1}=\frac{15a}{14} \end{gathered}$$
So the center of mass of left lamina is at $\left(\overline{x_1},\overline{y_1}\right)=\left(\frac{-11a}{28},\frac{15a}{14}\right).$

Step 4. x coordinate Center of mass of the right lamina

substituting $\rho (x,y)=ky$in the formula of the center of mass $\overline{x}$for the right lamina.

$$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ {\overline{x}}_2=\frac{{\int }_0^a{\int }_0^{-x+2a}xkydydx}{{\int }_0^a{\int }_0^{-x+2a}kydydx} \\ {\overline{x}}_2=\frac{{\int }_0^a{\left[\frac{y^2}{2}\right]}_0^{-x+2a}x dx}{k{\int }_0^a{\left[\frac{y^2}{2}\right]}_0^{-x+2a}dx} \\ \overline{{\overline{x}}_2}=\frac{{\int }_0^a(x^3-4a{x}^2+4a^{2}x)dx}{{\int }_0^a(x^2-4ax+4a^2)dx} \\ {\overline{x}}_2=\frac{{\left[\frac{x^4}{4}-\frac{4a{x}^3}{3}+\frac{4a^2{x}^2}{2}\right]}_0^a}{{\left[\frac{x^3}{3}-\frac{4a{x}^2}{2}+4a^{2}x\right]}_0^a} \\ {\overline{x}}_2=\frac{\frac{a^4}{4}-\frac{4a^4}{3}+\frac{4a^4}{2}}{\frac{a^3}{3}-\frac{4a^3}{2}+4a^3} \\ {\overline{x}}_2=\frac{11a}{28} \end{gathered}$$

Step 5. y coordinate the Center of mass of the right lamina

substituting $\rho (x,y)=ky$in the formula of the center of mass $\overline{y}$for the right lamina.

$$\begin{gathered} \begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\end{array} \\ {\overline{y}}_2=\frac{{\int }_0^a{\int }_0^{-x+2a}{y}^{2}kdydx}{{\int }_0^a{\int }_0^{-x+2a}kydydx} \\ {\overline{y}}_2=\frac{k{\int }_0^a{\left[\frac{y^3}{3}\right]}_0^{-x+2a}dx}{k{\int }_0^a{\left[\frac{y^2}{2}\right]}_0^{-x+2a}dx} \\ {\overline{y}}_2=\frac{{\int }_{-a}^0\left[\frac{-x^3+6a{x}^2-12a^{2}x+8a^3}{3}\right]dx}{{\int }_{-a}^0\left[\frac{x^2-4ax+4a^2}{2}\right]dx} \\ {\overline{y}}_2=\frac{{\left[\frac{-\frac{x^4}{4}+2a{x}^3-6a^2{x}^2+8a^{3}x}{3}\right]}_0^a}{{\left[\frac{\frac{x^3}{3}-2a{x}^2+4a^{2}x}{2}\right]}_0^a} \\ {\overline{y}}_2=\frac{\left(\frac{-\frac{a^4}{4}+2a^4-6a^4+8a^4}{3}\right)}{\left(\frac{\frac{a^3}{3}-2a^3+4a^3}{2}\right)} \\ {\overline{y}}_2=\frac{15a}{14} \end{gathered}$$

So the center of mass of right lamina is at $\left(\overline{x_2},\overline{y_2}\right)=\left(\frac{11a}{28},\frac{15a}{14}\right).$

Step 6. Center of mass of composition of the lamina.

The Center of mass of composition of the left and right lamina is following. 

$\begin{array}{rl}\overline{x}& =\frac{m_1{\overline{x}}_1+m_2{\overline{x}}_2}{m_1+m_2}\\ \overline{x}& =\frac{m(-\frac{11}{28}a)+m\left(\frac{11}{28}a\right)}{m+m}\\ \overline{x}& =0\\ \overline{y}& =\frac{m_1{\overline{y}}_1+m_2{\overline{y}}_2}{m_1+m_2}\\ \overline{y}& =\frac{m\left(\frac{15}{14}a\right)+m\left(\frac{15}{14}a\right)}{m+m}\\ \overline{y}& =\frac{15}{14}a\end{array}$

So the center of mass of the lamina is at $\left(\overline{x},\overline{y}\right)=\left(0,\frac{15a}{14}\right).$