Question

Use the lamina from Exercise 64, but assume that the density is proportional to the distance from the x-axis.

Short answer

The Center of mass of the lamina is at$\left(\overline{X},\overline{Y}\right)=\left(0,\frac{58}{15}\right).$

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. x coordinate Center of mass of the horizontal lamina 

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

role="math" localid="1650347825085" $$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \begin{array}{rl}{\overline{x}}_1& =\frac{{\int }_{-3}^3{\int }_4^{6}xkydydx}{{\int }_{-3}^3{\int }_4^{6}kydydx}\\ {\overline{x}}_1& =\frac{k{\int }_{-3}^3{\left[\frac{y^2}{2}\right]}_4^{6}xdx}{k{\int }_{-3}^3{\left[\frac{y^2}{2}\right]}_{-3}^{6}dx}\end{array} \\ \begin{array}{rl}{\overline{x}}_1=& \frac{2k{\int }_{-3}^{3}xdx}{2k{\int }_{-3}^{3}dx}\\ {\overline{x}}_1=& \frac{2k\left({\left[\frac{x^2}{2}\right]}_{-3}^3\right)}{2k\left([x{]}_{-3}^3\right)}\\ {\overline{x}}_1& \begin{array}{rl}=0& \end{array}\end{array} \end{gathered}$$

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

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

role="math" localid="1650348806680" $$\begin{gathered} \begin{array}{rl}\overline{y}& =\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA}\end{array} \\ \overline{y_1}=\frac{{\int }_{-3}^3{\int }_4^{6}ykydydx}{k{\int }_{-3}^3{\int }_4^{6}kydydx} \\ \overline{y_1}=\frac{k{\int }_{-3}^3{\left[\frac{y^3}{3}\right]}_{-4}^6}{k{\int }_{-3}^3{\left[y^2\right]}_{-4}^6}dx \\ \overline{y_1}=\frac{76k{\int }_{-3}^{3}dx}{15k{\int }_{-3}^{3}dx} \\ \overline{y_1}=\frac{76k[x{]}_{-3}^3}{15k[x{]}_{-3}^3} \\ \overline{y_1}=\frac{76}{15} \end{gathered}$$

So the center of mass of horizontal lamina is$\left(\overline{x_1},\overline{y_1}\right)=\left(0,\frac{76}{15}\right).$

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

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

role="math" localid="1650348765861" $$\begin{gathered} \overline{x}=\frac{{\iint }_{\mathrm{\Omega }}x\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \overline{x_2}=\frac{{\int }_{-1}^1{\int }_0^{4}xkydydx}{k{\int }_{-1}^1{\int }_0^{4}kydydx} \\ \overline{x_2}=\frac{k{\int }_{-1}^1{\left[\frac{y^2}{2}\right]}_0^{4}xdx}{k{\int }_{-1}^1{\left[\frac{y^2}{2}\right]}_0^{4}dx} \\ \overline{x_2}=\frac{8k{\int }_{-1}^{1}xdx}{8k{\int }_{-1}^{1}dx} \\ \overline{x_2}=\frac{8k{\left[\frac{x^2}{2}\right]}_{-1}^1}{8k[x{]}_{-1}^1} \\ \overline{x_2}=0 \end{gathered}$$

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

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

$$\begin{gathered} \overline{y_2}=\frac{{\iint }_{\mathrm{\Omega }}y\rho (x,y)dA}{{\iint }_{\mathrm{\Omega }}\rho (x,y)dA} \\ \overline{y_2}=\frac{{\int }_{-1}^1{{\int }_0^4}_{\mathrm{\Omega }}ykydydx}{{\int }_{-1}^1{\int }_0^{4}kydydx} \\ \overline{y_2}=\frac{k{\int }_{-1}^1{\left[\frac{y^3}{3}\right]}_0^{4}dx}{k{\int }_{-1}^1{\left[\frac{y^2}{2}\right]}_0^4}dx \\ \overline{y_2}=\frac{{\displaystyle \frac{64}{3}}{\int }_{-1}^{1}dx}{8{\int }_{-1}^{1}dx} \\ \overline{y_2}=\frac{8[x{]}_{-1}^1}{3[x{]}_{-1}^1} \\ \overline{y_2}=\frac{8}{3} \end{gathered}$$

So the center of mass of vertical lamina is$\left(\overline{x_2},\overline{y_2}\right)=\left(0,\frac{8}{3}\right).$

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

Considering the mass of each lamina is m then the Center of mass of composition of the lamina is following.

$\begin{array}{rl}\overline{x}& =\frac{m_1{\overline{x}}_1+m_2{\overline{x}}_2}{m_1+m_2}\\ \overline{x}& =\frac{m\left(0\right)+m\left(0\right)}{m+m}\\ \overline{x}& =0\\ \overline{y}& =\frac{m_1{\overline{y}}_1+m_2{\overline{y}}_2}{m_1+m_2}\\ \overline{y}& =\frac{m\left(\frac{76}{15}\right)+m\left(\frac{8}{3}\right)}{m+m}\\ \overline{y}& =\frac{\left(\frac{116}{15}\right)}{2}\\ \overline{y}& =\frac{58}{15}\end{array}$

So the center of mass of the lamina is at$\left(\overline{X},\overline{Y}\right)=\left(0,\frac{58}{15}\right).$