Q. 61 - Calculus

Question

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

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 $\left(\overline{x},\overline{y}\right)=\left(\frac{5}{6},\frac{5}{6}\right).$

Step 1. Given information.

Given lamina is a composition of rectangles and density is constant.

Step 2. center of mass of individual lumina.

Consider lumina ${\Omega }_1, {\Omega }_2,\& {\Omega }_3.$

The graph state that the center of mass of ${\Omega }_1$is $\left(\overline{x_1},\overline{y_1}\right)=\left(\frac{1}{2},\frac{3}{2}\right).$

Similarly center of mass of ${\Omega }_2$is $\left(\overline{x_2},\overline{y_2}\right)=\left(\frac{1}{2},\frac{1}{2}\right).$

center of mass of ${\Omega }_3$is $\left(\overline{x_3},\overline{y_3}\right)=\left(\frac{3}{2},\frac{1}{2}\right).$

Step 3. 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.  

$$\begin{gathered} \overline{x}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3} \\ \overline{x}=\frac{m\frac{1}{2}+m\frac{1}{2}+m\frac{3}{2}}{m+m+m} \\ \overline{x}=\frac{m\frac{5}{2}}{3m}=\frac{5}{6} \end{gathered}$$

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.  

$$\begin{gathered} \overline{y}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3} \\ \overline{y}=\frac{m\frac{3}{2}+m\frac{1}{2}+m\frac{1}{2}}{m+m+m} \\ \overline{y}=\frac{m\frac{5}{2}}{3m}=\frac{5}{6} \end{gathered}$$

So the center of mass of total lamina is at $\left(\overline{x},\overline{y}\right)=\left(\frac{5}{6},\frac{5}{6}\right).$