Q. 59 - Calculus

Question

Let $\Omega$be a lamina in the xy-plane. Suppose $\Omega$is composed of two non-overlapping lamin ${\Omega }_1$and ${\Omega }_2$, as follows:

Show that if the masses and centers of masses of ${\Omega }_1$and ${\Omega }_2$are $m_1$and  $m_2,$and $\left({\overline{x}}_1,{\overline{y}}_1\right)\text{ and }\left({\overline{x}}_2,{\overline{y}}_2\right)$respectively, then the center of mass of $\Omega$is $\left(\overline{x},\overline{y}\right),$where

$\overline{x}=\frac{m_1{\overline{x}}_1+m_2{\overline{x}}_2}{m_1+m_2}\text{ and }\overline{y}=\frac{m_1{\overline{y}}_1+m_2{\overline{y}}_2}{m_1+m_2}$

Step-by-Step Solution

Answer

center of mass $\overline{x} \& \overline{y}$is the ratio of the sum of linear moment of the mass about the y-axis and x-axis respectively of both regions to the sum of both masses.

So the center of mass of $\Omega$is $\left(\overline{x},\overline{y}\right)=\left(\frac{m_1\overline{x_1}+m_2\overline{x_2}}{m_1+m_2},\frac{m_1\overline{y_1}+m_2\overline{y_2}}{m_1+m_2}\right).$

Step 1. Given information.

The Center of mass of ${\Omega }_1$is $m_1$at $\left(x_1,y_1\right).$

The Center of mass of ${\Omega }_2$is $m_2$at $\left(x_2,y_2\right).$

 The Center of mass of $\Omega$is at $\left(\overline{x},\overline{y}\right).$

Step 2. moment of the mass.

The x-coordinate of the center of mass $\overline{x}$of ${\Omega }_1$is $\overline{x_1}=\frac{M_y}{m_1}.$

So linear moment of the mass about the y-axis in the region ${\Omega }_1$is $M_{y1}=m_1\overline{x_1}.$

The y-coordinate of the center of mass $\overline{y}$of ${\Omega }_1$is $\overline{y_1}=\frac{M_x}{m_1}.$

So linear moment of the mass about the x-axis in the region ${\Omega }_1$is $M_{x1}=m_1\overline{y_1}.$

The x-coordinate of the center of mass of ${\Omega }_2$is $\overline{x_2}=\frac{M_y}{m_2}.$

So linear moment of the mass about the y-axis in the region ${\Omega }_2$is $M_{y2}=m_2\overline{x_2}.$

The y-coordinate of the center of mass of ${\Omega }_2$is $\overline{y_2}=\frac{M_x}{m_2}.$

So linear moment of the mass about the x-axis in the region ${\Omega }_2$is $M_{x2}=m_2\overline{y_2}.$

Step 3. Center of mass.

center of mass $\overline{x}$is the ratio of the sum of linear moment of the mass about the y-axis of both regions to the sum of both masses.

$$\begin{gathered} \overline{x}=\frac{M_{y1}+M_{y2}}{m_1+m_2} \\ \overline{x}=\frac{m_1\overline{x_1}+m_2\overline{x_2}}{m_1+m_2} \end{gathered}$$

center of mass $\overline{y}$is the ratio of the sum of linear moment of the mass about the y-axis of both regions to the sum of both masses.

$\overline{y}=\frac{m_1\overline{y_1}+m_2\overline{y_2}}{m_1+m_2}$

So center of mass is $\left(\overline{x},\overline{y}\right)=\left(\frac{m_1\overline{x_1}+m_2\overline{x_2}}{m_1+m_2},\frac{m_1\overline{y_1}+m_2\overline{y_2}}{m_1+m_2}\right).$