Q. 60 - Calculus

Question

Let $\Omega$be a lamina in the xy-plane. Suppose $\Omega$is composed of n non-overlapping laminæ ${\Omega }_1, {\Omega }_2,\dots , {\Omega }_n.$Show that if the masses of these laminæ are $m_1,m_2,\dots ,m_n$and the centers of masses are $\left({\overline{x}}_1,{\overline{y}}_1\right),\left({\overline{x}}_2,{\overline{y}}_2\right),\dots ,\left({\overline{x}}_n,{\overline{y}}_n\right),$then the center of mass of $\Omega$is $\left(\overline{x},\overline{y}\right),$where

$\overline{x}=\frac{{\sum }_{k=1}^n{m}_k{\overline{x}}_k}{{\sum }_{k=1}^n{m}_k}\text{ and }\overline{y}=\frac{{\sum }_{k=1}^n{m}_k{\overline{y}}_k}{{\sum }_{k=1}^n{m}_k}.$

Step-by-Step Solution

Answer

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

So the center of mass of $\Omega$is $\overline{x}=\frac{{\sum }_{k=1}^n{m}_k{\overline{x}}_k}{{\sum }_{k=1}^n{m}_k}\text{ and }\overline{y}=\frac{{\sum }_{k=1}^n{m}_k{\overline{y}}_k}{{\sum }_{k=1}^n{m}_k}.$

Step 1. Given information.

lumina $\Omega$composed of n non-overlapping laminæ ${\Omega }_1, {\Omega }_2,\dots , {\Omega }_n.$

masses of lumina ${\Omega }_1, {\Omega }_2,\dots , {\Omega }_n$are $m_1,m_2,\dots ,m_n$with a center of masses at $\left({\overline{x}}_1,{\overline{y}}_1\right),\left({\overline{x}}_2,{\overline{y}}_2\right),\dots ,\left({\overline{x}}_n,{\overline{y}}_n\right).$

center of mass of $\Omega$is $\overline{x}=\frac{{\sum }_{k=1}^n{m}_k{\overline{x}}_k}{{\sum }_{k=1}^n{m}_k}\text{ and }\overline{y}=\frac{{\sum }_{k=1}^n{m}_k{\overline{y}}_k}{{\sum }_{k=1}^n{m}_k}.$

Step 2. moment of the mass.

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

So linear moment of the mass about the y-axis of ${\Omega }_1$is $M_y=m_1\overline{x}$

similarly, a linear moment of the mass about the y-axis of ${\Omega }_2, {\Omega }_3,\dots , {\Omega }_n$are following.

$$\begin{gathered} M_y=m_2\overline{x} \\ {M}_y=m_3\overline{x} \\ ⋮ \\ {M}_y=m_n\overline{x} \end{gathered}$$

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

So linear moment of the mass about the x-axis of ${\Omega }_1$is $M_x=m_1\overline{y}$

similarly, a linear moment of the mass about the x-axis of ${\Omega }_2, {\Omega }_3,\dots , {\Omega }_n$are following.

$$\begin{gathered} M_x=m_2\overline{y} \\ {M}_x=m_3\overline{y} \\ ⋮ \\ {M}_x=m_n\overline{y} \end{gathered}$$

Step 3. Center of mass.

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. 

$$\begin{gathered} \overline{x}=\frac{m_1{\overline{x}}_1+m_2{\overline{x}}_2+\cdots +m_n{\overline{x}}_n}{m_1+m_2+\cdots +m_n} \\ \overline{x}=\frac{{\sum }_{k=1}^n{m}_k{\overline{x}}_k}{{\sum }_{k=1}^n{m}_k} \end{gathered}$$

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. 

$$\begin{gathered} \overline{y}=\frac{m_1{\overline{y}}_1+m_2{\overline{y}}_2+\cdots +m_n{\overline{y}}_n}{m_1+m_2+\cdots +m_n} \\ \overline{y}=\frac{{\sum }_{k=1}^n{m}_k{\overline{y}}_k}{{\sum }_{k=1}^n{m}_k} \end{gathered}$$

So the center of mass of $\Omega$is $\overline{x}=\frac{{\sum }_{k=1}^n{m}_k{\overline{x}}_k}{{\sum }_{k=1}^n{m}_k}\text{ and }\overline{y}=\frac{{\sum }_{k=1}^n{m}_k{\overline{y}}_k}{{\sum }_{k=1}^n{m}_k}.$