Chapter 8.1, Problem 41E

To determine

Tofind the centre of mass.

Answer to Problem 41E

The centre of the mass is found to be $\left(\frac{4}{3},0\right)$ .

Explanation of Solution

Given information:

Suppose we have a finite collection of masses in the co-ordinate plane, the mass $m_k$ located at the point $\left(x_k,y_k\right)$ as shown in the figure.

  

The region bounded by the lines $y=x,y=-x,x=2$ with constant density $\delta$ .

The domain is found to be an isosceles triangle. The area is given by

  $A=\frac{1}{2}\cdot 4\cdot 2=4$

Let $X$ and $Y$ be the co-ordinates of the centre. Then by definition,

  $X=\frac{1}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}xdx}{\displaystyle \underset{-x}{\overset{x}{\int }}dy}=\frac{2}{A}\cdot {\displaystyle \underset{0}{\overset{2}{\int }}{x}^{2}dx}\frac{2}{A}\cdot \frac{8}{3}=\frac{4}{3}$

Which can be also inferred from the center of the triangle at the intersection of the median lines.

  $Y=0$ since they are symmetry.

In other words, the domain of integration over $dy$ is symmetric and the integrand is even.

Therefore, the centre of the mass is found to be $\left(\frac{4}{3},0\right)$ .