Question

Find the position of center of mass of the uniform planner section shown in figure with respect to the origin (O)

Answer 1

Mass of each part =$m$

Now, $X_{cm}=\frac{m.\frac{a}{2}+m.\frac{3a}{2}+m.\frac{a}{2}}{m+m+m}=\frac{5a}{6}$

$Y_{cm}=\frac{m.\frac{a}{2}+m.\frac{a}{2}+m.\frac{3a}{2}}{m+m+m}=\frac{5a}{6}$

$\therefore$ Center of mass at $(\frac{5a}{6},\frac{5a}{6})$