Question

From a square sheet of uniform density, a portion is removed shown shaded in figure. Find the centre of mass of the remaining portion if the side of the square is a.

Answer 1

We filled the removed portion with $\lambda$ as well as $-\lambda$ density.
COM of $\Delta$ will be at centroid
$2x+x=\frac{a}{2}$
$3x=\frac{a}{2}$
$x=\frac{a}{6}$
$2x=\frac{a}{3}$
COM of $\Delta$ will be at distance $\frac{a}{3}$ from centre of square.
Mass of square = $\lambda a^2$
mass of $\Delta$= $-\lambda \frac{1}{2}a\times \frac{a}{2}=\frac{-a^2\lambda }{4}$
distt of COM
$x_{com}=\frac{M_{sq}{x}_{sq}+M_{\Delta }{x}_{\Delta }}{M_{sq}+M_{\Delta }}$
$=\frac{\lambda a^2\left(\frac{a}{2}\right)-\frac{\lambda a^2}{4}\left(\frac{a}{2}+\frac{a}{3}\right)}{\lambda a^2-\frac{a^2\lambda }{4}}$
$=\frac{\frac{a}{2}-\frac{1}{2}\left(\frac{5a}{6}\right)}{\frac{3}{4}}$
$=\frac{a-\frac{5a}{6}}{\left(\frac{3}{2}\right)}=\frac{a}{6\times \frac{3}{2}}=\frac{a}{9}$