Question

A uniform rectangular thin sheet $\text{ABCD}$ of mass $M$ has length $a$ and breadth $b$, as shown in the figure. If the shaded portion $\text{HBGO}$ is cut-off, the coordinates of the centre of mass of the remaining portion will be:

• A. $(\frac{5a}{3},\frac{5b}{3})$
• B. $(\frac{3a}{4},\frac{3b}{4})$
• C. $(\frac{2a}{3},\frac{2b}{3})$
• D. $(\frac{5a}{12},\frac{5b}{12})$

Answer 1

The correct option is D $(\frac{5a}{12},\frac{5b}{12})$

With respect to point $O$, the CM of the cut-off potion will have coordinates, $(\frac{a}{4},\frac{b}{4})$

$\therefore$ With respect to point $D$, the CM of the cut-off potion will have coordinates, $[(\frac{a}{2}+\frac{a}{4}),(\frac{b}{2}+\frac{b}{4})]$

So, With respect to point $D$, the coordinates of CM of the cut-off portion are, $(\frac{3a}{4},\frac{3b}{4})$

Let the mass of the removed portion is $m$ and mass per unit area of the sheet is $\sigma$.

$\Rightarrow \sigma =\frac{M}{ab}=\frac{m}{\frac{a}{2}\times \frac{b}{2}}$

$\Rightarrow m=\frac{M}{4}$

$\Rightarrow x_{\text{CM}}=\frac{MX-mx}{M-m}$

$=\frac{M\times \frac{a}{2}-\frac{M}{4}\times \frac{3a}{4}}{M-\frac{M}{4}}=\frac{5a}{12}$

and $y_{\text{CM}}=\frac{MY-my}{M-m}$

$=\frac{M\times \frac{b}{2}-\frac{M}{4}\times \frac{3b}{4}}{M-\frac{M}{4}}=\frac{5b}{12}$

So, coordinates of CM of the remaining portion are,
$\frac{5a}{12},\frac{5b}{12}$

Hence, $\left(D\right)$ is the correct answer.