Question

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

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

Answer 1

The correct option is B $(\frac{5a}{12}.\frac{5b}{12})$

The given rectangular thin sheet ABCD can be drawn as shown in the figure below,

Here,
Area of complete lamina , $A_1=ab$
Area of shaded part of lamina $=\frac{a}{2}\times \frac{b}{2}=\frac{ab}{4}$
$(x_1,y_1)$ = coordinates of centre of mass of complete lamina $=(\frac{a}{2},\frac{a}{2})$
$(x_2,y_2)=$ coordinates of centre of mass of shaded part of lamina $=(\frac{3a}{4},\frac{3a}{4})$
$\therefore$ Using formula for centre of mass, we have
$X_{CM}=\frac{A_1{x}_1-A_2{x}_2}{A_2-A_2}$
$=\frac{ab\left(\frac{a}{2}\right)-\frac{ab}{4}\left(\frac{3a}{4}\right)}{ab-\frac{ab}{4}}=\frac{\frac{8a^{2}b-3a^{2}b}{16}}{\frac{3ab}{4}}=\frac{5a}{12}$
Similalry, $Y_{CM}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}$
$=\frac{ab\left(\frac{b}{2}\right)-\frac{ab}{4}\left(\frac{3b}{4}\right)}{ab-\frac{ab}{4}}=\frac{5b}{12}$