Question

Consider a rectangular plate of dimensions $a\times b$. If this plate is considered to be made up of four rectangles of dimensions $\frac{a}{2}\times \frac{b}{2}$ and if we remove one out of four rectangles. Find the position of centre of mass of the remaining system-

• A. $(\frac{-a}{12},\frac{b}{12})$
• B. $(\frac{a}{6},\frac{-a}{6})$
• C. $(\frac{-a}{6},\frac{-b}{6})$
• D. $(\frac{-a}{12},\frac{-b}{12})$

Answer 1

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


Using the relation for COM,

$X_{COM}=\frac{A_1{x}_1+A_2{x}_2}{A_1+A_2}......\left(1\right)$

Here,

$A_1=ab;A_2=-\frac{ab}{4}$

Taking $\text{O}$ as the origin.

$x_1=0;x_2=\frac{a}{4}$

Where,

$A_1$ = Area of complete rectangle

$A_2$ = Area of the small rectangle (cavity)

$x_1=$ position of COM of the complete rectangle and

$x_2=$ position of COM of the small rectangle (cavity)

From equation $\left(1\right)$ we get,

$X_{COM}=\frac{(ab\times 0)-\frac{ab}{4}\times \left(\frac{a}{4}\right)}{ab-\frac{ab}{4}}$

$X_{COM}=-\frac{a^{2}b}{12ab}=-\frac{a}{12}$

Similarly, for $Y_{COM},$

$Y_{COM}=\frac{A_1{y}_1+A_2{y}_2}{A_1+A_2}......\left(2\right)$

$y_1=0;y_2=\frac{b}{4}$

From eq. $\left(2\right)$ we get,

$Y_{COM}=\frac{(ab\times 0)-\frac{ab}{4}\times \left(\frac{b}{4}\right)}{ab-\frac{ab}{4}}$

$Y_{COM}=-\frac{b^{2}a}{12ab}=-\frac{b}{12}$

Therefore, the coordinates of center of mass of the remaining system is $(\frac{-a}{12},\frac{-b}{12})$.

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