Question

From a uniform square plate, of side $a$, with its centre at origin, the shaded portion is cut out and removed as shown in the figure.The coordinates of center of mass of the remaining portion is:

• A.
  $-\frac{a}{2},-\frac{a}{2}$
• B.
  $-\frac{a}{3},-\frac{a}{3}$
• C. $-\frac{a}{4},-\frac{a}{4}$
• D. $-\frac{a}{12},-\frac{a}{12}$

Answer 1

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

$A_1\to$ Total area $=a^2$

$A_2\to$ Area of the removed portion $=\frac{a^2}{4}$

As the lamina is homogeneous with uniform thickness, the coordinates of $\text{COM}$ of remaining portion is given by,

$X_{COM}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}=\frac{a^2\left(0\right)-\frac{a^2}{4}\left(\frac{a}{4}\right)}{\frac{3a^2}{4}}$

$=-\frac{a^3}{16}\times \frac{4}{3a^2}=\frac{-a}{12}$

Similarly,

$Y_{COM}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}=\frac{0-\frac{a^2}{4}\left(\frac{a}{4}\right)}{\frac{3a^2}{4}}=\frac{-a}{12}$

Hence, the co-ordinates of $\text{COM}$ of the remaining portion are:

$(-\frac{a}{12},-\frac{a}{12})$

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