Question

Four particles, each of mass m, are placed at the corners of a square of side a as shown in the figure. The position vector of the centre of mass is

• A. $a(\hat{i}+\hat{j})$
• B. $\frac{a}{2}(\hat{i}+\hat{j})$
• C. $a(\hat{i}-\hat{j})$
• D. $\frac{a}{2}(\hat{i}-\hat{j})$

Answer 1

The correct option is B

$\frac{a}{2}(\hat{i}+\hat{j})$

The (x, y) co-ordinates of the masses at O, A, B and C respectively are $(x_1=0,y_1=0),(x_2=a,y_2=0),(x_3=a,y_3=a)$ and $(x_4=0,y_4=a)$.
The (x,y) co-ordinates of the centre of mass are given by
$x_{CM}=\frac{m\times 0+m\times a+m\times a+m\times 0}{m+m+m+m}=\frac{a}{2}$
$y_{CM}=\frac{m\times 0+m\times 0+m\times a+m\times a}{m+m+m+m}=\frac{a}{2}$
Hence, the position vector of the centre of mass is $\frac{a}{2}(\hat{i}+\hat{j})$, which is choice (b).