Question

Three particles each of mass m, are placed at the corners of an equilateral triangle of side a, as shown in the figure. The position vector of the centre of mass is

• A. $\frac{a}{2}(\hat{i}+\frac{\hat{j}}{\sqrt{3}})$
• B. $\frac{a}{2}(3\hat{i}+\hat{j})$
• C. $\frac{a}{2}(3\hat{i}+\sqrt{3}\hat{j})$
• D. $\frac{a}{2}(3\hat{i}+\frac{\hat{j}}{\sqrt{3}})$

Answer 1

The correct option is A

$\frac{a}{2}(\hat{i}+\frac{\hat{j}}{\sqrt{3}})$

The (x, y) co-ordinates of the masses at O, A and B respectively are $(x_1=0,y_1=0),(x_2=a,y_2=0)$ and $(x_3=\frac{a}{2},y_3=\frac{a\sqrt{3}}{2})$.
Therefore, the (x,y) co-ordinates of the centre of mass are
$x_{CM}=\frac{m\times 0+m\times a+m\times \frac{a}{2}}{m+m+m}=\frac{a}{2}{y}_{CM}=\frac{m\times 0+m\times 0+m\times \frac{a\sqrt{3}}{2}}{m+m+m}=\frac{a}{2\sqrt{3}}$
$\therefore$ Position vector of center of mass is $\frac{a}{2}(\hat{i}+\frac{\hat{j}}{\sqrt{3}})$.
Hence, the correct choice is (a).