Question

Three particles, each of mass m, are placed at the corners of a right-angled triangle as shown in the figure. If OA = a and OB = b, the position vector of the centre of mass is

• A. $\frac{1}{3}(a\hat{i}+b\hat{j})$
• B. $\frac{1}{3}(a\hat{i}-b\hat{j})$
• C. $\frac{2}{3}(a\hat{i}+b\hat{j})$
• D. $\frac{2}{3}(a\hat{i}-b\hat{j})$

Answer 1

The correct option is A

$\frac{1}{3}(a\hat{i}+b\hat{j})$

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 $(x_3=0,y_3=b)$.
The (x,y) co-ordinates of the centre of mass are given by
$x_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{m\times 0+m\times a+m\times 0}{m+m+m}=\frac{a}{3}{y}_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{m\times 0+m\times 0+m\times b}{m+m+m}=\frac{b}{3}$
The position vector of the centre of mass is $x_{CM}\hat{i}+y_{CM}\hat{j}$ $=\frac{a}{3}\hat{i}+\frac{b}{3}\hat{j}=\frac{1}{3}(a\hat{i}+b\hat{j})$, which is choice (a).