Question

Three particles, each of mass $m$ are placed at the corners of a right angled triangle as shown in figure. If $OA=a$ and $OB=b,$ the position vector of the centre of mass is (Here $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ axes respectively).

• 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

Answer: (A)

The $(x,y)$ co-ordinates of the masses at $O,A$ and $B$
respectively are (refer to figure as given in the question)
$\left(x_1=0,y_1=0\right),\left(x_2=a,y_2=0\right)$ and $\left(x_3=0,y_3=b\right)$
The $(x,y)$ co-ordinates of the centre of mass are
$X_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}=\frac{m\times 0+m\times a+m\times 0}{m+m+m}=\frac{a}{3}$
$Y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}=\frac{m\times 0+m\times 0+m\times b}{m+m+m}=\frac{b}{3}$
The position vector of the centre of mass is
${\vec{R}}_{CM}=X_{CM}\hat{i}+Y_{CM}\hat{j}=\frac{a}{3}\hat{i}+\frac{b}{3}\hat{j}=\frac{1}{3}(a\hat{i}+b\hat{j})$