Question

Three particles of masses $2\text{kg}$, $3\text{kg}$ and $5\text{kg}$ are placed at the three vertices $A,B,C$ of a right angled triangle respectively. Co-ordinates of the particles are $A(0,3),B(0,0)$ and $C(3,0)$. Find the position of centre of mass of the particles.

• A. $(1,1)$
• B. $(2,1)$
• C. $(1.5,0.6)$
• D. $(1.5,1.6)$

Answer 1

The correct option is C $(1.5,0.6)$


$m_1=2\text{kg}$ having co-ordinates $(x_1,y_1)=(0,3)$
$m_2=3\text{kg}$ having co-ordinates $(x_2,y_2)=(0,0)$
$m_3=5\text{kg}$ having co-oridnates $(x_3,y_3)=(3,0)$
Now, the co-oridnates of their $COM$ are
$x_{com}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{\left(2\right)\left(0\right)+\left(3\right)\left(0\right)+\left(5\right)\left(3\right)}{2+3+5}=\frac{15}{10}$
$x_{com}=1.5$

Similarly,
$y_{com}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_1}{m_1+m_2+m_3}$
$=\frac{\left(2\right)\left(3\right)+\left(3\right)\left(0\right)+\left(5\right)\left(0\right)}{2+3+5}=\frac{6}{10}$
$y_{com}=0.6$
Thus, position of $COM$ of the three particles is $(x_{com},y_{com})=(1.5,0.6)$