Question

Three particles of masses 0.50 kg, 1.0 kg and 1.5 kg are placed at the three corners of a right-angles triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in figure. Locate the centre of mass of the system taking O as origin

• A. (1.3 cm, 1.5 cm)
• B. (2.6 cm, 3cm)
• C. (0.66 cm, 0.75 cm)
• D. (2 cm, 2cm)

Answer 1

The correct option is A

(1.3 cm, 1.5 cm)

Let us take the $4.0cm$ line as the x-axis and the $3.0cm$ line as the Y-axis. The coordinates of the three particles are as follows:

$\begin{array}{ccc}m& x& y\\ 0.50kg& 0& 0\\ 1.0kg& 4.0cm& 0\\ 1.5kg& 0& 3.0\end{array}$

The x - coordinate of the centre of mass is
$x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{\left(0.50kg\right).0+\left(1.0kg\right).\left(4.0cm\right)+\left(1.5kg\right).0}{0.50kg+1.0kg+1.5kg}$
$=\frac{4kg.cm}{3kg}$ = $1.3cm$
The y-coordinate of the centre of mass is
$y=\frac{m_1{y}_1+m_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{\left(0.50kg\right).0+\left(1.0kg\right).0+\left(1.5kg\right)\left(3.0cm\right)}{0.50kg+1.0kg+1.5kg}$
$=\frac{4.5kg.cm}{3kg}=1.5cm$
Thus, the centre of mass is $1.3cm$ right and $1.5cm$ above the $0.5kg$ particle.