Question

Three particles of masses 0.50 kg, 1.0 kg and 1.5 kg are placed at the three corners of right-angled 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.​

Answer 1

AnSwer :-

Let us take the 4.0 cm line as the X-axis and the 3.0 cm line as the y-axis .

The coordinates of the there particles are as follows

m$_1$= 0.50 kg, (x$_1$,y$_1$) = (0,0)

m$_2$= 1kg, (x$_2$,y$_2$) = (4,0)

m$_3$= 1.5 kg, (x$_3$,y$_3$) = (0,3)

The x-coordinate of the centre of mass is

$\sf x = \dfrac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}}$

$\sf= \dfrac{(0.50kg)(0) + (1.0kg)(4.0cm) + (1.5kg)(0)}{0.5kg + 1.0kg + 1.5kg}$

$\sf = \dfrac{4kg.cm}{kg} = 1.3cm$

The y-coordinate of the centre of mass is

$\sf y = \dfrac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3}}{m_{1} + m_{2} + m_{3}}$

$\sf = \dfrac{(0.50kg)(0) + (1.0kg)(0) + (1.5kg)(3.0cm)}{0.50kg + 1.0kg + 1.5kg}$

$\sf = \dfrac{4.5kg.cm}{3kg} = 1.5cm$

Thus, the centre of mass is 1.3cm right and 1.5 cm above the 0.5 kg particle.