Question

Three particles of masses 1 kg,2kgand 2kg are placed at the corners of an equilateral triangle. The co-ordinates of A,B,C are(0,0)(2,0)and(x,y) ,the co-ordinates of their center of mass are

Answer 1

Given:

Mass of three particles:

$m_{1} =1kg$

$m_{2}=2kg$

$m_{3}=2kg$

The coordinates of the equilateral triangle are:

$A=(0,0)$

$B=(2,0)$

$C=(x,y)$

To find:

The center of mass of the system.

Solution:

The formula for the center of mass is:

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

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

Substitute the values in the formula of the center of mass in x-direction:

$x=\frac{1(0)+2(2)+2(x) }{1+2+2 }$

$x=\frac{4+2x}{5}$

Substitute the values in the formula of the center of mass in y-direction:

$y=\frac{1(0)+2(0)+2(y)}{1+2+2}$

$y=\frac{2y}{5}$

Thus, the coordinate of the center of mass of the system is $(\frac{4+2x}{5},\frac{2y}{5} )$