Obtain an expression for the position vector of centre of mass of a system n particles in two dimension.
Obtain an expression for the position vector of centre of mass of a system n particles in two dimension.
Suppose that there are three particles not lying in a straight line so they are taken in a plane.
Suppose the coordinates of three particles of masses $m_{1}, m_{2}$ and $m_{3}$ are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ in two dimension respectively.
If the centre of mass of this system is $(x, y)$ then $\mathrm{X}$-coordinate of the centre of mass, X-coordinate $=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$ and
$\text { Y-coordinate }=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}$
$\therefore$ Centre of mass,
$(\mathrm{X}, \mathrm{Y})=\left(\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}, \frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}\right)$
If all three particles having same mass, $m_{1}=m_{2}=m_{3}=m$ assume,
$\therefore x$-coordinate of centre of mass,
$\mathrm{X} =\frac{m x_{1}+m x_{2}+m x_{3}}{m+m+m}$
$=\frac{x_{1}+x_{2}+x_{3}}{3} \text { and }$
$\text { y-coordinate } \mathrm{Y} =\frac{m y_{1}+m y_{2}+m y_{3}}{m+m+m}$
$=\frac{y_{1}+y_{2}+y_{3}}{3}$
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