Obtain an expression for the position vector of centre of mass of a system of $n$ particles in one dimension.
Obtain an expression for the position vector of centre of mass of a system of $n$ particles in one dimension.
According to as shown in figure. Let the distances of the two particles be $x_{1}$ and $x_{2}$ respectively from origin $\mathrm{O} .$ Let $m_{1}$ and $m_{2}$ be respectively the masses of the two particles on $\mathrm{X}$-axis.
The centre of mass of the system is that point $\mathrm{C}$ which is at a distance $\mathrm{X}$ from $\mathrm{O}$, then $\mathrm{X}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} \quad\left[\mathrm{X}\right.$ is denoted by $r_{\mathrm{cm}}$ also $]$
where $\mathrm{X}$ is the mass-weighted mean of $x_{1}$ and $x_{2}$. If the two particles have the same mass $m_{1}=m_{2}=m$, then
$\mathrm{X}=\frac{m x_{1}+m x_{2}}{m+m}=\frac{x_{1}+x_{2}}{2}$
Hence, for two particles of equal mass the centre of mass lies exactly midway between them. Similarly, $n$ particles of masses $m_{1}, m_{2}, m_{3}, \ldots, m_{n}$ their respectively, distances $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ from origin on $\mathrm{X}$-axis then centre of mass of system of particles is given by,
$\mathrm{X} =\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}+\ldots m_{n} x_{n}}{m_{1}+m_{2}+m_{3}+\ldots m_{n}}$
$=\frac{\Sigma m_{i} x_{i}}{\Sigma m_{i}}$
but $\Sigma m_{i}=m_{1}+m_{2}+m_{3}, \ldots, m_{n}$
$\Sigma=m=$ total mass of system
$\therefore \frac{\Sigma m_{i} x_{i}}{M}$ (where $i=1,2,3, \ldots, n$ )
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