The total moment of mass about the centre of mass:

Let us consider a system of n particles of masses $m_1$, $m_2$, $\cdots$ $m_n$ having position vectors $\vec{r_1}$, $\vec{r_2}$, $\cdots$, $\vec{r_n}$ respectively with respect to the origin O as shown in the above Fig. 1.

Let P be the centre of mass of the system of n particle, with position vector $\vec{R}$, is given by

$\displaystyle{\vec{R}=\frac{1}{M}\sum_{i=1}^{n}m_i\vec{r_i}}\tag{1}$

Where $\displaystyle{M=\sum_{i=1}^{n}m_i}$ is the total mass of the above system.

Let us consider ith particle of mass $m_i$ having position vector $\vec{r_i}$ with respect to the origin $O$. Let $\vec{r_i}’$ be the position vector of the ith particle with respect to the centre of mass point $P$. So from Fig.1 we can write,

$\vec{R}+\vec{r_i}’=\vec{r_i}\tag{2}$

From equations (1) & (2) ,

$\displaystyle{\vec{R}=\frac{1}{M}\sum_{i=1}^{n}m_i(\vec{R}+\vec{r_i}’)}$

or, $M\vec{R}=\displaystyle{\sum_{i=1}^{n}m_i\vec{R}+\sum_{i=1}^{n}m_i\vec{r_i}’}$

or, $M\vec{R}=M\vec{R}+\displaystyle{\sum_{i=1}^{n}m_i\vec{r_i}’}$

or, $\displaystyle{\sum_{i=1}^{n}m_i\vec{r_i}’}=0\tag{3}$

So the sum of the products of mass and the position vectors of all the particles about the centre of mass is zero.