Derive an expression for the position vector of the center of mass of particle system.

Explanation

(i) Consider the point masses `m_(1), m_(2), m_(3),….m_(n)` whose position coordintes from orgin O along X-direction are `x_(1), x_(2), x_(3),…x_(n)` as shown in figure.

(ii) The equation for the X coordinate of the centre of mass is,
`x_(CM) = (sum m_(i)x_(i))/(sum m_(i))`
where `sum m_(i) = M`, is the total mass of all the particles.
(iii) Hence
`x_(CM) = (sum m_(i)x_(i))/(M)`
(iv) Similarly, the Y and Z coordinates of centre of mass can be written as,
`y_(CM) = (sum m_(i)y_(i))/(M) , z_(CM) = (sum m_(i)z_(i))/(M)`
(v) The position of the centre of mass of these masses is `(X_(CM) , Y_(CM), Z_(CM))`. In general, the position of center of mass in vector form can be expressed as,
`vec(r)_(CM) = (sum m_(i) vec(r)_(i))/(M)`
where `vec(r)_(CM) = x_(CM) hat(i) + y_(CM) hat(j) + z_(CM) hat(k)` is the position vector of the center of mass and `vec(r_(1)) = x_(i) hat(i) + y_(i) hat(j) + z_(i) hat(k)` is the position vector of the distributed point mass.