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

Explanation

(i) To find the centre of mass for a collection of n point masses, say ` m_(1) , m_(2) , m_(3) …..m_(n)` we have to first choose an origin and an appropriate coordinate system as shown in Figure .
(ii) Let ` x_(1) , x_(2) , x_(3… x_(n)` be the X-coordinates of the positions of these point masses in the X direction from the origin . 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)` is the total mass M of all the particels ` ( sum m_(i) = M)`
(iii) Hence ` x_(CM) = ( sum m_(i)x_(i))/(M)`
(vi) Similarly, we can also find y and z coordinates of the centre of mass for these distributed point masses as indicated in Figure.
` y_(CM) = ( sum m_(i) y_(i))/( M) `
` ( z_(CM) = ( sum m_(i)z_(i))/( M) `
(v) Hence, the position of centre of mass of these point masses in a Cartesian cooridnate system is ` ( x_(CM) y_(CM) , z_(CM))` . In general , the position of centre of mass can be written in a vector form as ,
` vecr_(CM) = ( Sum m_(i) vecr_(i))/ M`
Where ` vecr_(CM) = x_(CM)= hati + y_(CM) hatj + z_(CM) hatk ` is the position vector of the centre of mass and ` vecr_(i) = x_(i) hati + y_(i) hatj + z_(i) hatk` is the position vector of the distributed point mass : where ` hati , hatj and hatk ` are the unit vectors along X, Y and Z -axes respectively