Derive an expression for the position vector of the center of mass of

1.	Derive an expression for the position vector of the center of mass of particle system.
Answer» Solution :(i) To find the centreof mass for acollection of npoint masses, say ` m_(1) , m_(2) , m_(3) …..m_(n)`we haveto firstchoosean 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 thesepoint massesin theX direction from the origin . The equation forthe X coordinate of the centreof mass is , ` x_(CM) = ( sum m_(i) x_(i))/( sum m_(i)) ` Where ` sum m_(i)`is the totalmass M of all the PARTICELS ` ( sum m_(i) = M)` (iii)Hence ` x_(CM)= ( sum m_(i)x_(i))/(M)` (vi)Similarly,we canalsofindy and z coordinatesof the centreof mass for thesedistributedpointmassesas indicatedin Figure. ` y_(CM) = ( sum m_(i) y_(i))/( M) ` ` ( z_(CM) = ( sum m_(i)z_(i))/( M) ` (v) Hence,the position of centreof massof thesepointmassesin a CARTESIAN cooridnate system is ` ( x_(CM) y_(CM) , z_(CM))` . In general , the positionof centreof masscan be written ina vectorformas , ` vecr_(CM) = ( Sum m_(i) vecr_(i))/ M` Where` vecr_(CM) = x_(CM)= hati + y_(CM) hatj + z_(CM) hatk ` is THEPOSITION vectorof the centre of mass and` vecr_(i)= x_(i) hati + y_(i) hatj + z_(i) hatk`is the position vectorof the distributedpointmass : where ` hati , hatj and hatk ` are the unitvectorsalongX, Y and Z -axesrespectively

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