The particles of mass `m_(1),m_(2) and m_(3)` are placed on the vertices of an equilateral triangle of sides .a.. Find the centre of mass of this system with respect to the position of particle of mass `m_(1)`.

Explanation

If we place the particle of mass `m_(1)` at the origin (0, 0) and particle of mass `m_(2)` along the x-axis at distance of .a. from the origin at (a, 0) position, then the co-ordinate of particle of mass `m_(3)` are
`(acos60^(@),asin60^(@))=((a)/(2),sqrt(3)/(2)a)`
The position vector of the particle of mass `m_(1),m_(2)andm_(3)` are respectively
`vec(r_(1))=(0,0),vec(r_(2))=(a,0),vec(r_(3))=((a)/(2),(sqrt(3))/(2)a)`
The position vector of C.M. with respect to `m_(1)` particle is
`vec(r_(cm))=(m_(1)vec(r_(1))+m_(2)vec(r_(2))+m_(3)vec(r_(3)))/(m_(1)+m_(2)+m_(3))`
`=(m_(1)(0,0)+m_(2)(a,0)+m_(3)((a)/(2),(sqrt(3))/(2)a))/(m_(1)+m_(2)+m_(3))`
`=(((m_(2)a+(m_(3)a)/(2),(sqrt(3)m_(3)a)/(2))))/(m_(1)+m_(2)+m_(3))`
`vecr_(cm)=[((m_(2)+(m_(3))/(2))a)/(m_(1)+m_(2)+m_(3)),(sqrt(3)/(2)m_(3)a)/(m_(1)+m_(2)+m_(3))]`