Three identical spheres, each of mass M , are placed at the corners of a right angle triangle with mutually perpendicular sides equal to 2   m (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of centre of mass.

Step-by-step solution

First, recall the formula of the center of mass of n particle system, r c m = m 1 r 1 + m 2 r 2 + m 3 r 3 + . . . + m n r n m 1 + m 2 + m 3 + . . + m n ⇒ x com = M × 0 + M × 2 + M × 0 3 M = 2 3 ⇒ y com  = M × 0 + M + 2 + M × 0 3 M = 2 3 So the position vector of center of mass is, 2 3 i ^ + 2 3 j ^ .