2020 Phase-2 System of Particles and Rotational MotionCentre of mass of a system of particles Medium Question 26 of 100 in System of Particles and Rotational Motion

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.

A $\dfrac{4}{3}(\hat{i}+\hat{j})$
B $2(\hat{i}+\hat{j})$
C $(\hat{i}+\hat{j})$
D $\dfrac{2}{3}(\hat{i}+\hat{j})$

The answer is D.

Solution

With masses at the origin, at $(2,0)$ and at $(0,2)$ (all mass M): $x_{cm}=\dfrac{M(0)+M(2)+M(0)}{3M}=\dfrac{2}{3}$, similarly $y_{cm}=\dfrac{2}{3}$. So position vector of COM $=\dfrac{2}{3}(\hat i+\hat j)$.

Concept Centre of mass of a system of particles

System of Particles and Rotational Motion in past papers

100 questions from this chapter have appeared across 27 exam years.