Question

Three identical spheres of mass $M$ each are placed at the corners of an equilateral triangle of side $2m$. Taking one of the corner as the origin, the position vector of the centre of mass is:

• A. $\sqrt{3}(\hat{i}-\hat{j})$
• B. $\frac{\hat{i}}{\sqrt{3}}+\hat{j}$
• C. $\frac{\hat{i}+\hat{j}}{3}$
• D. $\hat{i}+\frac{\hat{j}}{\sqrt{3}}$

Answer 1

The correct option is D $\hat{i}+\frac{\hat{j}}{\sqrt{3}}$

Position vector of center of mass will be position vector of the $centroid$,

which is on the angle bisector of the angle between any two $adjacent$ sides and at a distance of $\frac{2}{3}\times (side\times \frac{\sqrt{3}}{2})$

from that vertex.

Taking the vertex at the origin and one side along the $x-axis$

we get the unit vector as $\hat{i}Cos\left({60}^0/2\right)+\hat{j}Sin{30}^0=\frac{1}{2}(\sqrt{3}\hat{i}+\hat{j})$ angle of an $equilateral$ triangle is ${60}^0$

multiplying it by the distance $\frac{2}{\sqrt{3}}$, we get the vector as $\hat{i}+\frac{1}{\sqrt{3}}\hat{j}$