Q. Three identical spheres of mass $M$ each are placed at the corners of an equilateral triangle of side 2$m$. Taking one of the corner as the origin, the position vector of the centre of mass is
A) $\sqrt{3}$ ($\widehat{\text{i}}$-$\widehat{\text{j}}$)
B) $\widehat{\frac{\text{i}}{\sqrt{3}}}$ +$\widehat{\text{j}}$
C) $\frac{\widehat{\text{i}}\text{+}\widehat{\text{j}}}{3}$
D) $\widehat{\text{i}} +$ $\widehat{\frac{\text{j}}{\sqrt{3}}}$
Explanation
(b) The $x$ coordinate of centre of mass is
[image:7_image159.emf]
$\overline{x} = \frac{\sum_{}^{}{m_{i}x_{i}}}{\sum_{}^{}m_{i}}$
$= \frac{m \times 0 + m \times 1 + m \times 2}{m + m + m} = 1$
$\overline{y} = \frac{\sum_{}^{}{m_{i}y_{i}}}{\sum_{}^{}m_{i}}$
$= \frac{m \times 0 + m(2\sin{60{^\circ}) + m \times 0}}{m + m + m}$
$\overline{y} = \frac{\sqrt{3}m}{3m} = \frac{1}{\sqrt{3}}$
Position vector of centre of mass is $\left( \widehat{\mathbf{i}} + \frac{\widehat{\mathbf{j}}}{\sqrt{3}} \right)$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.