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 corners as the origin, the position vector of the centre of mass is
A) $\sqrt{3}(\widehat{\mathbf{i}} - \widehat{\mathbf{j}})$
B) $\frac{\widehat{\mathbf{i}}}{\sqrt{3}} + \widehat{\mathbf{j}}$
C) $\widehat{\mathbf{i}} + \widehat{\mathbf{j}}/3$
D) $\widehat{\mathbf{i}} + \widehat{\mathbf{j}}/\sqrt{3}$
Explanation
(c) $x_{CM} = \frac{\sum_{}^{}{m_{i}x_{i}}}{\sum_{}^{}m_{i}}$, Refer to figure
$= \frac{M \times 0 + M \times 1 + M \times 2}{M + M + M} = 1$
[image:7_image150.emf]
$y_{CM} = \frac{\sum_{}^{}{m_{i}y_{i}}}{\sum_{}^{}m_{i}}$
$= \frac{M \times 0 + M\left( 2\sin 60{^\circ} \right) + M \times 0}{M + M + M}$
$= \frac{\sqrt{3}M}{3M} = \frac{1}{\sqrt{3}}$
$\therefore$ Position vector of centre of mass is $\left( \widehat{\mathbf{i}} + \frac{1}{\sqrt{3}}\widehat{\mathbf{j}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.