Q.
Two balls each of mass $m$ are placed on the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side 1m. A ball of mass 2$m$ is placed at vertex $C$. The centre of mass of this system from vertex $A$ (located at origin) is
A) $\left( \frac{1}{2}m,\frac{1}{2}m \right)$
B) $\left( \frac{1}{2}m,\sqrt{3}\ m \right)$
C) $\left( \frac{1}{2}m,\frac{\sqrt{3}}{4}m \right)$
D) $\left( \frac{\sqrt{3}}{4}m,\frac{\sqrt{3}}{4}m \right)$
Explanation
(c) The centre of mass is given by
[image:7_image86.emf]
$\overline{x} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}}$
$\overline{x} = \frac{m \times 0 + m \times 1 + 2m \times \left( \frac{1}{2} \right)}{m + m + 2m}$
$\overline{x} = \frac{2m}{4m} = \frac{1}{2}m$
$\overline{y} = \frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3}}{m_{1} + m_{2} + m_{3}}$
$\overline{y} = \frac{m \times 0 + m \times 0 + 2m \times \frac{\sqrt{3}}{2}}{m + m + 2m}$
$= \frac{\sqrt{3}}{4}$m
$\therefore$ Centre of mass is $\left( \frac{1}{2}m,\frac{\sqrt{3}}{4}m \right)$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.