question_answer

Two balls each of mass m are placed on the vertices A and B of an equilateral triangle ABC of side\[1\text{ }m\]. A ball of mass \[2m\] 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)  \[\left( \frac{\sqrt{3}}{4}m,\frac{\sqrt{3}}{4}m \right)\]

Correct Answer: C

Solution :

The centre of mass is given by \[\bar{x}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[\bar{x}=\frac{m\times 0+m\times 1+2m\times \left( \frac{1}{2} \right)}{m+m+2m}\] \[\bar{x}=\frac{2\,m}{4\,m}=\frac{1}{2}m\] \[\bar{y}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[\bar{y}=\frac{m\times 0+m\times 0+2m\times \sqrt{3}/2}{m+m+2m}=\frac{\sqrt{3}}{4}m\] \[\therefore \] Centre of mass is \[\left( \frac{1}{2}m,\,\frac{\sqrt{3}}{4}m \right).\]