question_answer

Three identical spheres of mass m each are placed at the corners of an equilateral triangle of side 1m. Taking one of the corners as the 2 origin, the position vector of centre of mass is

A)  \[\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}\]                  

B)  \[\frac{1}{2}\hat{i}+\frac{1}{2\sqrt{3}}\hat{j}\]

C)  \[\hat{i}+\hat{j}\]                          

D)  \[2\hat{i}-\hat{j}\]

Correct Answer: B

Solution :

                \[{{X}_{CM}}=\frac{\Sigma {{m}_{i}}{{x}_{i}}}{\Sigma {{m}_{i}}}=\frac{m\times 0+m\times 1+m+\frac{1}{2}}{m+m+m}\] \[=\frac{3m/2}{3m}=\frac{1}{2}\] \[{{Y}_{cm}}=\frac{\Sigma {{m}_{i}}{{y}_{i}}}{\Sigma {{m}_{i}}}=\frac{m\times 0+m\times 1\times \sin 60+m\times 0}{3m}\] \[=\frac{\sqrt{3}m/2}{3m}=\frac{1}{2\sqrt{3}}\] So, position vector of centre of mass is\[\left( \frac{1}{2}\hat{i}+\frac{1}{2\sqrt{3}}\hat{j} \right)\]