question_answer

Four particles of masses m, 2m, 3m and 4m are arranged at the corners of a parallelogram with each side equal to a and one of the angle between two adjacent sides as\[60{}^\circ \]. The parallelogram lies in the \[x-y\] plane with mass m at the origin and 4m on the \[x-\]axis. The centre of mass of the arrangement will be located at

A)  \[\left( \frac{\sqrt{3}}{2}a,0.95a \right)\]

B)  \[\left( 0.95a,\frac{\sqrt{3}}{4}a \right)\]

C)  \[\left( \frac{3a}{4},\frac{a}{2} \right)\]

D)  \[\left( \frac{a}{2},\frac{3a}{4} \right)\]

Correct Answer: B

Solution :

\[\overline{x}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}+{{m}_{4}}{{x}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\] \[=\frac{0+\left( 2m\times \frac{a}{2} \right)+\left( 3m\times \frac{3a}{2} \right)+(4m\times a)}{m+2m+3m+4m}\] \[=\frac{ma+4.5ma+4ma}{10m}=\frac{9.5ma}{10m}=0.95a\] \[\overline{y}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}+{{m}_{4}}{{y}_{4}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+{{m}_{4}}}\] \[=\frac{(m\times 0)+(2m+a\sqrt{3}/2)+(3m\times a\sqrt{3}/2)+(4m\times 0)}{m+2m+3m+4m}\] \[=\frac{\sqrt{3}am+\sqrt{3}\times 1.5ma}{10m}=\frac{2.5\sqrt{3}am}{10m}=\frac{\sqrt{3}a}{4}\] \[\therefore \]Centre of mass is at\[\left( 0.95a,\frac{\sqrt{3}a}{4} \right)\]