Question

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) $

Answer 1

The Correct Option is B

$ \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) $