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

Answer: (B)

$\left( 0.95a,\frac{\sqrt{3}}{4}a \right)$

Answer 2

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