Question

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

• A. ($\frac{\sqrt{3}}{2}a,0.95a$)
• B. ($0.95a,\frac{\sqrt{3}}{4}a$)
• C. $M{R}^2=2I$
• D. ($\frac{a}{2},\frac{3a}{4}$)

Answer 1

The correct option is B ($0.95a,\frac{\sqrt{3}}{4}a$)

Center of mass is given by:

COM $=\frac{\sum m_i{x}_i}{\sum m_i}$ where $m_i$ is mass of $i^{th}$ particle and $r_i$ is position of $i^{th}$ particle.

In X-axis :X(x-center of mass)

$=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+m_4{x}_4}{m_1+m_2+m_3+m_4}$

$=\frac{(m\times 0)+(2m\times a\cos 60)+(3m\times (a+a\cos 60))+(4m\times a)}{10m}$

$=\frac{\left(ma\right)\times \left(3ma\right)+\frac{3ma}{2}+4ma}{10m}=\frac{2ma+6ma+3ma+8ma}{20m}=\frac{19ma}{20ma}=0.95a$

In Y-axis=

$=\frac{(m\times 0)+(2m\times a\sin 60)+(3m\times \left(a\sin 60\right))+(4m\times 0)}{10m}=\frac{\sqrt{3}ma+\frac{3\sqrt{3}}{2}ma}{10m}=\frac{2\sqrt{3}ma+3\sqrt{3}ma}{20ma}=\frac{5\sqrt{3}ma}{20ma}=\frac{\sqrt{3}}{4}a$

Therefore Center of mass=$(0.95a,\frac{\sqrt{3}}{4}a$)