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 ${60}^0$. 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. $(\frac{\sqrt{3}}{2}a,0.95a)$
• B. $(\frac{9a}{10},\frac{\sqrt{3a}}{4})$
• C. $(\frac{3a}{4},\frac{a}{2})$
• D. $(\frac{a}{2},\frac{3a}{4})$

Answer 1

Answer: (B)

$(\frac{9a}{10},\frac{\sqrt{3a}}{4})$

Therefore the coordinates of centre of mass are,

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

$=\frac{M\left(0\right)+2M\left(\frac{a}{2}\right)+3M\left(\frac{3a}{2}\right)+aM\left(a\right)}{M+2M+3M+4M}$

$\Rightarrow x_{cm}=\frac{\frac{Ma}{2}+\frac{9Ma}{2}+\frac{8Ma}{2}}{10M}=\frac{9Ma}{10M}$

$\Rightarrow x_{cm}=\frac{9a}{10}$

and,

$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3+m_4{y}_4}{m_1+m_2+m_3+m_4}$

$\frac{M\left(0\right)+2M\left(\frac{\sqrt{3}a}{2}\right)+3M\left(\frac{\sqrt{3}a}{2}\right)+4M\left(0\right)}{M+2M+3M+4M}$

$y_{cm}=\frac{\frac{5\sqrt{3}Ma}{2}}{10M}=\frac{5\sqrt{3}Ma}{20M}$

$y_{cm}=\frac{\sqrt{3a}}{4}$

Therefore the coordinates of the centre of mass are,

$(x_{cm},y_{cm})=(\frac{9a}{10},\frac{\sqrt{3a}}{4})$