Question

Four particles of masses m, 2m, 3m and 4m are at the vertices of a parallelogram in x y plane with one of the adjacent angle 60 and smaller side a and larger side 2a. The mass m is at the origin and mass 4m on the x-axis. The centre of mass of the system is

• A. $(\frac{33a}{20};\frac{5\sqrt{3}a}{2})$
• B. $(\frac{a}{2};\frac{3a}{2})$
• C. $(0.95a;\frac{\sqrt{3a}}{4})$
• D. $(\frac{\sqrt{3a}}{2};0.82a)$

Answer 1

The correct option is A $(\frac{33a}{20};\frac{5\sqrt{3}a}{2})$

Coordinates of mass $m=(0,0)$

Coordinates of mass $4m=(2a,0)$

Coordinates of mass $2m=(acos{60}^o,asin{60}^o)=(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Coordinates of mass $3m=(\frac{a}{2}+2a+\frac{\sqrt{3}a}{2})=(\frac{5a}{2},\frac{\sqrt{3}a}{2})$

therefore, CM of the system will be given by:

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

$x=\frac{33}{20}a$

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

$y=\frac{5a\sqrt{3}}{2}$

Thus the CM of the system will be at $(\frac{33}{20}a,\frac{5a\sqrt{3}}{2})$