Question

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

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

Answer 1

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

Let $m_1=m,m_2=2m,m_3=3m,m_4=4m$


$\because m_1$is at origin
Position vector of$m_1,$$\overrightarrow{r_1}=0\hat{i}+0\hat{j}$
Position vector of$m_2,$$\overrightarrow{r_2}=a\cos {60}^{\circ }\hat{i}+a\sin {60}^{\circ }\hat{j}=\frac{a}{2}\hat{i}+\frac{a\sqrt{3}}{2}\hat{i}$
Position vector of$m_3,$
$\overrightarrow{r_3}=(a+a\cos {60}^{\circ })\hat{i}+a\sin {60}^{\circ }\hat{j}=\frac{3a}{2}\hat{i}+\frac{a\sqrt{3}}{2}\hat{j}$
Position vector of$m_4,$
$\overrightarrow{r_4}=a\hat{i}$

Position vector of com $\overrightarrow{r}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}+m_4\overrightarrow{r_4}}{m_1+m_2+m_3+m_4}$
Putting all values in above equation
$\overrightarrow{r}=(0.95a\hat{i}+\frac{\sqrt{3}}{4}a\hat{j})$