Question

The position vector of four identical masses of mass $1\text{kg}$ each are $\overrightarrow{r_1}=(\hat{i}+2\hat{j}+7\hat{k}),\overrightarrow{r_2}=(3\hat{i}+5\hat{j}+\hat{k}),$ $\overrightarrow{r_3}=(6\hat{i}+2\hat{j}+3\hat{k})$ and $\overrightarrow{r_4}=(2\hat{i}-\hat{j}+5\hat{k})$. Find the position vector of their centre of mass.

• A. $(2\hat{i}+3\hat{j}+4\hat{k})$
• B. $(4\hat{i}+3\hat{j}+2\hat{k})$
• C. $(3\hat{i}+4\hat{j}+2\hat{k})$
• D. $(3\hat{i}+2\hat{j}+4\hat{k})$

Answer 1

The correct option is D $(3\hat{i}+2\hat{j}+4\hat{k})$

All masses are equal (Given)
$m_1=m_2=m_3=m_4=m=1\text{k}g$
The position vector of $COM$ of the four particles is given by
${\overrightarrow{r}}_{com}=\frac{m_1\overrightarrow{r_1}+m_2\overrightarrow{r_2}+m_3\overrightarrow{r_3}+m_4\overrightarrow{r_4}}{m_1+m_2+m_3+m_4}$
Substitute the values, we get
${\overrightarrow{r}}_{com}=\frac{\left(1\right)(\hat{i}+2\hat{j}+7\hat{k})+\left(1\right)(3\hat{i}+5\hat{j}+\hat{k})+\left(1\right)(6\hat{i}+2\hat{j}+3\hat{k})+\left(1\right)(2\hat{i}-\hat{j}+5\hat{k})}{1+1+1+1}$
$=\frac{(12\hat{i}+8\hat{j}+16\hat{k})}{4}$
${\overrightarrow{r}}_{com}=(3\hat{i}+2\hat{j}+4\hat{k})$