Question

Seven particles each of mass $m$ are placed at the corners of a cube of side $b$. The co-ordinates of the centre of mass of the system is

• A. $(\frac{2b}{7},\frac{2b}{7},\frac{2b}{7})$
• B. $(\frac{3b}{7},\frac{3b}{7},\frac{3b}{7})$
• C. $(\frac{4b}{7},\frac{4b}{7},\frac{4b}{7})$
• D. $(\frac{5b}{7},\frac{5b}{7},\frac{5b}{7})$

Answer 1

The correct option is B $(\frac{3b}{7},\frac{3b}{7},\frac{3b}{7})$

We can assume one more mass $m$ kept at the empty corner. Then the centre of mass of the system will be at the centre of the cube.
$\overrightarrow{r}=\frac{b}{2}\hat{i}+\frac{b}{2}\hat{j}+\frac{b}{2}\hat{k}$
Now, $C.O.M$ of the original system will be,
${\overrightarrow{r}}_{C.O.M}=\frac{M\overrightarrow{r}-m\overrightarrow{r^{{}^{\prime }}}}{M-m}$
${\overrightarrow{r}}_{C.O.M}=\frac{8m\times (\frac{b}{2}\hat{i}+\frac{b}{2}\hat{j}+\frac{b}{2}\hat{k})-m\times (b\hat{i}+b\hat{j}+b\hat{k})}{8m-m}$
${\overrightarrow{r}}_{C.O.M}=\frac{3b}{7}\hat{i}+\frac{3b}{7}\hat{j}+\frac{3b}{7}\hat{k}$
Hence, the coordinates of $COM$ is $(\frac{3b}{7},\frac{3b}{7},\frac{3b}{7})$