Four particles of masses $m,\ 2m,\ 3m$ and $4m$ are arranged at the corners of a parallelogram with each inside equal to $a$ and one of the angle between two adjacent sides is $60{^\circ}$. 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

Question

Dr. Abubacker · Accepted Answer

(b) Let $m_{1} = m,\ m_{2} = 2m,\ m_{3} = 3m,\ m_{4} = 4m$
${\overrightarrow{r}}_{1} = 0\widehat{i} + 0\widehat{j}$
${\overrightarrow{r}}_{2} = a\cos{60\widehat{i} + a\sin{60\widehat{j} = \frac{a}{2}i + \frac{a\sqrt{3}}{2}\widehat{j}}}$
${\overrightarrow{r}}_{3} = (a + a\cos{60)\widehat{i} + a\sin{60\widehat{j} = \frac{3}{2}a\widehat{i} + \frac{a\sqrt{3}}{2}\widehat{j}}}$
${\overrightarrow{r}}_{4} = a\widehat{i} + 0\widehat{j}$
By substituting above value in the following formula
$\overrightarrow{r} = \frac{m_{1}{\overrightarrow{r}}_{1} + m_{2}{\overrightarrow{r}}_{2} + m_{3}{\overrightarrow{r}}_{3} + m_{4}{\overrightarrow{r}}_{4}}{m_{1} + m_{2} + m_{3} + m_{4}} = 0.95a\widehat{i} + \frac{\sqrt{3}}{4}a\widehat{j}$
So the location of centre of mass $\left\lbrack 0.95a,\frac{\sqrt{3}}{4}a \right\rbrack$

Explanation

Reviewed by Dr. Abubacker

Want more practice?