Question

Three particles of masses $1\text{kg}$ , $2\text{kg}$ and $3\text{kg}$ are placed at the vertices $A,B,C$ respectively of an equilateral triangle $ABC$ of edge $1\text{m}$ as shown in the figure. Find the position of the centre of mass of the system. (If $A$ is assumed to be the origin).

• A. $(\frac{2}{3},\frac{\sqrt{3}}{6})$
• B. $(\sqrt{\frac{3}{16}},\frac{2}{3})$
• C. $(\frac{\sqrt{3}}{6},\frac{2}{3})$
• D. $(\frac{2}{3},\sqrt{\frac{3}{16}})$

Answer 1

The correct option is A $(\frac{2}{3},\frac{\sqrt{3}}{6})$


Given that origin is assigned at mass $1\text{kg}$
Hence,coordinate of the vertices of triangle $ABC$ are:
$A=(0,0)$
$B=(1\cos {60}^{\circ },1\sin {60}^{\circ })=(\frac{1}{2},\frac{\sqrt{3}}{2})$
$C=(1,0)$

$\begin{array}{ccc}\text{mass}& x\left(\text{m}\right)& y\left(\text{m}\right)\\ m_A=1\text{kg}& 0& 0\\ m_B=2\text{kg}& \frac{1}{2}& \frac{\sqrt{3}}{2}\\ m_C=3\text{kg}& 1& 0\end{array}$

Applying the formula for coordinates of $\text{CM}$ for system:

$x_{\text{CM}}=\frac{m_A{x}_A+m_B{x}_B+m_C{x}_c}{m_A+m_B+m_C}$
$=\frac{(1\times 0)+(2\times \frac{1}{2})+(3\times 1)}{1+2+3}$
$\therefore x_{\text{CM}}=\frac{4}{6}=\frac{2}{3}$

$\&y_{\text{CM}}=\frac{m_A{y}_A+m_B{y}_B+m_C{y}_C}{m_A+m_B+m_C}$
$=\frac{(1\times 0)+(2\times \frac{\sqrt{3}}{2})+(3\times 0)}{1+2+3}$
$\therefore y_{\text{CM}}=\frac{\sqrt{3}}{6}$
Hence, position of $\text{CM}$ is represented by coordinates $(\frac{2}{3},\frac{\sqrt{3}}{6})$