Question

Three particles of masses $50\text{g},100\text{g}$ and $150\text{g}$ are placed at the vertices of an equilateral triangle of side $2\text{m}$ as shown in the figure. The $(x,y)$ co-ordinates of the centre of mass (in $\text{m})$ will be

• A. $(\frac{7}{6},\frac{\sqrt{3}}{2})$
• B. $(\frac{7}{12},\frac{\sqrt{3}}{4})$
• C. $(\frac{7}{3},\sqrt{3})$
• D. $(\frac{7}{12},\frac{\sqrt{3}}{2})$

Answer 1

The correct option is A $(\frac{7}{6},\frac{\sqrt{3}}{2})$

Let $(x_{cm},y_{cm})$ be the co-ordinates of the centre of mass of system of the three particles.


$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{50\left(0\right)+100\left(2\right)+150\left(1\right)}{300}$
$=\frac{7}{6}\text{m}$
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{50\left(0\right)+100\left(0\right)+150\left(\sqrt{3}\right)}{300}$
$=\frac{\sqrt{3}}{2}\text{m}$
$\therefore (x_{cm},y_{cm})=(\frac{7}{6},\frac{\sqrt{3}}{2})\text{m}$