Question

Three identical carrom coins, each of radius $5\text{cm}$ are placed touching each other on a horizontal surface such that an equilateral triangle is formed when the centres of the coins are joined. Find the coordinates of this centre of mass, if the origin is fixed at the centre of coin A, as shown in figure.

• A. $(5\text{cm},\frac{5}{\sqrt{3}}\text{cm})$
• B. $(5\sqrt{3}\text{cm},\frac{5}{\sqrt{3}}\text{cm})$
• C. $(2\text{cm},\frac{5}{\sqrt{3}}\text{cm})$
• D. $(5\text{cm},5\text{cm})$

Answer 1

The correct option is A $(5\text{cm},\frac{5}{\sqrt{3}}\text{cm})$


Centre of mass of each coin will be at its respective centre, so with respect to origin :-
Centre of mass of coin $A$ $(x_1,y_1)=(0,0)$
Centre of mass of coin $B$ $(x_1,y_2)=(2r,0)$
Centre of mass of coin $C$: $(x_3,y_3)=(r,\sqrt{3}r)$

Let $(x_{cm},y_{cm})$ be the position of the centre of mass of the system. Then
$x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$=\frac{m\left(0\right)+m\left(2r\right)+m\left(r\right)}{3m}$
$\Rightarrow x_{cm}=r=5\text{cm}$
$y_{cm}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$=\frac{m\left(0\right)+m\left(0\right)+m\left(\sqrt{3}r\right)}{3m}$
$=\frac{r}{\sqrt{3}}=\frac{5}{\sqrt{3}}\text{cm}$
$\therefore$ Position of centre of mass of the system, $(x_{cm},y_{cm})$ = $(5\text{cm},\frac{5}{\sqrt{3}}\text{cm})$