Question

Two balls each of mass $m$ are placed on the vertices $A$ and $B$ of an equilateral $\triangle A B C$ of side $1\, m$. A ball of mass $2\, m$ is placed at vertex $C$. The centre of mass of this system from vertex $A$ (located at origin) is

• A. $\left(\frac{1}{2} m , \frac{1}{2} m \right)$
• B. $\left(\frac{1}{2} m , \sqrt{3} m \right)$
• C. $\left(\frac{1}{2} m , \frac{\sqrt{3}}{4} m \right)$
• D. $\left(\frac{\sqrt{3}}{4} m , \frac{\sqrt{3}}{4} m \right)$

Answer 1

Answer: (C)

The centre of mass is given by

$x_{ CM }=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$
$x_{ CM }=\frac{m \times 0+m \times 1+2 m \times\left(\frac{1}{2}\right)}{m+m+2 m}$
$x_{ CM }=\frac{2 m}{4 m}=\frac{1}{2} m$
$y_{ CM }=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}$
$y_{ CM }=\frac{m \times 0+m \times 0+2 m \times \frac{\sqrt{3}}{2}}{m+m+2 m}$
$=\frac{\sqrt{3}}{4} m$
Hence, the centre of mass is $\left(\frac{1}{2} m , \frac{\sqrt{3}}{4} m \right)$.