Question

Two balls each of mass $m$ are placed on the vertices $A$ and $B$ of an equilateral $△ABC$ of side $1m$. A ball of mass $2m$ 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

$x_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$
$x_{CM}=\frac{m\times 0+m\times 1+2m\times \left(\frac{1}{2}\right)}{m+m+2m}$
$x_{CM}=\frac{2m}{4m}=\frac{1}{2}m$
$y_{CM}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1+m_2+m_3}$
$y_{CM}=\frac{m\times 0+m\times 0+2m\times \frac{\sqrt{3}}{2}}{m+m+2m}$
$=\frac{\sqrt{3}}{4}m$
Hence, the centre of mass is $\left(\frac{1}{2}m,\frac{\sqrt{3}}{4}m\right)$.