Question

A circular ring of mass $6\text{kg}$ and radius $a$ is placed such that its centre lies at the origin. Two particles of masses $2\text{kg}$ each are placed at the intersecting points of the circle with +ve x-axis and +ve y-axis. Then, the angle made by the position vector of centre of mass of entire system with x-axis is

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${\tan }^{-1}\left(\frac{4}{5}\right)$
• D. ${30}^{\circ }$

Answer 1

The correct option is A ${45}^{\circ }$

The centre of mass of the ring will lie at origin $(0,0)$ by symmetry
Let $Q$ is the position of centre of mass of all three particles.


$\tan \theta =\frac{y_{CM}}{x_{CM}}=\frac{m_1{y}_1+m_2{y}_2+m_3{y}_3}{m_1{x}_1+m_2{x}_2+m_3{x}_3}$

$=\frac{6\times 0+2\times 0+2\times a}{6\times 0+2\times a+2\times 0}$
$=1$
or $\theta ={45}^{\circ }$
Hence, option (a) is correct.