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A circular ring of mass 6 kg and radius a is placed such that its centre lies at the origin. Two particles of masses 2 kg each are placed at the intersecting points of the circle with positive X-axis and positive Y-axis. Then, the angle made by the position vector of centre of mass of entire system with X-axis is

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45°

60°

30°

$\tan^{- 1} (\frac{4}{5})$

answer is A.

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Detailed Solution

P is the position of centre of mass of particles at 2 and 3.
Q is position of centre of mass of all three particles.

$\tan θ = \frac{y_{CM}}{x_{CM}} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}$

$= \frac{6 \times 0 + 2 \times 0 + 2 \times a}{6 \times 0 + 2 \times a + 2 \times 0} = 1$ or $θ = 45^{\circ}$

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