A small bead of mass m is carried by a circular hoop having centre at $O$ and radius $\sqrt{2} m$ which rotates about a fixed vertical axis. The coefficient of friction between beed and hoop is $μ = 0.5$ . The maximum angular speed of the hoop for which the bead does not have relative motion with respect to hoop. $(g = 10 m / s^{2})$

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$\sqrt{10}$

$\sqrt{30}$

$\sqrt{60}$

$\sqrt{15}$

answer is A.

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Detailed Solution

The maximum angular speed of the hoop corresponds to the situations when the bead is just about to slide upwards. For the bead not to slide upwards.

$\cos 45^{\circ} = \frac{x}{r} \Rightarrow x = r \cos 45^{\circ} m ω^{2} (r \sin 45) \cos 45 - m g \sin 45 \leq μ N$

Where $N = m g \cos 45 + m ω^{2} (r \sin 45) \sin 45$

On Solving $ω \leq \sqrt{\frac{(1 + μ) g}{(1 - μ) r \sin 45}}$

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