A coin is placed on a disc. The coefficient of friction between the coin and the disc is $$\mu$$. If the distance of the coin from the center of the disc is $$r$$, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is :

A coin is placed on a rotating disc at a distance $$r$$ from the center. The coefficient of friction between the coin and the disc is $$\mu$$. We need to find the maximum angular velocity so the coin does not slip.

For the coin to move in a circle on the disc, it needs a centripetal force directed toward the center. The only horizontal force acting on the coin is the static friction between the coin and the disc.

The centripetal force required for circular motion is $$F_{centripetal} = m\omega^2 r$$, where $$m$$ is the mass of the coin and $$\omega$$ is the angular velocity. The maximum static friction force available is $$f_{max} = \mu N = \mu mg$$, where $$N = mg$$ is the normal force (since the disc is horizontal).

For the coin not to slip, the required centripetal force must not exceed the maximum friction, which gives $$m\omega^2 r \leq \mu mg$$. At the maximum angular velocity, the two forces are equal: $$m\omega_{max}^2 r = \mu mg$$.

The mass $$m$$ cancels from both sides, yielding $$\omega_{max}^2 = \dfrac{\mu g}{r}$$ and hence $$\omega_{max} = \sqrt{\dfrac{\mu g}{r}}$$.

The correct answer is Option (3): $$\sqrt{\dfrac{\mu g}{r}}$$.