JEE Physics 2024 Question with Solution

Given: A coin is at distance $r$ from the center of a rotating disc, and the coefficient of friction is $\mu$.

Find: The maximum angular velocity $\omega$ for which the coin does not slip.

The required centripetal force for circular motion is

$$F_c = m\omega^2 r$$

The maximum static friction available is

$$F_f = \mu m g$$

For the coin to remain at rest relative to the disc, friction must provide the centripetal force:

$$m\omega^2 r \leq \mu m g$$

Cancelling $m$,

$$\omega^2 \leq \frac{\mu g}{r}$$

Therefore,

$$\omega_{\max} = \sqrt{\frac{\mu g}{r}}$$

So the correct expression is $\sqrt{\frac{\mu g}{r}}$. This matches option B.

The solution's marks option C, but the solution working clearly gives $\sqrt{\frac{\mu g}{r}}$. Hence the keyed answer on the page is inconsistent with the derivation.

Given: The coin rotates with the disc at radius $r$.

Find: The limiting angular speed before slipping starts.

The forces on the coin are:

• Weight $mg$ downward
• Normal reaction $N$ upward
• Friction toward the center of the disc

Vertical forces balance, so friction is the only horizontal force available to produce centripetal acceleration. Thus,

$$f = m\omega^2 r$$

At the limiting condition of slipping,

$$f_{\max} = \mu N = \mu mg$$

Hence,

$$m\omega^2 r = \mu mg$$

Cancelling $m$,

$$\omega^2 = \frac{\mu g}{r}$$

Taking the positive square root,

$$\omega = \sqrt{\frac{\mu g}{r}}$$

Therefore, the coin does not slip as long as the angular velocity does not exceed $\sqrt{\frac{\mu g}{r}}$.