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JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 3)

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $$\omega$$. The coefficient of static friction between the bottom of the beaker and the surface of the disc is $$\mu$$. The beaker will revolve with the disc if :
$$R \le {{\mu g} \over {2{\omega ^2}}}$$
$$R \le {{\mu g} \over {{\omega ^2}}}$$
$$R \ge {{\mu g} \over {2{\omega ^2}}}$$
$$R \ge {{\mu g} \over {{\omega ^2}}}$$

Explanation

The force that prevents the beaker from sliding is the static friction force. For an object in uniform circular motion, the net force acting on the object (which is the friction force in this case) is equal to the centripetal force.

The static friction force is given by the normal force (which is equal to the weight of the object) multiplied by the coefficient of static friction, which is :

$$F_{\text{friction}} = \mu mg.$$

The centripetal force needed to keep an object moving in a circle of radius R at angular velocity ω is given by :

$$F_{\text{centripetal}} = mR\omega^2.$$

For the beaker not to slide off, the static friction force must be at least as large as the centripetal force. Therefore, we have :

$$\mu mg \geq mR\omega^2.$$

After canceling the mass m from both sides, we get :

$$R \leq \frac{\mu g}{\omega^2}.$$

So, the correct answer is Option B :

$$R \leq \frac{\mu g}{\omega^2}.$$

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