QJEEM24Phba65: JEE Main 2024 Physics - Class 11 JEEM24Phba65 Solution: A coin is placed on a disc. The coefficient of friction between …

When the coin is on the disc and the disc starts rotating, centrifugal force acts on the coin, trying to push it away from the center. The friction between the coin and the disc opposes this motion. For the coin to not slip, the frictional force must be equal to the centrifugal force acting on the coin.

The normal force (N) acting on the coin is equal to the weight of the coin, which can be represented as:

$N = mg$

where $m$ is the mass of the coin, and $g$ is the acceleration due to gravity.

The centrifugal force ( $f$ ) that acts on the coin due to the rotation of the disc is given by:

$f = m\omega^2r$

where $\omega$ is the angular velocity, and $r$ is the distance of the coin from the center of the disc.

Friction force (f) is also given by the formula:

$f = \mu N$

where $\mu$ is the coefficient of static friction between the coin and the disc.

For the coin to not slip, the centrifugal force must be equal to the frictional force, hence:

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

Dividing both sides by $mr$ , we get:

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

This equation shows that the maximum angular velocity ( $\omega$ ) that can be given to the disc to prevent the coin from slipping off depends on the coefficient of friction ( $\mu$ ), the acceleration due to gravity ( $g$ ), and the distance of the coin from the center ( $r$ ).