A coin is placed on a disc. The coefficient of friction between the coin and the disc is μ. 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 :

Solution

Physics sir · 3 steps

Static friction is the only force pulling the coin inward, so set its maximum equal to mω² r.

1. Who supplies the pull. To move in a circle with the disc, the coin needs a centripetal force mω² r towards the centre. The only horizontal force on it is static friction from the disc.
2. Friction has a limit. On a horizontal disc the normal force is N = mg, so static friction can be at most μ N = μ mg. The coin stays put only while the needed force does not exceed this.

   mω² r ≤ μ m g ⇒ ω ≤ √((μ g)/r)

3. So. The mass cancels, and the maximum angular velocity is ωₘₐₓ = √((μ g)/r), which is option (b).

The idea to keep. Whenever friction alone keeps something on a circle, write mω² r ≤ μ mg. The mass cancels, and a larger radius means a lower safe speed.

Answer from the NTA final answer key. Solution by PrepWiser.