JEE Main 2022 · 25 June · evening · Single correct · +4 / −1

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 ω. The coefficient of static friction between the bottom of the beaker and the surface of the disc is μ. The beaker will revolve with the disc if :

  1. (a) R ≤ (μ g)/(2ω ²)
  2. (b) R ≤ (μ g)/(ω ²)
  3. (c) R ≥ (μ g)/(2ω ²)
  4. (d) R ≥ (μ g)/(ω ²)
Answer and solution

Answer (b) R ≤ (μ g)/(ω ²)

Solution

Physics sir · 3 steps

Static friction is the only force pulling the beaker inward, so it must be able to supply mω² R.

  1. Who provides centripetal force. To move in a circle of radius R, the beaker needs an inward force mω² R. Gravity and the normal reaction are vertical, so only static friction from the disc acts horizontally and can supply it.
  2. Limit of friction. The normal reaction is N = mg, so static friction can be at most μ N = μ mg. The beaker stays with the disc as long as the friction it needs does not exceed this.

    mω² R ≤ μ m g ⇒ R ≤ (μ g)/ω²

  3. So. The mass cancels, and the beaker revolves with the disc when R ≤ (μ g)/ω², which is option (b).

The idea to keep. On a rotating platform, set the needed mω² r against the maximum friction μ mg. Since the needed force grows with r, the safe condition is always an upper limit on r.

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

Study Circular Motion

Asked about 70 times in JEE Main since 2020. This question is from 2022.