JEE Main 2026 · 21 January · evening · Single correct · +4 / −1

A large drum having radius R is spinning around its axis with angular velocity ω, as shown in figure.
The minimum value of ω so that a body of mass M remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass M as μ, is :

JEE Main PYQ figure
  1. (a) √(2g/(μ R))
  2. (b) √(g/(2μ R))
  3. (c) √((μ g)/R)
  4. (d) √(g/(μ R))
Answer and solution

Answer (d) √(g/(μ R))

Solution

Physics sir · 3 steps

The wall pushes inward and supplies the centripetal force; friction from that push must hold up the weight.

  1. Horizontal direction. The only horizontal force on the block is the normal reaction N from the wall, pointing towards the axis. It provides the centripetal force, so N = Mω² R.
  2. Vertical direction. Gravity Mg pulls the block down along the wall. Only static friction, acting upward along the wall, can balance it, and friction can be at most μ N.

    μ Mω² R ≥ Mg ⇒ ω² ≥ g/(μ R)

  3. So. The mass M cancels, and the minimum angular velocity is ωₘᵢₙ = √(g/(μ R)), which is option (d).

The idea to keep. In a rotor problem the normal force is the centripetal force, and friction is the force that fights gravity. Faster spin means a bigger normal force, so more friction is available.

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 2026.