Laws of Motion

JEE Main level, single correct, medium

129 of 210

Question 129 of 210

A 0.50.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 44 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is 55 rad/s. The coefficient of friction between the drum's inner wall surface and mass is _______. (Take g=10g = 10 m/s2^2)

Show answer and solution

Answer: (A) 0.10.1

The normal force NN provides the necessary centripetal force for the mass to move in a circle:

N=mω2RN = m \omega^2 R

For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force:

f≥mgf \ge mg

Since the maximum static friction is fmax=μNf_{max} = \mu N, we have:

μN≥mg\mu N \ge mg

μ(mω2R)≥mg\mu (m \omega^2 R) \ge mg

μ≥gω2R\mu \ge \dfrac{g}{\omega^2 R}

Substituting the given values g=10g = 10 m/s2^2, ω=5\omega = 5 rad/s, and R=4R = 4 m:

μ≥1052×4\mu \ge \dfrac{10}{5^2 \times 4}

μ≥10100\mu \ge \dfrac{10}{100}

μ≥0.1\mu \ge 0.1

The minimum coefficient of friction is 0.10.1.

Answer: 0.10.1

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Laws of motion, friction and pulleys simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Laws of Motion (Frictional force), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.