Physics · Horizontal Circular Motion

JEE Main 2026 — 2 April, Evening Shift — Question 2

A $0.5\mathrm{kg}$ mass is in contact against the inner wall of a cylindrical drum of radius $4\mathrm{m}$ 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 $5\mathrm{rad / s}$ . The coefficient of friction between the drum's inner wall surface and mass is (Take $\mathrm{g} = 10\mathrm{m / s}^2$ )

Option A:
0.1
Correct
Option B:
0.5
Option C:
0.7
Option D:
0.3

Answer: A

Step-by-step solution

$\mathrm{N} = \mathrm{m}\omega^{2}\mathrm{R}$ , $\mathrm{f}_{\mathrm{s}} = \mathrm{mg}$ , $\mathrm{mg}\leq \mu \mathrm{m}\omega^{2}\mathrm{R}$ $\Rightarrow \mu \geq \frac{\mathrm{g}}{\omega^{2}\mathrm{R}} = \frac{10}{5^{2}\times 4} = 0.1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Horizontal Circular Motion
Topic: Dynamics of circular motion