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Physics Question 27 – JEE-MAIN 2026

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A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 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 rad/s. The coefficient of friction between the drum's inner wall surface and mass is _______. (Take g=10 m/s2)

Analyze all forces acting on the mass: gravity, normal force, and static friction.

Step 1: Identify Forces and Conditions for Equilibrium✦ Active

The forces acting on the mass are its weight mg downwards, the normal force N horizontally towards the center, and the static friction force fs upwards. For the mass not to fall, the upward friction force must balance the downward gravitational force:

fs=mg
Step 2: Relate Normal Force to Centripetal Force○ Expand

The normal force N provides the necessary centripetal force for the circular motion:

N=mRω2

For the minimum rotational speed, the static friction is at its maximum value, which is related to the normal force by the coefficient of static friction μs:

fs=μsN
💡 Teacher's Secret Hint

Remember that the normal force is perpendicular to the surface, and in this case, it's horizontal, providing the centripetal force.

Step 3: Calculate the Coefficient of Friction○ Expand

Substitute the expressions for fs and N into the force balance equation from Step 1:

μs(mRω2)=mg

Solving for μs:

μs=mgmRω2=gRω2

Substitute the given values: g=10 m/s2, R=4 m, ω=5 rad/s.

μs=104×(5)2=104×25=10100=0.1
💡 Teacher's Secret Hint

Ensure units are consistent before calculation. The mass m cancels out, indicating the coefficient of friction is independent of the mass in this scenario.

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