JEE Mains · Physics · STD 11 - 4.2 friction
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/s\(^2\))
- A \(0.1\)
- B \(0.5\)
- C \(0.7\)
- D \(0.3\)
Answer & Solution
Correct Answer
(A) \(0.1\)
Step-by-step Solution
Detailed explanation
The normal force \(N\) provides the necessary centripetal force for the mass to move in a circle: \(N = m \omega^2 R\) For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force: \(f \ge mg\) Since the…
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