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 su…

Question

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).

Accepted Answer

(1) 0.1. Idea: the wall's normal force supplies the centripetal force, and friction against that wall is what holds the mass up. Radially: N=mω²r. Vertically, at the minimum speed friction is at its limit: μ N=mg. Substituting and cancelling m: μ ω²r=g μ=g/(ω²r)=(10)/(25×4)=(10)/(100)=0.1. The mass cancels, which is why a 0.5 kg block and a person would stick at the same speed — the principle behind the fairground rotor.

Step-by-step solution

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