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 )

Step-by-step solution

The normal force N provides the necessary centripetal force for the mass to move in a circle: N = m ^2 R For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force: f mg Since the maximum static friction is f_ max = N , we have: N mg (m ^2 R) mg g ^2 R Substituting the given values g = 10 m/s ^2 , = 5 rad/s, and R = 4 m: 10 5^2 4 10 100 0.1 The minimum coefficient of friction is 0.1 . Answer: 0.1