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)
- 0.1
- 0.5
- 0.7
- 0.3
Answer
(1) 0.1
Solution
Normal reaction from the wall provides the centripetal force:
N = m 2r
Friction (vertical) must balance the weight:
N mg ⇒ m 2 r mg
At the minimum speed the friction is limiting:
∴ = g 2 r = 10(5)2 4 = 10100 = 0.1
(The mass 0.5 kg cancels out.)
Answer: (1)