Question

person is in contact with inner wall of a vertical hollow cylinder of radius 2m without floor under his feet, remains in equilibrium without slipping down
when the cylinder is rotated about it's own vertical axis. If the coefficient of static friction between person and wall of cylinder is 0.4, the minimum
angular velocity of cylinder should be
(rad/sec)​

Answer 1

Explanation:

Given Person is in contact with inner wall of a vertical hollow cylinder of radius 2m without floor under his feet, remains in equilibrium without slipping down when the cylinder is rotated about it's own vertical axis. If the coefficient of static friction between person and wall of cylinder is 0.4, the minimum  angular velocity of cylinder should be

• Now radius of cylinder = 2 m
• So wall of cylinder μ = 0.4
• We need to find the minimum angular velocity of cylinder.
• So for equilibrium maximum friction will balance the force towards the ground Mg.
• So Mg = μN  
• Now N = Mω^2R
• So substituting we get
•  Mg = μ M ω^2 R
•    So ω = √g / μ R
•           = √9.8 / 0.4 x 2
•           = √9.8 / 0.8
•           = √12.25
•           = 3.5 rad / s

Reference link will be

https://brainly.in/question/12436847

Answer 2

Answer:

3.5rad/swill be the answer