A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 riv/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Answers
Mass of the man, m = 70 kg
Radius of the drum, r = 3 m
Coefficient of friction, μ = 0.15
Frequency of rotation, ν = 200 rev/min = 200 / 60 = 10 / 3 rev/s
The necessary centripetal force required for the rotation of the man is provided by the normal force (FN).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force
(f = μFN) acting upward.
Hence, the man will not fall until:
mg < f
mg < μFN = μmrω2
g < μrω2
ω > (g / μr)1/2
The minimum angular speed is given as:
ωmin = (g / μr)1/2
= ( 10 / (0.15 X 3) )1/2 = 4.71 rad s-1
Explanation:
Mass of the man, m = 70 kg
Radius of the drum, r = 3 m
Coefficient of friction, μ = 0.15
Frequency of rotation, ν = 200 rev/min = 200 / 60 = 10 / 3 rev/s
The necessary centripetal force required for the rotation of the man is provided by the normal force (FN).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force
(f = μFN) acting upward.
Hence, the man will not fall until:
mg < f
mg < μFN = μmrω2
g < μrω2
ω > (g / μr)1/2
The minimum angular speed is given as:
ωmin = (g / μr)1/2
= ( 10 / (0.15 X 3) )1/2 = 4.71 rad s-1