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 rev/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 ?
Given: The mass of the man standing inside a hollow cylinder is 70 kg , the radius of the hollow cylinder drum is 3 m , the speed of rotation of the drum is 200 rev / min and the coefficient of friction between the wall and the clothing is 0.15 .
The diagram of a cylinder showing the forces acting on the surface is shown below:
The required centripetal force provided by the horizontal reaction on the man is,
N = m R ω 2
The energy balance equation of the man on the cylinder is given as,
Weight = μ N m g = μ m R ω 2 min ω min = g μ R
By substituting the given values in the above equation, we get
ω min = 10 ( 0.15 ) ( 3 ) = 22.2 ≈ 5 s − 1
Thus, the minimum rotational speed of the cylinder is 5 s − 1 .
Given: The mass of the man standing inside a hollow cylinder is
The diagram of a cylinder showing the forces acting on the surface is shown below:
The required centripetal force provided by the horizontal reaction on the man is,
The energy balance equation of the man on the cylinder is given as,
By substituting the given values in the above equation, we get
Thus, the minimum rotational speed of the cylinder is
