A 60 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 4m rotating about its vertical axis with 180 rev./min. The coefficient of friction between the wall and his clothing is 0.20. Find 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? Assume g = 10 $$\text{m/s}^2$$​

Correct option is A

​$$\text{When the floor is suddenly removed, three forces act on the man:} \\1. \text{Frictional force } (f_s) \text{ due to the contact between the man and the drum's wall.} \\2. \text{Centripetal force } (F = m r \omega^2) \text{ due to the rotation of the drum.} \\3. \text{Weight of the man } (W = mg). \\\text{The frictional force } (f_s) \text{ must balance the man's weight in order to prevent him from falling. The frictional force is given by:} \\f_s = \mu \cdot N \\\text{where } N = mg \text{ is the normal force, which in this case equals the weight of the man.} \\\text{Balancing the vertical forces:} \\f_s = mg \\\mu m r \omega^2 = mg$$

$$\textbf{Given:} \\\bullet \text{Mass of the man: } m = 60 \, \text{kg} \\\bullet \text{Radius of the drum: } r = 4 \, \text{m} \\\bullet \text{Coefficient of friction between the man and the drum: } \mu = 0.20 \\\bullet \text{Gravitational acceleration: } g = 10 \, \text{m/s}^2$$

$$\mu r \omega^2 = g \\\omega^2 = \frac{g}{\mu r} \\\text{Substituting the given values:} \\\omega^2 = \frac{10}{0.2 \times 4} = 12.5 \\\omega = \sqrt{12.5} = 3.54 \, \text{rad/s}$$​​​​