Minimum Rotational Speed and Coefficient of Friction in Cylindrical Drum

To find the coefficient of friction ($\mu$) between the drum's inner wall surface and the mass, we analyze the forces acting on the mass $m$ as it rotates in a horizontal circle inside the cylindrical drum:

1. Normal Force ($N$): The wall exerts a normal force directed radially inwards toward the center of the cylinder, providing the necessary centripetal force for the circular motion: $N = m \omega^2 R$

2. Vertical Forces: For the mass to remain stuck to the wall without falling, the upward static friction force ($f_s$) must balance the downward gravitational force ($mg$): $f_s = mg$

3. Condition for Non-Slipping: The force of static friction cannot exceed the maximum static friction force ($f_{s, \text{max}} = \mu N$): $f_s \le \mu N$

   Substituting $f_s = mg$ and $N = m \omega^2 R$ into the inequality: $mg \le \mu m \omega^2 R$

   Dividing both sides by $m$: $g \le \mu \omega^2 R$

   For the minimum rotational speed $\omega_{\text{min}}$, the limiting condition is reached: $g = \mu \omega_{\text{min}}^2 R \implies \mu = \frac{g}{\omega_{\text{min}}^2 R}$

Given data:

• $g = 10\text{ m/s}^2$
• $R = 4\text{ m}$
• $\omega_{\text{min}} = 5\text{ rad/s}$

Substituting the given values into the formula: $\mu = \frac{10}{(5)^2 \times 4} = \frac{10}{25 \times 4} = \frac{10}{100} = 0.1$

Therefore, the coefficient of friction is $0.1$, which corresponds to Option A.