A \(0.5\) kg mass is in contact against the inner wall of a cylindrical drum of radius \(4\) m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is \(5\) rad/s. The coefficient of friction between the drum's inner wall surface and mass is _______. (Take \(g = 10\) m/s\(^2\))

The normal force \(N\) provides the necessary centripetal force for the mass to move in a circle:

\(N = m \omega^2 R\)

For the mass to remain stuck to the wall without falling, the upward frictional force must balance the downward gravitational force:

\(f \ge mg\)

Since the maximum static friction is \(f_{max} = \mu N\), we have:

\(\mu N \ge mg\)

\(\mu (m \omega^2 R) \ge mg\)

\(\mu \ge \dfrac{g}{\omega^2 R}\)

Substituting the given values \(g = 10\) m/s\(^2\), \(\omega = 5\) rad/s, and \(R = 4\) m:

\(\mu \ge \dfrac{10}{5^2 \times 4}\)

\(\mu \ge \dfrac{10}{100}\)

\(\mu \ge 0.1\)

The minimum coefficient of friction is \(0.1\).

Answer: \(0.1\)