A 0.5 kg mass is in contact against the inner wall of a cylindrical...

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: $0.1$