Solution

Let's consider a particle on the inner wall of the rotating cylinder at an angle $\alpha$ to the horizontal plane.
According to Newton's second law for the radial direction, the centripetal acceleration is provided by the normal reaction force $N$ and the radial component of gravity:
$$m \omega^2 R = N + mg \sin\alpha \implies N = m(\omega^2 R - g \sin\alpha)$$

For the particle not to slide down along the inner wall, the required static friction force must balance the tangential component of gravity ($mg \cos\alpha$).
To prevent sliding, this required friction force must not exceed the maximum possible static friction force ($\mu N$):
$$mg \cos\alpha \le \mu N$$

Substitute the expression for the normal force $N$:
$$mg \cos\alpha \le \mu m(\omega^2 R - g \sin\alpha)$$

According to the problem statement, the coefficient of friction is $\mu = 1$. The mass $m$ cancels out:
$$g \cos\alpha \le \omega^2 R - g \sin\alpha$$
$$\omega^2 R \ge g(\cos\alpha + \sin\alpha)$$
$$\omega^2 \ge \frac{g}{R}(\cos\alpha + \sin\alpha)$$

To ensure the particles do not slide at any point on the inner surface of the cylinder, this condition must hold true for the maximum possible value of the function $f(\alpha) = \cos\alpha + \sin\alpha$.
Using trigonometric properties, the maximum value of $(\cos\alpha + \sin\alpha)$ is $\sqrt{2}$ (which occurs when $\alpha = 45^\circ$).

Therefore, the minimum angular velocity required to hold the particles at all points is determined by:
$$\omega^2 \ge \frac{g}{R} \sqrt{2}$$
$$\omega_{\text{min}} = \sqrt{\frac{g\sqrt{2}}{R}}$$