A large drum having radius $R$ is spinning around its axis with angular velocity $\omega$, as shown in figure. The minimum value of $\omega$ so that a body of mass $M$ remains stuck to the inner wall of the drum, taking the coefficient of friction between the drum surface and mass $M$ as $\mu$, is :

Consider the mass $M$ in the frame of the rotating drum. The forces acting on the mass are:
1. Weight $Mg$ acting vertically downwards.
2. Normal reaction $N$ from the wall acting towards the center, which provides the centripetal force: $N = M\omega^2 R$.
3. Frictional force $f$ acting vertically upwards to balance the weight.
For the mass to remain stuck to the wall, the frictional force must balance the weight: $f = Mg$.
The maximum available static friction is $f_{max} = \mu N = \mu M\omega^2 R$.
To prevent slipping, we must have $f_{max} \ge Mg$.
$\mu M\omega^2 R \ge Mg$
$\omega^2 \ge \frac{g}{\mu R}$
$\omega \ge \sqrt{\frac{g}{\mu R}}$
The minimum value of $\omega$ is $\sqrt{\frac{g}{\mu R}}$.