A person is stuck to the inside of a large rotating drum of radius r. The coefficient of static friction between the person's clothes and the drum surface is mu. What is the minimum angular velocity o

For the person to remain stuck to the wall, the upward static frictional force ($f_s$) must balance the downward gravitational force (weight, $Mg$).

$$f_s \ge Mg$$

The normal force ($N$) exerted by the wall on the person provides the centripetal force required for circular motion.

$$N = M\omega^2 r$$

where $M$ is the mass, $\omega$ is the angular velocity, and $r$ is the radius of the drum.

The maximum static frictional force is given by $f_{s,max} = \mu N$, where $\mu$ is the coefficient of static friction. For the person not to slide down, the frictional force must be at least equal to the weight:

$$\mu N \ge Mg$$

Substitute $N = M\omega^2 r$ into the inequality:

$$\mu (M\omega^2 r) \ge Mg$$

To find the minimum angular velocity, we set the inequality to an equality:

$$\mu M\omega_{min}^2 r = Mg$$

Divide both sides by $M$:

$$\mu \omega_{min}^2 r = g$$

Solve for $\omega_{min}^2$:

$$\omega_{min}^2 = \frac{g}{\mu r}$$

Take the square root of both sides:

$$\omega_{min} = \sqrt{\frac{g}{\mu r}}$$

Comparing this with the given options, option (a) matches our result.

The final answer is $\text{(a)} \ \omega_{min} = \sqrt{\frac{g}{\mu r}}$.