Mass of man m = 70 kg

Radius of the cylindrical drum r = 3 m

Frequency of the vertical axis f = 200 rev/min

Coefficient of the friction \mu = 0.15

Step 1: Note down the forces acting on the man

Frictional force acting upward is balanced by the weight of the man acting down ward

f_r = mg .....................(1)

The normal force acting towards the central axis gives centripetal force.

N = F_c

N = \frac{mv^2}{r} = mr \omega^2 ........(2)

Step 2: Set up an equation for the minimum angular velocity

Frictional force F_r = \mu * N

f_r = \mu * (mr \omega^2) \because \text{ equation (2) }

mg = \mu * (mr \omega^2) \because \text{ equation (1) }

g = \mu * (r \omega^2)

\omega = \sqrt{\frac{g}{\mu * r}}

Step 3: Find the minimum angular speed at which the drum rotates

\omega = \sqrt{\frac{10}{0.15 * 3}}

\omega = 4.7 rad/sec

Hence, minimum angular speed at which the drum rotates \omega = 4.7 rad/sec