m = 70 kg m

The drum’s radius is 3 metres.

= 0.15 is the coefficient of friction between the wall and his clothes.

The number of revolutions per minute of a hollow cylindrical drum is 200/60 = 10/3 revolutions per second.

The normal N of the wall on the guy provides the needed centripetal force.

mv2/R = m2R N = mv2/R = m2R

The guy adheres to the drum’s wall while the floor spins. As a result, the frictional force acting vertically upwards balances the weight of the man (mg) acting downward.

If mg limits frictional force fe (N), the guy will not fall.

mg ≤ μN

mg ≤ μ (mω2R)

ω2 ≥ g/Rμg

As a result, given the cylinder’s minimum rotational speed,

ω2 = g/Rμ = 10/(0.15 x 3) = 22. 2

ω =√22.2 = 4.7 rad