A `70kg` man stands in contact against the inner wall of a hollow cylindrical drum of radius `3m` rotating about its verticle axis. The coefficient of friction between the wall and his clothing is `0.15` . What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Explanation

`m = 70 kg , r = 3 m`
`n = 200 "rpm" = (200)/(60) "rps" , mu = 0.15 , omega = ?`
The horizontal force N by the wall on the man provides the necessary force `= m r omega^(2)` the frictional force (f) in this case is vertically upwards opposing the weight (mg) of the man
After the floor is removed the man will remain stack to the wall when
`mg = f lt mu N` i.e. `mg lt mu m r omega^(2)` or `g lt mu r omega^(2)`
`:.` Minimum angular speed of rotation of the cylinder is `omega = sqrt((g)/(mu r)) = sqrt (10/(0.15 xx 3)) = 4 .7 rad//s`.