A `70kg` man stands in contact against the inner wall of a hollow cylindrical drum of radius `3m` rotating about its vertical 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

R = 3 m, `omega`= 200 rev/min = `2 xx 22/7 xx 200/60 rad/s`
= 440 / 21 rad/s
and `mu` = 0.15
As shown in the figure, the normal reaction (N) of the wall on the man acts in the horizontal direction towards the axis of the cylinder while the force of frction (f) acts vertically upwards. The required centripetal force will be provided by the horizontal reaction N of the wall on the men, i.e.,
N = `mv^(2)` /R = m`omega^(2)`R
The frictional force f acting verically upwards will be balanced by the weight of the man Hence, the man remians stuck to the wall after the floor of the man. Hence, the man remains stuck to the wall after the floor is removed it mg `le` `mu`m`omega^(2)`R
or g `le` `mu``omega^(2)`R
or `mu``omega^(2)`R `ge` g or `omega` `ge` g / R`mu`
Hence , for minimum rotational speed of the cylinder
`omega^(2)` = g / `mu`R = 10 / `0.15 xx 3` = 22.2
`omega` = `sqrt 22.2` = 4.7 `rad/s`