Question

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 in $rad/s$ of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed? Take $g=10m{s}^{-2}$

Answer 1

Radius of cylindrical drum $R=3m$
The horizontal force $N$ by the wall on the man provides the needed centripetal force; $N=mR{\omega }^2$
The frictional force $f$ (vertically upward) will be balanced by the weight of the man. Hence, the man remains stuck to the wall after the floor is removed if
$mg\le f_l\Rightarrow mg\le \mu N\Rightarrow mg\le \mu m{\omega }^{2}R$
$\Rightarrow g\le \mu {\omega }^{2}R\Rightarrow \mu {\omega }^{2}R\ge g$
$\Rightarrow {\omega }^2\ge \frac{g}{\mu R}\ge \frac{10}{0.15\times 3}\ge 22.22$
$\therefore$ minimum rotational speed required
$\omega =\sqrt{22.22}=4.7rad/s$