Question

A $70$ kg man stands against the inner wall of a hollow cylindrical drum the rotor of radius $3$ m rotating about its vertical axis the coefficient of friction between the wall and his clothing is $0.15$

Answer 1

the cylinder being vertical the normal reaction of wall man acts horizontally and provides centripetal force acts on

$R=mr{\omega }^2$

the frictional force acting upwards

$F=mg$

the man willl remain stuck to the wall after the floor is removed he will continue to rotate with the cylender

$\mu \ge \frac{F}{R}$

$F\le \mu R$

$mg\le \mu mr{\omega }^2$

${\omega }^2\ge \frac{g}{\mu R}$

$\omega \ge \sqrt{\frac{g}{\mu R}}$

the minimum angular speed of rotation ${\omega }_{min}=\sqrt{\frac{g}{\mu R}}$

$\mu =0.15,r=3m,g=9.8{ms}^{-2}$

${\omega }_{min}=\sqrt{\frac{9.8}{0.15\times 13}}rad{s}^{-1}$

$4.67=5rad{s}^{-1}$