Question

A person with a mass m stands in contact against a rough wall of cylindrical drum
(rotor) of radius R without floor. The rotor is rotating about its axis with angular
speed w rad/s. What is minimum coefficient of friction required such that person
does not slip down?
​

Answer 1

Given:

A person with a mass m stands in contact against a rough wall of cylindrical drum (rotor) of radius R without floor. The rotor is rotating about its axis with angular speed $\omega$ rad/s.

To find:

Minimum coefficient of friction needed?

Calculation:

First of all, look at the diagram !

For equilibrium of the person, the frictional force should be equal to the weight of the person:

$\therefore \: f = mg$

$\implies \: \mu N = mg$

• Here normal reaction will be provided by centrifugal force.

$\implies \: \mu ( \dfrac{m {v}^{2} }{R} )= mg$

$\implies \: \mu \bigg \{\dfrac{m {( \omega R)}^{2} }{R} \bigg \}= mg$

$\implies \: \mu {\omega}^{2} R = g$

$\implies \: \mu = \dfrac{g}{ {\omega}^{2} R}$

So, minimum coefficient of friction needed is:

$\boxed{ \bf\: \mu = \dfrac{g}{ {\omega}^{2} R} }$