Question

A person stands in contact against the wall of a cylindrical drum of radius $R$ rotating with an angular velocity $\omega$ . If the coefficient of friction between the wall and the person is $\mu$ , the minimum rotational speed of the cylinder which enables the person to remain stuck to the wall when the floor is suddenly removed is

• A. ${\omega }_{min}=\sqrt{\frac{g}{\mu R}}$
• B. ${\omega }_{min}=\sqrt{\frac{\mu R}{g}}$
• C. ${\omega }_{min}=\sqrt{\frac{2g}{\mu R}}$
• D. ${\omega }_{min}=\sqrt{\frac{gR}{\mu }}$

Answer 1

The correct option is A ${\omega }_{min}=\sqrt{\frac{g}{\mu R}}$

The person will remain stuck to the wall if the limiting static frictional force acting between floor and the person
must balance its own weight.


Normal reaction provides the necessary centripetal force to the person.

$\Rightarrow N=M{\omega }^{2}R$

${\left(f_s\right)}_{max}=\mu N=\mu M{\omega }^{2}R$

or, applying condition of vertical equilibrium

${\left(f_s\right)}_{max}=Mg$

$\Rightarrow \mu M{\omega }^{2}R=Mg$

When friction acting is maximum, that will correspond to minimum angular velocity of drum such that slipping is prevented.

$\Rightarrow {\omega }_{min}^2=\frac{g}{\mu R}$

$\therefore {\omega }_{min}=\sqrt{\frac{g}{\mu R}}$