Question

A person with a mass $Mkg$ stands in contact with a wall of a cylindrical drum of radius $r$ rotating with an angular velocity $\omega$. If the coefficient of friction between the wall and the clothing is $\mu$, then 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. $\sqrt{\frac{g}{\mu r}}$
• B. $\sqrt{\frac{\mu r}{g}}$
• C. $\sqrt{\frac{2g}{\mu r}}$
• D. $\sqrt{\frac{rg}{\mu }}$

Answer 1

The correct option is A $\sqrt{\frac{g}{\mu r}}$

Due to rotation of the cylinder centripetal force $F_c=M{\omega }^{2}r$ acts on the person.
This is due to the normal reaction provided by the inner surface of the rotating cylinder.
Weight Mg acts downward.
Frictional force $F_{friction}=N\mu =M{\omega }^{2}r\mu$ acts upward.
when the person is in equilibrium, weight $Mg$ is balanced by $F_{friction}$
$\therefore Mg=M{\omega }^{2}r\mu$
$\Rightarrow {\omega }^2=\frac{g}{\mu r}$
$\omega =\sqrt{\frac{g}{\mu r}}$