Question

A person with a mass of $Mkg$, 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 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. $\omega =\sqrt{\frac{g}{\mu r}}$
• B. $\omega =\sqrt{\frac{\mu r}{g}}$
• C. $\omega =\sqrt{\frac{2g}{\mu r}}$
• D. $\omega =\sqrt{\frac{gr}{\mu }}$

Answer 1

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

Given drum radius $r,$ angular velocity $=\omega ,$ coefficient of friction $=\mu$, mass of person is $M$

and person is in contact with inner wall of rotor.

From the Free Body Diagram:

The normal reaction provides the necessary centripetal fornce$\Rightarrow F_C-F_N=0..........\left(1\right)$

Centrifugal Force $F_C=m{\omega }^{2}r......\left(2\right)$

$\Rightarrow F_N=m{\omega }^{2}R$

Friction force $f$ is given by

$\Rightarrow f=\mu F_N$

Person remains stuck to the wall even when the floor is removed downward

weight of the person will be balance the friction force $f$ upwards

$\Rightarrow f\ge Mg$

$\Rightarrow \mu F_N\ge mg\Rightarrow \mu m{\omega }^{2}R\ge mg$

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

least value of angular speed

$\Rightarrow \omega =\sqrt{\frac{g}{R\mu }}$