Question

A person with a mass of $M\text{kg}$ 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 clothing 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. ${\omega }_{\text{min}}=\sqrt{\frac{g}{\mu r}}$
• B. ${\omega }_{\text{min}}=\sqrt{\frac{\mu r}{g}}$
• C. ${\omega }_{\text{min}}=\sqrt{\frac{2g}{\mu r}}$
• D. ${\omega }_{\text{min}}=\sqrt{\frac{rg}{\mu }}$

Answer 1

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

The person will remain stuck to the wall if the static force of friction $\mu Mr{\omega }^2$ is equal to or more than the weight $Mg$.
Hence $\mu Mr{\omega }^2\ge Mg$
$\Rightarrow {\omega }^2\ge \frac{g}{\mu r}$
i.e., $\omega \ge \sqrt{\frac{g}{\mu r}}$,
$\therefore {\omega }_{\text{min}}=\sqrt{\frac{g}{\mu r}}$
Hence, the correct answer is option (a).