Question

A person with a mass $M$ $kg$ 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 $\mu$, then the minimum rotational speed of the cylinder which enables the person to remain struck to the wall when the floor is suddenly removed will be

• 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. ${\omega }_{min}=\sqrt{\frac{rg}{\mu }}$

Answer 1

Answer: (A)

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

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, $\mu Mr{\omega }^2\ge Mg\Rightarrow {\omega }^2\ge \frac{q}{\mu r}$
ie $\omega \ge \sqrt{\frac{q}{\mu r}}$
so, ${\omega }_{min}=\sqrt{\frac{q}{\mu r}}$