Question

A person stands in contact against the inner wall of a rotor of radius r. The coefficient of friction between the wall and the clothing is $\mu$ and the rotor is rotating about vertical axis. The minimum angular speed of the rotor so that the person does not slip downward is

• A. $\sqrt{\frac{\mu g}{r}}$
• B. $\sqrt{\frac{\mu r}{g}}$
• C. $\sqrt{\frac{g}{\mu r}}$
• D. $\sqrt{\mu rg}$

Answer 1

The correct option is D $\sqrt{\mu rg}$

Normal force= Centripetal force acting on the person.

$N=m{\omega }^{2}r$

Frictional force,$f=\mu N=\mu m{\omega }^{2}r$

Weight of the person,$w=mgf$

$f=W\mu m{\omega }^{2}r=mg$ or

$W=g\mu r$

Required linear speed, $V=rw=rg\mu r=gr\mu$

Thus,

the angular speed $\omega =g\mu r$ and

linear speed $V=\sqrt{g}r\mu$

Option $D$ is correct.