question_answer

A person with a mass of M kg, stands in contact against the wall of a cylindrical drum of radius r rotating with an angular velocity co. If the coefficient of friction between the wall and the clothing is  then minimum rotational speed of  the cylinder which enables the person to remain stuck to the wall when the floor is suddenly removed is

A)  \[{{\omega }_{\min }}=\sqrt{\frac{g}{\mu r}}\]  

B)         \[{{\omega }_{\min }}=\sqrt{\frac{\mu r}{g}}\]

C)  \[{{\omega }_{\min }}=\sqrt{\frac{2g}{\mu r}}\]              

D)         \[{{\omega }_{\min }}=\sqrt{\frac{gr}{\mu }}\]

Correct Answer: A

Solution :

 We known,             \[\mu \times r=mg\] \[\therefore \]      \[\mu \times m\omega 2r=mg\] \[\Rightarrow \]   \[\omega =\sqrt{\frac{g}{ur}}\]