A person stands in contact against the inner surface of a cylindrical drum of radius (R ) rotating with angular velocity `omega` . The coefficient of friction between the inner surface of the minumum rotational speed of the cylinder , which enables the person to ramain stuck to the wall , when the platform on which the person was standing is suddenly removed ?

Options

A. `sqrt((mu R)/(g))`
B. `sqrt((g)/(mu R))`
C. `sqrt((2g)/(mu R ))`
D. `sqrt((rg)/(mu))`

Answer

B

Explanation

The person will remain stuck to the inner surface of the rotating cylinder ,
If the frictional force (F ) ` ge ` weight (Mg ) of the person the centrifugi force `MR omega^(2)`
` =` The normal reaction (N )
` therefore F= mu N = MR omega ^(2)`
for equilibrium ,`mu R omega ^(2) = Mg `
` therefore omega_(min) ^(2) = (g)/( mu R ) " "therefore omega _(min ) = sqrt((g)/(mu R) )`