A small bead of mass `m` is carried by a circular hoop having center at `O` and radius `sqrt 2m` which rotates about a fixed vertical axis. The coefficient of friction between beed and hoop is `mu=0.5`. The maximum angular speed of the hoop for which the bead does not have relative motion with respect to hoop. `(g=10m//s^(2))`

Options

A. `sqrt5`
B. ` sqrt10`
C. `sqrt15`
D. `sqrt30`

Answer

D

Explanation

The maximum angular speed of the hoop corresponds to the situations when the bead is just about to slide upwards . For the bead not to slide upwards .

cos `45^(@) = (x)/(r) implies x = "r cos" 45^(@)`
`m omega^(2) ("r sin" 45^(@) + m omega^(2) ( "r sin" 45^(@) ) "sin" 45^(@)`
On solving , `omega le sqrt((1 + mu ) g)/((1 - mu) " r sin" 45^(@))`