A small bead of mass m = 1 kg is free to move on a circular hoop. The circular hoop has centre at C and radius r = 1 m and it rotates about a fixed vertical axis. The coefficient of friction between bead 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, at the position shown in figure is : (Take `g = 10 m//s^(2)`)

Options

A. `(5sqrt(2))^(1//2)`
B. `(10sqrt(2))^(1//2)`
C. `(15sqrt(2))^(1//2)`
D. `(30sqrt(2))^(1//2)`

Explanation

The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards. The free body diagram of the bead is

For the bead not to slide upwards.
`m omega^(2)(r sin 454^(@))cos45^(@)- mg sin45^(@) lt mu N.....(1)`
where `N=mg cos 45^(@)+ m omega^(2)(r sin 45^(@)) sin 45^(@)......(2)`
From 1 and 2 we get.
`omega=sqrt(30sqrt(2)) rad//s`