A thin circ

Let the radius vector joining the bead with the centre makes an angle Θ, with the vertical downward direction.

OP = R = Radius of the circle

N = Normal reaction

The respective Vertical and Horizontal equations of forces can be written as:

Mg = NCosΘ …(1)

m/ω² = Nsin Θ …(2)

In triangle OPQ, we have,

sinΘ = I/R

→ I = RSin Θ …(3)

Substituting equation (3) in equation (1) , we have,

m(R SinΘ )ω² = NsinΘ

→ mRω² = N …(4)

Substituting equation 4 in equation 1, we have,

mg = mRω²Cos Θ

CosΘ = g/ Rω² …(5)

Since, CosΘ ≤ 1, the bead will remain at the lowermost point,

for g/ Rω²≤ 1,

i.e. ω≤(g/R)^1/2

For ω = (2g/R)^1/2 …(6)

On equation equations 5 and 6, we have

2g/R = g/RcosΘ,

CosΘ = 1/2

Θ = Cos^-1(0.5)

→ Θ = 60°