Answer :

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°


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