A thin circular loop of radius …

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ∆. Show that a small bead on the wire loop remains at its lowermost point for ∆ ≤ √(g/R).What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ∆ = √(2g/R)? {∆ is 'omega'} Solution - (https://i.stack.imgur.com/YuoQ6.jpg) I want to ask, how the solution is showing that for ∆ ≤ √(g/R) bead will remain at lower most point. The inequality is derived by assuming ∅(theta) to be angle with the vertical and there is nothing stating ∅ to be 0 as only then bead will be at lower most point. My another question is that, how the bead will rise to higher position after increasing ∆ to √(2g/R)? I think question is asking about the highest angle, bead will remain(i.e., bead will not come down) at this ∆.

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