A thin circular wire of radius `R` rotatites about its vertical diameter with an angular frequency `omega` . Show that a small bead on the wire remain at its lowermost point for `omegalesqrt(g//R)` . What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction.

Explanation

`Ncostheta=mg`

`Nsintheta=mromega^(2)`
`=m(Rsionomega)omega^(2)`
Dividing these two equation,
we get `costheta=(g)/(Romega^(2))`
:. `omega=sqrt((g)/(Rcostheta))`
At lowermost point, `theta=0^(@)`
`omega=sqrt((g)/(R))`
Substituting `omega=sqrt(2g//R)` in Eqs, `(i)` we have,
`costheta=(1)/(2)`
:. `theta=60^(@)`