A thin circular wire of radius `R` rotates 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 center to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction.

Explanation

Let the radius vector joining the bead to the centre of the wire make an angle `theta` with the vertical downward direction. If N is normal reaction . Then from fig. ltbRgt mg = Ncos `theta` … (i)
mr`omega^(2)` = N sin`theta` … (ii)
or m(R sin`theta` )`omega^(2)` = N sin`theta`
or mR`omega^(2)` = N
From equation (i), mg = mR`omega^(2)`cos`theta`
or cos`theta` = g/ R`omega^(2)` ... (iii)
As `` `le` 1 , therefore, bead will remain at its lowermost point for
g/R`omega^(2)` `le` 1 or `omega` `le` `sqrt g/R`
When `omega` = `sqrt 2g/R` from equation (iii),
cos`theta` = g/R(R/2g) = 1/2
`therefore` `theta` = `60^(@)`.