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

Explanation

In Figure we have shown that radius vector joining the bead to the centre of the wire makes an angle `theta` with the downword direction . If N is normal reaction , then as is clear from the figure,
`"mg"=Ncostheta.....(i)`
m r `omega^(2)=Nsintheta.......(ii)`
or m `(Rsintheta)omega^(2)=Nsintheta" or m R "omega^(2)=N`
from (i) , mg = m R `omega^(2)costheta " or " costheta= (g)/(Romega^(2))........(iii)`

As `|costheta|lel` , therefore , bead will remain at its
lowermost point for `(g)/(Romega^(2))lel, " or " omegalesqrt((g)/(R))`
When `omega=sqrt((2g)/R)," from (iii)",costheta=(g)/(R)((R)/(2g))=(1)/(2)`
`therefore theta = 60^(@)`