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

Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

             
  N = Normal reaction
The respective vertical and horizontal equations of forces can be written as, 
mg = N Cosθ                         ... (i) 

mlω² = N Sinθ                       … (ii)
In ΔOPQ, we have
Sin θ = l / R 
l = R Sinθ                              … (iii)

Substituting equation (iii) in equation (ii), we get
 
m(R Sinθ) ω² = N Sinθ 

mR ω² = N                            ... (iv) 

Substituting equation (iv) in equation (i), we get
mg = mR ω² Cosθ 

Cosθ = g / Rω²                        ...(v) 

Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω² ≤ 1.
i.e., for                      ω ≤ (g / R)^1/2  
For ω = (2g / R)^1/2 
      ω² = 2g / R                  ...(vi) 
On equating equations (v) and (vi), we get
   = g / RCos θ  
 Cos θ = 1 / 2 
∴ θ = Cos^-1(0.5)  =  60⁰ , is the angle made by the radius vector joining the centre to the bead with the vertical downward direction.