Q3 : A thin circular loop of radius R  rotates about its vertical diameter with an angular frequency ω g / R .  What is the angle made by radius vector joining the centre to the bead with the vertical download direction for ω = 2 g / R ?  Neglect friction.  

 

Dear student

Let O be the centre of the circular loop and θ is the angle made by radius vector joining the centre to bead with vertical.

At P the normal reaction of the loop is along PO and its resolved components are
Ncosθ=mg Nsinθ=mrω2

where mrω2 is the centrifugal force on the bead and r=PR=Rsinθ where R is the radius of the circle so
Ncosθ=mgNsinθ=mRsinθ·ω2on dividing we getcosθ=gRω2
at lowest point θ=0o so ω=gR and thus if ω becomes higher this value bead rises. so bead remains at lowest position if ωgR.
angle when ω=2gR
cosθ=gR·R2gcosθ=12θ=60°

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