The Total solution for NCERT class 6-12
A thin circular loop of radius R rotatesabout its vertical diameter with an angular frequency ω. Show thata small bead on the wire loop remains at its lowermost point for.What is the angle made by theradius vector joining the centre to the bead with the vertical downwarddirection for ?Neglect friction.
Let the radius vector joining the bead withthe centre make an angle θ, with the vertical downward direction.
OP = R = Radius of thecircle
N = Normal reaction
Therespective vertical and horizontal equations of forces can be written as:
mg = Ncosθ … (i)
mlω2 = Nsinθ … (ii)
InΔOPQ, we have:
l = Rsinθ … (iii)
Substituting equation (iii) inequation (ii), we get:
m(Rsinθ) ω2 = Nsinθ
mR ω2 = N …(iv)
Substituting equation (iv) inequation (i), we get:
mg = mR ω2 cosθ
… (v)
For or
On equating equations (v) and (vi),we get: