Question

Question 5.40 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 ?Neglect friction.

Chapter Laws Of Motion Page 113

Answer 1

Hi______

hope u got all step easily

i can explain more last step ,
here
g/Rw^2=1
w^2=g/R
w=(g/R)^1/2
now us bionomial
w=2g/R............
we can neglect further term
hope it help u
if not msg me

Answer 2

see the figure :
Let the position of bead at P in such a way that radius joining with the bead to the centre of the wire makes ∅ angle with vertical.

Break in the components of Normal reaction ( N ) { see the figure }

in equilibrium condition:
mg = Ncos∅ __________(1)
mw²l = Nsin∅
where , w is angular frequency . and
l = Rsin∅ use it

mw²(Rsin∅) = Nsin∅
mw²R = N ____________(2)

from eqns (1) and (2)
mg = mw²Rsin∅
cos∅ = g/w²R
but we know, cos∅ ≤ 1
so, g/w²R ≤ 1
w ≤ √{g/R}
hence, a small bead on the wire remain at the lowermost point for w ≤ √{g/R}
hence, proved //
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angle made by radius vector joing the centre to the bead with vertical downward direction for w = √{2g/R} is ∅
cos∅ = g/w²R
= g/R{2g/R}
= 1/2
cos∅ = 1/2 = cos60°
∅ = 60°