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 ω ≤ √g/R. What is
the angle made by the radius vector joining the centre to the bead
with the vertical downward direction for ω = √2g/R? Neglect
friction.
Solution:
Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.
Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle
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 or ω2 = 2g / R .....(vi)
On equating equations (v) and (vi), we get:
2g / R = g / RCos θ
Cos θ = 1 / 2
∴ θ = Cos-1(0.5) = 600
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