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Laws of Motion

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Question : 40 of 40
Marks: +1, -0
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω\omega. Show that a small bead on the wire loop remains at its lowermost point for ω≤gR\omega \leq \sqrt{\frac{g}{R}} . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω=2gR\omega = \sqrt{\frac{2g}{R}} ? Neglect friction.
Solution:  
We have shown that radius vector joining the bead to the centre of the wire makes an angle θ with the vertical downward direction.
If N is normal reaction, then as is clear from the figure,
mg=N=cos⁡θ…(i)mg = N = \cos\theta \ldots (i)
mrω2=Nsin⁡θ…(ii)mr\omega^2 = N\sin\theta \ldots (ii)
or m(Rsin⁡θ)ω2=Nsin⁡θm(R\sin\theta)\omega^{2}=N\sin\theta or mRω2=NmR\omega^{2}=N
from (i),mg=mRω2cos⁡θ(i), mg = mR\omega^{2}\cos\theta
or cos⁡θ−gRω2\cos\theta - \frac{g}{R\omega^2}...(iii)
As ∣cos⁡θ∣≤1|\cos\theta| \leq 1, therefore, bead will remain at its lowermost point for
gRω2≤1,\frac{g}{R\omega^{2}} \leq 1, or ω≤gR\omega \leq \sqrt{\frac{g}{R}}
When ω=2gR,\omega = \sqrt{\frac{2g}{R}}, from (iii)
cos⁡θ=gR(R2g)=12\cos\theta = \frac{g}{R} \left(\frac{R}{2g}\right) = \frac{1}{2}
∴θ=60∘\therefore \theta = 60^{\circ}
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