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 $$ω ≤\dfrac{\sqrt{g}}{R}$$ . What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for $$ω =\dfrac{\sqrt{2g}}{R}$$ ? Neglect friction.

Asked by Sudhanshu | 1 year ago |  208

##### Solution :-

Let θ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction

mg = N cosθ —–(1)

mrω2 = Nsinθ —– (2)

(or) m (Rsinθ) ω= Nsinθ

mRω= N

Substituting the value of N in (1)

mg = mRω2cosθ

(or) cos θ = $$\dfrac{g}{Rω^2}$$ ———(3)

As l cosθ I ≤ 1 the bead will remain at the lowermost point

$$\dfrac{g}{Rω^2}$$≤ 1 or ω ≤ $$\dfrac{\sqrt{g}}{R}$$

For ω = $$\dfrac{\sqrt{2g}}{R}$$ , equation (3) becomes

cos θ = $$\dfrac{g}{Rω^2}$$

cos θ = ($$\dfrac{g}{R}$$) ($$\dfrac{R}{2g}$$) = $$\dfrac{1}{2}$$

θ = 60°

Answered by Sudhanshu | 1 year ago

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