A smooth wire of length 2 π r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed ω about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of ω 2 is equal to - (a) (√(3)g)/(2r) (b) (2g)/((r√(3))) (c) ((g√(3)))/(r) (d) 2g/r

Answer: (b). OPNmgr/2θ P is at distance r/2 from the axis, so the radius OP makes angle θ with the downward vertical where sin θ = 1/2, cos θ = √(3)/2. Normal reaction N acts along PO. Vertical: Ncos θ = mg. Horizontal: Nsin θ = mω² (rsin θ) ⇒ N = mω² r. mω² r · √(3)/2 = mg ⇒ ω² = (2g)/(√(3) r) So the answer is (b).

A smooth wire of length 2 π r is bent into a circle and kept in a… | JEE Main 2019 (12 Apr, Shift 2)

Q15JEE Main 2019 · 12 APR · SHIFT 2Difficult

Conical Pendulum & Funnel

A smooth wire of length 2 $\pi$ r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of $\omega$ 2 is equal to -

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Answer: (b)

P is at distance $\frac{r}{2}$ from the axis, so the radius OP makes angle $\theta$ with the downward vertical where $\sin\theta = \frac{1}{2}$ , $\cos\theta = \frac{\sqrt{3}}{2}$ .
Normal reaction N acts along PO. Vertical: $N\cos\theta = mg$ .
Horizontal: $N\sin\theta = m\omega^2 (r\sin\theta) \Rightarrow N = m\omega^2 r$ .
$m\omega^2 r \cdot \frac{\sqrt{3}}{2} = mg \Rightarrow \omega^2 = \frac{2g}{\sqrt{3}\,r}$
So the answer is (b).

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