A smooth wire of length 2π 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 ω2 is equal to :

Question

A smooth wire of length 2π 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 ω2 is equal to : (A) (g√3)/r (B) (√3g/2r) (C) 2g/r (D) 2g/(r √3)

Tardigrade Admin · Accepted Answer

N sin θ = m (r/2) ω2 ...(1) N cos θ = mg ...(2) tan θ = (r ω2/2g) (r/2 (√3r)2) = (r ω2/2g) ω2 = (2g/√3r)

$(g 3) / r$

$\frac{3 g}{2 r}$

$2 g / r$

$2 g / (r 3)$

$N s in θ = m \frac{r}{2} ω^{2} ... (1)$
$N cos θ = m g ... (2)$
$t an θ = \frac{r ω ^{2}}{2 g}$
$\frac{r}{2 \frac{3 r}{2}} = \frac{r ω ^{2}}{2 g}$
$ω^{2} = \frac{2 g}{3 r}$

(g3​)/r

2r3​g​

2g/r

2g/(r3​)

Nsinθ=m2r​ω2...(1) Ncosθ=mg...(2) tanθ=2grω2​ 223​r​r​=2grω2​ ω2=3​r2g​

$(g 3) / r$

$\frac{3 g}{2 r}$

$2 g / r$

$2 g / (r 3)$

$N s in θ = m \frac{r}{2} ω^{2} ... (1)$
$N cos θ = m g ... (2)$
$t an θ = \frac{r ω ^{2}}{2 g}$
$\frac{r}{2 \frac{3 r}{2}} = \frac{r ω ^{2}}{2 g}$
$ω^{2} = \frac{2 g}{3 r}$