# 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 value of $\omega ^{2}$ is equal to:Option 1)$\frac{\sqrt{3}g}{2r}$Option 2)$2g/(r\sqrt{3})$Option 3)$(g\sqrt{3})/r$Option 4)$2g/r$

S solutionqc

$N \sin \Theta =m\frac{r}{2}\omega ^{2}$

$N\cos \Theta =mg$

$\tan \Theta =\frac{r\omega ^{2}}{2g}\cdots \cdots (1)$

from geometry $\tan \Theta =\frac{r}{2.\frac{\sqrt{3}}{2}r}\cdots \cdots (2)$

equating (1) and (2)

$\omega ^{2}=\frac{2g}{\sqrt{3}r}$

Option 1)

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

Option 2)

$2g/(r\sqrt{3})$

Option 3)

$(g\sqrt{3})/r$

Option 4)

$2g/r$

Exams
Articles
Questions