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

Answers (1)
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

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