Question:
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 -
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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P is at horizontal distance \(r/2\) from the axis, so OP makes \(30^\circ\) with the vertical. Resolve the normal force: its vertical part balances weight, its horizontal part gives the centripetal force.
Updated On: Nov 11, 2026
- \(\frac{\sqrt{3}\,g}{2r}\)
- \(\frac{2g}{r\sqrt{3}}\)
- \(\frac{g\sqrt{3}}{r}\)
- \(\frac{2g}{r}\)
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The Correct Option is B
Solution and Explanation
The bead is at horizontal distance \(\frac{r}{2}\) from the axis, so with \(\phi\) the angle between OP and the vertical, \(\sin\phi = \frac{r/2}{r} = \frac{1}{2}\), \(\phi = 30^\circ\). The radius of its circular path is \(\frac{r}{2}\).
Forces: weight mg down and the normal force N along PO (towards the centre O).
Vertical: \(N\cos30^\circ = mg\).
Horizontal (centripetal): \(N\sin30^\circ = m\omega^2\frac{r}{2}\).
Dividing: \(\tan30^\circ = \frac{\omega^2 r}{2g}\), so \(\omega^2 = \frac{2g}{r}\tan30^\circ = \frac{2g}{\sqrt3\,r}\).
Answer: (2) \(\frac{2g}{r\sqrt{3}}\)
Forces: weight mg down and the normal force N along PO (towards the centre O).
Vertical: \(N\cos30^\circ = mg\).
Horizontal (centripetal): \(N\sin30^\circ = m\omega^2\frac{r}{2}\).
Dividing: \(\tan30^\circ = \frac{\omega^2 r}{2g}\), so \(\omega^2 = \frac{2g}{r}\tan30^\circ = \frac{2g}{\sqrt3\,r}\).
Answer: (2) \(\frac{2g}{r\sqrt{3}}\)
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