Question

A thin circular wire of radius R rotates about its vertical diameter with an angular frequency $\omega .$ Show that a small bead on the wire remains at its lowermost point $\omega \le \sqrt{glR}.$ What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega =\sqrt{2glR}?$ Neglect friction.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Let the angle be $\theta$.

balancing forces in vertical direction, $Ncos\theta =mg-\left(1\right)$

balancing forces in horizontal direction, $Nsin\theta =m{\omega }^{2}Rsin\theta$

$⟹N=m{\omega }^{2}R=2mg$

substituting in (1),

$2mgcos\theta =mg⟹cos\theta =0.5⟹\theta ={60}^0$