Question

A bead 'B' is threaded on a smooth circular wire frame of radius $r$. The radius at a certain position of the bead makes an angle $\theta$ with the vertical axis. The entire frame is rotated with an angular speed $\omega$ about the vertical axis. The distance of the bead from the vertical axis is $R$. If the bead is in equilibrium (with respect to the rotating frame), then identify the correct statements:

• A. $\omega =\sqrt{\frac{g\tan \theta }{R}}$
• B. $R=r\sin \theta$
• C. $N\sin \theta =m{\omega }^2\text{R}$
• D. $\omega =\sqrt{\frac{r}{g\cos \theta }}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\omega =\sqrt{\frac{g\tan \theta }{R}}$
B $R=r\sin \theta$
C $N\sin \theta =m{\omega }^2\text{R}$


The bead is in equilibrium with respect to the rotating wire frame, which is rotating about a vertical axis passing through its centre.
Hence a centrifugal force $F_{\text{centrifugal}}=m{\omega }^{2}r$ will appear to act in the outward direction, when observed from the rotating frame.

Applying the condition of equilibrium for bead as per FBD:
$N\sin \theta =m{\omega }^{2}R$ .....$\left(1\right)$
$N\cos \theta =mg$ .....$\left(2\right)$

Distance of bead from vertical axis
$R=r\sin \theta ............\left(3\right)$

Dividing $\left(1\right)$ by $\left(2\right)$:
$\frac{\sin \theta }{\cos \theta }=\frac{{\omega }^{2}R}{g}$
$\therefore \omega =\sqrt{\frac{g\tan \theta }{R}}$

Option a,b,c are correct.