Q3 : A thin circular loop of radius R  rotates about its vertical diameter with an angular frequency $\omega \le \sqrt{g/R.}$ What is the angle made by radius vector joining the centre to the bead with the vertical download direction for $\omega =\sqrt{2g/R?}$ Neglect friction.  

 

Dear student

Let O be the centre of the circular loop and $\theta$ is the angle made by radius vector joining the centre to bead with vertical.

At P the normal reaction of the loop is along PO and its resolved components are
$$\begin{gathered} N\cos \theta =mg \\ N\sin \theta =mr{\omega }^2 \end{gathered}$$

where $mr{\omega }^2$ is the centrifugal force on the bead and r=PR=$R\sin \theta$ where R is the radius of the circle so
$$\begin{gathered} N\cos \theta =mg \\ N\sin \theta =mR\sin \theta ·{\omega }^2 \\ on dividing we get \\ \cos \theta =\frac{g}{R{\omega }^2} \end{gathered}$$
at lowest point $\theta$=0^o so $\omega =\sqrt{\frac{g}{R}}$ and thus if $\omega$ becomes higher this value bead rises. so bead remains at lowest position if $\omega \le \sqrt{\frac{g}{R}}$.
angle when $\omega =\sqrt{\frac{2g}{R}}$
$$\begin{gathered} \cos \theta =\frac{g}{R}·\frac{R}{2g} \\ \cos \theta =\frac{1}{2} \\ \Rightarrow \theta ={60}^{°} \end{gathered}$$