A conical pendulum, a thin uniform rod of length $L$ and mass $M$, rotates uniformly about a vertical axis with angular velocity $\omega$ (the upper end of the rod is hinged). Find the angle $\theta$ between the rod and the vertical.

$$\begin{gathered} \mathrm{Choose} \mathrm{an} \mathrm{element} \mathrm{of} \mathrm{the} \mathrm{rod} \mathrm{of} \mathrm{width} \mathrm{dx} \mathrm{at} \mathrm{a} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{the} \mathrm{hinge}. \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{element}, dm=\frac{m}{𝓁} dx. \mathrm{The} \mathrm{centrifugal} \mathrm{force} \mathrm{on} \mathrm{this} \mathrm{element} \\ dF = \left(dm\right) {\omega }^2 (x \sin \theta ) \end{gathered}$$

$$\begin{gathered} \mathrm{Its} \mathrm{moment} \mathrm{of} \mathrm{force} \mathrm{about} \mathrm{the} \mathrm{hinge} \\ d\tau = dF \times x \cos \theta \\ =\left(dm\right) {\omega }^2 \left(x \sin \theta \right) \left(x \cos \theta \right) \\ =\left(\frac{m}{\mathcal{l}}dx\right){\omega }^2{x}^2 \left(\frac{\sin 2\theta }{2}\right) \\ =\frac{m{\omega }^2}{2\mathcal{l}} \sin 2 \theta x^2 dx ....\left(\mathrm{i}\right) \\ \mathrm{For} \mathrm{the} \mathrm{moment} \mathrm{of} \mathrm{force} \mathrm{of} \mathrm{whole} \mathrm{length} \mathrm{of} \mathrm{rod}, \mathrm{integrating} \left(\mathrm{i}\right) \\ \mathrm{\tau }=\frac{{\mathrm{m\omega }}^2}{2\mathcal{l}} \sin 2 \mathrm{\theta } {\int }_0^{\mathcal{l}}{\mathrm{x}}^2 \mathrm{dx} \\ =\frac{{\mathrm{m\omega }}^2{\mathcal{l}}^2}{6} \sin 2\mathrm{\theta } ....\left(\mathrm{ii}\right) \end{gathered}$$

In the rotating frame, apart from other forces the centrifugal force also act. For rotational equilibrium of the rod, we have $\Sigma \overrightarrow{\tau }=0$ Taking moment of all forces about hinge and putting their algebraic sum zero, we get

$$\begin{gathered} mg\frac{l}{2} \sin \theta = \frac{m{\omega }^2{\mathcal{l}}^2}{6}\sin 2\theta \\ or \cos \theta = \frac{3 g}{2{\omega }^2\mathcal{l}} \end{gathered}$$