Question

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). The angle $\theta$ between the rod and the vertical is ${\cos }^{-1}\left(\frac{xg}{2{\omega }^{2}l}\right)$. Find the value of $x$.

Answer 1

Let us consider the system in a frame rotating with the rod. In this frame, the rod is at rest and experiences not only the gravitational force $m\overrightarrow{g}$ and the reaction force $\overrightarrow{R}$, but also the centrifugal force ${\overrightarrow{F}}_{cf}$.
In the considered frame, from the condition of equilibrium i.e. $N_{0z}=0$
or, $N_{cf}=mg\frac{l}{2}\sin \theta$ .....(1)
where, $N_{cf}$ is the moment of centrifugal force about $O$.
To calculate $N_{cf}$, let us consider an element of length $dx$, situated at a
distance $x$ from the point $O$. This element is subjected to a horizontal pseudo force
$\left(\frac{m}{l}\right)dx{\omega }^{2}x\sin \theta$.
The moment of this pseudo force about the axis of rotation through the point $O$ is
$d{N}_{cf}=\left(\frac{m}{l}\right)dx{\omega }^{2}x\sin \theta x\cos \theta$
$=\frac{m{\omega }^2}{l}\sin \theta \cos \theta x^{2}dx$
So, $N_{cf}={\int }_0^l\frac{m{\omega }^2}{l}\sin \theta \cos \theta x^{2}dx=\frac{m{\omega }^2{l}^2}{3}\sin \theta \cos \theta$ .....(2)
It follows from equations (1) and (2) that,
$\cos \theta =\left(\frac{3g}{2{\omega }^{2}l}\right)$
or, $\theta ={\cos }^{-1}\left(\frac{3g}{2{\omega }^{2}l}\right)$