Question

A thin horizontal uniform rod $AB$ of mass $M$ and length $l$ can rotate freely about a vertical axis passing through its end $A$. At a certain moment the end $B$ starts experiencing a constant force $F$ which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane. Find the angular velocity of the rod as a function of its rotation angle $\phi$ counted relative to the initial position.

• A. $\omega =\sqrt{\frac{6F\sin \phi }{Ml}}$
• B. $\omega =\sqrt{\frac{3F\sin \phi }{Ml}}$
• C. $\omega =\sqrt{\frac{6F\cos \phi }{Ml}}$
• D. None of these

Answer 1

The correct option is A $\omega =\sqrt{\frac{6F\sin \phi }{Ml}}$

Let the rod be deviated through an angle $\phi$ from its initial position at an arbitrary instant of time, measured relative to the initial position in the positive direction. From the equation of the increment of the mechanical energy of the system.
$\Delta T=A_{ext}$
or, $\frac{1}{2}I{\omega }^2=\int N_{z}d\phi$
or, $\frac{1}{2}\frac{M{l}^2}{3}{\omega }^2={\int }_0^{\phi }Fl\cos \phi d\phi =Fl\sin \phi$
Thus, $\omega =\sqrt{\frac{6F\sin \phi }{Ml}}$