Question

A uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$ the rod makes an angle $\theta$ with it (see figure). To find $\theta$, equate the rate of change of angular momentum (direction going into the paper) $\frac{m{l}^2}{12}{\omega }^2\sin \theta \cos \theta$ about the centre of mass $\left(CM\right)$ to the torque provided by the horizontal and vertical forces $F_H\text{and}{F}_V$ about the $CM$. The value of $\theta$ is then such that:

• A. $\cos \theta =\frac{2g}{3l{\omega }^2}$
• B. $\cos \theta =\frac{g}{2l{\omega }^2}$
• C. $\cos \theta =\frac{g}{l{\omega }^2}$
• D. $\cos \theta =\frac{3g}{2l{\omega }^2}$

Answer 1

The correct option is D $\cos \theta =\frac{3g}{2l{\omega }^2}$

$\text{Vertical force,}{F}_V=mg$

$\text{Horizontal force = Centripetal force}$

$\Rightarrow F_H=m{\omega }^2\frac{l}{2}\sin \theta$

Torque due to vertical force
${\tau }_V=mg\frac{l}{2}\sin \theta$

Torque due to horizontal force
${\tau }_H=m{\omega }^2\frac{l}{2}\sin \theta \frac{l}{2}\cos \theta$

${\tau }_{net}=\frac{dL}{dt}$

$\Rightarrow mg\frac{l}{2}\sin \theta -m{\omega }^2\frac{l}{2}\sin \theta \frac{l}{2}\cos \theta =\frac{m{l}^2}{12}{\omega }^2\sin \theta \cos \theta$

$\Rightarrow \cos \theta =\frac{3}{2}\frac{g}{{\omega }^{2}l}$

Hence, option $\left(D\right)$ is correct.