$A$ 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 $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

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