A uniform rod of length '$$\ell$$' 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) $$\frac{m\ell^2}{12}\omega^2 \sin\theta$$ 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: