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:

$$F_v = mg$$

$$F_H = m a_{\text{cm}} = m \omega^2 \left(\frac{l}{2}\sin\theta\right)$$

Torque about the centre of mass due to forces acting at the pivot:

$$\tau_{\text{cm}} = F_H \cdot \left(\frac{l}{2}\cos\theta\right) - F_v \cdot \left(\frac{l}{2}\sin\theta\right)$$

$$\tau_{\text{cm}} = \left(m \omega^2 \frac{l}{2}\sin\theta\right)\left(\frac{l}{2}\cos\theta\right) - (mg)\left(\frac{l}{2}\sin\theta\right)$$

Equating to the given rate of change of angular momentum (taking magnitudes in direction):

$$\frac{m l^2}{12} \omega^2 \sin\theta \cos\theta = (mg)\left(\frac{l}{2}\sin\theta\right) - \left(m \omega^2 \frac{l}{2}\sin\theta\right)\left(\frac{l}{2}\cos\theta\right)$$

$$\frac{ml^2}{12}\omega^2 \sin\theta \cos\theta + \frac{ml^2}{4}\omega^2 \sin\theta \cos\theta = \frac{mgl}{2}\sin\theta$$

$$\frac{ml^2}{3}\omega^2 \sin\theta \cos\theta = \frac{mgl}{2}\sin\theta$$

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