A uniform rod of mass 1.2 kg and length 0.90 m hangs from a smooth pivot at its upper end…

Step 1. The torques. The weight gives $Mg\dfrac{L}{2}\sin\theta \approx 5.40\ \text{N m} \times \theta$. The spring is stretched by $L\theta$, pulls with $kL\theta$ and acts at $L$: $kL^2\theta = 48.60\ \text{N m} \times \theta$. In all $C = 54.00\ \text{N m}$ per radian.

Step 2. Period (a). $I = \dfrac{ML^2}{3} = 0.324\ \text{kg m}^2$, so $\omega^2 = \dfrac{C}{I} = \dfrac{54.00\ \text{N m}}{0.324\ \text{kg m}^2} = 166.7\ \text{s}^{-2}$, $\omega = 12.910\ \text{rad/s}$ and

Step 3. Speed of the end (b). The rod's angular speed is greatest as it passes the vertical: $\Omega = \omega\theta_0 = 0.6455\ \text{rad/s}$. The lower end moves at $\Omega L = 0.581\ \text{m/s}$.

Step 4. Equal simple pendulum (c). It needs $g/\ell = \omega^2$, so $\ell = \dfrac{10\ \text{m/s}^2}{166.7\ \text{s}^{-2}} = 6.00\ \text{cm}$.

Step 5. Spring force (d). The largest stretch is at release, $L\theta_0 = 0.0450\ \text{m}$, so $F = 60\ \text{N/m} \times 0.0450\ \text{m} = 2.70\ \text{N}$.