Rod with spring released, 90 degree rotation

Assumed geometry

Rod $AC$ is 0.600 m long: $A$ is 0.480 m above the pivot $B$, $C$ is 0.120 m below it. The fixed support $D$ is at the level of $B$, 0.350 m horizontally from $B$. The rod turns through $90^\circ$ so that it becomes horizontal and $C$ moves towards $D$.

            A
            |   0.48 m
            |
   D .......B (pivot)
     <0.35>  |   0.12 m
            C

Work-energy principle: $T_1 + V_1 = T_2 + V_2$ with $T_1 = 0$.

Spring

$$\begin{aligned} \text{Position 1: } CD_1 &= \sqrt{0.350^2 + 0.120^2} = 0.3700\ \text{m}, & x_1 &= 0.3700 - 0.150 = 0.2200\ \text{m}\\ \text{Position 2: } CD_2 &= 0.350 - 0.120 = 0.2300\ \text{m}, & x_2 &= 0.2300 - 0.150 = 0.0800\ \text{m} \end{aligned}$$ $$V_{e1} = \tfrac12(400)(0.2200)^2 = 9.68\ \text{J}, \qquad V_{e2} = \tfrac12(400)(0.0800)^2 = 1.28\ \text{J}$$

Gravity

The centre $G$ of the rod is $0.300$ m from $A$, i.e. $0.480 - 0.300 = 0.180$ m above $B$ initially, and $0.180$ m beside $B$ (level with it) at the end. So $G$ falls $0.180$ m:

$$W_g = mgh = 4(9.81)(0.180) = 7.06\ \text{J}$$

Moment of inertia about B

$$I_B = \frac{1}{12}(4)(0.6)^2 + 4(0.18)^2 = 0.2496\ \text{kg m}^2$$

Energy equation

$$0 + V_{e1} + V_{g1} = \tfrac12 I_B\omega^2 + V_{e2} + V_{g2}, \quad V_{g1}-V_{g2} = W_g \;\Rightarrow\; \tfrac12I_B\omega^2 = 9.68 - 1.28 + 7.06 = 15.46\ \text{J}$$ $$\omega = \sqrt{\frac{2(15.46)}{0.2496}} = 11.13\ \text{rad/s}$$

Answer: $\omega \approx 11.13$ rad/s (counter-clockwise, towards the spring support).