Mechanism of Mechanics.

Use conservation of energy: $\Delta K + \Delta U_g + \Delta U_s = 0.$ First, for the kinetic energy term, consider the rotation of the rod about its CM and the motion of the CM. The CM is located $d = 450/2 - 150 = 75$ mm to the left of the pivot. $K = K_{rot} + K_{tr} = \tfrac12I\omega^2 + \tfrac12mv_{CM}^2 \\ = \tfrac12m(\tfrac1{12}\ell^2 + \tfrac12d^2)\omega^2 = 0.03375\ \omega^2;$ The gravitational potential energy increases because the CM is raised vertically over a distance $d$ : $\Delta U_g = mgd = 2.25\ \text{J}.$ The spring was first stretched and ends up compressed. In the initial situation, the length of the spring is $\sqrt{150^2 + 200^2} = 250$ mm; in the final situation, $200 - 150 = 50$ mm. Thus $\Delta U_s = \tfrac12 k \Delta u^2 = 150\left((50-100)^2 - (250-100)^2\right) = -3.00\ \text{J}.$ Thus we have $0.03375\ (\omega'^2-\omega_0^2) + 2.25 - 3.00 = 0, \\ \omega'^2 = \omega_0^2 + \frac{0.75}{0.03375}, \\ \omega' = \sqrt{4^2 + 22.222} = \boxed{6.18241}\ \text{rad/s}.$