Rod and spring, angular velocity and reaction

Data and assumptions

$L = 2.5$ m, $m=15$ kg, $k = 300$ kN/m, compression $x = 1$ in $= 0.0254$ m. Pivot $O$ is 0.5 m from end $B$ (taken as the end resting on the spring), so the centre $G$ is $d = 1.25 - 0.5 = 0.75$ m from $O$ on the free (long) side.

  spring               O                  G
  B ====================o================== A
   |<---- 0.5 m ---->|<------ 2.0 m ------>|

The spring pushes the short end up, so the long side swings down and $G$ falls by $d$ on reaching the vertical. (If the spring pressed the long end instead, the spring energy would be only 96.8 J, less than the 110.4 J needed to lift $G$ through 0.75 m, and the rod could not reach the vertical.)

$$I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(2.5)^2 + 15(0.75)^2 = 16.250\ \text{kg m}^2$$

Angular velocity (work-energy)

$$V_e = \tfrac12kx^2 = \tfrac12(3\times10^5)(0.0254)^2 = 96.77\ \text{J}, \qquad W_g = mgd = 15(9.81)(0.75) = 110.36\ \text{J}$$ $$T_1 + U_{1\to2} = T_2:\quad 0 + 96.77 + 110.36 = \tfrac12 I_O\omega^2$$ $$\omega = \sqrt{\frac{2(207.14)}{16.250}} = 5.05\ \text{rad/s}$$

Reaction at the pivot in the vertical position

At the vertical, $G$ is directly below $O$. The weight passes through $O$, so $\sum M_O = 0$ and $\alpha = 0$: there is no tangential acceleration and no horizontal reaction. $G$ has only the normal (centripetal) acceleration $\bar a_n = \omega^2 d$ directed up towards $O$:

$$\bar a_n = \omega^2 d = (5.05)^2(0.75) = 19.12\ \text{m/s}^2$$ $$\sum F_y = m\bar a_n:\quad R - mg = m\omega^2 d$$ $$R = 15(9.81 + 19.12) = 434\ \text{N}$$

Answer: $\omega \approx 5.05$ rad/s; reaction at $O$ is vertical, $R \approx 434$ N upward (horizontal component zero).