Rod pivoted at O, spring compressed 25 mm

Data and choice of method

Energy method (work-energy): the rod starts from rest and the spring and gravity do work; the pivot reaction does none.

$L = 1.5$ m, $m = 15$ kg, $k = 300$ kN/m $= 3\times10^5$ N/m, compression $x = 0.025$ m. Pivot $O$ is 0.3 m from $B$, so the centre $G$ of the rod is

$$d = \frac{L}{2} - 0.3 = 0.75 - 0.3 = 0.45\ \text{m from } O \text{ on the } A \text{ side}$$

  B ===O=====G========== A
     0.3   0.45
                       ^ spring

Moment of inertia about $O$:

$$I_O = \frac{1}{12}mL^2 + md^2 = \frac{1}{12}(15)(1.5)^2 + 15(0.45)^2 = 5.850\ \text{kg m}^2$$

Position 1 (horizontal, at rest) and position 2 (vertical)

The spring is released, pushing end $A$ up, so the rod swings with $A$ up and $G$ rises by $d$ in reaching the vertical. $T_1 = 0$.

• Elastic energy released by the spring: $V_e = \tfrac12 kx^2 = \tfrac12(3\times10^5)(0.025)^2 = 93.75$ J
• Gain in potential energy of the rod: $V_g = mgd = 15(9.81)(0.45) = 66.22$ J

$$T_1 + V_1 = T_2 + V_2 \;\Rightarrow\; 0 + V_e = \tfrac12 I_O\omega^2 + V_g$$ $$\tfrac12 I_O\omega^2 = 93.75 - 66.22 = 27.53\ \text{J}$$ $$\omega = \sqrt{\frac{2(27.53)}{5.850}} = 3.07\ \text{rad/s}$$

Answer: $\omega \approx 3.07$ rad/s (counter-clockwise, with $A$ moving upward) as the rod passes the vertical.