Rod and spring, angular velocity at vertical

Impulse-momentum principle for a rigid body: the momenta of all particles of the body at time $t_1$, plus the impulses of the external forces acting between $t_1$ and $t_2$, equal the momenta at time $t_2$. The momenta of the particles reduce to a linear momentum vector $m\bar{\vec v}$ at the mass centre and an angular momentum $\bar I\omega$ about it:

$$\begin{aligned} m\bar{\vec v}_1 + \sum\int \vec F\,dt &= m\bar{\vec v}_2\\ \bar I\omega_1 + \sum\int M_G\,dt &= \bar I\omega_2 \end{aligned}$$

The moment equation may also be written about a fixed point $O$ (or about any point, with the appropriate momentum terms): $H_{O1} + \sum\int M_O\,dt = H_{O2}$. In scalar form the principle gives three equations (two for the linear part, one for the rotational part). It is the best method when forces are given as functions of time, or when a time interval (rather than a distance) is involved, as in impact problems.

 (momenta)1   +   (impulses)   =   (momenta)2

   m v1  .--.         F dt          m v2  .--.
   ---> (G) I w1  +   ------>  =    ---> (G) I w2
        '--'                             '--'

Numerical part

Data and choice of method

This part asks for a speed after a change of position, so the work-energy method is used: the rod starts from rest, the spring and gravity do work, and 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)

When the rod is released, the spring pushes end $A$ up, so the rod swings with $A$ moving up and $G$ rises by $d$ by the time the rod is 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.