To solve this problem, we need to calculate the angular speed of the system after the collision. The system involves two rods, with one initially horizontal and falling under gravity.

Key steps:

1. Moment of inertia (I): Each rod is thin and uniform with length $l$ and mass $m = 4.24 \, \text{kg}$. The moment of inertia of the horizontal rod about the pivot is: $I_{\text{horizontal}} = \frac{1}{3} m l^2$

   Similarly, the moment of inertia of the vertical rod (after collision) is the same: $I_{\text{vertical}} = \frac{1}{3} m l^2$

   Combined, the total moment of inertia after collision becomes: $I_{\text{total}} = 2 \times \frac{1}{3} m l^2 = \frac{2}{3} m l^2$

2. Initial energy of the horizontal rod: The horizontal rod starts from rest and rotates under gravity. Its center of mass falls a distance of $\frac{l}{2}$. The initial gravitational potential energy is: $U = m g \frac{l}{2}$ This potential energy is entirely converted to rotational kinetic energy just before collision: $\frac{1}{2} I_{\text{horizontal}} \omega_{\text{initial}}^2 = m g \frac{l}{2}$

   Substituting $I_{\text{horizontal}} = \frac{1}{3} m l^2$: $\frac{1}{2} \left(\frac{1}{3} m l^2\right) \omega_{\text{initial}}^2 = m g \frac{l}{2}$ Solving for $\omega_{\text{initial}}$: $\omega_{\text{initial}} = \sqrt{\frac{3 g}{l}}$

3. Angular momentum conservation during collision: The collision is perfectly inelastic, so angular momentum is conserved. Before the collision, only the horizontal rod contributes angular momentum. After the collision, both rods rotate together.

   Angular momentum before collision: $L_{\text{initial}} = I_{\text{horizontal}} \omega_{\text{initial}}$

   Angular momentum after collision: $L_{\text{final}} = I_{\text{total}} \omega_{\text{final}}$

   Using conservation of angular momentum: $I_{\text{horizontal}} \omega_{\text{initial}} = I_{\text{total}} \omega_{\text{final}}$ Substituting $I_{\text{horizontal}} = \frac{1}{3} m l^2$ and $I_{\text{total}} = \frac{2}{3} m l^2$: $\frac{1}{3} m l^2 \cdot \sqrt{\frac{3 g}{l}} = \frac{2}{3} m l^2 \cdot \omega_{\text{final}}$

   Simplifying: $\omega_{\text{final}} = \frac{\sqrt{\frac{3 g}{l}}}{2}$

4. Substitute values: Given $g = 9.8 \, \text{m/s}^2$, $l = 0.129 \, \text{m}$: $\omega_{\text{final}} = \frac{\sqrt{\frac{3 \cdot 9.8}{0.129}}}{2}$

   Compute: $\sqrt{\frac{3 \cdot 9.8}{0.129}} = \sqrt{227.9} \approx 15.1$ $\omega_{\text{final}} = \frac{15.1}{2} \approx 7.55 \, \text{rad/s}$

Final Answer:

The angular speed of the rods right after the collision is 7.55 rad/s.

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