For the above question I have proceeded in following way:
However I am unable to proceed and am unable to use $\theta$ in any further equation. I think I am getting the concept wrong somewhere.
For the second part I have proceeded in the following way: Is the approach correct?
For the kinetic energy calculation. Calling
$$ \cases{ p_g = (\xi, \eta)\\ \vec n = (\sin\theta,-\cos\theta)\\ p_1 = p_g+a\sin\phi\vec n\\ p_2 = p_g-a\sin\phi\vec n\\ } $$
Decomposing the kinetic energy into rotational plus translational we have
$$ T = \frac 12 L\left((\dot \theta+\dot \phi)^2+(\dot \theta-\dot\phi)^2\right)+\frac m2 \dot p_1^2 + \frac m2 \dot p_2^2 $$
Here $L = \frac {1}{12}m(2a)^2$ and $p_1,p_2$ are the mass center for the two rods respectively.
$$ T = m \left(\dot \xi^2+\dot\eta^2+a^2 \left(\cos ^2(\phi)+\frac{1}{3}\right) \dot\phi^2+a^2 \left(\sin ^2(\phi)+\frac{1}{3}\right) \dot\theta^2\right) $$