Why is the rotational kinetic energy term for the point mass of kinetic energy for a double pendulum not included in the Lagrangian equation?
\begin{align}
T&=\frac{m_1+m_2}2\ell_1^2\dot\theta_1^2+\frac{m_2}2\ell_2^2\dot{\theta}_2^2+m_2\ell_1\ell_2\dot{\theta_1}\dot{\theta}_2\cos\left(\theta_1-\theta_2\right),\\
U&=-\left(m_1+m_2\right)\ell_1g\cos\theta_1-m_2\ell_2g\cos\theta_2
\end{align}
Best Answer
It is there, it's just hidden by the change of coordinates. Written in Cartesian coordinates, the kinetic energy is $$ T=\frac12m_1\dot{x}_1^2+\frac12m_1\dot{y}_1^2+\frac12m_2\dot{x}_2^2+\frac12m_2\dot{y}_2^2+\frac12I_1\dot\theta_1^2+\frac12I_2\dot\theta_2^2\tag{1} $$ where the last term is the rotational kinetic enregy.
If you let \begin{align} x_1&=\frac12\ell_1\sin\theta_1\\ y_1&=-\frac12\ell_1\cos\theta_1\\ x_2&=\ell_2\left(\sin\theta_1+\frac12\sin\theta_2\right)\\ y_2&=-\ell_2\left(\cos\theta_1+\frac12\cos\theta_2\right), \end{align} then take the appropriate derivatives, you'll find that $T$ defined in (1) is equivalent to the $T$ defined in your problem.