# Rotational kinetic energy in double pendulum

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}

• I would argue that a point isn't oriented and can't rotate. But there are two terms that have units of inertia times the square of the rotation speed in your kinetic energy. Commented Jul 15, 2017 at 4:30

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.
• @Paras: I've fixed the missing factors of 1/2 in $x_1,\,y_1$ & made distinct the masses & moments of inertia. As far as I know, the difference between the compound and simple pendulum are the moment of inertia, $I$ (the compound including the parallel axis theorem). It should work out correctly. Commented Dec 21, 2014 at 15:30