# 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. – Orion Yeung Jul 15 '17 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. – Kyle Kanos Dec 21 '14 at 15:30