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Suppose I have a double pendulum with continuous distribution of mass, where $(x_1 , y_1) $ is the center of mass of first pendulum and $(x_2 , y_2)$ for second pendulum. I want to write lagrangian for this system, so I need to find the energy. My question: is the kinetic energy equal to $$\frac{m_1}{2} \left( \dot{x_1} ^2 + \dot{y_1} ^2 \right) + \frac{m_2}{2} \left( \dot{x_2} ^2 + \dot{y_2} ^2 \right) + \frac{I_1}{2} \omega_1 ^2 + \frac{I_2}{2} \omega_2 ^2$$? Or maybe, since the first pendulum only rotates around the stationary point on the top, the first term should be omited, and only energy assiociated with rotational motion should be taken under account?

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There's nothing wrong per se about what you have for the kinetic energy term. But you're going to need to express some of the variables in terms of each other.

There's a fairly decent derivation of the kinetic energy term here

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  • $\begingroup$ Yes, I understand that expressing $x_i , y_i $ in terms of angles would be the next step. It's just I've found a source in which the author omits the first term, allegedly due to the fact that it the first pendulum only rotates around the stationary point. I found this explanation weird, hence my question. I'll look into this derivation, thank you very much. $\endgroup$ – G.Fil Jul 9 '17 at 11:15

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