# Double Compound Pendulum: why use inertia about the center of mass for bottom pendulum?

I'm trying to wrap my head around the kinetic energy of a double compound pendulum, like the one shown in the Wikipedia article on double pendulums.

I know for computing the kinetic energy of the first (top) pendulum you can either use the angular velocity with the moment of inertia calculated from the end of the rod (i.e., $\frac{1}{2}I_{1,end}\dot\theta_1^2$), OR the linear plus angular velocity about the center of mass (i.e., $\frac{1}{2}M_1 v_1^2+\frac{1}{2}I_{1,cm}\dot\theta_1^2$). This makes sense to me because this is basically equivalent to applying the parallel axis theorem to change the moment of inertia from the center of mass to the end of the top pendulum.

However, it's not so clear to me why this works for the second (bottom) pendulum. The talk section of the article says it's a lot simpler to model the kinetic energy of the bottom rod by summing the linear plus angular velocities about the rod's center of mass (i.e., $\frac{1}{2}M_2 v_2^2+\frac{1}{2}I_{2,cm}\dot\theta_2^2$) where the position of the center of mass is a function of the two generalized coordinates, $\theta_1$ and $\theta_2$.

Can somebody please explain to me why this works? I can understand computing the linear kinetic energy about the center of mass, but it's unclear why the rotational energy is also chosen to be computed at the center of mass. Is this also just an application of the parallel axis theorem, but about some rotational axis which is more difficult to imagine?

All help appreciated! Thanks!

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If you define kinetic energy as $KE =\frac{1}{2} m (\vec{v}_G \cdot \vec{v}_G) + \frac{1}{2} I_G (\vec{\omega} \cdot \vec{\omega} )$ where $v_G$ is the linear velocity of the CG and $I_G$ is the mass moment of inertia about the CG, and then transform the quantities to the handle of the rigid body (i.e. the pin) then you will get what Wikipedia has.

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Thanks for the answer! Let me see if I understand what you're saying... When transforming the quantities (linear and rotational KE) to the handle you would need to subtract out as much from the linear term as you would add to the rotational term when applying the parallel axis theorem. Otherwise you'd be double counting some of the energy. Am I on the right track? –  JRintheSEA Apr 29 '13 at 15:04
Also, is CG the same as CM? –  JRintheSEA Apr 29 '13 at 15:05
Yes, CG is center of gravity, or center or mass CM. –  ja72 Apr 29 '13 at 19:28
By definition kinetic energy is summed at the CM because that is where one expresses Newton's & Euler's laws of motion. –  ja72 Apr 29 '13 at 21:07