# Issue when calculating the Lagrangian of a physical double pendulum When reading the solution to calculating the kinetic energy of the system I see that:

$$T_1 = \frac{1}{2}\int_0^L (r\dot\theta_1)^2 \frac{M}{l}dr$$

which is just $$T = \frac{1}{2} I \omega^2$$, where $$I = \int r^2 dm = \int_0^L r^2 \frac{M}{l} dr$$. But I am not sure why the $$r$$ and $$\dot\theta_1$$ are grouped in the first expression (which will be relevant in my next question).

For the second rod it is said that "a velocity offset" is applied $$T_2 = \frac{1}{2} \int_0^L \left[ (l\dot\theta_1 \cos\theta_1 + r \dot\theta_2 \cos\theta_2)^2 + (l\dot\theta_1 \sin\theta_1 + r \dot\theta_2 \sin\theta_2)^2 \right] \frac{M}{l} dr$$

I am having a hard time understanding how this expression is taking into account the rotational and translational components of the kinetic energy.

I know that $$l\dot\theta_1 \cos\theta_1$$ is the x component of the velocity of the end of the first rod and similarly $$l\dot\theta_1 \sin\theta_1$$ is the y component of that point.

But the integral is not calculating $$I$$, it seems to be $$\int (\dot{x}^2 + \dot{y}^2) \, dm$$ which does not make much sense to me because that does not take the form to calculate $$I$$ (also $$\omega^2$$ is missing for it to be the rotational KE), nor is it calculating the translational kinetic energy of the CM. So what is this calculating and why does it make sense that it is the kinetic energy of the second rod?

• The kinetic energy is $T=\dfrac{1}{2}mv_{cm}^{2}+\dfrac{1}{2}I _{cm}\omega ^{2}$
– Eli
Aug 15, 2022 at 14:43

The kinetic energy for an arbitrary distribution of matter is defined as $$\int \frac{1}{2}(\dot x^2+\dot y^2) dm$$ This formula can always be used. IF you have a rigid body then you can simplify this further to $$\frac{1}{2} m v_{\mathrm{CoM}}^2 + \frac{1}{2} I_{\mathrm{CoM}} \omega^2$$ In the case of a rigid object, such as one of the arms of the pendulum, those two quantities are equivalent. So the choice between them is one of computational convenience.