Lagrangian of an inverted pendulum on a moving cart

So I have been trying to derive the equations of motion of the inverted physical pendulum in a cart, but I seem to be confused about the derivation of its Kinetic Energy. I know this physical system is very popular and while I have searched and searched I couldn't find an answer to my question anywhere. So I divided the kinetic energy into the cart's and pendulum's: $$T = T_C + T_P$$

The cart's one is pretty straight forward $$T_C = 1/2 M \dot{x}^2$$, where I am denoting $$x$$ the horizontal coordinate of the cart's point mass.

My trouble is now with the pendulum's Kinetic Energy. I would assume I would have to sum the translational energy of the pivot point $$T_{pivot}=1/2 m \dot{x}^2$$ to the rotational energy of the pendulum $$T_{rot} = 1/2 I \dot{\theta}^2$$, where $$I$$ is the moment of inertia of the pendulum with respect to the pivot point (Note: the angle $$\theta$$ i chose is with respect to the upper vertical, unlike in the image up there).

With this I got: $$\mathcal{L} = \frac{1}{2}(M+m) \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 - mgl\cos\theta$$

And therefore the equations of motion: $$(M+m) \ddot{x} = F(t)$$ $$I \ddot{\theta} - mgl \sin\theta = 0$$

These equations, though, seem too simple compared to the equations I have seen out there for this problem. I would really appreciate if someone could point out my mistakes.

• Your position vector to the CM is $\vec R= [ x+l\cos(\theta),l\sin(\theta)]^T$ Thus the kinetic energy will be?
– Eli
May 7 '20 at 14:36
• @Eli I considered the translational kinetic energy of the pivot and not the CM, maybe that was a wrong assumption. In that case I would just get the $T=1/2 m (\dot{x_{CM}}^2 + \dot{y_{CM}}^2)$ for the pendulum? or would I have to add a rotational kinetic energy? (I was thinking maybe with moment of inertia relative to the center of mass) May 7 '20 at 15:38
• no you have to take the CM velocity for translation and for rotation $I=I_{cm}+m\,l^2$
– Eli
May 7 '20 at 16:02
• @Eli Thank you for your help, Eli. So basically I should consider the translational kinetic energy of the CM and the rotational relative to the pivot point? May 7 '20 at 16:26

I had the same question and after reading some definitions, I've got the answer: The kinetic energy of a rigid body which has planar motion is always

$$T=T_{Gtranslate}+T_{rotate/G}$$

or

$$T=1/2Mv^2_G+1/2l_G\omega^2$$

where $$G$$ is the center of mass. So in this pendulum you have to calculate $$v_G=\sqrt{\dot{x_G}^2+\dot{y_G}^2}$$ and $$\omega=\dot{\theta}$$ and. Then the kinetic energy will be

$$T=\frac{1}{2}M(\dot{x_G}^2+\dot{y_G}^2)+\frac{1}{2}I_G\dot{\theta}^2 + T_{cart}$$

There is a paper from MIT 2.003SC course which has the same solution: http://bit.ly/PendulumonACart

First, that the Lagrangian will have a term containing $$F(t)$$, or you will not get $$(M+m)\ddot{x}=F(t)$$. Second that if $$F(t)$$ is an explicit function of $$t$$, then Lagrangian will also be the explicit function of $$t$$ and then you have to consider more general form of Euler-Langrangian equation. For that refer to https://physics.stackexchange.com/a/437198/203041.

• That function $F(t)$ is a non conservative force (therefore not included in the lagriang, I believe) acting along the x-axis on the cart. I use the generalized forces which I believe would be $F(t)$ for the $x$ coordinate and $0$ for the $\theta$ coordinate. Am I wrong in doing this? May 7 '20 at 15:42