# Inverted pendulum on a cart - Lagrangian without moment of inertia?

I am modeling the inverted pendulum on a moving cart using Lagrangian methods. I see most examples model the pendulum's kinetic energy as a sum of translational and rotational components (using a $$I\dot\theta^2$$ term), but I also encounter examples that only use a sum the translational energies (see diagram and text snippet).

Is one method preferred over another? What assumptions would factor into the choice? Comparing both, I arrived at EOM for each with what seemed like a non-trivial difference - namely the $$mL^2\dot\theta^2$$ being smaller by a factor of two in the translational-only approach.  Update - I have added details of my model's kinetic energy for feedback:

The pendulum bob's position vector and resulting squared-velocity: $$\vec p= \begin{bmatrix} x+l\sin(\theta) \\ l \cos(\theta) \end{bmatrix}$$

$$v^2 = (\dot x + l\dot\theta \cos(\theta))^2 + \dot\theta^2 l^2 \sin^2(\theta)$$

KE: $$KE=\frac{1}{2}M\dot x + \frac{1}{2}m v^2 + \frac{1}{2}I\dot\theta^2$$

Which becomes: $$KE=\frac{1}{2}M\dot x + \frac{1}{2}m(\dot x^2+2l\dot x\dot\theta \cos(\theta)+l^2\dot\theta^2) + \frac{1}{2}m l^2 \dot\theta^2$$

I suspect I have too many terms here.

• You seem to be adding the $\frac{1}{2}I\dot\theta^2$ into your KE equation (the second to last equation you wrote) for no reason. This is why you have an extra term (you added it in by hand for seemingly no reason). The kinetic energy of the bob is already contained in your $\frac{1}{2}mv^2$ term.
– hft
Jun 2 at 0:41

The MMOI of the point mass $$m$$ about its center of mass is zero.

You only need $$K = \tfrac{1}{2} M \dot{x}^2 + \tfrac{1}{2} m (\vec{v} \cdot \vec{v})$$

You could evaluate $$K$$ at the pivot point as

$$K = \tfrac{1}{2} M \dot{x}^2 + \tfrac{1}{2} m \dot{x}^2 + \tfrac{1}{2} I \dot{\theta}^2$$

where $$I = m \ell^2$$ and the result would be the same.

Take a careful look at the last term in the cited text. It is of the form,

$$\frac{1}{2}ml^2\dot\theta^2 = \frac{1}{2}I\dot\theta^2$$

Which is exactly the kinetic energy of a point mass in pure rotation and I have reason to believe you missed that factor of one-half since you obtained different results. The convenience of using the moment of inertia is related to how one choose coordinates. Note that the mixed term (has $$\dot y\dot\theta$$) would otherwise be easy to miss and then there wouldn't be any coupling of the motion. i.e. would could write $$L(\theta,y) = L_1(\theta) + L_2(y)$$ and that is very unexpected.

• Thanks for the reply! I added my KE model to the post, which sums translational and rotational energies for the pendulum bob. This effectively doubles the l^2 term after simplifying, compared to a 'translational-only' model. Do I have an error here? Dec 26, 2020 at 17:33

Your approach has some overlap in the translational and rotational terms for the mass $$m$$.

To completely separate translation and rotation, you have to break down kinetic energy as: $$KE = \text{translation of the center of mass} + \text{rotation around the center of mass}.$$ If you model your entire rigid body as a massless rod connecting two point masses $$m$$ and $$M$$, then the position of the center of mass is at $$CM = \frac1{m+M}\left(m\begin{bmatrix} x+l\sin\theta \\ l\cos\theta \end{bmatrix} + M\begin{bmatrix} x \\ 0 \end{bmatrix}\right) = \begin{bmatrix} x+\frac{m}{m+M}l\sin\theta \\ \frac{m}{m+M}l\cos\theta \end{bmatrix}$$ and you can differentiate this to get velocity $$\vec{v}$$ of the center of mass.

Now considering rotation of this rigid body around its center of mass, even though the angle is different, angular velocity is again $$\dot\theta$$. Moment of inertia of the rigid body around its center of mass is $$I = \frac{mM}{m+M}l^2$$ so we get $$KE = \frac12(m+M)\vec{v}^2 + \frac12I\dot\theta^2$$ which reproduces the result from the first approach.