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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.