Equation of motion of a pendulum modeled as a cylinder

[diegoferigo]

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Pendulum System

Modeling with Newton Laws

1. Inertia of a cylinder rotating around its center of mass

\begin{equation} I_{com} = \frac{1}{12} m (3 r^2 + L^2) \end{equation}

2. Express inertia in the pivot point using the parallel axis theorem

\begin{aligned} I{pivot} &= I{com} + m \left( \frac{L}{2} \right)^2 \\ &= \dots = \frac{m}{12} (4L^2+3r^2) \end{aligned}

3. Apply the second Newton Law in its rotational form

\begin{aligned} \tau{net} = I{pivot} \ddot\theta = \left( \frac{L}{2} \right) m g sin(\theta) \end{aligned}

\begin{aligned} \ddot\theta = \dots = \frac{6g}{4L + 3 r^2 / L} sin(\theta) \end{aligned}

Modeling with Lagrange Equations

1. Calculate the lagrangian of the system

\begin{equation} K = \frac{1}{2} m v^2 + \frac{1}{2} I_{com} \dot\theta^2 \end{equation}

\begin{equation} U = \frac{1}{2} mg \left( \frac{L}{2} \right) cos(\theta) \end{equation}

\begin{aligned} L &= K - U \\ &= \frac{1}{2} m v^2 + \frac{1}{2} I{com} \dot\theta^2 - \frac{1}{2} mg \left( \frac{L}{2} \right) cos(\theta) \\ &= \frac{1}{2} m \left( \frac{L}{2} \right)^2 \dot\theta^2 + \frac{1}{2} I{com} \dot\theta^2 - \frac{1}{2} mg \left( \frac{L}{2} \right) cos(\theta) \end{aligned}

2. Apply the Euler-Lagrange equation

\begin{equation} \frac{d}{dt} \frac{\partial L}{\partial \dot\theta} - \frac{\partial L}{\partial \theta} = 0 \end{equation}

Calculate the partial derivatives:

\begin{equation} \frac{\partial L}{\partial \dot\theta} = \left( m \left( \frac{L}{2} \right)^2 + I_{com} \right) \dot\theta \end{equation}

\begin{equation} \frac{\partial L}{\partial \theta} = mg \left( \frac{L}{2} \right) sin(\theta) \end{equation}

Substituting the derivatives and $I_{com}$:

\begin{aligned} \ddot\theta = \dots = \frac{6g}{4L + 3 r^2 / L} sin(\theta) \end{aligned}

Calculate the system dynamics with Euler integration

\begin{aligned} \dot\thetak &= \dot\theta{k-1} + \ddot\theta_k \Delta t \\ \thetak &= \theta{k-1} + \dot\theta_k \Delta t \end{aligned}

[DanielePucci]

@diegoferigo what are the above equations needed for?

[diegoferigo]

@DanielePucci I implemented a test that compares three toy environments:

• Simple pendulum running inside Ignition Gazebo
• Simple pendulum running inside PyBullet
• Pendulum from the above equations

Most of the pendulum analysis you can find online and in the literature do not consider the pendulum as a cylinder but as a point mass, and they derive its inertia accordingly. Though, simulators typically read a URDF file that, if done properly, has the inertial element of a given body matching its visual appearance. The pendulum URDF of this repository has a cilynder-shaped body, and in this issue I recapped the steps to derive the equations to integrate the pendulum system under this condition.

This is necessary to have a (almost) perfect match with the simulation running in the considered physics engines. The only difference is the integration method used. For this simple comparison, a relatively small integration step together with Euler integration was good enough to provide a fair comparison.

This test allows to avoid regressions in reproducibility. I kept this issue open because it would be nice to add also the friction in the picture. Moreover, I would exclude the development of an example with contacts because in this case physics engine behave quite differently and creating an analytical model would be rather involved.

[diegoferigo]

Marking this issue as backlog since the variant with friction will be not implemented in the near future.