Exemple: Simple Pendulum (Angular parameterization)

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Carlos Adir edited this page Oct 29, 2021 · 1 revision

Let's suppose that we have a pendulum of mass m, which describes a circular route of radius a that starts at an angle theta0 with speed 0.

Initial conditions

As we start with an angle theta0 and 0 as speed, we have

$\theta (t = 0) = \theta_{0} \ \ \ \ \ \ \ \ \dot{\theta}(t=0) = 0$

System's energy

We have that the kinetic energy is given by

$\bg_white E_{k} = \dfrac{1}{2}mL^2 \cdot \dot{\theta}^2$

The potential energy is given by

$E_{p} = -mgL(1-\cos \theta)$

So, the total energy is

$E = \dfrac{1}{2}mL^2 \cdot \dot{\theta}^2 - mgL(1-\cos \theta)$

In a matricial form we get the total energy just like

$E = \dfrac{1}{2} \cdot \left[\dot{\theta}\right]^T \cdot \underbrace{ \left[mL^2\right]}_{\left[M\right]}\cdot \left[\dot{\theta}\right] + \underbrace{\left[mgL(-1+\cos \theta)\right]}_{\left[C \right ]}$

The Lagrangian

Now we compute the lagrangian

$\begin{align*} L_{\theta} & = \dfrac{d}{dt}\left(\dfrac{\partial E}{\partial \dot{\theta}}\right) - \dfrac{\partial E}{\partial \theta} \\ & = \dfrac{d}{dt}\left(mL^2 \cdot \dot{\theta}\right) - \dfrac{\partial}{\partial \theta}\left(-mgL \cdot (1-\cos \theta)\right) \\ & = mL^2 \cdot \ddot{\theta} + mgL \cdot \sin \theta \end{align*}$

As we don't have external force, our equation is only L_theta = 0:

$mL^2 \cdot \ddot{\theta} + mgL \cdot \sin \theta = 0$

In a matricial form we get:

$\left[mL^2 \right] \cdot \left[\ddot{\theta}\right] + \left[ mgL \cdot \sin \theta\right] = 0$