Tag

Without air resistance

To apply Equations (??), we will treat it as a simple pendulum with no friction, a massless rod and no air resistance. The pendulum will consist of a rod with a bob at one end and freely rotating around a central point. We will define this central point as being the origin in Cartesian coordinates. Say the rod is of fixed length $l$ metres. The pendulum bob will be of mass $m$ kilograms. As you can see in Figure , we have defined $\theta$ as the angle between the rod and the positive $x$ -axis going counter-clockwise. Hence the $x$ and $y$ coordinates of the pendulum bobs are:

\begin{aligned} x &= l\cos{\theta} \\ y &= l\sin{\theta}. \end{aligned}

Hence:

\begin{aligned} \dot{x} &= -l\dot{\theta}\sin{\theta} \\ \dot{y} &= l\dot{\theta}\cos{\theta}. \end{aligned}

As for velocity squared, it is defined as:

\begin{aligned} v^2 &= \dot{x}^2+\dot{y}^2 \\ &= l^2\dot{\theta}^2\sin^2{\theta} + l^2\dot{\theta}^2\cos^2{\theta} \\ &= l^2\dot{\theta}^2. \end{aligned}

Hence $v=l\dot{\theta}$ . The kinetic energy is:

\begin{aligned} T &= \dfrac{mv^2}{2} \\ &= \dfrac{ml^2\dot{\theta}^2}{2}. \end{aligned}

As for potential energy, given this bob is moving solely under the force of gravity:

\begin{aligned} V &= mgy \\ &= mgl\sin{\theta}. \end{aligned}

Hence:

\begin{aligned} L &= T - V \\ &= \dfrac{ml^2\dot{\theta}^2}{2} - mgl\sin{\theta}. \end{aligned}

$\dfrac{\partial L}{\partial \dot{\theta}}$ is hence:

\begin{aligned} \dfrac{\partial L}{\partial \dot{\theta}} &= ml^2 \dot{\theta}. \end{aligned}

And its time derivative is:

\begin{aligned} \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}} = ml^2 \ddot{\theta}. \end{aligned}

As for the $\dfrac{\partial L}{\partial \theta}$ term it is:

\begin{aligned} \dfrac{\partial L}{\partial \theta} &= -mgl\cos{\theta}. \end{aligned}

So Equations (??) becomes:

\begin{aligned} ml^2 \ddot{\theta} + mgl\cos{\theta} &= 0. \end{aligned}

Dividing by $ml^2$ and taking the second term to the right-hand side yields:

\begin{align} \ddot{\theta} &= -\dfrac{g}{l}\cos{\theta}. \tag{1} \end{align}

Equation (1) has, as its highest order derivative, a second-order derivative. It also only involves ordinary derivatives (not partial derivatives). Hence it is a second-order ordinary differential equation (ODE). Second-order ODEs are typically obtained in both Newtonian and Lagrangian mechanics. The number of second-order ODEs obtained for a given system depends on the number of degrees of freedom of the system.

Technically speaking it is possible to analytically solve Equation (1), but it requires the use of non-standard functions called elliptic integrals. If you would like to check out numerical solutions to this equation, see this page.

© Brenton Horne. I do not claim to be infallible, so feel free to open issues or pull requests on GitHub if you notice flaws or things that can be improved.
I favour Oxford spelling, so if you notice I generally use different spelling to Australian spelling, that is probably why.
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Without air resistance

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