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The differential equation that gives the equation of motion of a pendulum where:

  • $m$ is the mass
  • $L$ is the distance between the pivot and the body's centre of mass
  • $g$ is the acceleration due to gravity
  • $I$ is the moment of inertia of the body about the pivot

is given by:

$\displaystyle\frac{\partial^2 \theta}{\partial t^2} + \left(\frac{mgL}{I}\right) \sin\left(\theta\right) = 0$

Here, we're going to neglect air resistance and friction.

I plugged this equation into Wolfram Alpha and the solution seems to be:

$\displaystyle\theta = 2\,\text{am}\left(\frac{t + \omega_0}{2} \sqrt{\theta_0 + \frac{2mgL}{I}} \,\,\bigg| \,\,\frac{4mgL}{2mgL + I\theta_0}\right)$

where $\text{am}\left(x, y\right)$ is the Jacobi amplitude function.

But, plugging the numbers, the units don't cancel out and they aren't in the right order. Now, my question becomes: Is this the correct equation of motion? If not, what is it? Do units not matter when plugging into the equation of motion?

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The equation is fine and the units work out!

I think your issue is with the units of $\omega_0$ and $\theta_0$. Though $\omega$ is usually a frequency and $\theta$ is usually an angle, here they both have to be unitful. $\omega_0$ has to have units of time (since we add $t+\omega_0$) and $\theta_0$ appears in $m g L+ I \theta_0$, and so must have units, where "[x]" means "the units of x": $$\left[\theta_0 \right]=\left[ \frac{m g L}{I}\right]=\frac{\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}^2}\mathrm{m}}{\mathrm{kg}\cdot\mathrm{m}^2}=\frac{1}{\mathrm{s}^2}$$

With that plugged in, everything works out. Both $x$ and $y$ in $\mathrm{am}(x,y)$ are unitless.

$\omega_0$ and $\theta_0$ are just constants of integration.

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