What is the Improved Solution for a Non-Small Angle Simple Pendulum?

[Ackbach]

Here is this week's POTW:

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In freshman calculus-based physics, you studied the simple pendulum, and possibly the physical pendulum. Typically, in such problems, you used the small-angle approximation to simplify the resulting differential equation for $\theta$, the angle the pendulum makes with the vertical. Let us improve on this solution by not allowing the small-angle approximation. For the simple pendulum, reduce to a single integral the differential equation for $\theta$. You may obtain an implicit result, and you may neglect air resistance and any other second-order effects such as the Coriolis effect. Assume the length of the pendulum is $\ell$, $\theta(0)=\theta_0$, and $\dot{\theta}(0)=0$.

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[Ackbach]

Congratulations to topsquark for his correct solution! Actually, he solved the physical pendulum, not the simple pendulum, but that's ok. Here is his solution:

Spoiler

This problem seems to me to be intractable if we do it the straightforward way. But hey, I'm a Physicist and have a trick up my sleeve.

Let's work this out using the work-energy theorem. I'm using the physical pendulum for the equations but the diagram below is for the simple pendulum.

\(\displaystyle W_{nc} = \Delta KE + \Delta PE\)

There are no non-conservative forces present, so \(\displaystyle W_{nc} = 0\)

\(\displaystyle 0 = \Delta KE + \Delta PE\)

The KE is easy:
\(\displaystyle \Delta KE = \left ( \frac{1}{2} \right ) I \dot {\theta} ^2 - \left ( \frac{1}{2} \right ) I \dot {\theta _0} ^2\)

Define the zero of the potential energy to be at the bottom of the swing of the pendulum and the height of the pendulum at \(\displaystyle \theta _0\) to be h. Using the diagram below, we have that
\(\displaystyle cos(\theta) = \frac{L - h}{L} \implies h = L(1 - cos(\theta))\).

*Note that for the physical pendulum L is actually the radial distance from the pivot to the center of mass.

So the PE becomes:
\(\displaystyle \Delta PE = mgL(1 - cos(\theta)) - mgL(1 - cos( \theta _0)) \)

So back to the energy equation:
\(\displaystyle 0 = \left [ \left ( \frac{1}{2} \right ) I \dot {\theta} ^2 - \left ( \frac{1}{2} \right ) I \dot {\theta _0} ^2 \right ] + \left [ mgL(1 - cos(\theta)) - mgL(1 - cos( \theta _0)) \right ] \)

Noting that \(\displaystyle \dot{ \theta _0} = 0\) we get:
\(\displaystyle 0 = \left ( \frac{1}{2} \right ) I \dot {\theta} ^2 + mgL(cos(\theta _0) - cos(\theta))\)

or
\(\displaystyle 0 = \dot{\theta}^2 + \frac{2mgL}{I} (cos(\theta_0) - cos(\theta))\)

\(\displaystyle \dot{\theta} = \sqrt{\frac{I}{2mgL} (cos(\theta) - cos(\theta_0))}\)

Just for comparison, for a simple pendulum, \(\displaystyle I = mL^2\), where L is now the length of the string on the pendulum, giving:
\(\displaystyle \dot{\theta} = \sqrt{\frac{L}{2g} (cos(\theta) - cos(\theta_0))}\)

Back to the physical pendulum. This equation is separable:
\(\displaystyle \dot{\theta} = \frac{d \theta}{dt} = \sqrt{\frac{2gmL}{I} (cos(\theta) - cos(\theta_0))}\)

\(\displaystyle dt = \sqrt{\frac{I}{2mgL}}\left [ cos(\theta) - cos(\theta_0)) \right ] ^{-1/2} ~d \theta \)

\(\displaystyle \int_{t_0}^t dt = \int _{\theta _0}^{\theta} \sqrt{\frac{I}{2mgL}}\left [ cos(\theta) - cos(\theta_0)) \right ] ^{-1/2} ~d \theta \)

\(\displaystyle t - t_0 = \int _{\theta _0}^{\theta} \sqrt{\frac{I}{2mgL}}\left [ cos(\theta) - cos(\theta_0)) \right ] ^{-1/2} ~d \theta \)

Since \(\displaystyle t_0 = 0\) we have, finally:
\(\displaystyle t = \sqrt{\frac{I}{2mgL}} \int _{\theta _0}^{\theta} \left [ cos(\theta) - cos(\theta_0)) \right ] ^{-1/2} ~d \theta \)

This integral has no closed form, but note that if \(\displaystyle \theta = 0\) we can put the integral in terms of an incomplete elliptic integral of the first kind.