Gaussian Rod Pendulum - Corrected

Mass, center of mass, and moment of inertia

$m = \frac{\sqrt{\pi}}{2} \\ r_C = \frac{1}{\sqrt{\pi}} \\ I = \frac{\sqrt{\pi}}{4}$

Let $\theta$ be the angle from the downward vertical.

Kinetic and potential energies (potential is relative to rest position):

$E = \frac{1}{2} I \, \dot{\theta}^2 \\ U = m g \, (r_C - r_C \cos \theta)$

Lagrangian:

$L = E - U = \frac{1}{2} I \, \dot{\theta}^2 - m g \, (r_C - r_C \cos \theta)$

Evaluating equations of motion:

$\frac{\partial{L}}{\partial{\dot{\theta}}} = I \, \dot{\theta} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = I \, \ddot{\theta} \\ \frac{\partial{L}}{\partial{\theta}} = -mg \, r_c \, \sin \theta$

Equating:

$I \, \ddot{\theta} = -mg \, r_c \, \sin \theta \\ \frac{\sqrt{\pi}}{4} \, \ddot{\theta} = -\frac{10 \sqrt{\pi}}{2} \,\frac{1}{\sqrt{\pi}} \, \sin \theta \\ \ddot{\theta} = -\frac{20}{\sqrt{\pi}} \, \sin \theta$

Small angle approximation:

$\ddot{\theta} = -\frac{20}{\sqrt{\pi}} \, \sin \theta \\ \ddot{\theta} \approx -\frac{20}{\sqrt{\pi}} \, \theta$

This corresponds to SHM with the following angular frequency:

$\omega^2 = \frac{20}{\sqrt{\pi}} \\ \omega = \sqrt{\frac{20}{\pi^{1/2}}}$

The solution by @Steven Chase involves the usage of Lagrangian mechanics. I present a slightly different method as follows.

The mass of the rod about the hinge can be computed according to the relation:

$m = \int_{0}^{\infty} e^{-r^2}dr = \frac{\sqrt{\pi}}{2}$

The radial location of the center of mass can be found as such:

$r_c = \frac{1}{m}\int_{0}^{\infty}re^{-r^2}dr = \frac{1}{\sqrt{\pi}}$

The moment of inertia of the rod about the hinge can be found as such:

$I = \int_{0}^{\infty} r^2e^{-r^2}dr = \frac{\sqrt{\pi}}{4}$

Now, say the pendulum makes an angle $\theta$ with the vertical at an arbitrary instant of time. At this general instant, the kinetic and potential energy of the rod can be computed as such. Note that the X-axis is taken as the zero PE level.

$T = \frac{1}{2}I\dot{\theta}^2$ $V = -mgr_c \cos{\theta}$

Since there are no dissipative forces acting on the system, energy is conserved and we can say:

$E_{total} = T + V$

Here, $E_{total}$ is constant at all instants of time. Differentiating the above expression with respect to time gives:

$0 = \left(I\ddot{\theta} + mgr_c \sin{\theta}\right)\dot{\theta}$

Since $\dot{\theta}$ cannot be zero, this gives the equation of motion:

$I\ddot{\theta} + mgr_c \sin{\theta} = 0$

From here, on applying the small-angle approximation, one gets:

$I\ddot{\theta} + mgr_c \theta = 0$

Which finally leads to:

$\omega^2 = \frac{mgr_c}{I} = \frac{20}{\sqrt{\pi}}$