Skewed Dumbbell

Let $\theta$ be the angle between $OA$ and $Y'$ .

Kinetic energy:

$\mathcal{T} = \frac{1}{2} m (1-p)^2 L^2 \dot{\theta}^2 + \frac{1}{2} m p^2 L^2 \dot{\theta}^2$

Potential energy:

$\mathcal{V} = - m g (1-p)L \cos \theta + m g p L \cos \theta$

Lagrangian:

$\mathcal{L} = \mathcal{T} - \mathcal{V} = \frac{1}{2} m (1-p)^2 L^2 \dot{\theta}^2 + \frac{1}{2} m p^2 L^2 \dot{\theta}^2 + m g (1-p)L \cos \theta - m g p L \cos \theta$

Equation of motion:

$\frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}} = \frac{\partial{\mathcal{L}}}{\partial{\theta}}$

Evaluating and plugging in numbers results in:

$\ddot{\theta} = \frac{10p - 5}{1 - 2p + 2 p^2} \sin \theta$

For small oscillations:

$\ddot{\theta} \approx \frac{10p - 5}{1 - 2p + 2 p^2} \theta$

Angular frequency and period of small $\theta$ oscillations:

$\omega^2 = \frac{4 \pi^2}{T^2} =- \frac{10p - 5}{1 - 2p + 2 p^2}$

Period squared expression:

$T^2 = \frac{4 \pi^2 (1 - 2p + 2 p^2) }{5 - 10 p}$

Final integral:

$\int_0^{0.49} T^2 \, dp = \pi^2\frac{ 100 (50) \ell n (50) + 50^2 - 1}{10 (50^2)}$

Below is a plot of $T^2$ vs. $p$ . The period is smallest for $p = 0$ , meaning that the entire rod is below the origin. The period asymptotically approaches infinity as $p$ approaches $0.5$ . When $p = 0.5$ , the torques are balanced, resulting in no angular acceleration (you can see this in the theta double-dot equation above as well). If the system is initialized with $\theta \neq 0$ and $\dot{\theta} = 0$ , the pendulum never moves, which is effectively the same thing as oscillating with an infinite period.