T-Junction Oscillation

The combined gravitational potential energy of the two side masses is constant. It can therefore be neglected.

Kinetic energy, potential energy, and Lagrangian:

$T = \frac{3}{2} m L^2 \dot{\theta}^2 \\ V = - m g L \cos \theta \\ \mathcal{L} = T - V = \frac{3}{2} m L^2 \dot{\theta}^2 + m g L \cos \theta$

Equation of motion:

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

This results in:

$3 m L^2 \ddot{\theta} = - mg L \sin \theta \\ 3 L \ddot{\theta} = - g \sin \theta$

For small angles:

$3 L \ddot{\theta} = - g \theta \\ \ddot{\theta} = - \frac{g}{3 L} \theta$

This corresponds to simple harmonic motion with the following angular frequency:

$\omega^2 = \frac{g}{3 L}$

To compute the time period for large oscillations, probably the easiest thing to do is just run a numerical simulation using the following equation for the double-dot term.

$3 L \ddot{\theta} = - g \sin \theta$

Another way is to use conservation of energy:

$|V(\theta) - V(\theta_0)| = \frac{3}{2} m L^2 \dot{\theta}^2 = \frac{3}{2} m L^2 \Big( \frac{d \theta}{dt} \Big)^2$

Re-arrange to obtain an expression for $dt$ in terms of $\theta$ and $d \theta$ , and integrate this expression to derive the time period.

For small initial angle $\theta$ of a compound pendulum , we can find the period of the pendulum by first finding the equivalent length of pendulum given by:

$L_{eq} = \frac I{mR}$

where $I$ and $m$ are the moment of inertia and $m$ of the pendulum respectively, and $R$ is the distance between the pivot point and the center of mass. Then the angular frequency is given by:

$\omega^2 = \frac g{L_{eq}}$

For this problem, let the mass of each of the three balls be $m$ . Then the moment of initial of the pendulum is $I = 3mL^2$ .

Using the pivot point be the origin and the coordinates of center of mass be $C(x_c, y_c)$ , where $y$ increases vertically down. Then $x_c = \dfrac {mL-mL+0}{3m} = 0$ and $y_c = \dfrac {0+0+mL}{3m} = \dfrac L3$ . Therefore $R = \dfrac L3$ .

Then $L_{eq} = \dfrac {3mL^2}{3m\cdot \frac L3} = 3L$ and $\omega^2 = \dfrac g{3L} \implies \alpha = \boxed 3$ .