Dynamics of Constrained Charged Particles - 2

Let $\theta$ be the angle of the rod below the horizontal. Let $y_0$ be the vertical position of the middle ball. The position and velocity of the ball on the right are:

$x = L \cos \theta \\ y = y_0 - L \sin \theta \\ \dot{x} = -L \sin \theta \dot{\theta} \\ \dot{y} = \dot{y}_0 - L \cos \theta \dot{\theta}$

Since $\dot{x}$ and $\dot{y}$ are zero in the beginning, and $\theta = 0$ as well, we know that $\dot{\theta} = \frac{v}{L}$ in the beginning. The kinetic energy, potential energy, and Lagrangian are:

$T = m L^2 \dot{\theta}^2 + m \dot{y_0}^2 - 2 m L \cos \theta \dot{\theta} \dot{y_0} + \frac{1}{2} m_0 \dot{y_0}^2 \\ V = \frac{k q^2}{2 L \cos \theta} \\ \mathcal{L} = T - V = m L^2 \dot{\theta}^2 + m \dot{y_0}^2 - 2 m L \cos \theta \dot{\theta} \dot{y_0} + \frac{1}{2} m_0 \dot{y_0}^2 - \frac{k q^2}{2 L \cos \theta}$

Equations of motion:

$\frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{\dot{y_0}}} = \frac{\partial{\mathcal{L}}}{\partial{y_0}} \\ \frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}} = \frac{\partial{\mathcal{L}}}{\partial{\theta}}$

Crunching out the math results in the following coupled linear system:

\[\begin{pmatrix} 2m + m_0 & -2 m L \cos \theta \\ -2 m L \cos \theta & 2 m L^2 \\

\end{pmatrix}

\begin{pmatrix} \ddot{y}_0 \\ \ddot{\theta} \\

\end{pmatrix} = \begin{pmatrix} -2 m L \sin \theta \, \dot{\theta}^2 \\ -\frac{k q^2}{2L} \sec \theta \tan \theta \\

\end{pmatrix} \]

Initialize variables appropriately and solve the linear system for the double-dot terms on every time step. Numerical integration yields a period of motion of $\theta$ $T_{\theta} \approx 22.03$ .

The plot below shows $y_0$ and $\theta$ (in degrees) over time from $t = 0$ to $t = 100$ . You can see that $\theta$ oscillates sinusoidally, while $y_0$ increases steadily with some small oscillation on top.

Interestingly, the period of motion of the $x$ coordinate of the balls seems to be half that of $\theta$ and $y$ . Recall that $\theta$ is a time sinusoid (or at least a quasi-sinusoid). Since $x$ is the cosine of $\theta$ , its period is half that of $\theta$ . Since $y$ is the sine of $\theta$ , its period is equal to that of $\theta$ . You can go to Wolfram Alpha and ask it to expand the cosine of a sine and the sine of a sine to see this.

A slightly different solution from that of @Steven Chase

Let the $y$ coordinate of the vertical ball be $s$ . Let $\theta$ be the angle between any rigid rod and the Y-axis.

Coordinates of the ball on the right:

$x_1 = L \sin{\theta}$ $y_1 = s -L \cos{\theta}$

Coordinates of the ball on the left:

$x_2 = -L \sin{\theta}$ $y_2 = s -L \cos{\theta}$

Kinetic energy of the system:

$\mathcal{T} = \frac{m_o \dot{s}^2}{2} + \frac{m}{2}\left(\dot{x}_1^2 + \dot{y}_1^2\right)+\frac{m}{2}\left(\dot{x}_2^2 + \dot{y}_2^2\right)$

$\mathcal{T} = 2 \dot{s}^2 + 4 \dot{s} \dot{\theta} \sin{\theta} + 4 \dot{\theta}^2$

Potential Energy of the system:

$\mathcal{V} = \frac{1}{4 \sin{\theta}}$

Lagrange's equation with respect to $s$ is:

$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{s}}\right)=0$ $\implies \dot{s} + \dot{\theta} \sin{\theta}=C$

Where $C$ is an arbitrary constant. The above equation is the generalised momentum conservation law.

The initial conditions of the system are:

$\theta(0) = \frac{\pi}{2}$ $\dot{\theta}(0) = -\frac{v}{L}$ $s(0)=0$ $\dot{s}(0) = v$

Applying initial conditions gives: $C=\frac{1}{4}$ $\implies \dot{s} + \dot{\theta} \sin{\theta}=\frac{1}{4} \ \dots(2)$

Since total energy is conserved:

$\mathcal{T}+\mathcal{V} = C_2$

Applying initial conditions gives $C_2=0.5$ $\mathcal{T}+\mathcal{V} = \frac{1}{2}$

Eliminating $\dot{s}$ from the energy equation using (2) and simplifying:

$\left(4-2\sin^2{\theta}\right) \dot{\theta}^2 + \frac{1}{8} + \frac{1}{4 \sin{\theta}}=\frac{1}{2}$

$\dot{\theta}^2 = \frac{3\sin{\theta}-2}{8\sin{\theta}\left(4-2\sin^2{\theta}\right)}$

Recall that the minimum separation is:

$d = \frac{8}{3}$

That corresponds to: $\theta_o=\arcsin(2/3)$

So when the motion of the system starts, $\theta$ reduced. This means:

$\dot{\theta} =- \sqrt{\frac{3\sin{\theta}-2}{8\sin{\theta}\left(4-2\sin^2{\theta}\right)}}$

Rearranging:

$\int_{\theta_o}^{\pi/2} \sqrt{\frac{8\sin{\theta}\left(4-2\sin^2{\theta}\right)}{3\sin{\theta}-2}} \ d\theta = \int_{0}^{T/4} dt$

We know that $\theta= \pi/2$ is the mean position while $\theta = \theta_o$ is the extreme position. The motion from the mean to extreme position in an oscillation takes a quarter of the total time period. Hence:

$\boxed{T = 4\int_{\theta_o}^{\pi/2} \sqrt{\frac{8\sin{\theta}\left(4-2\sin^2{\theta}\right)}{3\sin{\theta}-2}} \ d\theta}$