Dipole and Ring Dynamics

Consider the particle to be at any arbitrary point on the ring. Its coordinates can be parameterised in polar coordinates as such:

$x = \cos{\theta}$ $y = \sin{\theta}$

The potential energy of the particle due to the di-pole is:

$V(\theta) = \frac{1}{4\pi}\left(\frac{1}{\sqrt{\cos^2{\theta} + \left(\sin{\theta} - 0.1\right)^2}} - \frac{1}{\sqrt{\cos^2{\theta} + \left(\sin{\theta} + 0.1\right)^2}}\right)$

The kinetic energy of the particle is:

$T = \frac{\dot{x}^2 + \dot{y}^2}{2} = \frac{\dot{\theta}^2}{2}$

From this point, the problem can be easily solved using Lagrangian mechanics and numerical integration. However, I present a semi-analytical approach as follows.

In order to understand how the particle moves, it is important to first understand that since there are no dissipative forces acting on the system, energy is conserved. This means:

$T + V(\theta) = V_{\mathrm{init}}+T_{\mathrm{init}}$

Here, the initial kinetic energy is zero as the particle is initially at rest and the initial potential energy can be found by substituting $\theta = \pi/6$ in the expression for the potential energy. This gives:

$T + V(\theta) = V_{\mathrm{init}}$

• So, if the particle on the ring has to speed up, it must speed up at the expense of potential energy.
• Moreover, the potential energy can never exceed its initial value as that would result in negative kinetic energy which is impossible. Therefore, the particle's motion lies between two values of $\theta$ such that these conditions remain satisfied at all times.
• Using a bit of elementary trigonometry, it is easy to see that the potential energy takes the same value when $\theta = \frac{\pi}{6}$ , $\theta = \frac{5\pi}{6}$ , and $\theta = \frac{-7\pi}{6}$ .
• Therefore the particle either oscillates between $\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}$ or $\frac{-7\pi}{6} \le \theta \le \frac{\pi}{6}$ .
• To confirm the range of the particle's motion, a plot of the potential energy vs. $\theta$ is shown as follows.

In the plot, $\theta$ ranges between $\frac{-7\pi}{6} \le \theta \le \frac{5\pi}{6}$ . The black dashed line is $V_{\mathrm{init}}$ . The potential energy intersects the black line at three points just as mentioned earlier. The three points being $\theta = \frac{\pi}{6}$ , $\theta = \frac{5\pi}{6}$ , and $\theta = \frac{-7\pi}{6}$ .

It can be seen that in the interval $\frac{-7\pi}{6} \le \theta \le \frac{\pi}{6}$ , the particle's potential energy is lower than or equal to the initial potential energy. Therefore, the range $\theta$ is:

$\frac{-7\pi}{6} \le \theta \le \frac{\pi}{6}$

Now, having established the bounds of motion, it is important to know that the particle moves from one extreme position to another in half a time period. Using all this information and the energy conservation principle, we get:

$T = V_{\mathrm{init}} - V(\theta)$ $\implies \frac{\dot{\theta}^2}{2} = V_{\mathrm{init}} - V(\theta)$ $\implies \dot{\theta} = -\sqrt{2\left(V_{\mathrm{init}} - V(\theta) \right)}$

An extra negative sign is introduced as since time increases, $\theta$ decreases. Let the time period be $T_p$ Separating variables and integrating:

$\int_{0}^{T_p/2} dt = -\int_{\pi/6}^{-7 \pi/6} \frac{d\theta}{\sqrt{2\left(V_{\mathrm{init}} - V(\theta)\right)}}$ $\boxed{T_p = 2\int_{-7 \pi/6}^{\pi/6} \frac{d\theta}{\sqrt{2\left(V_{\mathrm{init}} - V(\theta)\right)}} \approx 68.4807}$