Determining if a particle is bounded or unbounded with motion in polar coordinates

Question

_{(source: gyazo.com)}

Firstly, I determined the equation for the particles path for $k <-mV^2a^2$ as $u = {Ae^{\sqrt{\alpha}\theta}+Be^{-\sqrt{\alpha}\theta}}$ where $ \alpha = 1 + \dfrac{k}{mV^2a^2},$ $u = 1/r$

I have two questions;

(1) Could I solve for the constants $A,B$ with the information given? (It is not given that at r = a, $\theta = 0$, so I figured no)

(2) I'm not sure how to determine if the motion is bounded or unbounded in polar coordinates - For instance, if we were dealing with $r(t)$ then I'd take $t \to \infty$ and see what happens, but we have theta here, how can we deal with theta? In this case with $k <-mV^2a^2$ it is bounded (if we take $\theta \to \infty$)

Answer 1

You can solve for $A$ and $B$ by using $r(0) = a$, $\theta(0) = 0$ and evaluating $\dot{r}$ and $\dot{\theta}$ in your solution and using the initial conditions for velocity. You will need an equation for velocity as a vector in polar coordinates.

Furthermore, while you don't have $\theta(t)$ explicitly, it is a useful exercise to consider what happens to $u$ (or $r$) as $\theta$ varies over the whole range $(0 \dots 2\pi)$. If $u \to 0$ anywhere, then that would be an example of the particle escaping to infinity.

Answer 2

Define the potential energy $$U(r) = -\int_{\infty}^r F(\bar r)\cdot(-d \bar r) = -\int_r^\infty F(\bar r)\cdot d \bar r$$ with the condition that it is zero at infinity.

If the initial overall energy of the particle $E_0 = U(a) + \frac m2 V^2$ is larger than zero the particle will escape. (The only exception is the case that $k<0$ and the path is directed towards the origin. In that case the solution breaks down in finite time.)

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Note: In the above answer I assume that $V$ is really the velocity tangent to the particle path and not the velocity in angular direction.