Can a particle with non-zero angular momentum pass through the center of a spherical potential? Suppose you have a particle of mass $m$ moving in a potential $V(r) = -\frac{k}{r^2}$, with $r^2 = x^2+y^2+z^2$ and $k > 0$. Since the angular momentum $l$ is conserved, the particle will move in a plane: Let's use polar coordinates $(r, \theta)$. The expression for the energy is then:
$$
E = \frac12m\dot{r}^2+\left(\frac{l^2}{2m}-k\right)\frac1{r^2}
$$
Suppose $l < \sqrt{2mk}$. Then the effective potential is of the form $V_{eff} = -\frac{a}{r^2}$ with $a > 0$.
This implies that no matter what the initial energy $E$ of the particle is, it can and will at some point pass through $r=0$. Indeed, if you find the trajectory $r = r(\theta)$, you get things that look like either $r = \cos \theta$ or like $r = \sinh \theta$, which are $0$ at some point.
I don't understand how this can be. When the particle is very close to $r=0$, its momentum has to pointed towards $r=0$, but this would imply that $\vec{r}$ and $\vec{p}$ are parallel, and so $l$ would be $0$! How can this be? Doesn't conservation of angular momentum prevent the particle from reaching the center, no matter how strong the potential is?
 A: The constraint on angular momentum is $l=mr^2\dot{\theta}=\text{constant}$. As $r\rightarrow0$, we have $\dot{\theta}\rightarrow\infty$. This does cause the kinetic energy to diverge, but the potential energy blows up to $-\infty$, so the total energy stays constant.
A: $\vec r$ and $\vec p$ cannot be parallel. If you take a basis $\vec e_r, \vec e_\theta$, the coordinates of $\vec r$ and $\vec p$ are : 
$$\vec r = \begin{pmatrix} r\\ 0 \end{pmatrix}, \quad  \vec p = \begin{pmatrix} m \dot r\\ m  r \dot \theta\end{pmatrix} \tag{1}$$
The angular momentum $l = m r^2 \dot \theta$ is constant, and here, we will suppose that it is different of zero $l \neq 0$ (otherwise your question is trivial), so we may write :
$$\vec r = \begin{pmatrix} r\\ 0 \end{pmatrix}, \quad  \vec p = \begin{pmatrix} m \dot r\\ \frac{l}{r}\end{pmatrix} \tag{2}$$
You see that it is not possible, for any finite value of $r$, including $r=0$, that $\vec r$ and $\vec p$ are parallel, because $\frac{l}{r} \neq 0$ (For $r=0$, we may say that $\frac{l}{r} = \pm \infty$, but $\pm\infty \neq 0$)
[EDIT]
A different point of view is using the variable $u = \frac{1}{r}$. The equations of movement are then : 
$$ \frac{d^2u}{d \theta^2} + u = \frac{2mk}{l^2}u \tag{3}$$
If $l^2 > 2mk$, there is a solution $u = A ~cos (\omega \theta) + B sin (\omega \theta)$, with $\omega^2 = 1 -  \frac{2mk}{l^2}$
If $l^2 < 2mk$, there is a solution $u = A ~ch (\omega \theta) + B sh (\omega \theta)$, with $\omega^2 =    \frac{2mk}{l^2} - 1$
If $l^2 = 2mk$, there is a solution $u = A + B \theta$
