Effective potential of radial equation of Hydrogen In QM, the study of the hydrogen atom, why does it follow that for the radial equation given as $$-\frac{\hbar^2}{2m}\frac{d^2 u}{d r^2} + \bigg( -\frac{e^2}{4 \pi \epsilon_0} \frac{1}{r} + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} \bigg)u = Eu,$$ the term $\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$ is an effective potential that tends to keep the wave function with $l \neq 0$ away from the origin?
 A: The term
$$V_\mathrm{centripetal}=\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}$$
is an effective potential because it appears in the radial equation in the same way (coefficient of $u$) as the Coulomb potential
$$V_\mathrm{Coulomb}=-\frac{e^2}{4\pi\epsilon_0r}$$
The Coulomb potential is attractive for opposite charges which you can see from the minus sign in the equation. Taking only the Coulomb potential into account energy gets smaller the smaller $r$ is.
The effective potential on the other hand is always positive making it repulsive (energy gets smaller for large $r$). Specifically if you get close to the origin, $r\to 0$ the centripetal potential $V_\mathrm{centripetal}\to +\infty$. In physics, systems tend to minimize their energy, so a position where the potential energy is very large is not attractive for the system and you can say that the centripetal potential is "keeping the wave function away from the origin". 
In a classical mechanics analogy, a potential like this would result in a force $F=-\frac{dV}{dr}>0$, i.e. a force which is positive, pushing the "particle" away from the origin.
Since the centripetal potential is $\propto r^{-2}$ and the Coulomb potential $\propto r^{-1}$, at small $r\to 0 $ the centripetal potential will dominate the total potential $V_\mathrm{eff}=V_\mathrm{Coulomb}+V_\mathrm{centripetal}$. At large $r\to\infty$ the Coulomb potential will dominate. In-between there is an optimum $r$ at which the energy gets minimized. In a picture:

