Expression of the potential in the hydrogen atom while solving radial part of the wave function
The 'centrifugal' term appears when the coordinates are changed from $(x,y,z)$ to spherical $(r, \theta, \varphi)$. This is not to say that it is artificial, or a mathematical trick, but is a consequence of the fact that the electron has orbital angular momentum, and this is made clear by changing coordinate systems.
To summarize: the equation is $H\varphi(\pmb{r})=E\varphi(\pmb{r})$ where $\displaystyle H=-\frac{\hbar^2}{2\mu}\Delta +V(r)$. Changing to spherical coordinates produces,
$$\Delta =-\frac{1}{r}\frac{\partial ^2}{\partial r^2}+\frac{1}{r^2}\left(\frac{\partial ^2}{\partial \theta^2}+\frac{1}{\tan(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{\sin^2(\theta)}\frac{\partial^2}{\partial \varphi^2} \right)$$
which means that eigenfunctions of $\varphi(\pmb{r})$ are variables in $r, \theta, \varphi$
The second term is almost the same as that for $\pmb{L^2}$ and substituting for this makes the Hamiltonian
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r+ \frac{1}{2\mu r^2}\pmb{L^2}+V(r)$$
As $\pmb{L^2}\varphi(\pmb{r})=l(l+1)\hbar^2\varphi(\pmb{r})$  then
$$\displaystyle H=-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{\partial ^2}{\partial r^2}r + \frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$$
The potential is unchanged and still $V(r) = -e^2/r$ and the Hamiltonian is still a function of $r$ but now depends on variable $l$ thus the eigenvalues must depend on $l$.
The centrifugal term is so called because the corresponding force, the negative slope, always repels the electron from the centre and labelled a centrifugal potential or barrier. The term is always positive or zero and can change the shape of the potential the electron experiences and thus forms an effective potential written as  $\displaystyle V_{eff}=\frac{l(l+1)\hbar^2}{2\mu r^2}+V(r)$. When $l&gt;0$ the centrifugal term dominates at small $r$ causing $V_{eff}$ to be repulsive. As larger $r$ the potential becomes attractive.