Background Information
The lagrangian of a particle in a central force field $V(r)$ is $$ L=\frac12m(\dot r^2+r^2\dot\theta^2+r^2\sin^2\theta\dot\varphi^2)-V(r). $$ The particle must move in a plane, so the coordinate system can always be chosen such that $\dot\varphi=0$. The motion is restricted to be 2D. Then $$ L=\frac12m(\dot r^2+r^2\dot\theta^2)-V(r).\tag{1} $$ The Euler-Lagrange Equation for $r$ gives $$ m(\ddot r-r\dot \theta^2)+V'(r)=0.\tag{2} $$ With some trick one derives Binet equation, but the formula itself is not the point of this post.
Canonical variable $\theta$ doesn't appear in Lagrangian (1), so the canonical momentum is conserved, too. Set it to be $$ p_\theta=mr^2\dot\theta=l=\text{const}.\tag{3} $$ The Lagrangian becomes $$ L=\frac12m\dot r^2+\frac{l}{2mr^2}-V(r). $$ The problem is now 1D. Applying Euler-Lagrange Equation again, we will get $$ m\ddot r+\frac{l}{mr^3}+V'(r)=0.\tag{4} $$
Question
Eq (2) is not consistent with Eq (4), since Eq(3) holds. What's wrong? Thanks for answering.
