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I'm sure I've made a silly mistake here, so I would be very grateful if someone could help me clear it up! Here is my reasoning:

The Lagrangian in a spherical potential is $$ \mathcal{L}=\frac{m\mathbf{v}^2}{2}-U(r) = \frac{m}{2}(\dot{r}^2 +r^2\dot{\theta}^2+r^2\sin^2(\theta)\dot{\phi}^2)-U(r). $$ Now $\frac{\partial\mathcal{L}}{\partial\phi}=0$, so the corresponding momentum $p_\phi=\frac{\partial\mathcal{L}}{\partial\dot{\phi}}=mr^2\sin^2(\theta)\dot{\phi}$ is constant in time. This means that the Lagrangian can instead be written $$ \mathcal{L} =\frac{m}{2}(\dot{r}^2 +r^2\dot{\theta}^2+\underbrace{[r^2\sin^2(\theta)\dot{\phi}]}_{p_\phi}\dot{\phi})-U(r) =\frac{m}{2}(\dot{r}^2 +r^2\dot{\theta}^2)-U(r)+\frac{d}{dt}(\frac{m}{2}p_\phi \phi). $$ Since $p_\phi$ is constant the last term is a total derivative and so can be omitted completely, and the Lagrangian may be written as $$ \mathcal{L}’= \frac{m}{2}(\dot{r}^2 +r^2\dot{\theta}^2)-U(r) $$ The Lagrangian now has $\frac{\partial\mathcal{L}’}{\partial\theta}=0$ which was not true for the original one. Something has clearly gone wrong, but where?

Also, following the same logic for $\theta$ gives that $$ \mathcal{L}’’= \frac{m\dot{r}^2}{2}-U(r) $$ Which is also wrong - the effective potential energy has the wrong form.

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The main point is that one is not allowed to use EOM in the Lagrangian. For such problems one should instead form (minus) the Routhian $$-R(r,\dot{r};\theta,\dot{\theta};\phi,p_{\phi})~=~L - p_{\phi}\dot\phi~=~ \frac{m}{2}\left(\dot{r}^2 + r^2\dot{\theta}^2\right) \color{Red}{-}\frac{p_{\phi}^2}{2mr^2\sin^2\theta} -U(r) $$ by Legendre transforming the cyclic coordinate $\phi$.

Warning: To forget the minus sign in red is a common mistake, cf. e.g. this and this related Phys.SE posts. The minus sign is important because it is a centrifugal (rather than a centripetal) potential.

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