In polar, just remember these\begin {align*} \vec {r} & =\rho \hat {e}_{\rho }\\ d\vec {r} & =\hat {e}_{\rho }d\rho +\hat {e}_{\phi }\rho d\phi \\ \vec {v} & =\frac {d\vec {r}}{dt}\\ & =\hat {e}_{\rho }\frac {d\rho }{dt}+\hat {e}_{\phi }\rho \frac {d\phi }{dt}\\ \frac {d}{dt}\hat {e}_{\rho } & =\dot {\phi }\hat {e}_{\phi }\\ \frac {d}{dt}\hat {e}_{\phi } & =-\dot {\phi }\hat {e}_{\rho } \end {align*}
Given \(\vec {r}=\rho \hat {e}_{\rho }\), then \begin {align*} \vec {v} & =\dot {\rho }\hat {e}_{\rho }+\rho \frac {d}{dt}\hat {e}_{\rho }\\ & =\dot {\rho }\hat {e}_{\rho }+\rho \dot {\phi }\hat {e}_{\phi } \end {align*}
And similarly for \(\vec {a}\). \[ \vec {a}=\left (\ddot {\rho }-\rho \dot {\phi }^{2}\right ) \hat {e}_{\rho }+\left ( \rho \ddot {\phi }+2\dot {\rho }\dot {\phi }\right ) \hat {e}_{\phi }\]
This is much better than the alternatives.
In Cylindrical\begin {align*} d\hat {e}_{\rho } & =\hat {e}_{\phi }d\phi \\ d\hat {e}_{\phi } & =-\hat {e}_{\rho }d\phi \\ d\hat {e}_{z} & =0 \end {align*}
\(dr\) is different coordinates
Cartessian\[ dr=\hat {e}_{x}dx+\hat {e}_{y}dy+\hat {e}_{z}dz \] Cylindrical\[ dr=\hat {e}_{\rho }d\rho +\hat {e}_{\phi }\rho d\phi +\hat {e}_{z}dz \] Spherical\[ dr=\hat {e}_{r}dr+\hat {e}_{\theta }rd\theta +\hat {e}_{\phi }r\sin \theta d\phi \] \(v\) is different coordinates
Use these for finding Lagrangian.
In Cartessian\[ \vec {v}=\dot {x}\hat {e}_{x}+\dot {y}\hat {e}_{y}+\dot {z}\hat {e}_{z}\] Polar\[ \vec {v}=\dot {\rho }\hat {e}_{\rho }+\rho \dot {\phi }\hat {e}_{\phi }\] Spherical\begin {align*} \vec {v} & =\dot {\rho }\hat {e}_{\rho }+\rho \dot {\theta }\hat {e}_{\theta }+\rho \sin \theta \dot {\phi }\hat {e}_{\phi }\\ \nabla V\left (\rho ,\theta ,\phi \right ) & =\hat {e}_{\rho }V_{r}+\hat {e}_{\theta }\frac {1}{\rho }V_{\theta }+\hat {e}_{\phi }\frac {1}{\rho \sin \theta }V_{\phi } \end {align*}