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$$L=\frac{1}{2}mv^2=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$$

$$L=\frac{1}{2}mv^2=\frac{1}{2}m(\dot{r}^2+r^2\dot{φ}^2)$$

I dont get this part. $$\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{φ}}}\right)-\frac{\partial{L}}{\partial{{φ}}}=0 \longrightarrow \ddot{φ}+\frac{2}{r}\dot{r}\dot{φ}=0$$

Shouldn't the derivative of the Lagrangian w.r.t. $φ$ be zero instead of this $$\frac{2}{r}\dot{r}\dot{φ},$$ because the Lagrangian doesn't contain any $φ$, thus derivative w.r.t. this should be zero.

Source: http://en.wikipedia.org/wiki/Action_(physics)

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closed as off-topic by AccidentalFourierTransform, ACuriousMind, user36790, Hritik Narayan, Gert May 8 '16 at 23:51

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  • $\begingroup$ Hint: $r$ is a function of time. $\endgroup$ – Rajath Krishna R May 7 '16 at 9:06
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Indeed the Lagrangian is independent of $\phi$. However, the partial derivative w.r.t $\dot{\phi}$,

\begin{equation} \frac{\partial L}{\partial\dot{\phi}}=mr^2\dot{\phi}, \end{equation}

contains $r$ and $\dot{\phi}$, both depending on time $t$.

Therefore, you need product rule to compute the total time derivative. That's why you have two terms. They come from differentiating $\partial{L}/\partial\dot{\phi}$.

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