Mass attached to elongated spring and rotating around fixed point Let there be a point mass $m$ attached to a spring of elastic constant $k$, with its other end attached to a fixed point $O$, free to rotate around it. The natural length of the spring is $L_0$, but in the initial state it has length $L_0$, and is set in motion with a speed $v_0$ perpendicular to the spring. There is no friction. 
I need to find the speed of the mass when the spring is at length $L_0$ again using Lagrangian mechanics. I also need to discuss if any motion takes place in the case that $\frac{k L_{0}^{2}}{m}<3 v_{0}^{2}$.
Before writing anything down, I have thought that by conservation of angular momentum, it be true that $2L_0 mv_0 = L_0 m v$, and therefore $v=2 v_0$. I am not sure if this is right, since the presence of a spring makes me think this has to be more complicated than that. How would one go on about solving this using Lagrangian mechanics? I don't know how to define the velocity of the mass, which would be necessary if I want to define the kinetic energy of the system. The potential energy could be easily expressed in terms of the elongation, so I can define the length of the spring at any moment as $L(t) = L_0 + x(t)$ and therefore:
$$V = \frac{1}{2}kx(t)^2$$
Edit: after following @Tobi7's advice, I got two equations:
$$m\ddot{x} = m \dot{\varphi}^2 x + k(L_0-x)$$
$$2m\dot{\varphi}(t) x \dot{x} + mx^2 \ddot{\varphi}^2 = 0$$
But I don't know how to proceed to find out the speed of the mass when $x = L_0$. I imagine that, when $x=L_0$, $\ddot{x}=0$ (the restoring force from the spring will be at its minimum when the spring has its natural length). Solving for $\dot{\varphi}^2$ in the first equation gives $\dot{\varphi} = 0$ which I assume is not the case.
 A: An importat concept of classical mechanics are degrees of freedom, the independent parameters that define the configuration of the system. In your case, the mass performs a circular motion, so choosing an angle $\varphi$ seems appropriate. The other DOF is due to the spring, which someone might denote as $x$. The position of the mass with regards to the origin is 
$$\boldsymbol{r} = \begin{pmatrix} 
x \cos \varphi \\
x \sin \varphi 
\end{pmatrix}.$$
The derive the kinetic energy $T$ of the system, you need to calculate the time derivative of $\boldsymbol{r}$:
$$\dot{\boldsymbol{r}} = \begin{pmatrix}
\dot{x} \cos\varphi - x \sin\varphi \dot{\varphi} \\
\dot{x} \sin\varphi + x \cos\varphi \dot{\varphi}
\end{pmatrix}.$$
The kinetic energy is simply
$$T = \frac{1}{2} m \dot{\boldsymbol{r}} \cdot \dot{\boldsymbol{r}}.$$
From now on, you have to take the usual steps to obtain the equations of motion, i.e. define the Lagrangian $\mathcal{L} = T - V$ and plug $\mathcal{L}$ into the Euler-Langrange equation. If $x$ is counted from the origin onwards, the potential of the spring should be  $V(x) = 1/2 k (x - L_0)^2$.
A hint to solve your problem: The velocity perpendicular to the mass is $\dot{\varphi}$.
