# Find lagrangian of a very specific model with two masses

This is just a homework problem on classical mechanics with respect to the lagrangian perspective. Below I quote the problem.

Two masses $m_1$ and $m_2$ are conected by an ideal string $l$ that passes through a hole in a table. The mass $m_1$ is above the table (so $m_2$ is hang on the other extreme of the string) and there are no friction in the problem. We have that $m_2$ moves only vertically. $(a)$ What is the initial velocity of $m_1$ such that $m_2$ keep hold still a distance $z$ from the table. $(b)$ If $m_2$ varies vertically in small changes there will be small oscillations effects. Using Euler-Lagrange equations determine the small oscillations period.

For the first part I just note that the tension is the same in the entire string then $mg = F_{cp}$ where this $F_{cp}$ is the so called centripetal force then $m_2g = m_1v^2/(l-z)$. So $v^2 = (m_2/m_1)g(l-z)$. This is the initial velocity such that the system maintains in an equilibrium. But the second part is of some trouble. To construct the lagrangian in a classical way ($T-U$ for all classical-mechanical systems) we will have the kinetic energies

$$T_1 = \frac{m_1\dot{z}^2}{2}+\frac{m_1(l-z)^2\dot{\theta}^2}{2}$$

$$T_2 = \frac{m_2\dot{z}^2}{2}$$

and the potential energy with respect to the table

$$U_2 = -m_2gz$$

Then we construct the lagrangian function

$$\mathcal{L} = \frac{(m_1+m_2)}{2}\dot{z}^2 + \frac{m_1(l-z)^2}{2}\dot{\theta}^2 + m_2gz$$

and the part of the small oscilations I could not get because it is clear that we will have something with the $\dot \theta =$ constant. But then I'm having trouble with the effective potential because the second derivative is becoming zero. Can someone help me with any hints? Thanks!

Find first the equations of motion for $z$ and $\theta$. In particular: $$\frac{d}{dt}\frac{\partial L}{\partial \dot\theta}-\frac{\partial L}{\partial \theta}=\frac{d}{dt}\frac{\partial L}{\partial \dot\theta}=0$$ so that $$p_\theta=\frac{\partial L}{\partial \dot\theta}= m_1(l-z)^2\dot\theta\, .\tag{1}$$ is constant. Insert (1) into the EOM for $z$: \begin{align} (m_1+m_2)\ddot{z}+m_1(l-z)\dot{\theta}^2-m_2g&=0\, ,\\ (m_1+m_2)\ddot{z}-\left(m_2g-\frac{p_\theta^2}{m_1(l-z)^3}\right)&=0\, . \tag{2} \end{align} The equilibrium position occurs when $$-\frac{dV_{eff}}{dz}\vert _{z=z_0}=0=m_2g-\frac{p_\theta^2}{m_1(l-z_0)^3}$$ from which you can find $z_0$ and recover the small oscillation by expanding (2) about this point to leading order in $\Delta z=z-z_0$.