Lagrangian of an elastic pendulum

Question

I'm trying to understand the way my teacher found the Lagrangian of an elastic pendulum.

Given a spring pendulum connected to the origin, the equilibrium point is $(0,0,\frac{-mg}{k})$.

The length of the relaxed spring is $0$.

The Lagrangian with respect to the equilibrium point is $$ L = \frac{1}{2}m \dot r^2 - \frac{k}{2} r^2. $$

I don't understand why the gravitational potential energy wasn't taken into account.

Can it be related to the given initial conditions? The mass starts at the equilibrium point with velocity $v$ to the right ($x$ direction).

Answer 1

I don't understand why the gravitational potential energy wasn't taken into account.

The reasoning goes as follows. Suppose we have $U_g(x)= m g x$ and $U_e(x)= \frac{1}{2} k x^2$ where $x=0$ is at the unstretched position. Then the minimum total potential is found at $\frac{d}{dx}(U_g(x)+U_e(x))=0$. Call that minimum point $x_0=-gm/k$.

Now, we can make a coordinate transform $x \rightarrow X - x_0$ and we can write the total potential as $$U_g(X) + U_e(X) = \frac{k}{2} X^2-\frac{g^2 m^2}{2 k}$$ but since the constant term drops out of all of the equations of motion we can drop it from the Lagrangian without changing anything so we can get a simplified potential $$U_s(X)=\frac{k}{2} X^2 = U_e(X)$$

So the reason that we neglect the gravitational potential energy is that if we set our coordinate $X$ to be zero at the minimum potential (the equilibrium point) then the only effect of the gravitational potential is a constant offset, which can be dropped.

Answer 2

I don't think that the Lagrangian that your teacher found is correct ?

you have two generalized coordinate die spring deflection $~r(t)~$ and the angle $~\varphi(t)~$ \begin{align*} &\text{stating with the position vector}\\ &\mathbf{R}= \left[ \begin {array}{c} r\sin \left( \varphi \right) \\ r\cos \left( \varphi \right) \end {array} \right]\\ &\text{the kinetic energy}\quad, T=\frac{m}{2}\mathbf{v}\cdot \mathbf{v}\quad,\mathbf{v}=\mathbf{\dot{R}}\\ & \text{the potential energy}\quad, U=-m\,g\,\mathbf R_y-\frac{k}{2}\,r^2\\\\ &\text{the Lagrangian} \\ &\mathcal{L}=T+U = \frac{1}{2}\left[m \left( {{\it pr}}^{2}+{r}^{2}{p\varphi }^{2} \right) -2\,mgr\cos \left( \varphi \right) -k{r}^{2} \right] \end{align*}

with EL you obtain the EOM's

\begin{align*} & {\ddot{r}}+{\frac {-m\,r{\dot\varphi }^{2}+m\,g\cos \left( \varphi \right) +k r}{m}} =0\\ &\ddot\varphi -{\frac {-2\,\dot\varphi \,{\dot r}+g\sin \left( \varphi \right) }{r}} =0 \end{align*}

put the time derivative in the EOM's equal zero and solve for $~r~,\varphi~$ you obtain the equilibrium state

$$r_0=-\frac{m\,g}{k}~,\varphi_0=0$$

thus:

the Lagrangian from the equilibrium state is:

$$\mathcal{L}=\mapsto\mathcal{L}(r=r-r_0,\varphi)$$