# Modelling a Falling Slinky w/ Lagrangian

1. Oct 9, 2011

### oxidized

1. The problem statement, all variables and given/known data

Hi everyone! This is not actually a homework problem, but I thought it was similar to one so I am putting it here.

Basically I was watching this youtube video of a falling slinky and I decided I wanted to try modelling it with physical equations:

The problem I have posited is as follows:

A slinky is suspended in the air so that its top is at height $h$. The slinky has a uniform mass distribution and its total mass is $m$. Assume the slinky perfectly follows Hooke's law and has spring constant $k$. The equilibrium length of the slinky is $l_{eq}$. This experiment is done on Earth so gravity is just $g$.

Derive two equations. One that gives the distance of the top of the slinky to ground wrt time, and another that gives the distance of the bottom of the slinky to ground wrt time.

First question I have is: did I pose the problem correctly? Tell me if I am doing something wrong. I think it is good to ask for the position of the top and bottom of the slinky, although an alternative setup would be to ask for the bottom and the total length of the slinky.

2. The attempt at a solution

I have chosen two generalized coordinates $q_t$ and $q_b$ which are the distance of the top and bottom of the slinky to the ground, respectively.

Then the center of mass is: $q_c = \frac{1}{2} (q_t+q_b)$

So the kinetic energy is: $T = \frac{1}{2} m \dot{q_c}^2 = \frac{m}{8} (\dot{q_t}+\dot{q_b})^2$

Potential energy is
$V = mgq_c + \frac{1}{2} k (l_{total} - l_{eq})^2$
$V = \frac{mg}{2}(q_t+q_b) + \frac{1}{2} k (q_t-q_b - l_{eq})^2$

So Lagrangian is
$\cal{L} = T - V$
$\cal{L} = \frac{m}{8} (\dot{q_t}+\dot{q_b})^2 - \frac{mg}{2}(q_t+q_b) - \frac{1}{2} k (q_t-q_b - l_{eq})^2$

Euler Lagrange Eqns are:
$\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_i}} = \frac{\partial \cal{L}}{\partial q_i}$

Solving:
$\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_t}} = \frac{m}{4}(\ddot{q_t}+\ddot{q_b})$
$\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_b}} = \frac{m}{4}(\ddot{q_t}+\ddot{q_b})$
$\frac{\partial \cal{L}}{\partial q_t} = -\frac{1}{2}mg - k(q_t-q_b-l_{eq})$
$\frac{\partial \cal{L}}{\partial q_b} = -\frac{1}{2}mg + k(q_t-q_b-l_{eq})$

So
$\frac{m}{4}(\ddot{q_t}+\ddot{q_b}) = -\frac{1}{2}mg + k(q_t-q_b-l_{eq})$
$\frac{m}{4}(\ddot{q_t}+\ddot{q_b}) = -\frac{1}{2}mg - k(q_t-q_b-l_{eq})$

This is where I run into a road block. What do I do from here. If I add the two equations I get:
$\frac{m}{2}(\ddot{q_t}+\ddot{q_b}) = -mg$
$\ddot{q_c} = - g$

Wow great! I could have derived that in two lines with Newtonian mechanics. I want an separate equations for $q_t$ and $q_b$, but I am not sure how to separate $\ddot{q_t}$ and $\ddot{q_b}$ from my Euler-Lagrange eqns.

Questions:
1. What am I doing wrong here? Or is it impossible to get $q_t$ and $q_b$ from my setup?

2. Is using a Lagrangian the best way to go about solving this problem? Is there a much easier way that I am missing?

Thanks!
(edit: fixed a minus sign, but it doesn't help me solve the problem)

Last edited by a moderator: Sep 25, 2014
2. Oct 10, 2011

### ehild

It is a very interesting experiment! But you can not model the slinky with two point masses and a massless spring, I am afraid. It is an extended elastic body, and a wave starts in it when dropped. As the old man said, the bottom end needs some time till it gets know that the upper part has been released. The bottom part is in equilibrium and the tension is equal to its weight. The upper part starts to fall when released as it has tension downward and gravity. As it moves, it changes the tension gradually along its length. It takes some time when this change of tension reaches the very bottom. It looks that the bottom ring gets know that it should fall only when the slinky has almost its equilibrium shape - with almost no tension. So the bottom would not fall till there is an upward tension higher than the mass of the last ring. That means the speed of propagation of a disturbance in the slinky is slow, compared to its fall.

ehild

3. Oct 10, 2011

### diazona

Apparently their explanation in the video is kind of iffy - you don't need to invoke the bottom end not knowing that the top has been released.

Anyway, the point is that according to the numerical results in that post, the oscillations aren't symmetric in the CM frame, so you can't assume that $q_c = \frac{1}{2}(q_t + q_b)$.

4. Oct 10, 2011

### I like Serena

Welcome to PF, rusty!

I'm not so sure you can model the slinky this way (as ehild and diazona said), but I can point out a mistake you made.

I'm afraid your kinetic energy is wrong.
Each mass has its own kinetic energy. Those 2 kinetic energies must be added.
So:
$${\cal T} = \frac{1}{2} \frac m 2 \dot{q_t}^2 + \frac{1}{2} \frac m 2 \dot{q_b}^2$$
(You can only use the center of mass if the body is stiff.)

Btw, you can define an extra boundary condition:
$$k l_0 = k(q_{t,0} - q_{b,0}) = mg$$

Oh, and besides adding your 2 final equations, you should also subtract them...

To answer your other question, I believe the Lagrangian is the preferred method here.

Last edited: Oct 10, 2011
5. Oct 10, 2011

Hey all,