(adsbygoogle = window.adsbygoogle || []).push({}); 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 [itex]h[/itex]. The slinky has a uniform mass distribution and its total mass is [itex]m[/itex]. Assume the slinky perfectly follows Hooke's law and has spring constant [itex]k[/itex]. The equilibrium length of the slinky is [itex]l_{eq}[/itex]. This experiment is done on Earth so gravity is just [itex]g[/itex].

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 [itex]q_t[/itex] and [itex]q_b[/itex] which are the distance of the top and bottom of the slinky to the ground, respectively.

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

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

Potential energy is

[itex]V = mgq_c + \frac{1}{2} k (l_{total} - l_{eq})^2[/itex]

[itex]V = \frac{mg}{2}(q_t+q_b) + \frac{1}{2} k (q_t-q_b - l_{eq})^2[/itex]

So Lagrangian is

[itex]\cal{L} = T - V[/itex]

[itex]\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[/itex]

Euler Lagrange Eqns are:

[itex]\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_i}} = \frac{\partial \cal{L}}{\partial q_i}[/itex]

Solving:

[itex]\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_t}} = \frac{m}{4}(\ddot{q_t}+\ddot{q_b})[/itex]

[itex]\frac{\partial}{\partial t}\frac{\partial \cal{L}}{\partial \dot{q_b}} = \frac{m}{4}(\ddot{q_t}+\ddot{q_b})[/itex]

[itex]\frac{\partial \cal{L}}{\partial q_t} = -\frac{1}{2}mg - k(q_t-q_b-l_{eq})[/itex]

[itex]\frac{\partial \cal{L}}{\partial q_b} = -\frac{1}{2}mg + k(q_t-q_b-l_{eq})[/itex]

So

[itex]\frac{m}{4}(\ddot{q_t}+\ddot{q_b}) = -\frac{1}{2}mg + k(q_t-q_b-l_{eq})[/itex]

[itex]\frac{m}{4}(\ddot{q_t}+\ddot{q_b}) = -\frac{1}{2}mg - k(q_t-q_b-l_{eq})[/itex]

This is where I run into a road block. What do I do from here. If I add the two equations I get:

[itex]\frac{m}{2}(\ddot{q_t}+\ddot{q_b}) = -mg [/itex]

[itex]\ddot{q_c} = - g [/itex]

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

Questions:

1. What am I doing wrong here? Or is it impossible to get [itex]q_t[/itex] and [itex]q_b[/itex] 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)

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# Modelling a Falling Slinky w/ Lagrangian

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