Rubber Physics (Part 5)

Classical Mechanics Level 4

We refer back to the model of a one-dimensional chain from part 4 .

The chain consists of N N chain links, each having a length of l . l. This time we assume that the first chain link is fixed, while on the end of the chain a force of f ⃗ = f e ⃗ x \vec f = f \vec e_x acts. The outer force aligns the chain links in the + x +x direction so that the chain assumes its minimum potential energy when extended to its maximum length r ⃗ = N l e ⃗ x . \vec r = N l \, \vec e_x.

How much energy Δ E \Delta E does it take to flip a single chain link in the opposite direction?

Δ E = − N l f \Delta E = -N l f Δ E = − 2 l f \Delta E = -2 l f Δ E = − l f \Delta E = -l f Δ E = l f \Delta E = l f Δ E = 2 l f \Delta E = 2 l f Δ E = N l f \Delta E = N l f

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jan 13, 2018

The potential energy of the chain depends linear on the distance vector r ⃗ = x e ⃗ x \vec r = x \vec e_x : E pot = − r ⃗ ⋅ f ⃗ = − ∑ i = 1 N l ⃗ i ⋅ f ⃗ = − ∑ i = 1 N s i l f E_\text{pot} = - \vec r \cdot \vec f = - \sum_{i=1}^N \vec l_i \cdot \vec f = - \sum_{i = 1}^N s_i l f with the orientation vectors of the chain links, l ⃗ i = s i l e ⃗ x \vec l_i = s_i l \vec e_x , and the corresponding sign, s i = ± 1 s_i = \pm 1 . If all signs s i s_i are positive, the potential energy takes the value E pot = E pot min = − N l f E_\text{pot} = E_\text{pot}^\text{min} = - N l f If we flip a sigle chain link, we get E pot = − ( N − 1 ) l f + N l f = − N l f + 2 l f = E pot min + Δ E ⇒ Δ E = 2 l f \begin{aligned} E_\text{pot} &= -(N-1) l f + N l f = - N l f + 2 l f \\ &= E_\text{pot}^\text{min} + \Delta E \\ \Rightarrow \quad \Delta E &= 2 l f \end{aligned}

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