The Elastic Catenary Problem
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How to solve the elastic catenary problem?
Elastic Catenary
A catenary that accounts for elastic energy can be obtained as a perturbation of the rigid catenary.
1 Inelastic Solution Form
Consider the functional:
\[E = \int \sqrt{x'(t)^2 + y'(t)^2} \, y(t) \, dt\]The Lagrangian is:
\[L = \sqrt{x'(t)^2 + y'(t)^2} \, y(t)\]Substituting into the Lagrange equations gives:
\[\frac{d}{dt} \frac{y' y}{\sqrt{x'(t)^2 + y'(t)^2}} = \sqrt{x'(t)^2 + y'(t)^2}\] \[\frac{x' y}{\sqrt{x'(t)^2 + y'(t)^2}} = C_1\]Solving:
\[y = a \cosh t\] \[x = a t - b\]2 Considering Elastic Energy
First, consider the case where the rope is inextensible, as above. If the rope has elastic energy, then the coordinates of any point on the rope are merely corrections to the inextensible case — the result of competition between gravitational potential energy and elastic energy.
Let the changes in the rope be $\epsilon_x(t), \epsilon_y(t)$. The infinitesimal elements and coordinates before and after the correction can be written as:
Before correction:
\[y = a \cosh t\] \[x = a t - b\] \[ds_0 = a \cosh t \, dt\]After correction:
\[x = a \left( t + \epsilon_x(t) \right) - b\] \[y = a \left( \cosh t + \epsilon_y(t) \right)\] \[\begin{aligned} ds &= a \sqrt{ \cosh^2 t + \epsilon_x'(t)^2 + \epsilon_y'(t)^2 + 2 \epsilon_x'(t) + 2 \epsilon_y'(t) \sinh t } \, dt \end{aligned}\]The change in energy can be written as:
\[\begin{aligned} \Delta E &= \Delta E_p + \Delta E_k \\ &= \int \rho g a^2 \epsilon_y(t) \cosh(t) \, dt + K \int \left( \frac{ds - ds_0}{ds_0} \right)^2 ds_0 \\ &= \int \rho g a^2 \epsilon_y(t) \cosh(t) \, dt + K \int \frac{ \left( \sqrt{ \cosh^2 t + \epsilon_x'(t)^2 + \epsilon_y'(t)^2 + 2 \epsilon_x'(t) + 2 \epsilon_y'(t) \sinh t } - \cosh t \right)^2 }{ \cosh t } \, dt \\ &= \rho g a^2 \int \left[ \cosh(t) \, \epsilon_y(t) + K_1 \frac{ \left( \sqrt{ \cosh^2 t + \epsilon_x'(t)^2 + \epsilon_y'(t)^2 + 2 \epsilon_x'(t) + 2 \epsilon_y'(t) \sinh t } - \cosh t \right)^2 }{ \cosh t } \right] dt \end{aligned}\]where $K_1 = \dfrac{K}{\rho g a^2}$.
It can be seen that the Lagrangian has the form:
\[\begin{aligned} L\left( \epsilon_y'(t), \epsilon_x'(t), \epsilon_y(t), \epsilon_x(t), t \right) &= \cosh(t) \, \epsilon_y(t) \\ &\quad + K_1 \frac{ \left( \sqrt{ \cosh^2 t + \epsilon_x'(t)^2 + \epsilon_y'(t)^2 + 2 \epsilon_x'(t) + 2 \epsilon_y'(t) \sinh t } - \cosh t \right)^2 }{ \cosh t } \\ &= \cosh(t) \, \epsilon_y(t) + K_1 \frac{ \left( \sqrt{ \cosh^2 t + \phi } - \cosh t \right)^2 }{ \cosh t } \end{aligned}\]where $\phi = \epsilon_x’(t)^2 + \epsilon_y’(t)^2 + 2 \epsilon_x’(t) + 2 \epsilon_y’(t) \sinh t$.
At the same time, we satisfy:
\[\frac{d}{dt} \frac{\partial L}{\partial \epsilon_y'} = \frac{\partial L}{\partial \epsilon_y}\] \[\frac{d}{dt} \frac{\partial L}{\partial \epsilon_x'} = \frac{\partial L}{\partial \epsilon_x}\]After variation:
\[\frac{\partial L}{\partial \phi} \frac{\partial \phi}{\partial \epsilon_y'} = 2 \frac{\partial L}{\partial \phi} (\sinh t + \epsilon_y') = C_1 + \sinh t\] \[\frac{\partial L}{\partial \phi} \frac{\partial \phi}{\partial \epsilon_x'} = 2 \frac{\partial L}{\partial \phi} (1 + \epsilon_x') = C_2\]From the above two equations:
\[\epsilon_y' = (A - 1) \sinh t + (A \sinh t + B) \epsilon_x' + B\]This is the general form when elastic energy is considered. When $\epsilon_y = 0, \epsilon_x = 0$, the above must also hold, giving:
\[A = 1, \quad B = 0\]Thus:
\[\epsilon_y' = \sinh t \, \epsilon_x'\]Substituting into the second equation and simplifying:
\[\epsilon_x'(t) = C \cdot \cosh t\]Finally, solving:
\[\epsilon_x(t) = C \cdot \sinh t\] \[\epsilon_y(t) = C \cdot \frac{\sinh^2 t}{2}\]