What is the potential energy for a large spring that is bent into an arbitrary shape? The solution should be of the form $V=\int (…) ds $
EDIT:
I have reduced this problem to the problem of the 2D elastica, or a bent elastic rod (no stretching or twisting). (Following Equation 10 in https://www2.eecs.berkeley.edu/Pubs/TechRpts/2008/EECS-2008-103.pdf and many other sources):
The energy for this system is $$E[\theta(s)] = \int_0^Lds \frac1{R^2} =\int_0^Lds \left(\frac{d \theta}{ds}\right)^2 \tag{1}$$
Where $R$ is the radius of curvature, $R d\theta = ds$.
Varying the energy using Euler-Lagrange equations gives $\frac{d\theta}{ds}=constant$. This means the solution is just some section of a circle. This can't be right and does not allow for a positional boundary condition on the endpoint.
To fix this issue, consider Equation 22 of https://www2.eecs.berkeley.edu/Pubs/TechRpts/2008/EECS-2008-103.pdf, which seems to be the form of energy I am looking for, and allows for two additional positional boundary conditions for the endpoint.
$$\theta'' + \lambda_1 \cos(\theta) +\lambda_2 \sin(\theta)=0\tag{2}$$
An outline/resources on the solution of (2) would be appreciated (assuming this is the equation I am looking for!).
Also, (2) reduces to (1) in the limit that $\lambda_1=\lambda_2=0$. Is it ever physically meaningful to consider (1) or is (1) just a completely incorrect/meaningless equation?
Best Answer
In general, you have to find the potential for individual components of the spring and sum them up over the trajectory.[Integrate] Remember you do not have to increase k as you are not physically disconnecting the components. And the equation of the shape matters [because through this you will have the displaced height from the equilibrium] My approach! (Based on your constrained process) is given below, $$V=\frac{1}{2}ky^{2}$$ $$dV=kydy$$ $$dy=\frac{dV}{ky}----(I)$$ let assume the equation of the shape is given by y=f(x), y is the vertical displacement from the equilibrium. $${\dot{y}}=f'(x)=\frac{dy}{dx}$$ Thus, $$dy=f'(x)dx----(II)$$ From eq-(I) & (II) $$dx=\frac{dV}{kf(x)f'(x)}$$ As we know, $$ds=\sqrt{(dx)^{2}+(dy)^{2}}$$ Substituting dx and dy, $$ds=\frac{dV}{kf(x)}\sqrt{1+\left(\frac{dy}{dx}\right)^{-2}}$$ That implies, $$dV=\frac{ky}{\sqrt{1+(\dot{y})^{-2}}}ds$$ Therefore, $$V=k\int_{s_{1}}^{s_{2}}\left(\frac{y}{\sqrt{1+(\dot{y})^{-2}}}\right)ds$$