Potential energy of a bent spring

Question

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 $

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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?

Answer 1

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, $$V=k\int_{s_{1}}^{s_{2}}\left(\frac{y}{\sqrt{1+(\dot{y})^{-2}}}\right)ds$$

Answer 2

I am not sure whether I have understood your question correctly, but if it is going to help you, I could outline a derivation of equation (2), which after the application of a trigonometric identity, can be converted into pendulum's equation.

Fix a 2D inertial, orthonormal coordinate system $O\,\vec{i}\,\vec{j}$

Parametrize the curve by arc-length: $$\vec{r}(s) = x(s)\,\vec{i} \,+\, y(s)\,\vec{j}$$ where $s \in [0, L]$ is the arc-length parameter. An arc-length parametrization is characterized by the identity $$\left\|\frac{d\vec{r}}{ds}\right\|^2 \,=\,\Big(\,\frac{dx}{ds}\,\Big)^2+\, \Big(\,\frac{dy}{ds}\,\Big)^2 \,=\, 1$$ which is a constraint. Then, after denoting by $R = R(s)$ radius of curvature at distance $s$ from the origin of the curve, $$\frac{1}{R^2} \,=\, \left\|\frac{d^2\vec{r}}{ds^2}\right\|^2 \,=\, \left(\frac{d^2x}{ds^2}\right)^2 \,+\, \left(\frac{d^2y}{ds^2} \right)^2$$ The length of the curve is fixed, and on top of that usually the endpoints are also fixed: one end is at the origin $O$ of the coordinate system, the other end is at positive coordinate position $(a, \,b) \,=\, a\,\vec{i} + b\,\vec{j}$ which yields the constraints $$\vec{r}(0) \,=\, \vec{0}\,\,\text{ and }\,\,\vec{r}(L) \,=\, a\,\vec{i} + b\,\vec{j}$$ Consequently, the constraint Lagrangian is $$\int_0^{L} \, \frac{1}{2}\,\left\|\frac{d^2\vec{r}}{ds^2}\right\|^2\,ds \,+\, \lambda \, \big(\,\vec{i}\cdot\vec{r}(L) - a\,\big) \,+\, \mu \, \big(\,\vec{j}\cdot\vec{r}(L) - b\,\big)\,+\, \nu\,\left(\,\left\|\frac{d\vec{r}}{ds}\right\|^2 - 1 \,\right)$$ which, by Newton-Leibniz, can be written as $$\int_0^{L} \, \frac{1}{2}\,\left\|\frac{d^2\vec{r}}{ds^2}\right\|^2\,ds \,+\, \lambda \, \Big(\,\int_{0}^{L}\, \Big(\, \vec{i}\cdot\frac{d\vec{r}}{ds}\,\Big)\,ds - a\,\Big) \,+\, \mu \, \Big(\,\int_{0}^{L}\, \Big(\, \vec{j}\cdot\frac{d\vec{r}}{ds}\,\Big)\,ds - b\,\Big)\,+\, \nu\,\left(\,\left\|\frac{d\vec{r}}{ds}\right\|^2 - 1 \,\right)$$ an then $$\int_0^{L} \,\left(\, \frac{1}{2}\,\left\|\frac{d^2\vec{r}}{ds^2}\right\|^2 \,+\, \lambda\, \Big(\, \vec{i}\cdot\frac{d\vec{r}}{ds}\,\Big) \,+\, \mu \, \Big(\, \vec{j}\cdot\frac{d\vec{r}}{ds}\,\Big)\,\right)\,ds\,+\, \nu\,\left(\,\left\|\frac{d\vec{r}}{ds}\right\|^2 - 1 \,\right) \,-\, \lambda\,a - \mu\, b$$ where the constant terms can be removed, yielding the constraint Lagrangian $$\int_0^{L} \,\left(\, \frac{1}{2}\,\left\|\frac{d^2\vec{r}}{ds^2}\right\|^2 \,+\, \lambda\, \Big(\, \vec{i}\cdot\frac{d\vec{r}}{ds}\,\Big) \,+\, \mu \, \Big(\, \vec{j}\cdot\frac{d\vec{r}}{ds}\,\Big)\,\right)\,ds\,+\, \nu\,\left(\,\left\|\frac{d\vec{r}}{ds}\right\|^2 - 1 \,\right)$$ In order the reduce the order of the derivatives, substitute $$\vec{u}(s) \,=\, \frac{d\vec{r}}{ds}(s)$$ and the Lagrangian is now $$\int_0^{L} \,\left(\, \frac{1}{2}\,\left\|\frac{d\vec{u}}{ds}\right\|^2 \,+\, \lambda\, \Big(\, \vec{i}\cdot\vec{u}\,\Big) \,+\, \mu \, \Big(\, \vec{j}\cdot\vec{u}\,\Big)\,\right)\,ds\,+\, \nu\,\left(\,\left\|\vec{u}\right\|^2 - 1 \,\right)$$ One can easily remove the last holonomic constraint by introducing $\theta = \theta(s)$ such that $$\vec{u}(s) \,=\, \cos\theta(s)\,\vec{i}\,+\, \sin\theta(s)\,\vec{j}$$ which turns the Lagrangian into $$\int_0^{L} \,\left(\, \frac{1}{2}\,\left(\,\frac{d\theta}{ds}\,\right)^2 \,+\, \lambda\, \cos(\theta) \,+\, \mu \, \sin(\theta)\,\right)\,ds$$ The Euler-Lagrange equation is $$\frac{d^2\theta}{ds^2}\,=\, -\,\lambda \,\sin(\theta) \,+\, \mu\, \cos(\theta)$$ By certain trigonometric identities, one can find $\omega$ such that if one makes the substitution $\varphi \,=\, \theta + \omega$ the equation becomes something like
$$\frac{d^2\varphi}{ds^2} \,=\, -\, \Big(\,\sqrt{\lambda^2 + \mu^2}\,\Big)\,\cos(\varphi)$$

Answer 3

I think you are asking about what is called an an elastica.

The energy is given by $$ E=\int \frac 12 (\lambda \kappa^2 + \nu \tau^2) ds $$ where $s$ is the arc length, $\lambda$ and $\nu$ are constants. The functions $\kappa$ and $\tau$ are the curvature and torsion of the curve ${\bf r}(s)$ that are defined by the Serret-Frenet equations: $$ \frac{d{\bf t}}{ds}=\kappa {\bf n}\\ \frac{d{\bf n}}{ds}=-\kappa {\bf n}+\tau {\bf b}\\ \frac{d{\bf b}}{ds}= - \tau{\bf n} $$ Here $$ {\bf t}= \frac{d{\bf r}}{ds} $$ is the unit tangent, and the normal ${\bf n}$ and binormal ${\bf b}$ vectors are implicitly defined by these equations.

There is a large literature --- starting from Euler who defined the notion of an elastic rod --- on these equations and their solutions