| course |
course_year |
question_number |
tags |
title |
year |
Variational Principles |
IB |
79 |
IB |
2016 |
Variational Principles |
|
Paper 2, Section II, C |
2016 |
A flexible wire filament is described by the curve $(x, y(x), z(x))$ in cartesian coordinates for $0 \leqslant x \leqslant L$. The filament is assumed to be almost straight and thus we assume $\left|y^{\prime}\right| \ll 1$ and $\left|z^{\prime}\right| \ll 1$ everywhere.
(a) Show that the total length of the filament is approximately $L+\Delta$ where
$$\Delta=\frac{1}{2} \int_{0}^{L}\left[\left(y^{\prime}\right)^{2}+\left(z^{\prime}\right)^{2}\right] d x$$
(b) Under a uniform external axial force, $F>0$, the filament adopts the shape which minimises the total energy, $\mathcal{E}=E_{B}-F \Delta$, where $E_{B}$ is the bending energy given by
$$E_{B}[y, z]=\frac{1}{2} \int_{0}^{L}\left[A(x)\left(y^{\prime \prime}\right)^{2}+B(x)\left(z^{\prime \prime}\right)^{2}\right] d x$$
and where $A(x)$ and $B(x)$ are $x$-dependent bending rigidities (both known and strictly positive). The filament satisfies the boundary conditions
$$y(0)=y^{\prime}(0)=z(0)=z^{\prime}(0)=0, \quad y(L)=y^{\prime}(L)=z(L)=z^{\prime}(L)=0$$
Derive the Euler-Lagrange equations for $y(x)$ and $z(x)$.
(c) In the case where $A=B=1$ and $L=1$, show that below a critical force, $F_{c}$, which should be determined, the only energy-minimising solution for the filament is straight $(y=z=0)$, but that a new nonzero solution is admissible at $F=F_{c}$.