Wire basket shape for maximum Moment of Inertia

Question

Attempted to find the meridian shape for optimum moment of inertia with a minimum mass of wire constraint and like to share in the iso-perimetric problem in the plane.

Assuming constant density per unit length

$$ \int y^2 ds - \lambda^2 \int ds= \int( y^2-\lambda^2) ds ,\; {ds}/{dx}=\sqrt{1+y^{'2}}\tag1$$

Lagrangian is

$$ ( y^2-\lambda^2) \sqrt{1+y^{'2}}$$

Using Euler-Lagrange equations on the Lagrangian we obtain

$$( y^2-\lambda^2) \sqrt{1+y^{'2}} -y'\cdot ( y^2-\lambda^2)\dfrac{y'}{\sqrt{1+y^{'2}}} = c^2 $$

$$\dfrac{y^2-\lambda^2}{\sqrt{1+y^{'2}}} = c^2 \tag2 $$

To eliminate $c^2$ raising to second order with one differentiation with respect to arc

$$\dfrac{1}{c^2} =\dfrac{\sec \phi}{y^2- \lambda^2} = \dfrac{ \sec \phi \tan \phi\;k(s)}{ 2 y \sin \phi } $$

$$ \text{curvature} \;k(s)= \dfrac{d \phi}{ds} , \quad k(s)= \dfrac{ 2 y \cos ^2\phi}{c^2} \tag3 $$

changing $y=r$ as symbol for radius in cylindrical coordinates and since $ \sin \phi = \dfrac{dr}{ds},$

$$\sec^2 \phi \sin \phi\; d\phi=\sec \phi \tan \phi\; d\phi=\dfrac{2 r dr}{c^2} \tag4 $$

Integrating

$$ \sec\phi =\dfrac{r^2}{c^2}+\text{const.} \tag 5 $$

and initial condition $ \phi=0, r = r_1$

$$ \sec \phi -1 = \dfrac{r^2-r_1^2}{c^2} \tag 6 $$

which is the ODE for meridians of the basket. For the given graph initial conditions are at vertex $ \phi_{i}=0, r_i=0.8 $. Maximum curvature at vertex is $ k_{max}=\dfrac{2r_1}{c^2}$.

Please comment about correctness and or whether the curve is already known.

Answer 1

From the above maximum curvature for other $c$ values are

$$ k_{max_1}=2 r_1/c^2$$

Other curves can be generated and basket profiles are shown plotted:

Answer 2

You formulate $$ I[y]=\int L(y,y')ds=\int (y^2-λ^2)\sqrt{1+y'^2}\,dx $$ with the start position fixed at the rotation axis and the other end free.

The Euler-Lagrange equations are $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$ \begin{align} 0=\frac{d}{dt}\pd{L}{y'}-\pd{L}{y} &=\frac{d}{dt}\left((y^2-λ^2)\frac{y'}{\sqrt{1+y'^2}}\right)-2y\sqrt{1+y'^2} \\ &=-\frac{2y}{\sqrt{1+y'^2}}+(y^2-λ^2)\frac{y''}{\sqrt{1+y'^2}^3} \end{align} Separating the derivatives gives $$ 0=\frac{y'y''}{1+y'^2}-\frac{2yy'}{y^2-λ^2} \\ (y^2-λ^2)=C\sqrt{1+y'^2} $$ As for $y=0$ at the rotation axis the left side is negative, so must be the constant $C=-c^2$ $$ λ^2=c^2\sqrt{1+y'^2}+y^2 $$