relation between moment of inertia about center of mass and curvature energy for a smooth plane curve of finite fixed length.

Question

Given an inextensible rope of fixed length $L$ and uniform linear density, lying in a plane, mathematically represented by a smooth plane curve $(x(s),y(s))$, parameterized in arc length $s$, of fixed finite length $L$, its moment of inertia about its center of mass (axis perpendicular to the plane) is given as $$I_{cm} = \int_0^L ((x(s)-x_{cm})^2 + (y(s)-y_{cm})^2) ds$$ Let $k(s)$ be its curvature, and consider curvature energy $CE$ given by the expression $$CE = \int_0^L k^2(s) d{s}$$

We know that $I_{cm}$ and $CE$ are uniquely determined by $k(s)$.

My question is, for a curve of fixed length, what kind of relation exists between $I_{cm}$ and $CE$ for a smooth plane curve of fixed length? Is it that, the greater the value of $CE$ then lower the value of $I_{cm}$?

useful Reference : Question

Answer 1

There is no general relationship.

Consider the following three curves:

1. Straight line
2. Spiral
3. Sinusoidal wave (with very small amplitude and high frequency)

Curve 1 will have smaller CE but larger $I_{cm}$ than Curve 2. But Curve 3 will have bother larger CE and larger $I_{cm}$ than Curve 2.