Position/velocity of a fixed point as a function of rotation, on a rolling body with arbitrary curvature. I'm trying to find how the curvature of a rolling shape influences its rolling behavior, and in particular the velocity of the center of mass (or of some fixed point on the shape).

Consider a rigid body with arbitrary curvature, such as the egg in the figure, that rolls without slipping The rolling velocity $\frac{d\alpha}{dt}$ is determined by the angular momentum and moment of inertia around the contact point
$$
L = I(\alpha)\frac{d\alpha}{dt}.
$$
The angular momentum around the contact point is
$$
I(\alpha) = I_0 + m r^2
$$
where $I_0$ and $m$ are constants.
I think we should be able to get $\dot{r} = \frac{ds}{dt} + r\frac{d\alpha}{dt}$, that is, the velocity of the contact point itself plus the rotational velocity of the center-of-mass about the contact point.
The first part is simple, as we can use the constant angular momentum (see above) to solve for $d\alpha$ and then plug that into the definition for curvature:
$$
\kappa = \frac{d\alpha}{ds} \\
\kappa = \frac{L dt}{I(\alpha)ds} \\
\frac{ds}{dt} = \frac{L}{I(\alpha)\kappa} = \frac{LR}{I(\alpha)}.
$$
We can then plug this into
$$
\dot{r} = \dot{s} + \dot{\alpha} = \frac{LR}{I(\alpha)} + \frac{L}{I(\alpha)} \\
 = \frac{L R}{I_0 + mr^2} + \frac{L}{I_0 + mr^2}
$$
Question
However, I'm a bit stumped on the relation between $R$ and $r$. How do we get:
$$ \frac{\partial r}{\partial R}$$
?
 A: Assume we have the following set-up:

*

*a convex loop of length $2\pi$:, parametrised by arclength $s\in S^1=\mathbb{R}/2\pi\mathbb{Z}$, i.e., an angle diffeomorphism $\psi\in\operatorname{Diff}(S^1)$ (basically your $\alpha$).  The condition $\int_{S^1}e^{i\psi(s)}\,\mathrm{d}s=0$ ensures we have a simple closed loop and $\psi$ a diffeomorphism ensures strict convexity for rolling motion, and let's just choose orientation so that $\kappa=\psi'>0$;

*a linear mass density function $\rho\colon S^1\to\mathbb{R}^+$, which we will assume have total mass $1$ to save some typing; and

*we will only consider variation of $\psi$ keeping the total length ($2\pi$) and the mass distribution along the curve constant.

You can consider other variants (e.g., lemina, varying perimeter) in similar ways but you will need to take care of, e.g., converting between different coordinate systems.
Then the location of the centre of mass is
$$
\mathbf{r}(s_0)=
\mathbf{t}(s_0)\int_{S^1}\rho(s)\int_{s_0}^s\cos(\psi(\sigma)-\psi(s_0))\,\mathrm{d}\sigma\,\mathrm{d}s
+\mathbf{n}(s_0)
\int_{S^1}\rho(s)\int_{s_0}^s\sin(\psi(\sigma)-\psi(s_0))\,\mathrm{d}\sigma\,\mathrm{d}s
$$
with respect to the (tangent,normal) frame at $s_0$, i.e., the components are
\begin{align*}
T(s_0)&=\int_{S^1}\rho(s)\int_{s_0}^s\cos(\psi(\sigma)-\psi(s_0))\,\mathrm{d}\sigma\,\mathrm{d}s\\
&=\int_{s_0}^{s_0+2\pi}\int_{s}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\cos(\psi(s)-\psi(s_0))\,\mathrm{d}s
\\
N(s_0)&=\int_{S^1}\rho(s)\int_{s_0}^s\sin(\psi(\sigma)-\psi(s_0))\,\mathrm{d}\sigma\,\mathrm{d}s\\
&=\int_{s_0}^{s_0+2\pi}\int_{s}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\sin(\psi(s)-\psi(s_0))\,\mathrm{d}s
\end{align*}
Now take small $C^1$-variation $\delta\psi$ of $\psi$, fixing the gauge $\delta\psi(s_0)=0$ (so $\mathbf{t}(s_0)$ and $\mathbf{n}(s_0)$ are kept constant) and also being a closed curve $\int_{S^1} e^{i(\psi+\delta\psi)}=0$.
\begin{align*}
\delta T(s_0)&=\int_{s_0}^{s_0+2\pi}\int_s^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,(-\sin(\psi(s)-\psi(s_0)))(\delta\psi(s)-\delta\psi(s_0))\,\mathrm{d}s+o(\lVert\delta\psi\rVert)
\\
\delta N(s_0)&=\int_{s_0}^{s_0+2\pi}\int_{s}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\cos(\psi(s)-\psi(s_0))(\delta\psi(s)-\delta\psi(s_0))\,\mathrm{d}s
+o(\lVert\delta\psi\rVert)
\end{align*}
Since $\psi(s_0)=0$ and $\psi'=\kappa$, such $\delta\psi$ are of course completely determined by the $C^0$-small $\delta\kappa$
\begin{align*}
\delta T(s_0)&=\int_{s_0}^{s_0+2\pi}\int_{s}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,(-\sin(\psi(s)-\psi(s_0)))\int_{s_0}^s\delta\kappa(s')\,\mathrm{d}s'\,\mathrm{d}s+o(\lVert\delta\kappa\rVert)
\\
\delta N(s_0)&=\int_{s_0}^{s_0+2\pi}\int_{s}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\cos(\psi(s)-\psi(s_0))\int_{s_0}^s\delta\kappa(s')\,\mathrm{d}s'\,\mathrm{d}s
+o(\lVert\delta\kappa\rVert)
\end{align*}
So we have the functional derivatives
\begin{align*}
\frac{\delta T(s_0)}{\delta\kappa(s)}&=\int_{s_0}^s\sin(\psi(s')-\psi(s_0))\int_{s'}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\mathrm{d}s'
\\
&=\int_{s_0}^s\left(\int_{s'}^{s_0+2\pi}\rho\right)\sin\left(\int_{s_0}^{s'}\kappa\right)\,\mathrm{d}s'
\\
\frac{\delta N(s_0)}{\delta\kappa(s)}&=\int_s^{s_0+2\pi}\cos(\psi(s')-\psi(s_0))\int_{s'}^{s_0+2\pi}\rho(\sigma)\,\mathrm{d}\sigma\,\mathrm{d}s'\\
&=\int_s^{s_0+2\pi}\left(\int_{s'}^{s_0+2\pi}\rho\right)\cos\left(\int_{s_0}^{s'}\kappa\right)\,\mathrm{d}s'
\end{align*}
Of course, we also have
$$
R=\frac1\kappa,\quad\mathbf{R}=R\mathbf{n}.
$$
