This should be straightforward. The only thing unusual about this transformation from $(\theta, L(\theta)) \to (x,y)$ is that the transformation can be parameterized using one variable. But, it seems to me, that the math should still work. It's still just moving to a rotating system (similar to this problem and here). My answer seems reasonable too. 1.Am I trying to do something that is not possible? 2.Am I making a mistake? Is there an easier way to do this?
$L(\theta)\equiv L-R\theta$
$$\begin{eqnarray} x(\theta,L(\theta) ) & = & R \sin(\theta) + (L - R \theta) \cos(\theta)\\ y(\theta,L(\theta) ) & = & -R \cos(\theta) + (L - R \theta) \sin(\theta) \end{eqnarray}$$
$\equiv$
$$\begin{eqnarray} x(\theta,L(\theta)) & = & R \sin(\theta) + L(\theta) \cos(\theta)\\ y(\theta,L(\theta)) & = & -R \cos(\theta) + L(\theta) \sin(\theta) \end{eqnarray}$$
The physics behind this transformation comes from this problem.
Transforming $g_{cartesian} \to g_{\text{this rotating system}}$
$g_{\theta \theta} = L(\theta)^2 \to g^{\theta \theta} = \frac{1}{L(\theta)^2}$
$g_{rr}=1$
$\Gamma^{\theta} _{\theta \theta}= \frac{-R}{L^2}$
$\Gamma^{\theta} _{\theta L}= \frac{-R}{L}$
$F^r =0$
$F^{\theta}=\ddot{\theta}^2 - \frac{R}{L^2} \omega^2 -\frac{2R}{L} \omega^2=\ddot{\theta}^2-\omega^2(\frac{R}{L^2}+\frac{2R}{L})$
The units in the middle term don't match up.
Any help would be appreciated.