{\bar \kappa}(t) = \kappa(s(t)),$$ where $s(t)$ and $\dot s(t)$ are solutions of a differential equation to be specified later. The motion of the bicycle on the curve $\C$ is now found by substituting (\[eqn\_for\_v\]), (\[eqn\_for\_v\_dot\]), (\[eqn\_for\_bar\_kappa\]) and $\tau = \kappa'(s) \dot s$ in (\[eq:original\_system\]): $$\label{eq:constrained_system}
\bar{M} \begin{bmatrix} \ddot \varphi \\ \|\sigma^\prime\| \ddot {s} +
\frac{(\sigma')\trans \sigma''}{\|\sigma'\|} \dot{s}^2 
\end{bmatrix} = 
\bar F + \begin{bmatrix} 0 \\1/m \end{bmatrix} f\;,$$ where $\bar M=M|_{\bar \kappa =\kappa(s)}$ and $$\bar F = \left(F+B \begin{bmatrix} 1 \\ 0
\end{bmatrix}\kappa'(s) \dot s
\right)\bigg|_{\bar \kappa =\kappa(s), v=\|\sigma'(s)\|\dot s}.$$ Note that we have used the control input $\tau$ to make the curve $\C$ invariant, so in  we are left with one control input, the thrust force $f$. One can check that system  is a Lagrangian control system with input $f$, i.e., it has the form  with $q=
(\varphi,s)\in \cQ =S^1 \times [\Re]_T$, $G = [0 \ \ 1/m]\trans$, and $$L(q,\dot q) = \frac 1 2  [\dot
  \varphi \ \ \|\sigma^\prime(s)\| \dot{s}] \bar{M}(q) \begin{bmatrix}
  \dot \varphi \\ \|\sigma^\prime(s)\| \dot{s}
\end{bmatrix}- g h \cos \varphi.$$ Since the control force $f$ enters nonsingularly in the $\ddot s$ equation, we can define a feedback transformation for $f$ in  such that $\ddot s=u$, where $u$ 