Table of Contents
Unforced System
Euler-Lagrange
Conder the mechanical system given by
with moment of inertia $J$, mass $m$. This single body system has $Q_{free}=\mathbb{R}^3\times SO(3)$; however due to the imposed contraints, the configuration manifold $Q=\mathbb{S}^1\times\mathbb{R}$, which is an embedded submanifold of $Q_{free}$ through the following mapping
\[ \mathbb{S}^1\times\mathbb{R}\ni(\theta,x)\to(\mathbf{R},(x,0,0))\in SO(3)\times\mathbb{R}^3\]
with
\[\mathbf{R} = \begin{bmatrix}
\cos\theta & -\sin\theta & 0 \
\sin\theta & \cos\theta & 0 \
0 & 0 & 1 \
\end{bmatrix}.
\]
We can define our body and spatial angular velocities as
\[
\hat{\Omega}=\mathbf{R}^\top\dot{\mathbf{R}}=\hat{\omega}=\dot{\mathbf{R}}\mathbf{R}^\top=\dot{\theta}\begin{bmatrix}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 0 \
\end{bmatrix},
\]
which are equal because the local and global out of plane axes align. With this, we can define our kinetic energies as
\[
\begin{align}
KE_{trans}(t) &= \frac{1}{2}\mu(B)\Vert\dot{r}(t)\Vert_{\mathbb{R}^3}^2 = \frac{1}{2}\dot{x}^2\notag\\
KE_{rot}(t) &=
\frac{1}{2}\mathbb{G}_{\mathbb{R}^3}(\mathbb{I}_c(\Omega(t)),\Omega(t))=
\frac{1}{2}J\dot{\theta}^2\notag\\
KE(t) &= \frac{1}{2}(m\dot{x}^2(t)+J\dot{\theta}^2(t))\notag
\end{align}
\]
We can construct our Lagrangian $L(t)=KE(t)$, and evaluate the unforced Euler-Lagrange equations
\[
\frac{d}{dt}\left(\frac{\partial L}{\partial v^i}\right)-\frac{\partial L}{\partial q^i}=0
\]
to get
\[
\begin{align}
m\ddot{x}(t)&=0\notag\
J\ddot{\theta}(t)&=0\notag
\end{align}
\]
the solutions of which are
\[
\begin{align}
x(t) &= c_1 t + c_2\notag\
\theta(t) &= c_3 t + c_4\notag
\end{align}
\]
Geodesics
Consider the unit cylinder as defined by $C=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=1\}\subset\mathbb{R}^3$. Next, let $M=\mathbb{R}^2\backslash\{(0,0)\}$ with the following diffeomorphism
\[
\begin{align}
\phi:M&\to C\notag\
(\theta,r)&\to(\cos\theta,\sin\theta,\log(r))\notag.
\end{align}
\]
We can now construct our differentials as
\[
\begin{align}
dx &= \frac{\partial x}{\partial \theta}d\theta+\frac{\partial x}{\partial r}dr=-\sin\theta d\theta \notag\
dy &= \frac{\partial y}{\partial \theta}d\theta+\frac{\partial y}{\partial r}dr=\cos\theta d\theta \notag\
dr &= \frac{\partial r}{\partial \theta}d\theta+\frac{\partial r}{\partial r}dr= dr \notag\
\end{align}
\]
and perform a coordinate transform on the standard Euclidean Riemannian metric to get
\[
\begin{align}
\mathbb{G}&=dx\otimes dx+dy\otimes dy+dz\otimes dz\notag\
&=d\theta\otimes d\theta+\frac{1}{r^2}dr\otimes dr\notag.
\end{align}
\]
We can calculate the Christoffel symbols by
\[
\Gamma^k_{ij}=\frac{1}{2}\mathbb{G}^{kl}\left(\frac{\partial G_{il}}{\partial x^j}+\frac{\partial G_{jl}}{\partial x^i}-\frac{\partial G_{ij}}{\partial x^l}\right)
\]
with the only nonzero Christoffel symbol being
\[
\Gamma^r_{rr}=\frac{1}{2}r^2(-2r^{-3})=-\frac{1}{r}.
\]
Finally, with the following formula for the geodesic equations
\[
\ddot{x}^k(t)+\Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t)=0
\]
we get
\[
\begin{align}
\ddot{r}(t)-\frac{1}{r}\dot{r}^2(t)&=0\notag\
\ddot{\theta}(t)&=0\notag.
\end{align}
\]
The solutions have the following form
\[
\begin{align}
r(t)&=c_1e^{c_2 t}\notag\
\theta(t)&=c_3 t+c_4\notag,
\end{align}
\]
which, after begin transformed to the coordinates of $C$ become
\[
\begin{align}
x&=\cos(c_3 t + c_4)\notag\
y&=\sin(c_3 t + c_4)\notag\
z&=c_3 t + c_4\notag\
\end{align}
\]
This example has shown that the solutions to the unforced Euler-Lagrange equations align with the geodesic equations of the configuration manifold.
Forced System
Based on the Lagrange D’Alembert principle, we know that for the forced system
\[
\frac{d}{dt}\left(\frac{\partial L}{\partial v^i}\right)-\frac{\partial L}{\partial q^i}=Q_i
\]
where \(Q_i\) represents the generalized force in the \(q_i\) degree-of-freedom. For the above example, assume that \(F^1\) is applied along the \(x\) direction, and \(F^2\) is applied along the $\theta$ direction. Then, we have that
\[
m\ddot{r}(t)=F^1(t)\
J\ddot{\theta}(t)=F^2(t).
\]
Let $\gamma(t)=(r(t),\theta(t))$ represent a solution to the mechanical system. In the unforced case, we had that
\[
\nabla_{\gamma’(t)}\gamma’(t)=0,
\]
exactly because it was a geodesic. In the forced case, we have that
\[
\nabla_{\gamma’(t)}\gamma’(t)=\mathbb{G}^\sharp(F(t),\gamma’(t)).
\]
This means that the “geometric acceleration” in the local frame of the configuration manifold is exactly the force being applied to the system. Consider the following examples
Force Independent of State
In this case, we apply the following force to the system
\[
\begin{align}
F^1(t) &= \alpha\in\mathbb{R}\notag\
F^2(t) &= \sin(\omega t)\in\mathbb{R}\notag
\end{align}
\]
We then have
\[
\begin{align}
r(t) &= \frac{\alpha}{2m}t^2+v_0 t + x_0\notag\
\theta(t) &= -\frac{\sin(\omega t)}{J\omega^2}+\dot{\theta}_0t+\theta_0\notag
\end{align}
\]