The 2D case

The dynamics of a 2D rigid body that moves in the plane can be handled as a special case of a 3D body. Let $ {\cal A}\subset {\mathbb{R}}^2$ be a 2D body, expressed in its body frame. The total external forces acting on $ {\cal A}$ can be expressed in terms of a two-dimensional total force through the center of mass and a moment through the center of mass. The phase space for this model has six dimensions. Three come from the degrees of freedom of $ SE(2)$, two come from linear velocity, and one comes from angular velocity.

The translational part is once again expressed as

$\displaystyle {F}(u) = {d{D}\over dt} = {m}{\ddot p}.$ (13.107)

This provides four components of the state transition equation.

All rotations must occur with respect to the $ z$-axis in the 2D formulation. This means that the angular velocity $ \omega$ is a scalar value. Let $ \theta $ denote the orientation of $ {\cal A}$. The relationship between $ \omega$ and $ \theta $ is given by $ {\dot \theta}=
\omega$, which yields one more component of the state transition equation.

At this point, only one component remains. Recall (13.92). By inspecting % latex2html id marker 155763
$ (\ref{eqn:momeq})$ it can be seen that the inertia-based terms vanish. In that formulation, $ \omega_3$ is equivalent to the scalar $ \omega$ for the 2D case. The final terms of all three equations vanish because $ \omega_1 = \omega_2
= 0$. The first terms of the first two equations also vanish because $ {\dot \omega}_1 = {\dot \omega}_2 = 0$. This leaves $ {N}_3(u) = I_{33}
{\dot \omega}_3$. In the 2D case, this can be notationally simplified to

$\displaystyle {N}(u) = {d{E}\over dt} = \frac{d(I\omega)}{dt} = I \frac{d\omega}{dt} = I {\dot \omega},$ (13.108)

in which $ I$ is now a scalar. Note that for the 3D case, the angular velocity can change, even when $ {N}(u) = 0$. In the 2D case, however, this is not possible. In both cases, the moment of momentum is conserved; in the 2D case, this happens to imply that $ \omega$ is fixed. The sixth component of the state transition equation is obtained by solving (13.108) for $ {\dot \omega}$.

The state transition equation for a 2D rigid body in the plane is therefore

$\displaystyle {\dot p}_1$ $\displaystyle = v_1$ $\displaystyle \qquad {\dot v}_1$ $\displaystyle = {F}_1(u)/{m}$    
$\displaystyle {\dot p}_2$ $\displaystyle = v_2$ $\displaystyle \qquad {\dot v}_2$ $\displaystyle = {F}_2(u)/{m}$ (13.109)
$\displaystyle {\dot \theta}$ $\displaystyle = \omega$ $\displaystyle \qquad {\dot \omega}$ $\displaystyle = N(u)/I .$    

Steven M LaValle 2012-04-20