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

$${F}(u) = {d{D}\over dt} = {m}{\ddot p}.$$ (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 $(\ref{eqn:momeq})$ 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

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

$${\dot p}_1 = v_1 \qquad {\dot v}_1 = {F}_1(u)/{m}$$

$${\dot p}_2 = v_2 \qquad {\dot v}_2 = {F}_2(u)/{m}$$ (13.109)

$${\dot \theta} = \omega \qquad {\dot \omega} = N(u)/I .$$