13.3.2.1 Motion of a single particle

Consider the case of a single particle of mass $m$ that moves in ${\cal W}= {\mathbb{R}}$. The force becomes a scalar, $f \in {\mathbb{R}}$. Let $q(t)$ denote the position of the particle in ${\cal W}$ at time $t$. Using this notation, acceleration is ${\ddot q}$, and Newton's second law becomes $f = m {\ddot q}$. This can be solved for ${\ddot q}$ to yield

$${\ddot q}= f / m .$$ (13.50)

If $f$ is interpreted as an action variable $u$, and if $m = 1$, then (13.50) is precisely the double integrator ${\ddot q} = u$ from Example 13.3. Phase variables $x_1 = q$ and $x_2 = {\dot q}$ can be introduced to obtain a state vector $x = (q,{\dot q})$. This means that for a fixed $u$, the motion of the particle from any initial state can be captured by a vector field on ${\mathbb{R}}^2$. The state transition equation is

$$\begin{split}{\dot x}_1 &= x_2 \\ {\dot x}_2 &= \frac{u}{m}, \end{split}$$ (13.51)

in which $x_1 = q$, $x_2 = {\dot q}$, and $u = f$. Let $U = [-f_{max},f_{max}]$, in which $f_{max}$ represents the maximum magnitude of force that can be applied to the particle. Forces of arbitrarily high magnitude are not allowed because this would be physically unrealistic.

Now generalize the particle motion to ${\cal W}= {\mathbb{R}}^2$ and ${\cal W}= {\mathbb{R}}^3$. Let $n$ denote the dimension of ${\cal W}$, which may be $n=2$ or $n=3$. Let $q$ denote the position of the particle in ${\cal W}$. Once again, Newton's second law yields $f = m {\ddot q}$, but in this case there are $n$ independent equations of the form $f_i = m {\ddot q}_i$. Each of these may be considered as an independent example of the double integrator, scaled by $m$. Each component $f_i$ of the force can be considered as an action variable $u_i$. A $2n$-dimensional state space can be defined as $x = (q,{\dot q})$. The state transition equation for $n=2$ becomes

$${\dot x}_1 = x_3 \qquad {\dot x}_3 = u_1/m$$ (13.52)

$${\dot x}_2 = x_4 \qquad {\dot x}_4 = u_2/m ,$$

and for $n=3$ it becomes

$${\dot x}_1 = x_4 \qquad {\dot x}_4 = u_1/m$$

$${\dot x}_2 = x_5 \qquad {\dot x}_5 = u_2/m$$ (13.53)

$${\dot x}_3 = x_6 \qquad {\dot x}_6 = u_3/m .$$

For a fixed action, these equations define vector fields on ${\mathbb{R}}^4$ and ${\mathbb{R}}^6$, respectively. The action set should also be bounded, as in the one-dimensional case. Suppose that

$$U = \{ u \in {\mathbb{R}}^n \;\vert\; \Vert u\Vert \leq f_{max} \} .$$ (13.54)

Now suppose that multiple forces act on the same particle. In this case, the vector sum

$$F = \sum f$$ (13.55)

yields the resultant force over all $f$ taken from a collection of forces. The resultant force $F$ represents a single force that is equivalent, in terms of its effect on the particle, to the combined forces in the collection. This enables Newton's second law to be formulated as $F = m{\ddot q}$. The next two examples illustrate state transition equations that arise from a collection of forces, some of which correspond to actions.

Example 13..7 (Lunar Lander)  

Figure 13.8: There are three thrusters on the lunar lander, and it is under the influence of lunar gravity. It is treated as a particle; therefore, no rotations are possible. Four orthogonal forces may act on the lander: Three arise from thrusters that can be switched on or off, and the remaining arises from the acceleration of gravity.

Using the Newton-Euler model of a particle, an example will be constructed for which $X = {\mathbb{R}}^4$. A lunar lander is modeled as a particle with mass $m$ in a 2D world shown in Figure 13.8. It is not allowed to rotate, implying that ${\cal C}= {\mathbb{R}}^2$. There are three thrusters on the lander, which are on the left, right, and bottom of the lander. The forces acting on the lander are shown in Figure 13.8. The activation of each thruster is considered as a binary switch. Each has its own associated binary action variable, in which the value $1$ means that the thruster is firing and 0 means the thruster is dormant. The left and right lateral thrusters provide forces of magnitude $f_l$ and $f_r$, respectively, when activated (note that the left thruster provides a force to the right, and vice versa). The upward thruster, mounted to the bottom of the lander, provides a force of magnitude $f_u$ when activated. Let $g$ denote the scalar acceleration constant for gravity (this is approximately $1.622$ m/s$^2$ for the moon).

From (13.55) and Newton's second law, $F = m{\ddot q}$. In the horizontal direction, this becomes

$$m \ddot{q_1} = u_l f_l - u_r f_r ,$$ (13.56)

and in the vertical direction,

$$m \ddot{q_2} = u_u f_u - m g .$$ (13.57)

Opposing forces are subtracted because only the magnitudes are given by $f_l$, $f_r$, $f_u$, and $g$. If they were instead expressed as vectors in ${\mathbb{R}}^2$, then they would be added.

The lunar lander model can be transformed into a four-dimensional phase space in which $x = (q_1,q_2,{\dot q}_1,{\dot q}_2)$. By replacing $\ddot{q}_1$ and $\ddot{q}_2$ with $\dot{x}_3$ and $\dot{x}_4$, respectively, (13.56) and (13.57) can be written as

$$\displaystyle\strut \dot{x}_3 = \frac{1}{m} (u_l f_l - u_r f_r)$$ (13.58)

and

$$\displaystyle\strut \dot{x}_4 = \frac{u_u f_u}{m} - g .$$ (13.59)

Using $\dot{x}_1 = x_3$ and $\dot{x}_2 = x_4$, the state transition equation becomes

$${\dot x}_1 = x_3 \qquad {\dot x}_3 = \frac{1}{m} (u_l f_l - u_r f_r)$$

$${\dot x}_2 = x_4 \qquad {\dot x}_4 = \frac{u_u f_u}{m} - g,$$ (13.60)

which is in the desired form, ${\dot x}= f(x,u)$. The action space $U$ consists of eight elements, which indicate whether each of the three thrusters is turned on or off. Each action vector is of the form $(u_l,u_r,u_u)$, in which each component is 0 or $1$. $\blacksquare$

The next example illustrates the importance of Newton's third law.

Example 13..8 (Pendulum)  

Figure 13.9: The pendulum is a simple and important example of a nonlinear system.

A simple and very important model is the pendulum shown in Figure 13.9. Let $m$ denote the mass of the attached particle (the string is assumed to have no mass). Let $g$ denote the acceleration constant due to gravity. Let $L$ denote the length of the pendulum string. Let $\theta$ denote the angular displacement of the pendulum, which characterizes the pendulum configuration. Using Newton's second law and assuming the pendulum moves in a vacuum (no wind resistance), the constraint

$$m L {\ddot \theta}= - m g \sin \theta$$ (13.61)

is obtained. A 2D state space can be formulated in which $x_1 = \theta$ and $x_2 = {\dot \theta}$. This leads to

$$\begin{split}{\dot x}_1 &= x_2 \\ {\dot x}_2 &= -\frac{g}{L} \sin x_1 , \end{split}$$ (13.62)

which has no actions (the form of (13.62) is ${\dot x}= f(x)$).

A linear drag term $k L {\dot \theta}$ can be added to the model to account for wind resistance. This yields

$$m L {\ddot \theta}= - m g \sin \theta - k L {\dot \theta},$$ (13.63)

which becomes

$$\begin{split}{\dot x}_1 &= x_2 \\ {\dot x}_2 &= -\frac{g}{L} \sin x_1 - \frac{k}{m} x_2 \end{split}$$ (13.64)

in the state space form.

Now consider applying a force $u_f$ on the particle, in a direction perpendicular to the string. This action can be imagined as having a thruster attached to the side of the particle. This adds the term $u_f$ to (13.63). Its sign depends on the choice of the perpendicular vector (thrust to the left or to the right). The state transition equation ${\dot x}= f(x,u)$ then becomes

$$\begin{split}{\dot x}_1 &= x_2 \\ {\dot x}_2 &= -\frac{g}{L} \sin x_1 - \frac{k}{m} x_2 + \frac{1}{mL} u_f . \end{split}$$ (13.65)

$\blacksquare$

Although sufficient information has been given to specify differential models for a particle, several other concepts are useful to introduce, especially in the extension to multiple particles and rigid bodies. The main idea is that conservation laws can be derived from Newton's laws. The linear momentum (or just momentum) $d$ of the particle is defined as

$$d = m {\dot q}.$$ (13.66)

This is obtained by integrating $f = m {\ddot q}$ with respect to time.

It will be convenient when rigid-body rotations are covered to work with the moment of momentum (or angular momentum). A version of momentum that is based on moments can be obtained by first defining the moment of force (or torque) for a force $f$ acting at a point $q \in W$ as

$$n = q \times f ,$$ (13.67)

in which $\times$ denotes the vector cross product in ${\mathbb{R}}^3$. For a particle that has linear momentum $d$, the moment of momentum $e$ is defined as

$$e = q \times d .$$ (13.68)

It can be shown that

$$\frac{de}{dt} = n ,$$ (13.69)

which is equivalent to Newton's second law but is expressed in terms of momentum. For the motion of a particle in a closed system, the linear momentum and moment of momentum are conserved if there are no external forces acting on it. This is essentially a restatement of Newton's first law.

This idea can alternatively be expressed in terms of energy, which depends on the same variables as linear momentum. The kinetic energy of a particle is

$$T = \frac{1}{2} m {\dot q}\cdot {\dot q},$$ (13.70)

in which $\cdot$ is the familiar inner product (or dot product). The total kinetic energy of a system of particles is obtained by summing the kinetic energies of the individual particles.