Consider the case of a single particle of mass that moves in . The force becomes a scalar, . Let denote the position of the particle in at time . Using this notation, acceleration is , and Newton's second law becomes . This can be solved for to yield
(13.51) |
Now generalize the particle motion to and . Let denote the dimension of , which may be or . Let denote the position of the particle in . Once again, Newton's second law yields , but in this case there are independent equations of the form . Each of these may be considered as an independent example of the double integrator, scaled by . Each component of the force can be considered as an action variable . A -dimensional state space can be defined as . The state transition equation for becomes
(13.52) | ||||
(13.53) | ||||
(13.54) |
Now suppose that multiple forces act on the same particle. In this case, the vector sum
Using the Newton-Euler model of a particle, an example will be constructed for which . A lunar lander is modeled as a particle with mass in a 2D world shown in Figure 13.8. It is not allowed to rotate, implying that . 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 means that the thruster is firing and 0 means the thruster is dormant. The left and right lateral thrusters provide forces of magnitude and , 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 when activated. Let denote the scalar acceleration constant for gravity (this is approximately m/s for the moon).
From (13.55) and Newton's second law, . In the horizontal direction, this becomes
The lunar lander model can be transformed into a four-dimensional phase space in which . By replacing and with and , respectively, (13.56) and (13.57) can be written as
(13.58) |
(13.59) |
(13.60) |
The next example illustrates the importance of Newton's third law.
A simple and very important model is the pendulum shown in Figure 13.9. Let denote the mass of the attached particle (the string is assumed to have no mass). Let denote the acceleration constant due to gravity. Let denote the length of the pendulum string. Let 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
(13.61) |
A linear drag term can be added to the model to account for wind resistance. This yields
Now consider applying a force 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 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 then becomes
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) of the particle is defined as
(13.66) |
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 acting at a point as
(13.67) |
(13.68) |
(13.69) |
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
Steven M LaValle 2020-08-14