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OF A PENDULUM
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Consider the pendulum of Fig. 31.6. As the pendulum is moving in the direction shown, there are two forces acting to oppose its motion. These forces, which act tangentially to the circle of motion, are (1) the gravitational force mg sin 13 and (2) the friction in the pivot, which we suppose to be proportional to the tangential velocity of the mass, BR(d&dt). We shall assume the air resistance to be negligible and the rod to be of negligible mass. Application of Newton s second law gives d20 -mR- = mg sin9 + BR$ dt2
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EXAMPLES OF NONLINEAR SYSTEMS
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FIGURE 31-6 Forces acting on pendulum.
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Rearrangement leads to -+Dg+oisinB = 0 where D = m
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B _ d29 dt2
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(31.9)
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This equation resembles the equation for the motion of the spring-mass-damper system. However, the presence of the term involving sin 8 makes the equation nonlinear. Equation (31.9) has the following form in phase coordinates:
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de i -= d t da, -= -ozsin 8 - 06 dt
(31.10)
and a phase diagram would be a plot of angular velocity 8r versus position 8. At this point, we can gain some insight by simple analysis of Eq. (31.10) without actually obtaining a solution. Referring for the moment back to the spring-mass:damper system, we saw that the system ceased to oscillate when the point X = X = 0 was reached. That is, all curves stopped at the origin of Fig. 3 1 S. The reason for this is quite clear; when X = X = 0 is substituted into Eqs. (31.3), there is obtained
dX di -E-Z o dt dt
Since neither X nor X is changing with time, the motion ceases. Further examination of Eqs. (31.3) shows that X = X = 0 is the only point at which both dX/dt and dildt are zero. Thus, we see that the mass will come to rest only when the situation of zero displacement and zero velocity is reached.
NONLINEAR CONTROL
Now we perform a similar analysis on Eqs. (31.10). we are asking the following question: At what point or points in the phase plane (0 versus 8 diagram) do both dl-lldt and dO/dt become zero From the first of these equations, we see that this can happen* only when 8 = 0. Using this result in the second equation, it can be seen that it is also necessary that sin8 = 0 Equation (3 1.11) is satisfied at any of the points 8 = n7r where it is a positive or negative integer or zero. However, from a physical standpoint, we can really distinguish between only two of these points, which we take as 8 = 0 and 0 = n. Thus, the positions 8 = 0, 27r, 47r, -27r, etc., all look the same to us; i.e., the pendulum is hanging straight down. Similarly, the points 8 = 7r, 37r, etc., all correspond to the pendulum standing straight up. Thus, the analysis leads to the conclusion that the motion will cease when the pendulum comes to rest in either of the positions shown in Fig. 31.7. In addition, it is clear from Eqs. (31.10) that, if the pendulum stops at any other point, the motion continues. Of course, this analysis agrees with our physical intuition. However, we expect to find a distinction between the stability characteristics of the two equilibrium points, since the position at r is likely to be hard to attain and maintain. This distinction will be explored in more detail in Chap. 32. (31.11)
*The reader should not be lulled into a false sense of security at this. point. It would be wise to disregard the fact that dtVdt and ~9 are, in fact, the same quantity; 13 should be thought of as a coordinate in the phase plane, and dO/dt as the rate of change with time of the other coordinate. The virtue of making this distinction will become clear in the next example, a chemical reactor.
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