Classical Mechanics I
Classical Mechanics I PHYS 3201
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Date Created: 11/02/15
Phys 3201 Special Topic Spring 2005 Satellite Maneuver Free Return Trajectory These notes are a supplement to the textbooks Chapter 8 The same physies which ecs plains planetary motion can be used to maneuver rockets and satellites in useful ways The free return trajectory is a kind of insurance policy early in the mission the ship is put in a trajectory which will automatically return to Earth just in case it becomes dif cult or impossible to re the rockets For example consider the situation where the ship square is halfway between the Earth diamond and the moon circle We want to put it into a trajectory which will wind around the moon and rendezvous with Earth To keep things simple were going to ignore the gravitational pull of the Earth all pictures are represented in the frame of reference of the moon which sits at the origin Now the rst thing to realize is that the Earth is going to move from its present spot Let7s suppose that it will move to the point indicated by the new diamond on the next picture 1 i O 05 7 0 7 O D O 7 05 7 7 1 1 5 i l l l i i 2 1 5 1 05 0 0 5 We can use all our results for the Kepler problem to deduce the desired trajectory This wouldnt be true if we also included the gravitational pull ofthe Earth First of all7 let7s plan to put the ship in an elliptical trajectory with minimum kinetic energy at the rendezvous point This makes some sense we want the ship to be going as slowly as possible for a re entry The orbital equation is for origin at the rightmost focus and major axis along the line 6 0 04 T T 1 600 19 The ship initially lies a distance RZ from the moon and at angle 7139 7 around from the positive major axis7 so 04 Bi 16cos7rz The rendezvous point lies a distance Rf from the moon and at an angle 7139 around from the positive major axis so i a 7 These equations can be inverted to nd the needed values of E and a After a little algebra one nds Rf E i Rf 7 R Rf 7 R cos 7 and Oz Rf 1 7 6 From these we readily determine the total energy and angular momentum of the orbit L Mk0 k E a 62 7 1 where k GmM Here m is the rocket ship7s mass and M is the moon7s mass and M is the reduced mass Finally we can use these to determine the velocity vector we need to give the ship to put it into this trajectory If we write Vi1r r19 9 then L mRZve 7 m 2 2 i Gm E 7 3 1 119 which can be solved for 119 and then 1 i L v9 7 2 GmM 7 EltE R gt7 The following picture shows the results free return trajectory perfect hitl Phys 3201 Special Topic Spring 2005 Chaos The Butter y Effect and Lyapunov Exponents The hallmark of chaos is sensitivity with respect to initial conditions A chaotic system is perfectly deterministic yet in a practical sense unpredictable If you consider the evolution of the system using two slightly different initial conditions7 the two diverge exponentially fast As an example7 here is a plot of the s coordinate versus time for the Lorenz equations Two time series are shown on the same graph7 with initial conditions differing by about only one part in 10 million l X l m y 15 h U1 01 l I l l l l quoto 10 20 30 4o 50 t The break point77 appears suddenly7 at about t 17 The next gure tracks the difference between these two trajectories Shown is a semilog plot of la 7 ml versus time 1 O O 1 O 20 3O 4O 50 t You can see that the initial separation is very small of course and that its growth is not monotonic But on average the graph follows a pretty respectable straight line until is suddenly levels of7 right about the break point77 time A straight line on a semilog plot corresponds to exponential growth The exponential growth doesn7t last forever7 of course7 since Ax can never be greater than the difference between the largest and smallest values of x on the attractor The Lyapunov exponent is the growth rate of Am We can t an exponential to the data lAzl lAz0l and empirically determine the Lyapunov exponent A Taking logarithms of both sides yields ln lAzl ln le0l At so that the slope of the line is equal to A From the graph7 you can see that Ax grows from about 10 7 to about one in about a time 157 which gives a rough estimate of A N ln10 7 7 ln1 N 15 so that the distance grows by a factor of about 27 in an interval of 1A 09 units of time Precisely the same ideas can be applied to iterative maps Consider the map you studied for homework 11 Phys 3201 Special Topic Spiing 2005 Notes on Nonlinear Oscillators These notes are a supplement to the textbooks Chapter 4 The main purpose is to introduce you to just a couple of the fascinating things that happen in nonlinear systems7 and to develop a simple but effective tool known as harmonic balance77 to quantitatively understand some of what happens The Jump Phenomenon Numerical Investigation We saw for the linear driven oscillator that the amplitude of the response was a continuous function of the various system parameters drive amplitude and frequency7 for example Something very different can happen when we consider driven nonlinear oscillators One new possibility is the jump phenomenon 7 where the response amplitude changes suddenly at some critical value of the control parameter this is our rst example of a bifurcation This behavior is accompanied by another new feature7 namely hysteresis Equation of motion Consider a plane pendulum7 in the presence of a driving torque ignore the damping force for now The equation of motion is Io 7mgLsin Acoswt 1 where I is the moment of inertia7 m is the bob mass7 L is the rod length7 g is the acceleration due to gravity7 and i5 is the angle the rod makes with respect to the vertical For small angle motion7 one can Taylor expand the sine term A 7 1 3 1 5 Sln i ig quot Keeping only the rst term yields the harmonic oscillator equation Here7 well keep the second term too7 and see what effect the nonlinearity has on the resulting motion Our equation becomes7 after dividing through by I7 and including the usual damping term 2 w02 7 a g Fooswt 2 where woz ngL and F AI Jump Phenomenon Sweep Up 2 i i i i i 20 005 01 015 02 025 03 035 F Figure 1 Numerical Simulations Here are the results of numerical simulations of this equation with parameter values 7 005 we 11 w 10 and F 6 0035 lnitially with the system in equilibrium and at rest the forcing amplitude is zero As 1 very slowly ramp up F the response amplitude grows At rst the response amplitude is linearly proportional to F but as the nonlinearity becomes important the growth is a bit faster than linear Suddenly at a critical value of F the response jumps to a large value As F continues to increase the response increases but now relatively slowly What happens if we lower the driving force There is another jump point but it is somewhat lower than the rst one The result of the numerical simulation is shown on the next graph Finally take a look at the two gures superimposed There is a hysteresis loop for values Jump Phenomenon Sweep Down 2 i i i i i 20 005 01 015 02 025 03 035 F Figure 2 of F between the two jump points7 the system response can have two different amplitudes Which one the system Chooses depends on the initial conditions This is an example of a rather common feature of nonlinear systems7 namely coercistmg attractors This is impossible in a linear system 20 005 01 015 02 025 03 035 F Figure 3 The Method of Harmonic Balance This is an approximate method which is similar to the Fourier series method we used for the linear oscillator We can use it to investigate periodic solutions The steps are 1 Assume the solution can be represented as a truncated Fourier series 2 Substitute the assumed solution into the equation of motion7 and expand each term as a Fourier series 3 Throw out any higher frequency harmonics which are not included in the original assumed solution 4 Balance coef cients for each Fourier term harmonic This leads to a set of algebraic equations 5 Solve the algebraic equations Let7s apply this method to the pendulum problem For simplicity7 ignore the damping term7 so that w02 tbs Fcoswt F F gt a1 a2 a3 a a conditio I Figure 4 Solving the cubic graphically v Assume a solution of this form gtt 1 cos wt Our goal is to determine the allowed values of a We have if 7aw2 cos wt 3 1 153 1 cos3 wt a3 cos wt 1 cos 3wtgt Substitute these into the differential equation and throw out the higher harmonic cos 3wt The remaining terms all oscillate as cos wt This common factor can be factored out with the result 3 3 7a F 64 7aw2 w02a 7 l we2 7 w2a 7 gwozag F 3 This is a cubic equation for the amplitude 17 which may have either one or three real solutions Now7 it is possible to write down an exact formula for the solutions of this equation7 though the expression is pretty cumbersome Lets see what we can learn by trying to solve the cubic graphically7 by plotting the lefthand and righthand sides of the Eq3 vs 1 There are two different cases7 depending on the sign of wOZ 7 wZ7 shown in the gure We see that if LUZ gt woz there is a unique solution which is just like77 the linear oscillator problem However if LUZ lt woz we can have either one or three solutions depending on the value of Fl Suppose we imagine slowly increasing the drive amplitude F and measuring the response amplitude a If LUZ lt woz we get a surprising result Assume the system starts in the low amplitude state a2 in the gure 4 then as F increases from zero the response increases until F reaches a critical value F0 beyond which the solution branch a2 is missing What happens One possibility is that the system evolves to the solution branch with a a1 which is the only branch remaining But there is more if we now decrease F back past the value F0 we dont expect the system to jump back since nothing special happens to the solution branch al at that point This phenomenon where the output depends not just on the system parameters but on the previous history of the system is called hysteresis and always accompanies this kind of jump phenomenon We can calculate F0 by noting that it corresponds to the point where the cubic has a local maximum In fact F0 is equal to the local maximum of the cubic We locate the critical value of a by setting 3F6a 0 in Eq3 3 Aug 7 LL12 7 7w a2 0 8 so that at the critical point 8 a g 17 wZwg and the corresponding value of F0 is obtained by plugging this back into Eq3 The results of our thought experiment are summarized in gure Harmonic Balance and Fourier Analysis The approximation used above was a very simple application of the method called harmonic balance This method is closely related to the method of Fourier Analysis since we are looking for a periodic solution we can write 00 zt a0 Em cos nwt b sin nwt n1 Fourier Analysis allows one to nd an exact solution to linear equations with constant coef cients by turning the differential equation into a set of uncoupled linear algebraic equations one for each harmonic The problem encountered for nonlinear equations is that absta jump I Figure 5 Results of our thought experiment as we have seen the nonlinearities couple together these different harmonics7 leading to an in nite number of coupled nonlinear algebraic equations The method of harmonic balance simply truncates the Fourier series at some arbitrary usually low order7 and throws out any unwanted harmonics generated by the nonlinearities Although the approximation is thus uncontrolled it can be systematically improved by including more terms in the Fourier expansion The van der Pol Oscillator The van der Pol oscillator was originally1 studied as a model for an electronic device called a triode which can sustain voltage oscillations without bene t of any external driving input The van der Pol equation is z 5z2 i 1z wozx 0 4 You can also think of this equation as describing a mechanical oscillator with a nonlinear friction term which provides damping when x2 gt 1 7 and anti damping7 when x2 lt 1 If you numerically integrate the van der Pol equation7 you nd that there is one unstable equilibrium point at z z 0 All other initial conditions settle down7 after a transient7 1see Philosophical Magazine Volume 3 pages 6580 1927