YangQuan Chen, Ph.D., Director, MESA (Mechatronics, Embedded Systems and Automation) Lab

1. ME280 “Fractional Order Mechanics”Integer Order Analytical Mechanics andInteger Order Optimal Control Problems YangQuan Chen, Ph.D., Director, MESA (Mechatronics, Embedded Systems and Automation)Lab ME/EECS/SNRI/UCSolar, School of Engineering, University of California, Merced E: yqchen@ieee.org; or, yangquan.chen@ucmerced.edu T: (209)228-4672; O: SE1-254; Lab: Castle #22 (T: 228-4398) 11/12/2013 and 11/14/2013. Tue. Thu. 09:00-10:15, KL217 (Week-12)

2. Part 1: Integer-Order Analytical Mechanics • What’s the essence of “Analytical Mechanics”? • Why Lagrange is so great? • We can derive Newton’s second law in one slide! • Post-Newton Mechanics - D'Alembert，Euler，Lagrange，Hamilton，Jacobi，Gauss, Poisson • We focus on basic concepts • http://en.wikipedia.org/wiki/Analytical_mechanics ME280 "Fractional Order Mechanics" @ UC Merced

3. Acknowledgements • Internet, I picked up lots of forgotten “mechanics” • Dale E. Gary’s slides, NJIT Physics Dept. “Physics 430” • Mechanics, (3rd Edition) Volume 1 (Course of Theoretical Physics) by L. D. Landau and E.M. Lifshitz ME280 "Fractional Order Mechanics" @ UC Merced

4. Phys 205: Analytic Mechanics • UC Merced, Fall 2013 taught by Prof. Michael Scheibner • Subject matter: one semester course (subject to change). • Lagrange’s equations – constraints, generalized coordinates, D’Alamber’s principle • Variational methods – minimal action principle, Hamilton’s extrema principle, gauge transformations, Lagrange’s multiplicators • Symmetries and conservation laws – Noether’s theorem, time translation, Galilean relativity • Hamilton mechanics – canonical equations, canonical transformation, Liouville theorem, Hamilton- Jacobi theory, Poisson brackets • Central forces – Kepler’s laws, scattering processes • Oscillations – Green’s function, coupled oscillations • Rigid body – rotation, inertia tensor, Euler’s equations • Accelerated reference frames & relativistic mechanics – postulates of special relativity, time delation, Lorentz transformation ME280 "Fractional Order Mechanics" @ UC Merced

5. Classical Mechanics • First began with Galileo (1584-1642), whose experiments with falling bodies (and bodies rolling on an incline) led to Newton’s 1st Law. • Newton (1642-1727) then developed his 3 laws of motion, together with his universal law of gravitation. • This is where your previous experience in mechanics doubtless ends, but the science of mechanics does not end there, as you’ll see in this course. • Two additional, highly mathematical frameworks were developed by the French mathematician Lagrange (1736-1813) and the Irish mathematician Hamilton (1805-1865). • Together, these three alternative frameworks by Newton, Lagrange, and Hamilton make up what is generally called Classical Mechanics. • They are distinct from the other great forms of non-classical mechanics, Relativistic Mechanics and Quantum Mechanics, but both of these borrow heavily from Classical Mechanics. ME280 "Fractional Order Mechanics" @ UC Merced

6. Newton’s Three Laws • Law of Inertia • In the absence of forces, a particle moves with constant velocity v. • (An object in motion tends to remain in motion, an object at rest tends to remain at rest.) • Force Law • For any particle of mass m, the net force F on the particle is always equal to the mass m times the particle’s acceleration: F = ma. • Conservation of Momentum Law • If particle 1 exerts a force F21 on particle 2, then particle 2 always exerts a reaction force F12 on particle 1 given by F12 = - F21. • (For every action there is an equal and opposite reaction.) ME280 "Fractional Order Mechanics" @ UC Merced

7. Equivalence of First Two Laws • The Law of Inertia and the Force Law can be stated in equivalent ways. Obviously, if , then in the absence of forces • Thus, the velocity is constant (objects in motion tend to remain in motion) and could be zero (objects at rest tend to remain at rest). • The second law can be rewritten in terms of momentum: • In Classical Mechanics, the mass of a particle is constant, hence • So we can write . • In words, forces cause a change in momentum, and conversely any change in momentum implies that a force is acting on the particle. ME280 "Fractional Order Mechanics" @ UC Merced

8. Conservation of Momentum • This final result: says that the internal forces on a system of particles do not matter—they all cancel each other, so the change in total momentum of a system of particles is only due to external forces. • In particular, if there are no external forces, the momentum cannot change, i.e. it is constant. This is the principle of conservation of momentum: • This is one of the most important results in classical physics, and is true also in relativity and quantum mechanics. • We showed that Newton’s 3rd Law implies Conservation of Momentum. Conservation of Momentum implies Newton’s 3rd Law. Principle of Conservation of Momentum If the net external force Fext on an N-particle system is zero, the system’s total momentum P is constant. ME280 "Fractional Order Mechanics" @ UC Merced

9. Simplest example • Statement of Problem: • Two objects of masses m1 and m2 are subject to no external forces. Object 1 is traveling with velocity v when it collides with the stationary object 2. The two objects stick together and move off with common velocity v. Use conservation of momentum to find v in terms of v, m1 and m2. • Solution: • Conservation of momentum says the momentum before the collision must be the same as the momentum after the collision: • Since v1 = v, and v2 = 0(second object is stationary), we simply solve for v to find: ME280 "Fractional Order Mechanics" @ UC Merced

10. r1+dr dr r1 Kinetic Energy and Work • An obvious form of energy is energy of motion, or kinetic energy. We will use the symbol T, which is perhaps strange to you but is very much standard in Classical Mechanics. The kinetic energy of a particle of mass m traveling at speed v is defined to be: • Consider such a particle moving on some trajectory through space while its kinetic energy changes, on moving from position r1 to r1 + dr. We can take the time derivative of the kinetic energy, after writing , so that • But the first term on the right is the force . Thus, we can write the derivative of kinetic energy as • Finally, multiplying both sides by dt and noting that vdt = dr, we have Work-KE Theorem ME280 "Fractional Order Mechanics" @ UC Merced

11. Line Integrals and Work • The equation just derived is only valid for an infinitesimal displacement, but we can extend this to macroscopic displacements by integrating, to get: which says that the change in kinetic energy of a particle is equal to the sum of force (in the direction of the displacement) times the incremental displacement. • However, note that this is the displacement along the path of the particle. Such an integral is called a line integral. In evaluating the integral, it is usually possible to convert it into an ordinary integral over a single variable, as in the following example (which we will look at in a moment). • With the notation of the line integral where the last is a definition, defining the work done by F in moving from point 1 to point 2. Note that F is the net force on the particle, but we can also add up the work done by each force separately and write: ME280 "Fractional Order Mechanics" @ UC Merced

12. When Force Is Conservative? • Two conditions for a force to be conservative are • It turns out that there is an easy way to check whether a force has the second property, using a concept from vector calculus. It can be shown via a theorem called Stokes’ Theorem (which you will have seen if you have had the vector calculus course) that a force has the desired property, that the work it does is independent of the path, if and only if everywhere. The quantity is called the curl of F, or just “curl F,” or “del cross F.” • It follows the usual rules for the cross product. Conditions for a Force to be Conservative A force F acting on a particle is conservative if and only if it satisfies two conditions: 1. F depends only on the particle’s position r (and not on the velocity v, or the time t, or any other variable); that is, F = F(r). 2. For any two points 1 and 2, the work W(1  2) done by F is the same for all paths between 1 and 2. ME280 "Fractional Order Mechanics" @ UC Merced

13. Curl of F • Two find the curl of a vector, you form the matrix and find its determinant: • It may not be obvious that this being zero is equivalent to the condition that is path independent, but Stokes’ Theorem shows that it is. This gives a handy way to determine the path-independence property. ME280 "Fractional Order Mechanics" @ UC Merced

14. Potential Energy Potential Energy • The reason that forces meeting these conditions are called conservative is that, if all of the forces on an object are conservative we can define a quantity called potential energy, denoted U(r), a function only of position, with the property that the total mechanical energy is constant, i.e. is conserved. • To define the potential energy, we must first choose a reference point ro, at which U is defined to be zero. (For gravity, we typically choose the reference point to be ground level.) Then U(r), the potential energy, at any arbitrary point r, is defined to be • In words, U(r) is minus the work done by F when the particle moves from the reference point ro to the point r. ME280 "Fractional Order Mechanics" @ UC Merced

15. Example: A Cylinder Rolling Down an Incline y • Statement of the problem: • A uniform rigid cylinder of mass M, radius R rolls without slipping down a sloping track as shown in the figure. Use energy conservation to find its speed v when it reaches a vertical height h below its point of release. • Solution: • The cylinder is a rigid body, so we may ignore its internal potential energy. The external potential energy is where y is the vertical position of its CM measured from any convenient location. • The KE of a rotating (rolling) cylinder is where I is the moment of inertia, I = ½MR2, and w is the angular velocity of rolling. Because the cylinder is rolling without slipping, w = v/R. Thus, the final kinetic energy is • If the cylinder starts from rest, the change in kinetic energy is equal to minus the change in potential energy • So the final speed is ME280 "Fractional Order Mechanics" @ UC Merced

16. ME280 "Fractional Order Mechanics" @ UC Merced

17. Pierre-Louis Moreau de Maupertuis (1698 – 1759) Sir William Rowan Hamilton (1805 – 1865) Richard Phillips Feynman (1918 – 1988) • Variational principle • Maupertuis: Least Action Principle • Hamilton: Hamilton’s Variational Principle • Feynman: Quantum-Mechanical Path Integral Approach ME280 "Fractional Order Mechanics" @ UC Merced

18. Feynman’s teacher told him about the “Principle of Least Action”, one of the most profound results in physics. • When I was in high school, my physics teacher called me down one day after class and said, “You look bored, I want to tell you something Interesting”. Then he told me something I have always found fascinating. Every time the subject comes up I work on it. Richard Feynman http://www.damtp.cam.ac.uk/user/dt281/dynamics/two.pdf ME280 "Fractional Order Mechanics" @ UC Merced

19. y y2 y1 y = y(x) 2 1 x x1x2 Calculus of Variations: Two Examples • The calculus of variations involves finding an extremum (maximum or minimum) of a quantity that is expressible as an integral. • Let’s look at a couple of examples: • The shortest path between two points • Fermat’s principle (light follows a path that is an extremum) • What is the shortest path between two points in a plane? You certainly know the answer—a straight line—but you probably have not seen a proof of this—the calculus of variations provides such a proof. • Consider two points in the x-y plane, as shown in the figure. • An arbitrary path joining the points follows the general curve y = y(x), and an element of length along the path is • We can rewrite this (pay attention, because you will be doing this a lot in the next few chapters) as which is valid because Thus, the length is ME280 "Fractional Order Mechanics" @ UC Merced

20. Shortest Path Between 2 Points • Note that we have converted the problem from an integral along a path, to an integral over x: • We have thus succeeded in writing the problem down, but we need some additional mathematical machinery to find the path for which L is an extremum (a minimum in this case). Fermat’s Principle: • A similar but somewhat more interesting problem is to find the path light will take through a medium that has some index of refraction n1. You may recall that light travels more slowly through such a medium, and we define the index of refraction as n = c/v. where c is the speed of light in vacuum, and v is the speed of light in the medium. The total travel time is then • Here we are allowing the index of refraction to vary arbitrarily vs. x and y. ME280 "Fractional Order Mechanics" @ UC Merced

21. Variational Principles • Obviously, both problems on the previous slide are similar, and such cases arise in many other situations. • In our usual minimizing or maximizing of a function f(x), we would take the derivative and find its zeroes (i.e. the values of x for which the slope of the function is zero). These points of zero slope may be minima, maxima, or points of inflection, but in each case we can say that the function is stationary at those points, meaning for values of x near such a point, the value of the function does not change (due to the zero slope). • In analogy with this familiar approach, we want to be able to find solutions to these integrals that are stationary for infinitesimal variations in the path. This is called calculus of variations. • The methods we will develop are called variational methods, and a principle like Fermat’s Principle are called variational principles. • These principles are common, and of great importance, in many areas of physics (such as quantum mechanics and general relativity). ME280 "Fractional Order Mechanics" @ UC Merced

22. y y2 y1 2 1 y = y(x) (right) x x1x2 Euler-Lagrange Equation-1 • We are now going to discuss a variational method due to Euler and Lagrange, which seeks to find an extremum (to be definite, let’s consider this a minimum) for an as yet unknown curve joining two points x1 and x2, satisfying the integral relation • The function f is a function of three variables, but because the path of integration is y = y(x), the integrand can be reduced to a function of just one variable, x. • To start, let’s consider two curves joining points 1 and 2, the “right” curve y(x), and a “wrong” curve Y(x) that is a small displacement from the “right” curve, as shown in the figure. • We will write the difference between these curves as some function h(x). ME280 "Fractional Order Mechanics" @ UC Merced

23. Euler-Lagrange Equation-2 • There are infinitely many functions h(x), that can be “wrong,” but we require that they each be longer that the “right” path. To quantify how close the “wrong” path can be to the “right” one, let’s write Y = y + ah, so that • This is going to allow us to characterize the shortest path as the one for which the derivative dS/da = 0 when a = 0. To differentiate the above equation with respect to a, we need to evaluate the partial derivative via the chain rule so dS/da = 0 gives ME280 "Fractional Order Mechanics" @ UC Merced

24. Euler-Lagrange Equation-3 • We can handle the second term in the previous equation by integration by parts: but the first term of this relation (the end-point term) is zero because h(x) is zero at the endpoints. • Our modified equation is then • This leads us to the Euler-Lagrange equation • We come to this conclusion because the modified equation has to be zero for any h(x). ME280 "Fractional Order Mechanics" @ UC Merced

25. Euler-Lagrange Equation-4 • Let’s go over what we have shown. We can find a minimum (more generally a stationary point) for the path S if we can find a function for the path that satisfies • The procedure for using this is to set up the problem so that the quantity whose stationary path you seek is expressed as where is the function appropriate to your problem. Write down the Euler-Lagrange equation, and solve for the function y(x) that defines the required stationary path. • Let’s do a few examples to make the procedure clear. ME280 "Fractional Order Mechanics" @ UC Merced

26. Shortest Path Between Two Points • We earlier showed that the problem of the shortest path between two points can be expressed as • The integrand contains our function • The two partial derivatives in the Euler-Lagrange equation are: • Thus, the Euler-Lagrange equation gives us • This says that • A little rearrangement gives the final result: y2= constant (call it m2), so y(x) = mx + b. In other words, a straight line is the shortest path. ME280 "Fractional Order Mechanics" @ UC Merced

27. 1 x 2 y Example: The Brachistochrone • This is a famous problem in the history of the calculus of variations. Statement of the problem: • Given two points 1 and 2, with 1 higher above the ground, in what shape could we build a track for a frictionless roller-coaster so that a car released from point 1 would reach point 2 in the shortest possible time? See the figure, which takes point 1 as the origin, with y positive downward. • Solution: • The time to travel from point 1 to 2 is where , from kinetic energy considerations. • Since this depends on y, we will take y as the independent variable, hence • Our integral now becomes: • From the Euler-Lagrange equation: Notice that, since we are using y as the independent variable, we swap x and y ME280 "Fractional Order Mechanics" @ UC Merced

28. Solution, cont’d: • Since clearly and so . • Evaluating this derivative and squaring it for convenience, we have where the constant is renamed 1/2a for future convenience. • Solving for x we have: Finally, to get x we integrate: • It is not obvious, but this can be solved by the substitution y = a(1 -cosq), which gives • The two equations that give the path are then: in terms of q. ME280 "Fractional Order Mechanics" @ UC Merced

29. 1 2a a 2 3 • Solution, cont’d: • This curve is called a cycloid, and is a very special curve indeed. As you will show in the homework, it is the curve traced out by a wheel rolling (upside down) along the x axis. • Another remarkable thing is that the time it takes for a cart to travel this path from 23 is the same, no matter where 2 is placed, from 1 to 3! Thus, oscillations of the cart along that path are exactly isochronous (period perfectly independent of amplitude). ME280 "Fractional Order Mechanics" @ UC Merced

30. y(x) (x2,y2) y (x1,y1) x ds Another Example • Statement of the problem: • A surface of revolution is generated as follows: Two fixed points (x1, y1) and (x2, y2) in the x-y plane are joined by a curve y = y(x). [Actually, you’ll make life easier if you start out writing this as x = x(y).] The whole curve is now rotated about the x axis to generate a surface. Show that the curve for which the area of the surface is stationary has the form where xo and yo are constants. (This is often called the soap-bubble problem, since the resulting surface is usually the shape of a soap bubble held by two coaxial rings of radii y1 and y2.) • Solution: • What we have to minimize now is not a line, but rather a surface, dA = y dfds. • We could write this but then both Euler-Lagrange terms will be non-zero. Instead, follow the hint: ME280 "Fractional Order Mechanics" @ UC Merced

31. Solution, cont’d: • We identify so and • Squaring, we have • Solving for • Integrating • The statement of the problem wants y(x), so we have to solve for y, to get where we identify C = yo. ME280 "Fractional Order Mechanics" @ UC Merced

32. y y2 y1 2 1 y = y(x) (right) x x1x2 More Than Two Variables • Aside: Maximum and minimum vs. stationary — geodetics on surfaces. • When we discussed the shortest path between two points, we examined the situation in the figure below. • However, this is not the most general path. Here is one that cannot be expressed as a function y(x). • To handle this case, we need to consider a path specified by an independent variable u along the path, i.e. x = x(u), y = y(u) . • Now the length of the path is then • In general, the problem is • This may seem more difficult than the previous case, but in fact we just proceed in the same manner with the “wrong” paths ME280 "Fractional Order Mechanics" @ UC Merced

33. More Than Two Variables-2 • The stationary path is the one for which • Following the same procedure as before, we end up with two Euler-Lagrange equations that the function f must satisfy: • One of the great strengths of Lagrangian mechanics is its ability to deal with cartesian, cylindrical, spherical, and any other coordinate systems with ease. • In order to generalize our discussion, we write (x, y, z), (r, f, z) or (r, q, f) as generalized coordinates(q1, q2, …, qN). ME280 "Fractional Order Mechanics" @ UC Merced

34. The Lagrangian • We will use slightly different notation, and refer to our integrand function as the Lagrangian • Note that because the independent variable is t, we can use rather than . • We will then make the action integral stationary which requires that L satisfy ME280 "Fractional Order Mechanics" @ UC Merced

35. Advantages Over Newtonian Mechanics • We are now going to use the ideas of the previous lecture to develop a new formalism for mechanics, called Lagrangian mechanics, invented by Lagrange (1736-1813). There are two important advantages of the Lagrange formalism over that of Newton. • First, as we have seen, Lagrange’s equations take the same form for any coordinate system, so that the method of solution proceeds in the same way for any problem. • Second, the Lagrangian approach eliminates the forces of constraint. This makes the Lagrangian formalism easier to solve in constrained problems. • This module is the heart of advanced classical mechanics, but it introduces some new methods that will take getting used to. Once you master it, you will find it an extraordinarily powerful way to solve mechanics problems. ME280 "Fractional Order Mechanics" @ UC Merced

36. Lagrange’s Equations for Unconstrained Motion • Recall our general function f, that played such a large role in the Euler-Lagrange equation • To make the connection to mechanics, we are now going to show that a function called the Lagrangian, L= T – U, is the function that, when used in the Euler-Lagrange equation, leads to the equation of motion for a particle. • Here, T is the particle’s kinetic energy and U is the potential energy • Note that L is NOT the total energy, E = T + U. One can ask why the quantity T – U should give rise to the equation of motion, but there seems to be no good answer. • I do note, however, that the problem can be cast in terms of the total energy, which gives rise to Hamiltonian mechanics. ME280 "Fractional Order Mechanics" @ UC Merced

37. Lagrangian • Obviously, with the dependences of T on the x, y, z velocities, and U on the x, y, z positions, the Lagrangian depends on both, i.e. • Let’s look at the first two derivatives • But note that if we differentiate the second equation with respect to time we get • So you can see that the Lagrange equation is manifestly true for a free particle: • In Cartesian coordinates (so far) in three dimensions, we have: Note: this last equality is only true in an inertial frame. ME280 "Fractional Order Mechanics" @ UC Merced

38. The actual path that a particle follows between two points 1 and 2 in a given time interval, t1 to t2, is such that the action integral is stationary when taken along the actual path. Connection to Euler-Lagrange • If you compare the Lagrange equations with the Euler-Lagrange equation we developed in the previous chapter, you see that they are identical. • Since the Euler-Lagrange equation is the solution to the problem of stationarity of a path integral, we see that is stationary for the path followed by the particle. • This integral has a special name in Physics—it is called the action integral, and when it is a minimum it is called the principle of least action, although that is a misnomer in the sense that this could be a maximum, or even an inflection point. • The principle is also called Hamilton’s Principle: ME280 "Fractional Order Mechanics" @ UC Merced

39. Generalized Coordinates • We have now seen that the following three statements are completely equivalent: • A particle’s path is determined by Newton’s second law F = ma. • The path is determined by the three Lagrangian equations (at least in Cartesian coordinates). • The path is determined by Hamilton’s Principle. • Again, the great advantage is that we can prove that Lagrange’s equations hold for any coordinate system such that, for any value r = (x, y, z) there is a unique set of generalized coordinates (q1, q2, q3) and vice versa. • Using these, we can write the Lagrangian in terms of generalized coordinates as • In terms of these, we then have • These are true in any coordinate system, so long as the coordinates are measured in an inertial frame. ME280 "Fractional Order Mechanics" @ UC Merced

40. Example: Deriving Newton’s 2nd Law • For a single particle in two dimensions, in Cartesian coordinates, under some arbitrary potential energy U(x, y), the Lagrangian is • In this case, there are two Lagrange equations • The left side of each equation is just • The right side of each equation, in turn, is just • Equating these, we have Newton’s second law ME280 "Fractional Order Mechanics" @ UC Merced

41. Generalized Force and Momentum • Notice that the left side term in Cartesian coordinates gives the force • When this is expressed in terms of generalized coordinates (which, for example, could be angular coordinates) this term is not necessarily a force component, but it plays a role very like a force, and indeed is called the generalized force. • Likewise, the right side in terms of generalized coordinates plays the role of a momentum, and is called the generalized momentum. • Thus, the Lagrange equation can be read as generalized force = rate of change of generalized momentum. ME280 "Fractional Order Mechanics" @ UC Merced

42. angular momentum mr2w. centripetal acceleration rw2. torque. Example: A different coordinate system • Let’s repeat the previous example, but in polar coordinates. In this case, the potential energy is U(r, f), and you should be able to write down the kinetic energy directly without much thought as so the Lagrangian is just • In this case, there are two Lagrange equations • Inserting the above Lagrangian into these equations gives • The first of these equations says • The second equation is best interpreted by recalling the gradient in polar coordinates ME280 "Fractional Order Mechanics" @ UC Merced

43. N Free Particles • It should be obvious how to extend these ideas to a larger number of particles. In the case of two particles, at positions r1 and r2, for example, Newton’s second law says each component of each particle obeys the equations • In the Lagrangian formalism, we have the equivalent Lagrange equations • We could likewise write these in generalized coordinates as • An important example we will use repeatedly in Chapter 8 is to replace r1 and r2 with RCM = (m1r1 + m2r2)/(m1 + m2), and r = r1-r2. • For N particles, then, there are 3N Lagrange equations. ME280 "Fractional Order Mechanics" @ UC Merced

44. f h Constrained Systems Example • We have said that the great power of the Lagrange formalism is that constraint forces disappear from the problem. • Before getting into the general case, let’s do a familiar problem—instead of a free particle, let’s consider one that is tied to the ceiling, i.e. the simple pendulum. • The pendulum bob moves in both x and y, but it moves under the constraint • A perfectly valid way to proceed might be to eliminate the y variable, and express everything in terms of x, e.g. replace y with • However, it is much simpler to express the problem in terms of the natural coordinate f. • The first task is to write down the Lagrangian L = T – U in terms of f. Clearly, for this problem U = mgh = mgl(1 -cosf). Likewise, the kinetic energy is • The relevant Lagrange equation is and you can basically write down the solution Equivalent to G = Ia ME280 "Fractional Order Mechanics" @ UC Merced

45. Constrained Systems in General • Let’s number a system of N particles as a = 1, …, N. The positions of these N particles are ra. We say that the parameters q1, …, qn are a set of ngeneralized coordinates for the system if each position ra can be expressed as a function of q1, …, qn and possibly time t, and conversely each qi can be expressed in terms of ra and possibly t, • In addition, we require that the number n of the generalized coordinates is the smallest number that allows the system to be parametrized in this way. In three dimensions, the number of generalized coordinates for N particles is certainly no more than 3N, and for constrained systems is usually less. For a rigid body of, say 1023 particles, for example, the number of generalized coordinates is n = 6, three for the position of the center of mass, and three for the orientation. • For the case of the pendulum we discussed last time, there is one body (the pendulum bob), and two coordinates (x and y), but there is only one generalized coordinate, f, since r = (x, y) = (l sin f, lcosf). ME280 "Fractional Order Mechanics" @ UC Merced

46. f1 f2 xs=1/2 at2 x l y f a (given) Generalized Coordinates-2 • Consider the double pendulum, with two bobs confined to motion in a plane. • Now we have two particles, four coordinates (x1, y1, x2, y2), but only two generalized coordinates f1 and f2. • In these two examples, the transformation between Cartesian coordinates and generalized coordinates did not depend on time, but here is an example that does. • Consider a pendulum hanging from a car that is undergoing a constant acceleration a to the right. • Because Lagrange’s equation was derived assuming that the coordinates are defined in an inertial frame, we are not allowed to use coordinates defined in the frame of the accelerating car! • However, we can express them relative to the ground. • In this case, the conversion from Cartesian to generalized coordinates is • Generalized coordinates that do not depend on t are called natural. ME280 "Fractional Order Mechanics" @ UC Merced

47. Degrees of Freedom • The number of degrees of freedom of a system is the number of coordinates that can be independently varied, i.e. the number of “directions” a system can move in small displacements from any initial configuration. • A simple pendulum has one degree of freedom, while the double pendulum has two. A free particle has three, while a system of N free particles has 3N degrees of freedom, i.e. each particle has complete freedom. • When the number of degrees of freedom of a system of N particles is less than 3N, we say that the system is constrained. A system of free particles constrained to move in two dimensions has 2N degrees of freedom. Some further examples: a rigid body has 6 degrees of freedom, a bead on a wire has 1 degree of freedom, and a particle on a surface has 2 degrees of freedom. • In each of these examples, the number of degrees of freedom equals the number of generalized coordinates (and so the number of Lagrange equations that apply). • A system with this natural-seeming property is said to be holonomic. This course will only treat holonomic systems, which are easier to solve. ME280 "Fractional Order Mechanics" @ UC Merced

48. A Non-Holonomic System • You might think that a system that does not have this natural property must be rare and bizarrely complicated. However, there are some simple examples of a nonholonomic system. Here is one. • Imaging a rubber ball free to roll (but not slide or spin) on a 2-d surface. • Starting at position (x, y) on the 2-d surface, it can only move in two independent directions and you might think that only two coordinates are necessary to completely describe its configuration, the coordinates x and y of its center. • But consider the following—place the ball at the origin O, and paint a dot on its top. Then roll it a distance equal to its circumference c along the x axis, so that the dot is again on top. • Now roll it a distance c in the y direction to a point P where its dot returns to the top. • Finally, roll it along the hypotenuse back to the origin. Now the dot is not on top, even though its position is again at O. • Evidentally, the two coordinates x and y are not enough to uniquely specify the configuration. In fact, we need three more, the orientation of the ball. • So 5 coordinates are needed, even though the ball has only two degrees of freedom. Such a system is nonholonomic. ME280 "Fractional Order Mechanics" @ UC Merced

49. Proof of Lagrange’s Equations with Constraints • We are now ready to prove Lagrange’s equations for any holonomic system. We will prove it for one particle, but it can easily be extended to an arbitrary number • Let’s take a particle constrained to move on a surface, so that it has two degrees of freedom and hence two independent generalized coordinates q1 and q2. • There are two types of forces on the particle—forces of constraint (whatever forces are keeping the particle constrained), which we’ll denote Fcstr, and all other forces F. The key is that the forces Fcstr can do no work on the particle. Note that the Fcstr forces may not be conservative, but this does not matter, since the Lagrange equations are not going to include them. • We shall assume that the non-constraint forces do satisfy the second condition, at least, of conservative forces, i.e. that they can be derived from the gradient of a potential energy, U(r, t): • If all forces F are really conservative, then they do not depend on t, but we do not need to assume this. The total force on the particle is Ftot = Fcstr+ F. ME280 "Fractional Order Mechanics" @ UC Merced

50. Action Integral Stationary on Right Path • Consider a constrained particle that moves through two points r1 and r2 at times t1 and t2. We will denote r(t) as the position when the particle is on the “right” path and R(t) as the position along any neighboring “wrong” path. • For a small displacement e(t) between the right and wrong path, we have • Note that e(t) = 0 at the end points r1 and r2, since both paths go through these points. Note also that r(t) and R(t) are in the surface, so e(t) is also. We denote the action integral by taken along any path R(t) lying in the surface, and by So the corresponding integral taken along the right path r(t). • We wish to prove that the integral S is stationary when R(t) = r(t), i.e. when e(t) = 0. Another way to say this is that the difference in the integrals is zero to first order in e. • Now where • We can substitute and ME280 "Fractional Order Mechanics" @ UC Merced