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by: Dr. Simeon Wiza


Dr. Simeon Wiza
GPA 3.96


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Class Notes
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This 11 page Class Notes was uploaded by Dr. Simeon Wiza on Wednesday September 9, 2015. The Class Notes belongs to PHYS 505 at University of Washington taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/192447/phys-505-university-of-washington in Physics 2 at University of Washington.


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Date Created: 09/09/15
Lecture 6 The Methods of Lagrange H More on Coordinate Transformations and Constraints As we have discussed in previous lectures the laws of Newton are truly simple only in Cartesian coordinates in an inertial frame Here we want to generalize the discussion of the previous lectures to generalized coordinates to obtain a formulation of Newtonian dynamics valid even in noninertial frames We consider a general coordinate transformation described in terms of three holonomic ordinary functions 71 Zxj yjazjRjq1quotquotqf t q1qftgj q1qfthj q1qft 61 Here we start with 3N j 1 N coordinates describing our N particle system in 3 Cartesian dimensions with n constraints and we transform to f 3N n generalized coordinate q j 1 3N n f Note that we are explicitly assuming that by an appropriate choice of coordinates which are orthogonal to any constraint surfaces we have eliminated the constraints from the analysis We retum to the variational form at the end of Lecture 5 Eq 544 but without the constraint forces I 43 55 Oa 62 M2 K g where the forces included may be either conservative or dissipative We can deduce the form of the transformed force from the relation N QIM 2F 5Fr 63 11 jl We find the generalized force acting on the generalized coordinate q to be N a Q F 3 64 l 0 Physics 505 Lecture 6 1 Autumn 2005 so that the corresponding variational equation looks like 11 i f Q5 0 m 1m2 q Next consider the transformation of the acceleration term We start with M w a 1m a W W ql 6 6t and note that I 6 6511 6 71 6 11 at Then it follows from K Vl wga dag Ldag r a t2 aq dt aq J dt aq d 4 dt aq 39 r aq at that we can write Physics 505 Lecture 6 2 Autumn 2005 3 D amp d26 6r d ari d mj 2 39 mer mjrj39 dt aq dt aq dt aq 1d 5 1 6 mj mj 1 j l dt aq 6 where we have recognized the kinetic energy in this expression Finally we see that Eq 65 becomes the transformed version of Hamilton s principle d 8T 8T Q 6q 0 dtaql aql I 1 610 which we might have expected from the EulerLagrange equations Since the generalized coordinates were chosen to eliminate the constraints the variations of these variables the 56 can be chosen arbitrarily Hence Eq 610 can be satisfied for all such variations if and only if 1 a 6TQ dt aql aql z 611 This last relation is called Lagrange s equation of the first kind For the familiar case of Cartesian coordinates in an inertial frame T m 11712 2 Q1 Fl Eq 611 clearly becomes just Newton s second law On the other hand Eq 611 applies for both dissipative and conservative systems in all generalized coordinates both inertial and accelerated Two familiar and useful examples of generalized coordinates are cylindrical and spherical coordinates which are especially relevant when the constraints exhibit the corresponding symmetry structure For the former we have Physics 505 Lecture 6 3 Autumn 2005 I7 pbzi d7 dip3 pdqsq d22 ipbp45q 2z 61123 12 29159613 2 612 Tp2p2 222 m 2 q qfq 613 Lagrange s equations then read as we saw in Lecture 4 m5 p 2 Qp mp2q m2ppq5p2 Q 613 1712 Q In spherical coordinates we have A fzrr ddrfrd6 rsin6d q i r9 rsin6gz5qi 611 r 612 9q3 T 02 r292 r2 sin2 6 614 m 2 112 613422 6112 sinz 6126132 So in spherical coordinates Lagrange tells us to write the equations of motion as Physics 505 Lecture 6 4 Autumn 2005 m39r r6 2 rsin2 6952 Q d 2 2 2 m r6 r s1n6c056 4 gt m2r 9 r267 r2sin6c056q52Q9 615 2 2 mdt r s1n 695 2r sin2 695 2r2 sin6 cos66lq r2 sin2 Q The slick thing here is that all the extra terms which we obtained in Lecture 4 by careful analysis of the accelerating frame appear naturally from Lagrange s equation As usual we now look at the case for conservative forces taken to be defined by a scalar function of both the new and old coordinates U71 7NUCI1qf 616 so that at Q15q1 5q1 617 6 11 Now we can define a Lagrangian as we did in Lecture 5 and the equations of motion for a conservative system are the generalized version of Eq 522 modulo the sign to match our current notation L T U d 6L 6L 0 a 618 draqk aqk which is Lagrange s equation of the second kind Consider for example a central force potential in spherical coordinates U U r for which Lagrange s equations give Physics 505 Lecture 6 5 Autumn 2005 6r 2W9 r26l r2 sin6lcos6ql2 0 r2 sin2 constant moi 1 92 rsin26952 619 The last line is of course the conservation of the 2 component of angular momentum It is customary and useful to make the following definition 6T P 620 which is called the canonical momentum conjugate to the generalized coordinate q In Cartesian coordinates this is just the mass times the velocity but this is not the case more generally For example in spherical coordinates we have 6TW p ar 6T 2 p9 E mr 6 621 pa mr2 sin2 6q5 L2 This definition will be very helpful when we study the role of symmetries and conservation laws in the Lagrangian formalism It also leads naturally to the appearance of extra terms in Newton s laws inherent in accelerating frames We can rewrite Lagrange s equations of the first kind Eq 611 dp 6T dt aqj Q 622 The first term on the righthandside the position dependence of the kinetic energy is exactly the extra term we found earlier when transforming to curvilinear coordinates In a conservative system with U typically not a function of the velocities we have Physics 505 Lecture 6 6 Autumn 2005 LT U 6L Pj aqj a L 623 dt 6 where again we expect any peculiarities of the coordinates to appear as position dependence in the kinetic energy and contribute to the righthand side Finally we note that in a linearly accelerated frame I7 I70 72 0 we have gmmg Egzmmmz n quotm 624 The use of the canonical momentum preserves the simple form of Newton but with the correct answer ie including the acceleration of the frame Now let us reconsider again the question of constraints What if we cannot eliminate all constraints by our choice of generalized coordinates Or what if we want to leam something about the constraint forces Assume that we are dealing with A7 generalized coordinates and n constraints which we can always express in differential form ajkdqj atk5t0jlZVk1n 625 If a constraint is holonomic ie the constraint is integrable we can simplify the analysis by writing the constraint in terms of a point function of the generalized coordinates a aJk j k q1q1v 626 Physics 505 Lecture 6 7 Autumn 2005 Note that even if the constraint is nonholonomic either because we cannot integrate Eq 625 or because the constraint is really an inequality we can still proceed using Lagrange s method of undetermined multipliers as introduced in Lecture 5 Minimizing the action yields the familiar result 6L16L5 0 aqj dtaqj 61p 627 where the variations are now not all arbitrary ie only JV n can be varied freely To address this issue we introduce n undetermined parameters and express the non variation of the constraints under the Virtual variations of the coordinates as 27611qu 0 628 Combining Eqs 627 and 628 we have 2a 5 20 an ath Ink 1 629 for all values of j and where the indeX k is summed over Now after agreeing that we will eventually determine the lk so that the constraints are satis ed we treat the variations as independent Thus we now solve for JV n unknowns the q I and 2k using the 7 n equations d 6L 6L E6 6 lkajk qj qj aqu ark 0 630 As an example consider a point particle of mass m that slides without friction on the inside of a smooth paraboloid of revolution defined by p2 012 The paraboloid is oriented vertically in the earth s uniform gravitational field If the particle is initially moving in a horizontal trajectory with velocity v0 at radius p0 what must the magnitude of v0 be in order that the particle moves neither up nor down Clearly we Physics 505 Lecture 6 8 Autumn 2005 should choose cylindrical coordinates where the Lagrangian is L2302 p2 222 mgz 631 The constraint of moving on the speci ed surface is expressed by 2 aZ p 0 2pdpadz0 6 32 all 2 2pa21 0 131 a We note that the initial conditions include Z0 pg a Lagrange s equations Eq 6 30 yield mb mpq52 2p11 mpqu 0 2 mp2q5 mp0v0 C0nstant m z mgal1 633 2 zz prszz pr 2p a a a Clearly for no 2 motion we want 21 mg a from the third equation and 339 p2 p 0 from the last equations Thus it follows from the first two equations that AIZWCWLE 2 2p a 634 2 3 V0 2 p0 7g The two components of the constraint force in this special case are Physics 505 Lecture 6 9 Autumn 2005 fz all 2 mg 2p fp QM 0mg 635 a f 0 The interested student is encouraged to consider the general problem ie motion not confined to the horizontal plane and determine for example the general form for the constraint force Here we simply note that for total conserved energy E and conserved angular momentum L the Lagrange multiplier has the form 4E 8L mg 21 Z a ma a I 4pzjz 636 2 a Another class of related and interesting problems is associated with round bodies rolling on other round bodies The example of a cylinder rolling on another fixed cylinder is discussed in detail in Chapter 3 of FampW and will appear in a slightly differ form in HW IV We can see the relevant issues by returning to the problem of the hoop on the inclined plane we discussed in Lecture 5 see pages 12 and 13 In that problem there are formally 2 constraints rolling on an inclined plane which was built into the choice of coordinates x along the plane and rolling without slipping which leads to the constraint in Eq 536 Keeping this latter constraint in the Lagrange multiplier formalism allowed us to determine the tangential frictional force along the plane in Eq 543 On the other hand since we eliminated the other constraint by our choice of coordinates we did not determine the normal force to the inclined plane To connect to the class of problems being described here consider the case where we replace the inclined plane by the surface of a sphere ie the hoop rolls upright on the surface of a sphere where the starting point is the hoop at rest at the top of the sphere Since we typically want to determine when the hoop ies off the sphere we want to determine the normal force it vanishes as the hoop ies off and so we want to explicitly keep both constraints So consider a sphere of radius R1 and a hoop of radius RZ as in Figure 192 in FampW note that the figure looks the same even though the setup is slightly different Further we define other variables as in the figure r is the distance between the center of the sphere and the center of the hoop39 91 is the polar angle of the contact point between the sphere and the hoop 92 Physics 505 Lecture 6 10 Autumn 2005 is the angle through which the hoop has rotated since it started rolling down the sphere The constraint of rolling on the sphere is see Eq 1937 in FampW 1rrRl R2 0 637 The rolling without slipping constraint replacing Eq 536 looks like see Eq 1938 in FampW 9152 9162R161 R2 62 91 0 638 Note in particular the final term 12261 which might be surprising recall that the corresponding expression for the inclined plane was 6 x x r6 0 and we might have eXpected that we should just make the changes x gt R161 r6 gt R262 The last term accounts for the fact that we are using curvilinear coordinates and the hoop rotates as moves along the sphere even if it not rolling at all ie even if it is slipping with the same point always in contact with the sphere Another way to look at Eq 6 38 is that it guarantees that the point on the hoop that is instantaneously in contact with the sphere has vanishing instantaneous velocity v R1R26391 R292 0 which must be true if the point is not slipping on the sphere As discussed in FampW we can proceed to solve this problem using 2 Lagrange multipliers for Eqs 637 and 638 and the Lagrangian R2 L 2 0 r2612 mngOS91 639 ie find an extremal path for L 21 1 12 In particular we find that the multiplier corresponding to Eq 637 has the form see Eq 1954 in FampW modulo a minus sign In 21 24 7C0S61 640 Hence the normal force vanishes and the hoop ies off the sphere for cos 61 47 61 552 Physics 505 Lecture 6 11 Autumn 2005


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