CLASSICAL MECHANICS PHYS 7221
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Date Created: 10/13/15
Single and Double plane pendulum Gabriela Gonzalez 1 Introduction We will write down equations of motion for a single and a double plane pendulum following Newton s equations and using Lagrange s equations Figure 1 A simple plane pendulum left and a double pendulum right Also shown are free body diagrams for the forces on each mass 2 Newton s equations The double pendulum consists of two masses m1 and m2 connected by rigid weightless rods of length 1 and 2 subject to gravity forces and constrained by the hinges in the rods to move in a plane We choose a coordinate system with the origin at the top suspension point the x axis as a horizontal axis in the plane of motion and the y axis pointing down so that gravity forces have positive components The single plane pendulum a simpler case has a single particle hanging from a rigid rod 21 Constraints The simple pendulum system has a single particle with position vector r z y There are two constraints it can oscillate in the xy plane and it is always at a xed distance from the suspension point Mathematically 1 2 The double pendulum system has two particles N2 with position vectors r1 r2 each with components zhyhzi There are four constraints each particle moving in the x y plane and each rod having constant lengths These constraints can be expressed as 20 l M 21 0 3 22 0 4 lril l1 5 lrz Fil l2 6 These constraints are holonomic they are only algebraic relationships between the coor dinates not involving inequalities or derivatives In the single pendulum case we only have one particle N1 so we have 3N3 coor dinates Since we have two constraints m2 we are left with n3N m3 21 only one generalized coordinate This is the angular position of the pendulum 0 which we can use to write r lsin 0 cos 0 0 7 In the double pendulum We know there should be only two generalized coordinates since there are 3N6 coordinates and m4 constraints so n3N m6 42 We can nd expressions for r1 r2 in terms of two angles 01 02 r1 1sin01cos010 8 r2 r1lgsin02cos020 9 We can express velocity an acceleration vectors in terms of generalized coordinates For the single pendulum r lsin0cos0 10 139quotV 0cos07sin0 11 391 a l cos07sin07102sin0cos0 12 M7102 13 The velocity vector V is perpendicular to the position vector r which is the expression of the constraint lrl l constant We should recognize the tangential and centripetal acceleration terms proportional to the velocity and to the inverted radial directions re spectively For the double pendulum we derive the same expressions for the rst particle r1 1sin 01 cos 1 14 f1 V1 191cos 01 7sin 01 15 11 a1 11 1cos017sin017110fsin01cos01 16 zl lol 7 119 17 and the second particle r2 r1 2sin 02 cos 1 18 f2 V2 V1 1202cos027 sin02 19 142 a2 a1 2152cos 02 7 sin 02 7 2038111 027 COS 02 20 22 Forces In the single pendulum case the forces on the particle are gravity and tension Gravity is along the y direction or the direciton of gravitational acceleration g and the tension is pointing towards the origin along the direction of 7r 7r T FT mg77rmg 21 r In the double pendulum the forces on m1 are the tension in the two rods and gravity The tension in the upper rod is along the direction 7r1 the tension force on m1 due to the lower rod is along the direction r2 7 r1 so we can write the force F1 as 7r r 7r T T F1 T1T2m1giilrii1quot21quot1m1g 22 1391 1392 7 I39ll 1 2 The forces on m2 are the tension in the lower rod and gravity The tension on m2 is along the direction of 7r2 7 r1 1 2 1 1 7 T F2 T2 77129 i1quot2 1quot1 77129 23 lrz Pil 12 23 Equations of motion 231 Single Pendulum In the single pendulum case Newton s law is F mi Writing the two non trivial compo nents we have mi Fi rmg 24 ml lt cos 0 7 sin 0 7 02sin 0 cos 0 7Tsin 0 cos 0 mg0 1 25 We thus have two equations ml cos0 7 0 2 sin a iTsm0 26 iml lt sin002 cos0gt iTcos0mg 27 Notice that although we only have one generalized coordinate 0 we have two equa tions That is because the equations also have the magnitude of the tension as an unknown so we have two equations for two unknowns t9 and T The equations are not only coupled but are also non linear involving trigonometric functions ughl A common trick with expressions involving trig is to make use of trig identities like cos2 9 1 sin2 9 1 For example multiplying 43 by cos 0 and adding Eq 45 multiplied by ism 0 we obtain a simpler equation for Using these identities we can write the equations as z 7981110 28 mm T 29 We cannot solve Eq 28 analytically but we could either solve it numerically or in the small angle approximation There will be two constants of integration because it is a second order differential equation we can relate those constants to the initial position and velocity or to conserved quantities such as the total energy but not linear momentum nor angular momentum the forces and torques are not zerol Whichever way once we have a solution for 0t we can use it in Eq29 to solve for the other unknown the tension T Eq29 does not invovle derivatives of T so there are no new constants of integration for the problem 232 Double Pendulum In the double pendulum Newton s second law on each particle is F mii i T T miri Tllrl 70quot2 P1 77119 30 T mzrz T1quot2 P1 77129 31 Are these six equations each equation has three componsents for two coordinates 01 02 Again not quite the equations only have two non zero components in the Xy plane and we have four unknowns 01 02 T1 and T2 so we have four equations for four unknowns just as expected We write the equation of motion for the two particles split into their two components in the plane mlll l cos 01 7 a sin 01 7 7T1 sin 01 T2 sini92 32 7m111 sin01 cos 4 7T1 cos 01 T2 cos 02 mg 33 m2 zl l cos 01 7 110 sin 01 zz z cos 01 7 120 sin 02 7 7T2 sin 02 34 7m2 ht91 sin 91 10 COS 91 2amp2 sin 92 203 COS 92gt 7T2 COS 92 77129 35 We have then four differential equations for four unknowns 01 02 T1 T2 Trigonometric identities such as cos2 9 sin2 9 1 and sin 02 cos 01 7 cos 02 sin 01 sin02 7 01 can be used to write the equations of motion as M 7 T2m1sin02 7 01 7 gsin01 36 119 7 T1 m1 7 T2m1cos02 7 n 7 gcos01 37 zl l cosi92 7 01110 fsin02 7 01 12 7 79 sin 02 38 711 sin02 7 01110 fcos02 7 01 120 7 T2m2 7 9 cos 02 39 We now can use Eqns 3637 to make substitutions in Eqns 38 39 and use more trig identities to simplify these equations further mg 7 79 sin 02 7 Tgm1sin02 7 01 7 gsin 01 cosi92 7 01 7T1m1 7 T2m1COS02 7 t1 7 gCOSt91 Sint92 7 91 7 7T1m1 sin02 7 01 40 120 Tzmz 7 gcos 02 T2m1 sini92 7 01 7 gsin 01 sini92 7 01 7T1m1 7 T2m1cosi92 7 t1 7 gCOS 91 COS92 7 91 T2m2T2m17T1m1cosi92 7 91 41 The four equations of motion are then zl l 7 Tgm1sin02 7 01 7 gsin 01 42 110 7 T1 m1 7 T2m1COS02 7 01 7 gcos01 43 zz z 7 7T1m1 sin02 7 01 44 120 7 Tgmg T2 m1 7 T1 m1 cosi92 7 01 45 Since the equations do not have derivatives of T1 T2 the best way to cast these equa tions for analytical or numerical solution is to obtain two differential equations for 01 02 without T1 T2 terms and use their solutions in expressions for T1 T2 in terms of 01 02 and their derivatives We obtain such expressions for T1 T2 from Eqns 44 42 zz z T L 46 1 m1 sin02 7 01 l zl l 9 sin 01 T 47 2 m1 sin02 7 01 l We use these expressions in Eqn 43 110 7 Tlm1 7 TZm1cos02 7 01 7 gcos01 2amp2 1amp1 9 sin 91 sin z 7 01 7 sin02 7 01 cos02 7 01 7 9 cos 01 7110 sin02 7 01 zz z 11 cos02 7 01 9 sin 02 48 3 Lagrange s equations 31 Simple Pendulum We have one generalized coordinate 0 so we want to write the Lagrangian in terms of 0 and then derive the equation of motion for 0 The kinetic energy is T 12mv2 12mlzt92 using Eq 11 for the velocity The potential energy is the gravitational potential energy V 7mg 7mgl cos 0 Notice we can derive the gravitational force from the potential Fg 7VV mg0 1 mg but not the tension force on the pendulum that is a constraint force The Lagrangian is 1 L T 7 V 57711292 mgl cos0 49 The Lagrange equation is d 8L 8L 0 3 7 w 50 d 2 i 7 a ml 0 7 7m91sm0 51 771120 mgl sin0 52 z 7gsin0 53 This is of course the same equation we derived from Newton s laws Eq 28 We do not have however an equation to tell us about the tension similar to Eq 29 we need to use Lagrange multipliers to obtain constraint forces 32 Double Pendulum We need to write the kinetic and potential energy in terms of the generalized coordiantes 01 02 We already wrote velocity vectors in terms of the angular variables in Eqns 1519 Using those expressions the kinetic energy is 1 1 T m1v 1 577121 54 1 i 1 i i i i 57711110 5mg 110 130 211120102 cos02 7 01 55 1 i 1 i i i 5m1 m2l 0 57712130 771211120102 cos02 7 01 56 The potential energy is the gravitational potential energy V 7m19y1 7 m29y2 57 imlgll cos01 7 mggll cos01 Z2 cos02 58 7m1 717 m2gll COS 91 7 7712912 COS 92 The Lagrangian is then L T 7 V 1 2 2 1 2 2 577741 m2llt91 717 5777421202 71471211120102 COS02 7 91 m1 m2gl1 COS 91 m2gl2 COS 92 60 We begin calculating the terms needed for the Lagrange equation for 01 8L i i a m1 m2l 01 7712111202 COS02 7 91 61 1 1 8L m1 m2l 01 7712111202 COS02 7 91 77774211120 sin02 7 91 7774211129102 SlHltt92 7 91 8L A A i i 8701 771211120102 s1n02 7 91 7 m1 m2gl1 S111 91 63 Lagrange s equation for 01 is then 0 18 LL 7 dt 801 m1 717 m2lll2 2 COS02 7 91 77712111203 smog 7 01 m1 77121 sin 01 64 Similarly7 the Lagrange s equation for 02 is it 77121302 7712111201 COS02 7 91 65 802 d 8L 77121302 7712111201 COS02 7 91 m211129 sin02 7 91 7 771211120192 sin02 7 91 66 W 7771211120102 sin02 7 91 7 7712912 sin 92 67 2 0 i d 8L 8L 7 dt 802 m2lg 2 m2lll2 1 COS02 7 91 m211120 sin02 7 01 ngzz sin 02 68 We collect the two Lagrange equations of motion7 which are7 of course7 the same ones we got from Newton s law m1 m2ll 1 m2l2 2 COS02 7 91 77121203 sin02 7 91 7 m1 m2g sin 91 69 zg g 11 cos02 7 01 7110f smog 7 01 7 9 sin 02 70 Energy function Gabriela Gonzalez September 97 2006 Consider a Lagrangian describing a system of N particles The Lagrangian is a function of generalized coordinates qj Where the index j 1n 3N is the number of degrees of freedom of the system7 their time derivatives qj and time L Lqj7cij7 75 and Lagrange s equation in the presence of a dissipation function f and of forces that cannot be derived from a potential Where the index i 1N indicates the particle on which the force is acting d8L 8L7 afZFgem 387787739187739 8 The energy function is de ned as 8L h 47 7 L 2 Jaqj We want to calculate the time derivative of the energy function The time derivative of the Lagrangina function is dL 8L l 8L 8L a Eq a q a The time derivative of the energy function is dh 8L l d 8L dL 5 21787 q87j a 7 Zi39daL 8L leaL jqidtaq j 8t jqiaqj We use Lagrange s equation in the rst term to obtain dh 8L 8f 817 8L i 8L 7 2 a zn Jimi dt 8 8g 1 8 8t 7 8 8L 8f 8r if 7 iii i F39 1 at QJ8qjq7 z 8 The dissipation function is a homogeneous function of second degree with respect to qj so 8 2f Egg aqj We assume the constraints are holonomic7 and the particles7 positions rl are functions of only coordinates qj but not of their derivatives qj ri riqj t Thus7 8m 739 i 817 i Ejjquijm aqj gqjaq dI39l39 8m F939ltE Egt 8i The complete expression for the time derivative of the energy function is then db 7 8L ari a 7 7572f2ijriv zijrr t for f fqt a function of coordinates and time If we use a restricted version of the action principle7 keeping both q and p xed at the initial and nal times7 then the functions f can be a function of coordinates and momenta F Fqp7 25 Each such choice of function F will 77generate77 a change of coordinates if dt pq HdtPQ Hdt dF intpdq 7 Hdt 7 PdQ 7 H dt dF pdq 7 PdQ H 7 Hdt This is a restricting condition on the coordinates7 since the combination pdq 7 PdQ H 7 Hdt will not in general be an exact differential ie7 we would not be be able to nd an exact integral When integrated this expression gives us a function Fq Q t such that 8F 8F 8F 7 77P 7H 7H aq 1 8Q 875 A different and more useful way to View these equations is to think that anyquot function FqQ 25 will generate a coordinate transformation qp a 62 P where the change of coordinates are given by the equations 8F 7 8F 7 871 T p E T The equations of motion in either coordinate set will have the form of canonical equations of motion if H QPt Hqpt 8F8t where in the right hand side we use q 61Q7P7 t7FFqQ7P7 t7PQ7P7t7t The formalism can be derived in a straightforward manner for n generalized coordinates For example consider the generating function F qQ The coordinate change will be 8F g 7 7 P 8F 7 The new coordinate q is the old momentum p and the new momentum P is minus the old coordinate ql Just like in Lagrangian mechanics the generalized coordinates need not have units of length or be position coordinates in Hamiltonian mechanics the canonical momenta need not be related to linear momenta or velocities in fact the roles of coordinates and momenta are interchangeable up to a signl The solutions to the new equations of motion should of course be the same physical solutions as for the 77old77 problem T illustrate this point let s consider the Hamiltonian of a simple harmonic oscillator P qP P2 1 2 H 7 g 5kg The canonical equations of motion are q 8H8p pm and p iaHaq ikq Using the q equation into the p equation we obtain the SH0 equation k q Eq with the usual solution qt qo coswt gt with LUZ km If we perform the change of coordinates generated by F qQ the new Hamiltonian is the same as the old one because 8F8t O The change of coordinates is q 7Pp Q so the new Hamiltonian is 2 2 mg l 27 l 2 H72m2kPi2m2kP The new equations of motion are 8H kP Q 8P 8H p 7 m mm Again using the equation into the P equation we obtain a SHO equation Q 9 k m with the usual solution Qt Q0 coswt 110 and Pt 7Q0wk sinwt 110 and LUZ km We see that the 77old solution77 W mmww is the same as the 77new77 solution 7Pt QOwk sinwt die if 110 10 7r2 and Q0 quk The initial values of the coordinates are indeed the ones that tell us about differences in the choices for coordinates If the oscillator is a mass on a spring a physical statement of initial conditions could be 77start from rest with the spring stretched by a length A from its rest length From that physical description we have to come up with values for our constants of integration be those qo 10 or 620110 In this case the condition 77at rest77 is described by 10 0 but 110 7r2 while the initial spring length tells us that go A and Q0 Awk We know the Hamiltonian functions are the same under the transformation generated by F qQ because 8F8t 0 but the Lagrangian functions will not be the same L i L PQ 7 H i 2961 7 H PC 71961 41in dqpdt The Lagrangian functions differ by a total time derivative of the function iqp imqq This is notquot the generic transformation we learned about in Goldstein s problem 1 8 since it is a function of q q not just q canonical transformations are more general We can use Legendre transformations to nd generating functions of different pairs of variables other than q Q associated with Fq Q t For example if we want generating functions of q P that is using P instead of Q we know that ltIgtqP Fq Q PQ and dltIgt dFPQ pdqi PdQ H 7HdtPdQ QdP pdq QdP H 7Hdt so the equations for the coordinate transformations and the change in Hamiltonian is aqgt 53 aqgt aiqipy iQy EiHiH 3 Let us consider for example a change of coordinates generated by ltIgtqP t qP 7 fq t The new and old Hamiltonian functions are not the same now but 81 8f H H 7 H 7 7 l 8t 875 The coordinate transformation is 81 8f 7 p 7 7 p 5 61 5 61 81 Q q Only momenta are transformed While coordinates are the same The change in Lagrangian will be A 8 8 8 8 d any This is the case we studied in Goldstein s problem 1 8 and reVisited many times through the course These transformation includes the case of the electromagnetic gauge transfor mation in Problem 1 9 for example We have seen then that allquot of the transformations generated by arbitrary functions of coordinates fq t are canonical transformations generated by functions of the form ltIgtq P t qP 7 fq t a small set of the generic functions ltIgtq P t we could use
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