Question

given a double rigid body pendulum with an applied motor torque at the top of the pendulum, what would be our equation of motion for the change in angle using Lagrangian mechanics?

Answer 1

Single and Double plane pendulum Gabriela González 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. Om mg mg B. T2 m, m29 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 mi and m2, connected by rigid weightless rods of length 11 and 12, 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 1

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. 2.1 Constraints The simple pendulum system has a single particle with position vector r = (x, y, z). There are two constraints: it can oscillate in the (x,y) plane, and it is always at a fixed distance from the suspension point. Mathematically, 2 0 (1) (2) r 1. The double pendulum system has two particles (N=2) with position vectors rı, r2, each with components (Li, Yi, zi). There are four constraints: each particle moving in the x-y plane, and each rod having constant lengths. These constraints can be expressed as 21 0 (3) 22 0 11 12 (5) (6) r2 - ril 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 (N=1), so we have 3N=3 coor- dinates. Since we have two constraints (m=2), we are left with n=3N-m=3-2=1: only one generalized coordinate. This is the angular position of the pendulum 0, which we can use to write: r=1(sin , cos 0,0). (7) In the double pendulum We know there should be only two generalized coordinates, since there are 3N=6 coordinates, and m=4 constraints, so n=3N-m=6-4=2. We can find expressions for rı, r2 in terms of two angles 01, 02: ri = 1 (sin 1, cos 01,0) r2 = ri +12(sin 02, cos 62,0) (8) (9) We can express velocity an acceleration vectors in terms of generalized coordinates. For the single pendulum, 2

r = f = a (13) (sin 0, cos ) (10) r=v 10(cos e, - sin ) (11) lë(cos 0, - sin ) - 102 sin 0, cos ) (12) = löy - 102 The velocity vector v is perpendicular to the position vector r, which is the expression of the constraint r = 1 = 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 first particle ri (14) (15) ri = v1 11 (sin 01, cos 01) 116 cos 01, -sin 01) 110 cos 01, -sin 01) - 110 (sin 01, cos 01) 101 - 11671 r = aj (16) (17) and the second particle: r2 (18) ri +12(sin 02, cos 02) Vi +1262(cos 02, -sin 02) al + 1272 (cos 62, – sin 02) – 1202 (sin 62, cos (2) 12 = V2 r2 = a2 (19) (20) 2.2 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 -r: T F=T + mg = ir+mg (21) In the double pendulum, the forces on mı are the tension in the two rods, and gravity. The tension in the upper rod is along the direction -rı, the tension force on mı due to the lower rod is along the direction r2 - rı, so we can write the force F, as -ri r 2 - ri T T +T + mig = eri + (r2 - ri) + mig (22) ri r2 - ri The forces on m2 are the tension in the lower rod, and gravity. The tension on my is along the direction of -(r2 - ri): (r2 -r1) T2 F = T2 - + m2g = - (r2-ri) + m2g (23) r2-ril F1 = 117 3

2.3 Equations of motion 2.3.1 Single Pendulum In the single pendulum case, Newton's law is F = mi. Writing the two non-trivial compo- nents, we have T mr F = -r+mg 1 (24) ml (ölcos 0, - 1, – sin 6) – 72 (sin 0, cos e)) = -T(sin , cosa) + mg(0,1). (25) We thus have two equations: ml (ö cos 0 – ? sin ) -T sine (26) -ml (ö sin 0 + Ö'cos e) = -T cos 0 + mg (27) Notice that although we only have one generalized coordinate (O), 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, 0 and T. The equations are not only coupled, but are also non-linear, involving trigonometric functions (ugh!). A common trick with expressions involving trig is to make use of trig identities like cos? 0 + sin? 0 = 1. For example, multiplying 43 by cos 0, and adding Eq. 45 multiplied by - sin 0, we obtain a simpler equation for 0. Using these identities, we can write the equations as lē -9 sine (28) mlh2 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 zero!). Whichever way, once we have a solution for e(t), we can use it in Eq.29 to solve for the other unknown, the tension T. Eq.29 does not invovle derivatives of T, so there are no new constants of integration for the problem. 2.3.2 Double Pendulum In the double pendulum, Newton's second law on each particle is Fi = miri: T miii -r + (r2-r1) + mig T2 12 (30) m2r2 = T2 (r2 - r.) + m2g (31) 4

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 x,y 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: mi (61 i cos 01 – oſ sin 01) -Ti sin 6+T2 sin 02 (32) -malı (öz sin 61 + oſ cos 0 -Tcos 01 +T2 cos 02 +m19 (33) m2 -m2 (101 cos 61 – 110; sin 01 + 1272 cos 01 – 1203 sin 02. -T2 sin 02 (34) (lö sin 01 +1102 cos 61 + 1202 sin 02 + 1203 cos 02) = -T2 cos 02 + m2g. (35) We have then four differential equations, for four unknowns (01, 02, T1, T2). Trigonometric identities such as cos0 + sin? 0 = 1, and sin 62 cos 01 - cos 02 sin 01 sin(02 - 01) can be used to write the equations of motion as: lä (T2/mı) sin(02 - 01) - gsin 01 (36) 110 (T1/mi) - (T2/mı) cos(O2 - tu) - g cos 01 (37) 10 cos(02 – 01) + 110 sin(02 – 01) +1272 = -g sin 02 (38) -17, sin(02 – 01) +110 cos(02 – 01) + 1203 = (T2/m2) - g cos 02. (39) We now can use Eqns. 36,37 to make substitutions in Eqns. 38, 39, and use more trig identities to simplify these equations further: 172 = -9 sin 02 – ((T2/mı) sin(02 - 01) - gsin (1) cos(02 – 01) -(T/mı) - (T2/mı) cos(02 – tı) - g cos ) sin(02 - 01) = -(T1/mı) sin(02 - 01) (40) 1203 (T2/m2) - g cos 02 + ((T2/mı) sin(02-01) - gsin (1) sin(02-01) - (T1/mi) - (T2/mı) cos((2 – tı) - g cos 01) cos(02 - 01) = (T2/m2) + (T2/mı) - (T1/mı) cos(02 - 01) (41) The four equations of motion are then 18 (T2/mı) sin(02 - 01) - gsin 01 (42) 1,03 (T1/mı) - (T2/m1) cos(02 - 01) - g cos 1 (43) 120, -(T1/mı) sin(02-01) (44) (T2/m2) + (T2/mı) - (T1/mı) cos(02 - 01) (45) 5

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 T, T, from Eqns. 44, 42: 1272 T -mi (46) sin(02-01) 118: +gsin T2 (47) sin(02-01) mi We use these expressions in Eqn. 43: 1102 (T1/mı) - (T2/mı) cos(02-01) - gcos @ 1272 1,0 +gsin 01 - cos(02-01) - g cos sin(02-01) sin(02-01) 1272 +1,0 cos(02 - 01) + g sin 02 -110 sin(02 – 01) (48) 3 Lagrange's equations 3.1 Simple Pendulum We have one generalized coordinate, 0, so we want to write the Lagrangian in terms of 0,0 and then derive the equation of motion for 0. The kinetic energy is T = (1/2)mv2 = (1/2)m1202 (using Eq. 11 for the velocity). The potential energy is the gravitational potential energy, V = -mgy = -mgl cos 0. Notice we can derive the gravitational force from the potential, F, = -VV = mg(0,1) = mg, but not the tension force on the pendulum: that is a constraint force. The Lagrangian is L=T-V= + (49) The Lagrange equation is 0 (50) d al al dt ad ao d - (-mgl sin o) dt = ml?ö + mgl sin 0 = -9 sin (m1²0) (51) (52) (53) 10 6

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. 3.2 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. 15,19. Using those expressions, the kinetic energy is 1 T + (54) 2 1 žmaloz + 5m2 (1202 +1303 +24120102 cos(Oz – 01 (55) 1 1 (56) (57) (58) (59) (60) + 2 The potential energy is the gravitational potential energy; V -migyi – m2942 -migli cos 01 – m29(l1 cos 01 + 12 cos 02) -(mi + m2)glı cos 01 - magl2 cos 02 The Lagrangian is then L T-V 3(mı + m2)102 + z malzo +malı20102 cos(02 - 01) + (mi + m2)gli cos 0 + m2gl2 cos 02 We begin calculating the terms needed for the Lagrange equation for 01: al = (mi + m2) + malı1202 cos((2 – 01) 201 d al = (mi + m2)t+ m2l11202 cos(02 – 01) -malı10 sin(02 - 01) + mal1120102 sin(02 – 01) al = molil20,02 sin(02 - 01) - (mi + m2)glı sin 01 a Lagrange's equation for , is then d al al 0 dt að = (mı + m2) 10. + m2lılző, cos(02 – 01) -malıl202 sin(02 – 01) + (mı + m2)glı sin 01 (61) dt a (62) (63) (64) 7

Similarly, the Lagrange's equation for 02 is al a02 d al malğó2 + malıl20, cos((2 – 01) (65) dt ai2 m21202 + m2lı120 cos(62 – 01) +malıl20 sin(02 - 01) – mal1l20102 sin(02 – 01) = -mel1120102 sin(02 - 01) - m2gl2 sin 02 (66) al a02 (67) 0 dt a02 d al al a02 malzőz + m2lılző, cos(62 – 01) +malı120 sin(02 – 01) + m2gl2 sin 02 (68) We collect the two Lagrange equations of motion, which are, of course, the same ones we got from Newton's law: (mı + m2) + m2l272 cos(02 - 01) = 1272 +117 cos(02-01) malzó sin(02-01) -(mi + m2)g sin 01 (69) -110 sin(02 - 01) - g sin 02 (70) 8