Assignment 2. Tyler Shendruk October 8, Hamilton s Principle - Lagrangian and Hamiltonian dynamics.

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1 Assignent Tyler Shendruk October 8, Marion and Thornton Chapter 7 Hailton s Principle - Lagrangian and Hailtonian dynaics. 1.1 Proble 7.9 y z x l θ α Figure 1: A disk rolling down an incline plane. Fro the axis of the disk hangs a siple pendulu. Consider a disk of ass M and radius R that s rolling down a plane of angle α as shown in Fig. 1. Fro the axis of this disk is a siple pendulu of length l < R with a bob of ass. Consider the otion of the pendulu. We ll do this in z, θ coordinates as shown in Fig. 1. The Cartesian coordinates in thers of z and θ are x disk = z cos α y disk = z sin α ẋ disk = ż cos α ẏ disk = ż sin α. The Cartesian coordinates of the bob are just that of a siple pendulu but in the regerence frae of the disk i.e. x bob = x disk + l = z cos α + l y bob = y disk l cos θ = z sin α l cos θ ẋ bob = ż cos α + l cos θ ẏ bob = ż sin α + l 1

2 Also notice that the rotation angle φ of the disk is siply proportional to the distance travelled i.e. φ = z R If we take the oent of inertia about the centre of the disk to be I = MR then the energy and Lagrangian are then given to be U = U disk + U bob = Mgy disk + gy bob = gz M + sin α gl cos θ 1a T = T bob + T kin,disk + T rot,disk = [ẋ bob + ẏbob M ẋ + disk + ẏdisk 1 + I φ = [ ż cos α + l cos θ + ż sin α + l + M ż cos α + ż sin α + 1 MR ż R = ż M + + l + ż l cos α cos θ sin α + Mż 4 = ż 3 M + + l + ż l cos α + θ 1b L = T U = ż 3 M + + l + ż l cos α + θ + gz M + sin α + gl cos θ 1c Notice, since it s coe up ultiple ties the kinetic energy of the disk and the bob are T disk = T kin,disk + T rot,disk = M ż + M 4 ż = 3 4 Mż T bob = [ẋ bob + ẏbob = [ ż + l + ż l cos α cos θ sin α = [ ż + l + ż l cos α + θ Lagrange Equation for z 0 = z d dt ż = Mg sin α + g sin α d dt [ = M + g sin α z [ 3 ż 3 M + M + + l cos α + θ + l θ cos α + θ l sin α + θ

3 3 z M + + l θ cos α + θ l sin α + θ M + g sin α = 0 Lagrange Equation for θ 0 = θ d dt = ż l sin α + θ gl d [l + żl cos α + θ dt = ż l sin α + θ gl [l θ + zl cos α + θ żl sin α + θ 1. Proble 7.34 l θ + z cos α + θ + g = 0 3 y x M x x R y a Cartesian Coordinates y X x R r θ b Unconstrained Coordinates Figure : A particle sliding down a sooth circular wedge. Consider a particle of ass sliding down a sooth circular wedge of ass M. The particle denoted by subscript slides - not rolls. The wedge denoted by subscript M slides on a sooth horizontal surface. Of course, you can always chose whatever coordinates you want but soe are ore judicious than others. We will use X, r, θ fro Fig. b. We can related these unconstrained coordinates to the perhaps ore intuitive cartesian coordinates of Fig. a. Wedge: Particle: x M = X y M = 0 x = X + r cos θ y = r 3

4 ẋ = Ẋ + ṙ cos θ r ẏ = ṙ r cos θ Now this is really great because we can set up the energy in cartesian and then talk about the unconstrained coordinates after their set up i.e. U = gy = gr 4a T = 1 Mẋ M + 1 }{{} [ ẋ + ẏ }{{} wedge particle = 1 MẊ + 1 [Ẋ + Ẋṙ cos θ Ẋr + ṙ cos θ rṙ cos θ +r sin θ + ṙ sin θ + ṙr cos θ + r cos θ = 1 M + Ẋ + 1 [ ṙ cos θ + sin θ +r cos θ + sin θ + Ẋṙ cos θ Ẋ r = 1 M + Ẋ + 1 [ṙ + r + Ẋṙ cos θ Ẋ r L = T U = 1 M + Ẋ + 1 [ṙ + r + Ẋṙ cos θ Ẋ r 4b + gr 4c Now we can get the Lagrange equations for each of the two or three variables. When we recognize that r = R right off the bat then there is no need for a Lagrange ultiplier λ and so we can copletely solve the proble by the two Lagrange equations for θ and X. However, if we leave r as a variable and consider a third Lagrange equation then we require a λ that connects the. By doing this we will learn about the physical process that ensures r = R. Since r = R is what we wish to ensure by including λ, the function f is Notice, fx, θ, r = r R = 0. 5 Lagrange Equation for X f X = 0, f θ = 0, f r = 1 X d dt Ẋ + λ f X = 0 X d dt Ẋ = 0 6a 4

5 Lagrange Equation for θ Lagrange Equation for r θ d + λ f dt θ = 0 X d dt Ẋ = 0 r d dt ṙ + λ f r = 0 r d dt ṙ + λ = 0 6b We see that the equations for X and θ are no different than if we hadn t used the ultiplier but we have the extra equation for r which will give us the force of constraint that ensures r = R i.e. the force of the wedge and the particle on each other. So now let s find the equations of otion fro Eq. 6a and Eq. 6b later we ll find the force of constraint fro Eq. 6c. Lagrange Equation for X 0 = X d dt Ẋ = 0 d [M + dt Ẋ ṙ cos θ r + 0 = M + [ r Ẍ + cos θ ṙ θr ṙ r cos θ = M + [ r Ẍ + cos θ ṙ r θ + cos θ. 6c Applying the constraint r = R gives r = 0 = ṙ so that we can see 0 = M + θ Ẍ R + cos θ Ẍ = θ M + R + cos θ 7 Lagrange Equation for θ 0 = θ d dt = [0 + 0 Ẋṙ Ẋ r cos θ + gr cos θ d [0 + 1 dt 0 + r + 0 Ẋr + 0 [ r Ẋr = Ẋṙ Ẋ r cos θ + gr cos θ d dt = Ẋṙ Ẋ r cos θ + gr cos θ = ṙ rṙ r θ + Ẍr + Ẋṙ Ẋr cos θ [ Ẋ r + Ẋ + r [g cos θ r θ + Ẍ. 5

6 Again r is constant so this reduces to 0 = 0 + R [g cos θ R θ + Ẍ R θ = Ẍ + g cos θ θ = Ẍ + g cos θ R i.e. X and θ are given by two second order coupled ODEs. 8 Since we kept r as a variable in Eq. 4c we are already in the perfect position to find the Lagrange ultiplier λ fro Eq. 6c. 0 = r d dt ṙ + λ = [0 + r + 0 Ẋ + g d [0 + 1 ṙ dt Ẋ cos θ λ = r Ẋ + g r + Ẍ cos θ Ẋ + λ λ = r g + r + Ẍ cos θ Now using that r = R and ṙ = 0 = r we can say λ = r + g Ẍ cos θ 9 = F r where we have explicitly put a inus sign in to reind us that this constraining force, noral to the circular surface is centripetal and has a direction inward. We can use our solutions for θ and Ẍ to get F r in ters of θ and only by solving Eq. 7 and Eq. 8 for Ẍ. Ẍ = θ M + R + cos θ = Ẍ sin θ + g cos θ + R cos θ Ẍ Ẍ M + Ẍ M + Ẍ = g cos θ + R M + cos θ 1 1 cos θ = g cos θ + R M + M + cos θ M M + + M + cos θ = g cos θ + R M + cos θ Ẍ M + cos θ = g cos θ + R cos θ Ẍ = g cos θ + R cos θ M + cos θ 10 6

7 With Ẍθ, we can give F r to be F r = R + g Ẍ cos θ = R + g g cos θ + R cos θ M + cos θ MR = + Mg + R cos θ + g cos θ M + cos θ g cos θ + R cos θ M + cos θ = M cos MR θ + Mg M R + g F r = M cos θ 11 7