P235 Midterm Examination Prof. Cline

Transcription

1 P235 Mier Exaination Prof. Cline THIS IS A CLOSED BOOK EXAMINATION. Do all parts of all four questions. Show all steps to get full credit. 7: p, 30 October :(20pts) Consider a rocket fired vertically upwards fro the ground in a unifor vertical gravitation field g =9.81/s 2. The rocket, with initial ass I,ejectsα kg/sec of propellant at an exhaust velocity u /sec. Derive the equation giving thetiedependenceoftheverticalvelocityoftherocket. SOLUTION: When there is a vertical gravitational external field the vertical oentu is not conserved. In a tie the rocket ejects propellant d p with exhaust velocity u. Thus the oentu iparted to this propellant is dp p = ud p Therefore the rocket is given an equal and opposite increase in oentu The thrust F due to ejection of the exhaust is given by dp R =+ud p F = dp R = dp p Consider the total proble for the special case of vertical ascent of the rocket against a gravitational force F G = g. Then This can be rewritten as g + u d p = dv R g + uṁ p = v R The second ter coes fro the variable ass. Let the fuel burn be constant at ṁ = ṁ p = α where α>0. Then the equation becoes Since then Inserting this in the above equation gives Integration gives But the change in ass is given by That is Thus dv = dv = ³ g + α u d = α d α = ³ g α u d v = g α ( I )+u ln Z I d = α Z t 0 I = αt v = gt + u ln ³ I ³ I 1

2 2; [30pts] Consider a ass attracted by a force that is directed toward the origin and proportional to the distance fro the origin, F = k r. The ass is constrained to ove on the surface of a cylinder with the origin on the cylindrical axis as shown. The radius of the cylinder is R. a) Derive the Lagrangian and Hailtonian b) Explain if the Hailtonian equals the total energy and if it is conserved? c) Derive the equations of otion d) Show which variables, if any, are cyclic e) Derive the frequency of the otion along the axis of the cylinder. SOLUTION This proble considers the case where a ass is attracted by a force directed toward the origin and proportional to the distance fro the origin. and the ass is constrained to ove on the surface of a cylinder defined by x 2 + y 2 = R 2 a) Derive the Lagrangian and Hailtonian It is obvious to use cylindrical coordinates ρ, z, θ for this proble. Since the force is just Hooke s law F = k r the potential is the sae as for the haronic oscillator, that is That is, it is independent of θ. In cylindrical coordinates the velocity is Confined to the surface of the cylinder eans that Then the Lagrangian siplifies to U = 1 2 kr2 = 1 2 k(ρ2 + z 2 ) v 2 =. ρ 2 + ρ 2. θ 2 + ż2 ρ = R. ρ = 0 L = T U = 1 2 µ R 2. θ 2 + ż2 1 2 k(r2 + z 2 ) The generalized coordinates are θ, z and the generalized oenta are p θ = L θ. =. R2 θ p z = L = ż ż (b) (a) 2

3 H = T + U = p 2 θ 2R 2 + p2 z k(r2 + z 2 ) b) Explain if the Hailtonian equals the total energy and if it is conserved? Since the L t =0the Hailtonian is conservative, and since the transforation fro rectangular to cylindrical coordinates does not depend explicitly on tie, then the Hailtonian equals the total energy. c) Derive the equations of otion. This can be done directly for the Lagrange equations. That is, gives Λ z L = d L ż L z =0.. z + k z =0 Siilarly Λ θ L = d L L θ θ =0 gives d R2 θ =0 d) The cyclic variables The angular oentu about the axis of the cylinder is conserved. since p θ = constant. That is, θ is a cyclic variable. e)the frequency of sall oscillations.. z + k z =0 is the equation for siple haronic otion with angular frequency r k ω = It is interesting that this proble has the sae solutions for the z coordinate as the haronic oscillator while the θ coordinate oves with constant angular velocity. 3

4 Figure 1: The yo-yo coprises a falling disc unrolling fro an attached string attached to a fixed support. 3; (30pts) As part of the Halloween, and the post ier exa celebrations, Julieta, and Wendi, organize their colleagues to drape crepe ribbons fro the highest bridge in Wilson Coons. They plan to do this by dropping rolls of crepe ribbon after fastening the loose end to the parapet. Being very careful physicists they decide that first they need to know the tension in the ribbon to ensure that it does not tear when the roll drops, as well as the speed of the roll after falling 20 to ensure it does not hurt anyone. Assue that the rolls of ribbon have a ass M and radius a and that the thickness and ass of the unrolled ribbon is negligible copared with the ass of the roll, that is M and a are constant as it drops, thus the oent of inertia of the roll is I = 1 2 Ma2. a) Derive the Lagrangian for one roll of crepe dropping with the upper loose end of the crepe ribbon held fixed. b) Use Lagrange ultipliers to derive the equations of otion c) Derive the tension in the ribbon. d) Derive the vertical acceleration and final velocity of the roll after falling 20 assuing it starts with zero velocity. SOLUTION: a) Consider the roll of ribbon coprising a disc that has a ribbon wrapped around it with one end attached to a fixed support. The roll is allowed to fall with the ribbon unwinding as it falls as shown in the figure. This is identical to the yo-yo proble. Let us derive the equations of otion and the forces of constraint. The kinetic energy of the falling yo-yo is given by T = 1 2 ẏ I φ 2 = 1 2 ẏ a2 φ2 where is the ass of the disc, a the radius, and I = 1 2 a2 is the oent of inertia of the disc about its central axis. The potential energy of the disc is U = gy Thus the Lagrangian is L = 1 2 ẏ a2 φ2 + gy b) The one equation of constraint is g(y, φ) =y aφ =0 The two Lagrange equations are f y d f y 0 + λ g y =0 f φ d f φ 0 + λ g φ =0 with only one Lagrange ultiplier. Evaluating these gives the two equations of otion g ÿ + λ = a2 φ λa = 0 4

5 Differentiating the equation of constraint gives φ = ÿ a Inserting this into the second equation and solving the two equations gives Inserting λ into the two equations of otion gives λ = 1 3 g ÿ = 2 3 g φ = 2 g 3 a c) The generalized force of constraint is the tension in the ribbon F y = λ g y = 1 3 g and the constraint torque is N φ = λ g φ = 1 3 ga d) Thus we have that the ribbon reduces the acceleration of the disc in the gravitational field by one third. ÿ = 2 3 g Since the acceleration is unifor and initial velocity is zero, then the final velocity is given by Inserting nubers gives v = = 16.17/s v 2 =2ad = 4 3 gd 5

6 Figure 2: Spring pendulu having spring constant k and oscillating in a vertical plane. 4; [30pts] The P235W ier exa induces Jacob to try his hand at bungee juping. Assue Jacob s ass is suspended in a gravitational field by the bungee of unstretched length b and spring constant k. Besides the longitudinal oscillations due to the bungee jup, Jacob also swings with plane pendulu otion in a vertical plane. Use polar coordinates r, φ, neglect air drag, and assue that the bungee always is under tension. a; Derive the Lagrangian b; Deterine Lagrange s equation of otion for angular otion and identify by nae the forces contributing to the angular otion. c; Deterine Lagrange s equation of otion for radial oscillation and identify by nae the forces contributing to the tension in the spring. d; Derive the generalized oenta e; Deterine the Hailtonian and deterine Hailton s equations of otion. a: Derive the Lagrangian: This is identical to the spring pendulu proble. The syste is holonoic, conservative, and scleronoic. Introduce plane polar coordinates with radial length r and polar angle φ as generalized coordinates. The generalized coordinates are related to the cartesian coodinates by Therefore the velocities are given by y = r cos φ x = r sin φ ẏ = ṙ cos φ + r φ sin φ ẋ = ṙ sin φ r φ cos φ The kinetic energy is given by T = 1 2 ẋ 2 + ẏ 2 = 1 2 ³ ṙ 2 + r 2 φ2 The gravitational plus spring potential energies are U = gr cos φ + k (r b)2 2 where r 0 denotes the rest length of the spring. The Lagrangian thus equals L = 1 ³ 2 ṙ 2 + r φ2 2 + gr cos φ k (r b)2 2 b; Deterine Lagrange s equation of otion for angular otion and identify by nae the forces contributing to the angular otion; Forthepolarangleφ, the Lagrange equation Λ φ L =0gives d ³ r 2 φ = gr sin φ But the angular oentu p φ = r 2 φ, thus the equation of otion can be written as ṗ φ = gr sin φ 6

7 ³ Alternatively evaluating d r 2 φ gives r 2 φ = gr sin φ 2rṙ φ The last ter in the right-hand side is the Coriolis force caused by the tie variation of the pendulu length r. c; Deterine Lagrange s equation of otion for radial oscillation and identify by nae the forces contributing to the tension in the spring. For the radial distance r, the Lagrange equation Λ r L =0gives r = r φ 2 + g cos φ k (r r 0 ) This equation just equals the tension in the spring, i.e. F = r. The first ter on the right-hand side represents the centrifugal radial acceleration, the second ter is the coponent of the gravitational force, and the third ter represents Hooke s Law for the spring. For sall aplitudes of φ the otion appears as a superposition of haronic oscillations in the r, φ plane. In this exaple the orthogonal coordinate approach used gave the tension in the spring thus it is unnecessary to repeat this using the Lagrange ultiplier approach. d; Derive the generalized oenta: p r = L ṙ = ṙ p φ = L φ = r2 φ e; Deterine the Hailtonian and deterine Hailton s equations of otion. Since the L t = 0 the Hailtonian is conservative, and since the transforation fro rectangular to cylindrical coordinates does not depend explicitly on tie, then the Hailtonian equals the total energy. Thus H = T + U = p 2 φ 2r 2 + p2 r 2 gr cos φ + k 2 (r r 0) 2 Thus Hailton s equations of otion are given by the canonical equations ṗ φ = H θ = gr sin φ (c) Equations d and e give that ṗ r = H r = k (r b)+g cos φ + p2 φ r 3 φ = H = p φ p φ r 2 ṙ = H p r = p r (d) (e) (f) ṗ r = r = k (r b)+g cos φ + p2 φ r 3 = k (r b)+g cos φ + r φ 2 7