Goldstein 9-33 (3rd ed. 9.35) plus an additional part (b) (a) The Hamiltonian for this one dimensional system is H= p2 mk +2. 2m x (1)
We want to nd y (t), where y is the quantity x2 , using the Poisson bracket operator formalism. The formal solution is y
Goldstein 9-28 (3rd ed. 9.30) (a) Let H be a Hamiltonian for some dynamical system. If A and B are two constants of the motion that explicitly depend on time, then it must be true that A = [A, H ] = [H, A] , t and B = [B, H ] = [H, B ] . t We want to prov
Physics 409/Classical Mechanics Name ANSWERS
Test #5
19 November 2010
Answer all parts of each of the two questions. 50 points total. The points value of each part is indicated. Begin the answer to each question on a new sheet of paper. Clearly define all
Physics 409/Classical Mechanics Name ANSWERS
Test #4
4 November 2010
Answer all parts of each of the two questions. 50 points total. The points value of each part is indicated. Begin the answer to each question on a new sheet of paper. Clearly define all
Physics 409/Classical Mechanics Name ANSWERS
Test #3
19 October 2010
Answer all parts of each of the two questions. 25 points apiece. 50 points total. Begin the answer to each question on a new sheet of paper. Clearly define all coordinates and variables.
Physics 409/Classical Mechanics Name
Test #2
30 September 2010
Answer all parts of each of the two questions. 50 points total. The points value of each part is indicated. Begin the answer to each question on a new sheet of paper. Clearly define all coordi
Physics 409/Classical Mechanics Name ANSWERS
Test #1
16 September 2010
Answer all parts of each of the two questions. 50 points total. The points value of each part is indicated. Begin the answer to each question on a new sheet of paper. Clearly define al
Physics 409: Classical Mechanics 16 November
Fall 10
Lecture schedule: 18 November: Review class for Test #5 19 November: 9 am - Fuller Chapter Test #5 on Ch. 5 (excluding heavy top) 30 November - Ch. 9, mainly Sec. 5 and 6 (2nd ed.); Sec. 6 and 7 (3rd ed
Physics 409: Classical Mechanics 9 November
Fall 10
Lecture schedule: 11 November: Chap. 8, Sec. 8.2 16 November: Ch. 9, mainly Sec. 4 and 5 (2nd ed.); Sec. 5 and 6 (3rd ed.) 18 November: Review class for Test #5 (See reverse side for topics.) 19 November
Physics 409: Classical Mechanics 2 November
Fall 10
Lecture schedule: 4 November: Chapter Test #4 on Ch. 4 9 November: start Chap. 8, Sec. 8.1;(Skip symplectic notation in Sec. 8-1.) 11 November: Chap. 8, Sec. 8.2 Note: We will not cover Sec. 8-3, 8-4 or
Physics 409: Classical Mechanics 26 October
Fall 10
Lecture schedule: 28 October: Chap. 5, Sec. 5.5, 5.6 2 November: Chap. 5, Sec. 7; Review Session for Ch. 4, 3:30-6, Rm. 202 4 November: Chapter Test #4 on Ch. 4 For Test #4 you should know or be able to
Physics 409: Classical Mechanics 21 October Lecture schedule: 26 October: Ch. 5, Sec. 5.4 28 October: Chap. 5,Sec. 5.5, 5.6
Fall 10
Reading tips for 26 October: What is a principal axis transformation? What are the principal axes of a rigid body? What are
Physics 409: Classical Mechanics 28 September 2010
Fall 10
Lecture schedule: 28 September: Chapter 3; 30 September: Chapter Test #2 on Ch. 2 (more problems setting up and solving Lagrangians, including Lagrange multipliers, plus calculus of variations pro
Physics 409: Classical Mechanics Fall 10 21 September Lecture schedule: L8: 21 September Ch. 3, Sec. 1-3 (See typos and comments on other side.) L9: 23 September: Ch. 3, Sec. 3, 5, 6 L10: 28 September Ch.3 Sec. 7, 8 Review session for Ch. 2: 3:30-6, Rm202
Physics 409: Classical Mechanics Fall 10 14 September Lecture schedule:L7 - 14 September 2.4 (but not the version in GPS) 16 September: Chapter Test #1 on Chapter 1 L8 - 21 September: Chap. 3 Review session: Tuesday, 14 September, 3:30 pm Rm 202 Reading t
Physics 409: Classical Mechanics 7 September Lecture schedule: L6 - 9 September: Ch.2, Sec. 2.2, 2.1, 2.3,2.5, 2.6, 2.7(3rd Ed.) L7 - 14 September 2.4 (but not the version in GPS) 16 September: Chapter Test #1 on Chapter 1
Fall 10
Reading tips for 9 Sept.
Physics 409: Classical Mechanics 31 August Lecture schedule: 31 August- 2 September: Chapter 1 7 September: Finish Ch. 1, start Ch.2, Sec. 2.2
Fall 10
Reading tips for 2 Sept.: Focus on: The Principle of Virtual Work and D'Alembert's Principle What is a g
Physics 409: Classical Mechanics 24 August Lecture schedule: 24 August - 2 September: Chapter 1 Homework Set #1: Due Tuesday, 31 August (1) Chap.1, GPS Problem 1 (#4 in 2nd ed) (2) Chap.1, GPS Problem 2 (#5 in 2nd ed) (3) Chap.1, GPS Problem 3 (#6 in 2nd
Goldstein 9-39a (3rd ed. 9.39a) (1) USING CARTESIAN COORDINATES Using Cartesian canonical variables, the Hamiltonian for the Kepler problem is H= where p2 = p2 + p2 + p2 , x y z and r= p x2 + y2 + z 2 . mk r. r (3) (2) k p2 , 2m r (1)
The Laplace-Runge-Le
Poisson Brackets for the Heavy Top The rst thing to do is dene the canonical variables, i.e., the generalized coordinates and momenta for the problem. The Euler angles , , and are the generalized coordinates. The corresponding canonical momenta are found
Goldstein 9-37 (3rd ed. 9.37) Let r be the position vector of the bob of mass m. In spherical coordinates the components of r are x = r sin cos , (1) y = r sin sin , z = r cos , (2) (3)
where r = |r| =constant, is the polar angle between r and the z axis,
GPS 8-27 (2nd ed. 8.17) (a) Using the identity, sin 2 = 2 sin cos , the given Lagrangian can be rewritten as m 2 2 22 L= q sin t + 2qq sin t cos t + q , (1) 2 from which we nd the generalized momentum p p= L q The energy function h is h=q
= mq sin2 t +
Goldstein 8-15 (3rd ed. 8.25) This problem diers from 8-14 because the cylinder is now driven to rotate at a constant angular speed , and its kinetic energy is no longer variable. Since the cylinders angular motion is prescribed, we only need to consider
Goldstein 8-14 (3rd ed. 8.24) This view is looking down on the cylinder from above. The z axis is coming up out of the page through the center of the circular cylinder. The x axis is fixed in space. The x& axis is fixed in the cylinder. The angle measures
GPS Problem 8.20 (2nd ed. 8.10) Modied Planar Double Pendulum This problem is a variation of the planar double pendulum problem, illustratrated in Fig. 1.4. The solution to that problem, GPS 1.22 or 2nd ed. 1.20, will be useful here. There are two major m
GPS 8-17 or Goldstein 8-7 Direct route: From Problem 5.13 (2nd ed.) or 5.19 (3rd ed.) the Lagrangian of this pendulum is L= M 2 I 2 1 y M ly sin + ky 2 + M g [y + l cos ] . 2 2 2 L y L
(1)
The generalized momenta, py and p , are py = and p = = M y M l s