HW13_Prob2_GPS17_G07

HW13_Prob2_GPS17_G07 - GPS 8-17 or Goldstein 8-7 Direct...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 · y 2 Ml · y · θ sin θ + I 2 · θ 2 1 2 ky 2 + Mg [ y + l cos θ ] . (1) The generalized momenta, p y and p θ ,are p y = ∂L · y = M · y Ml · θ sin θ, (2) and p θ = ∂L · θ = I · θ Ml · y sin θ. (3) The energy function h for this system is h = · yp y + · θp θ L, (4) and after substituting Eqs.(1), (2), and (3) we obtain h = M 2 · y 2 Ml · y · θ sin θ + I 2 · θ 2 + 1 2 ky 2 Mg [ y + l cos θ ] , (5) so it is clear that h = E , the energy of this pendulum. To obtain the Hamil- tonian, we solve Eqs. (2) and (3) for · y and · θ , · y = Ip y /M + Bp θ I MB 2 , (6) and · θ = Bp y + p θ I MB 2 , (7) where B = l sin θ. (8) Recalling that for this pendulum I =4 Ml 2 / 3 , (9) we see that I MB 2 = Ml 2 (4 / 3 sin 2 θ ) , (10) and is, thus, never equal to zero. After substituting Eqs.(6) and (7) into Eq.(5), we obtain for the Hamiltonian
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

Page1 / 3

HW13_Prob2_GPS17_G07 - GPS 8-17 or Goldstein 8-7 Direct...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online