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

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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 H = 1 2 M ( Ip y + MBp θ ) 2 ( I MB 2 ) 2 B ( Ip y + MBp θ )( Bp y + p θ ) ( I MB 2 ) 2 + I 2
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