Goldstein_19_24_25

Goldstein_19_24_25 - Homework 9 8.19 8.24 8.25 Michael Good Nov 2 2004 8.19 The point of suspension of a simple pendulum of length l and mass m is

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Homework 9: # 8.19, 8.24, 8.25 Michael Good Nov 2, 2004 8.19 The point of suspension of a simple pendulum of length l and mass m is con- strained to move on a parabola z = ax 2 in the vertical plane. Derive a Hamilto- nian governing the motion of the pendulum and its point of suspension. Obtain the Hamilton’s equations of motion. Answer: Let x 0 = x + l sin θ z 0 = ax 2 - l cos θ Then T = 1 2 m ( ˙ x 0 2 + ˙ z 0 2 ) U = mgz 0 Solving in terms of generalized coordinates, x and θ , our Lagrangian is L = T - U = 1 2 m ( ˙ x 2 +2 ˙ xl cos θ ˙ θ +4 a 2 x 2 ˙ x 2 +4 ax ˙ xl ˙ θ sin θ + l 2 ˙ θ 2 ) - mg ( ax 2 - l cos θ ) Using L = L 0 + 1 2 ˜ ˙ qT ˙ q where ˙ q and T are matrices. We can see ˙ q = ± ˙ x ˙ θ ² T = ± m (1 + 4 a 2 x 2 ) ml (cos θ + 2 ax sin θ ) ml (cos θ + 2 ax sin θ ) ml 2 ² with 1
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L 0 = - mg ( ax 2 - l cos θ ) The Hamilitonian is H = 1 2 ˜ pT - 1 p - L 0 Inverting T by ± a b c d ² - 1 = 1 ad - bc ± d - b - c a ² with the algebra, 1 ad - bc = 1 m 2 l 2 (1 + 4 ax 2 ) - m 2 l 2 (cos θ + 2 ax sin θ ) 2 this is = 1 m 2 l 2 (sin 2 θ + 4 ax 2 - 4 ax cos θ sin θ - 4 a 2 x 2 sin 2 θ ) = 1 m 2 l 2 (sin 2 θ - 4 ax sin θ cos θ + 4 a 2 x 2 cos 2 θ ) which I’ll introduce, for simplicity’s sake, Y. = 1 m 2 l 2 (sin θ - 2 ax cos θ ) 2 1 m 2 l 2 Y So now we have T - 1 = 1 m 2 l 2 Y ± ml 2 - ml (cos θ + 2 ax sin θ ) - ml (cos θ + 2 ax sin θ ) m (1 + 4 a 2 x 2 ) ² T - 1 = 1 mY ± 1 - (cos θ + 2 ax sin θ ) /l - (cos θ + 2 ax sin θ ) /l (1 + 4 a 2 x 2 ) /l 2 ² I want to introduce a new friend, lets call him J J (cos θ + 2 ax sin θ ) Y (sin θ - 2 ax cos θ ) 2 So, T - 1 = 1 mY ± 1 - J/l - J/l (1 + 4 a 2 x 2 ) /l 2 ² Proceed to derive the Hamiltonian, H = 1 2 ˜ pT - 1 p - L 0 2
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we can go step by step, T - 1 p = 1 mY ± 1 - J/l - J/l (1 + 4 a 2 x 2 /l 2 ²± p x p θ ² = 1 mY ± p x - ( J/l ) p θ ( - J/l ) p x + (1 + 4 a 2 x 2 /l 2 ) p θ ² and ˜ pT - 1 p = 1 mY ( p 2 x - J l p θ p x - J l p θ p x + 1 + 4 a 2 x 2 l 2 p 2 θ ) the full Hamiltonian is H = 1 2 mY ( p 2 x - 2 J l p θ p x + 1 + 4 a 2 x 2 l 2 p 2 θ ) + mg ( ax 2 - l cos θ ) plugging in my Y and J H = 1 2 m (sin θ - 2 ax cos θ ) 2 ( p 2 x - 2 cos θ + 2 ax sin θ l p θ p x + 1 + 4 a 2 x 2 l 2 p 2 θ )+ mg ( ax 2 - l cos θ ) Now to find the equations of motion. They are ˙ x = ∂H ∂p x ˙ θ = ∂H ∂p θ ˙ p x = - ∂H ∂x ˙ p θ = - ∂H ∂θ The first two are easy, especially with my substitutions.
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This note was uploaded on 03/17/2012 for the course PHYS 202 taught by Professor Atkin during the Spring '12 term at Amity University.

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Goldstein_19_24_25 - Homework 9 8.19 8.24 8.25 Michael Good Nov 2 2004 8.19 The point of suspension of a simple pendulum of length l and mass m is

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