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hw7_scribe - Problem 1: Position of M: (x(t),x2 (t))...

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Unformatted text preview: Problem 1: Position of M: (x(t),x2 (t)) Position of m: (x(t) + l ∗ sin(θ),x2 (t) − l ∗ cos(θ)) Velocity of M: (x, 2xx) ˙ ˙ ˙ ˙ Velocity of m: (x + l ∗ cos(θ) ∗ θ,2 ∗ x ∗ x + l ∗ sin(θ) ∗ θ) ˙ ˙ 1 KE = 2 M VM + 1 m Vm 2 1 ˙ ˙ = 1 M ∗ (x2 + 4x2 x2 ) + 2 m ∗ ((x + l ∗ cos(θ) ∗ θ)2 + (2xx + l ∗ sin(θ) ∗ θ)2 ) ˙ ˙ ˙ ˙ 2 1 1 2 2 2 ˙2 = 2 (M + m) ∗ (1 + 4x ) ∗ x + 2 ml θ + mlxθ(cos(θ) + 2x ∗ sin(θ)) ˙ ˙˙ P E = M gx2 + mg (x2 − l ∗ cos(θ)) = (m + M )x2 − mglcos(θ) 2 2 Lagrangian: L = KE − P E Euler-Lagrange equations: 2x ∗ sin(θ) + 2x ∗ cos(θ)) ˙ 1 dL ˙ ˙2 dx = 2 (M + m)8xx + 2mlxθsin(θ ) − 2(m + M )gx d dL dL dt [ dq i ] − dq i = 0 ˙ dL 2 ˙ ˙ dx = (M + m)(1 + 4x )x + mlθ (cos(θ ) + 2x ∗ sin(θ )) ˙ d dL 2 2 ¨ ˙ ˙ x dt [ dx ] = (M +m)(8xx +(1+4x )¨)+mlθ (cos(θ )+2x∗sin(θ )+mlθ (−sin(θ )+ ˙ ¨ ˙ 4(M + m)xx2 +(M + m)(4x2 +1)¨ + mlθ(cos(θ)+2xsin(θ))+ mlθ2 (2xcos(θ) − ˙ x sin(θ)) + 2(M + m)gx = 0 yielding: And for theta: Energy: ¨ 2mlsin(θ)x2 + ml(2xsin(θ) + cos(θ))¨ + ml2 θ + mglsin(θ) = 0 ˙ x ˙ E = dL x + dL θ − L ˙ dx ˙ ˙ dθ dE d dL dL ¨ d dL ˙ dL ¨ ˙ ˙ dt = dt ( dx )x + dx x + dt ( dθ )θ + dθ θ − ˙˙ ˙˙ d d ˙ = [ dt ( dL ) − dL ]x + [ dt ( dL ) − dL ]θ ˙ ˙ dx ˙ dx dθ dθ =0 dL ˙ dx x − dL ˙ dθ θ − dL dx x ˙¨ − dL ¨ ˙θ dθ So energy is conserved. Hamiltonian: ˙ H = i p q i − L where p = dL dq ˙ Problem 8: x = −x ˙ y = 2y − 5x3 ˙ Mus prove that x,y: y = x3 is invariant: condition for invariance: vector eld is always tangent to your manifold Consider a curve with dx = −x, y = x3 ds dy = 3x2 dx = −3x3 = (2 − 5)x3 = 2y − 5x3 as required ds ds so the curve is a trajectory of the system implying that the vector eld is tangeant to this trajectory viewed as a manifold. 1 ...
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This note was uploaded on 02/28/2011 for the course MATH 140A taught by Professor Marsden during the Fall '09 term at Caltech.

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