Dirac structures and geometry of nonholonomic constraints
Dirac structures and geometry of nonholonomic constraints Dirac structures and geometry of nonholonomic constraints
Example Modeling constraints: a sleigh subject to strong viscous force perpendicular to the sleigh Equations for a curve t ↦−→ (x(t), y(t), ϕ(t), px(t), py (t), π(t)) (γ < 0) ˙x = px/m, ˙px = (γ/m)(py cos ϕ − px sin ϕ)(− sin ϕ) ˙y = py /m, ˙py = (γ/m)(py cos ϕ − px sin ϕ)(cos ϕ) ˙ϕ = π/I , ˙π = 0 KG (KMMF) Seminarium Geometryczne 13 X 2010 24 / 26
Example Initial momentum along the sleigh, π0/I = 1, ϕ0 = 0 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 black: solution with constraints red: γ/m = −3 blue: γ/m = −10 green: γ/m = −100 KG (KMMF) Seminarium Geometryczne 13 X 2010 25 / 26
- Page 5 and 6: Contents Dirac structure Dirac stru
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- Page 11 and 12: Dirac structure Definition A Dirac
- Page 13 and 14: Almost Dirac structure in mechanics
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- Page 19 and 20: Almost Dirac structures in mechanic
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- Page 25 and 26: Constrained Lagrangian system accor
- Page 27 and 28: Constrained Lagrangian system accor
- Page 29 and 30: Constrained Lagrangian system accor
- Page 31 and 32: Constrained Lagrangian system accor
- Page 33 and 34: Constrained Lagrangian system accor
- Page 35 and 36: My point of view Lagrangian dynamic
- Page 37 and 38: My point of view Back We treat ∆
- Page 39 and 40: My point of view Back We treat ∆
- Page 41 and 42: My point of view How to get ˜ ∆
- Page 43 and 44: My point of view How to get ˜ ∆
- Page 45 and 46: My point of view How to get ˜ ∆
- Page 47 and 48: My point of view The constrained ca
- Page 49 and 50: My point of view The constrained ca
- Page 51 and 52: My point of view ˜∆ KG (KMMF) Se
- Page 53 and 54: My point of view Conclusion The rel
- Page 55: Example Equations for a curve t ↦
Example<br />
Modeling <strong>constraints</strong>: a sleigh subject to strong viscous force<br />
perpendicular to the sleigh<br />
Equations for a curve t ↦−→ (x(t), y(t), ϕ(t), px(t), py (t), π(t)) (γ < 0)<br />
˙x = px/m, ˙px = (γ/m)(py cos ϕ − px sin ϕ)(− sin ϕ)<br />
˙y = py /m, ˙py = (γ/m)(py cos ϕ − px sin ϕ)(cos ϕ)<br />
˙ϕ = π/I , ˙π = 0<br />
KG (KMMF) Seminarium Geometryczne 13 X 2010 24 / 26
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