University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

169

Pages:

5
Author:

Jabin F.

Lagrangian of Different Systems Simple Pendulum (Polar Coordinates) T = 1/2 m(ṙ² + r²θ̇²) = 1/2 ml²θ̇² V = -mgl cosθ L = T - V = 1/2 ml²θ̇² + mgl cosθ EOM: d/dt(ml²θ̇) + mgl sinθ = 0 For small oscillations (sinθ ≈ θ): θ̈ + g/l θ = 0 SHM with ω = √(g/l) Simple Pendulum (Cartesian Coordinates) T = 1/2 m(ẋ² + ẏ²) = 1/2 m[ẋ² + (x/√(l² - x²))²ẋ²] = 1/2 mẋ²(l² / (l² - x²)) V = ±mg√(l² - x²) L = T - V = 1/2 mẋ²(l² / (l² - x²)) ± mg√(l² - x²) Lagrangian of Simple Harmonic Oscillator (SHO) 1D: T = 1/2 mx², V = 1/2 kx², L = 1/2 mx² - 1/2 kx² 2D Isotropic: T = 1/2 m(ẋ² + ẏ²), V = 1/2 k(x² + y²), L = 1/2 m(ẋ² + ẏ²) - 1/2 k(x² + y²) 2D Anisotropic: T = 1/2 m(ẋ² + ẏ²), V = 1/2 k1x² + 1/2 k2y², L = 1/2 m(ẋ² + ẏ²) - 1/2 k1x² - 1/2 k2y² Lagrangian of a Spring Pendulum T = 1/2 m(ṙ² + r²θ̇²) V = -mgr cosθ + 1/2 k(r - l)² L = T - V = 1/2 m(ṙ² + r²θ̇²) + mgr cosθ - 1/2 k(r - l)² Radial EOM: r̈ - rθ̇² - g cosθ + (k/m)(r - l) = 0 Angular EOM: θ̈ + 2(ṙ/r)θ̇ + (g/r) sinθ = 0 Spherical Pendulum T = 1/2 m(ṙ² + r²θ̇² + r²sin²θφ̇²) As r = l (constant) → ṙ = 0 T = 1/2 m(l²θ̇² + l²sin²θφ̇²) = 1/2 ml²(θ̇² + sin²θφ̇²) V = mgz = mglsinθ = mglcosθ L = T - V = 1/2 ml²(θ̇² + sin²θφ̇²) - mglcosθ pφ = ∂L/∂φ̇ = ml²sin²θφ̇ = constant sin²θφ̇ = constant