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

2024
Views:

492

Pages:

5
Author:

Jabin F.

#Lagrangian (L) L(qi, qi̇, t) = T(qi, qi̇, t) - V(qi, t) where qi = q₁, q₂, q₃, ..., qₙ Lagrangian is not uniquely defined for a system. So it doesn't have any physical significance. If L is the Lagrangian of a system, then we can define another equivalent Lagrangian L' for the system where: L' = L + dF(qi, t) / dt L and L' give the same EOM. So these two Lagrangians describe the same dynamics of the system. P-76: Lagrangian of a particle moving in a plane is given in cartesian coordinate as L = ẋẏ - x⁹ - y⁹. In polar coordinate the expression for the canonical momentum p_θ is: Solution: Firstly we have to convert in polar form as: x = r cos θ → ẋ = ṙ cos θ - r sin θ θ̇ y = r sin θ → ẏ = ṙ sin θ + r cos θ θ̇ L = (ṙ cos θ - r sin θ θ̇)(ṙ sin θ + r cos θ θ̇) - r⁹ cos⁹ θ - r⁹ sin⁹ θ L = (ṙ² - r²θ̇²)sin θ cos θ + rṙθ̇(cos² θ - sin² θ) - r⁹(cos⁹ θ + sin⁹ θ) L = (ṙ² - r²θ̇²)sin θ cos θ + rṙθ̇cos 2θ - r⁹ P-70: A bead of mass m moves along a fixed vertical circle of radius R. The bead is attached to the highest point of the circle with a light spring of spring constant k. If θ, shown in the figure, is taken as the generalized coordinate, what is the Lagrangian of the system? [Natural length of spring is R and gravitation potential energy is taken to be zero at the center of circle.] Solution: From first circle: ϕ = 9R cos θ → ϕ̇ = -9R sin θ θ̇ From second circle: R + h = l → h = R(cos θ - 1) = R(cos 2θ - 1) T = (1/2)m(ṙ² + R²θ̇²) V = mgh = mgR(cos 2θ - 1) L = T - V = (1/2)m(ṙ² + R²θ̇²) - mgR(cos 2θ - 1) L = (1/2)mṙ² + (1/2)mR²θ̇² + mgR(1 - cos 2θ) Continuation of the previous image: Now using these values, we can find out KE & PE: T = (1/2)m(ṙ² + R²θ̇²) = (1/2)m(9m²R²θ̇² + 4R²θ̇²) = 9mR²θ̇²/2 V = -mgh + (1/2)k(ϕ - R)² = -mgR(cos θ) + (1/2)k(R - 2R(1 - cos θ))² V = -mgR(cos θ) + (1/2)kR²(1 - cos θ)² V = -mgR(cos θ) + (1/2)kR²(1 - 2cos θ + cos² θ) V = -mgR(cos θ) + (1/2)kR²(1 - 2cos θ + 1 - sin² θ) V = -mgR(cos θ) + kR²(1 - cos θ - (1/2)sin² θ) L = T - V = 9mR²θ̇²/2 + mgR(cos θ) - kR²(1 - cos θ - (1/2)sin² θ)