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

2024
Views:

345

Pages:

5
Author:

Jabin F.

# PB with angular momentum -> same as CB {x, Lx} = a {x, Ly} = -z {x, Lz} = y {y, Lx} = -z {y, Ly} = x {y, Lz} = -x {z, Lx} = y {z, Ly} = -x {z, Lz} = x If PB of a component of a vector is evaluated with angular momentum, then in result we get another component of that vector. {x, Lx} = 0 {x, Ly} = z {x, Lz} = -y {y, Lx} = -z {y, Ly} = 0 {y, Lz} = x {z, Lx} = y {z, Ly} = -x {z, Lz} = 0 @Lagrangian of a system is L = 1/2 ml^2θ̇² + mgl cosθ If H be Hamiltonian of system then find value of ∂L/∂θ, ∂L/∂θ̇, H Solution: L = 1/2 ml^2θ̇² + mgl cosθ ∂L/∂θ = -mgl sinθ ∂L/∂θ̇ = ml²θ̇ Now H = ∂L/∂θ̇ θ̇ - L = ml²θ̇² - (1/2 ml^2θ̇² + mgl cosθ) = 1/2 ml²θ̇² - mgl cosθ P-29 P/a-10 --> {a,b,c} x {a,b,c} is equal to ? Solution {a,b,c} x {a,b,c} = {(a,a), (a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b), (c,c)}