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

2024
Views:

298

Pages:

5
Author:

Jabin F.

# Another form of Lagrangian EOM: d/dt (∂L/∂qi̇) - ∂L/∂qi = 0 L = T - V in electric circuit: qi → charge qi̇ → current generalized KE = Li̇²/2 d/dt (∂(T-V)/∂qi̇) - ∂(T-V)/∂qi = 0 d/dt (∂T/∂qi̇) - ∂T/∂qi - (-∂V/∂qi) = 0 d/dt (∂T/∂qi̇) - ∂T/∂qi = ∂V/∂qi Is derivative of potential is called force. So here we called Qi as generalized force. → Another form of Lagrangian EOM used when V is not known but actual force is given. → Relation b/w generalized potential and generalized force. 128 Lagrangian L = T - V in cartesian coordinate: L = 1/2m(ẋ² + ẏ² + ż²) - V(x,y,z) in spherical polar: L = 1/2m(ṙ² + r²θ̇² + r²sin²θφ̇²) - V(r,θ,φ) in plane polar: L = 1/2m(ṙ² + r²θ̇²) - V(r,θ) in cylindrical co-ordinate: L = 1/2m(ṙ² + r²θ̇² + ż²) - V(r,θ,z) P-78 A thin rod of mass 'M' and length 'L' is suspended from one end. The rod is free to rotate about its one end. If motion of the rod is not confined to one plane, its Lagrangian in spherical polar co-ordinate is: L = 1/2 ∫ dm(ṙ² + r²θ̇² + r²sin²θφ̇²) + ∫ dmgr cosθ Solution: Consider an element of rod as shown in figure. Motion of element is same as motion of bob in spherical pendulum. Since there is rod instead of string in spherical pendulum. Lagrangian of element: dL = dT - dV = 1/2 dm(ṙ² + r²θ̇² + r²sin²θφ̇²) - (-dmgr cosθ) = dm[1/2(ṙ² + r²θ̇² + r²sin²θφ̇²) + gr cosθ] P-78 For uniform rod, m = M/L ⇒ dm = M/L dr dL = 1/2(dm)r²(θ̇² + sin²θφ̇²) + (dm)gr cosθ On integrating r → 0 to r = L we get, L = 1/2 (M/L) ∫[r³(θ̇² + sin²θφ̇²) + 2gr cosθ] dr L = (M/6L)[r³(θ̇² + sin²θφ̇²) + 3gr² cosθ] L = (M/6)[L²(θ̇² + sin²θφ̇²) + 3gL cosθ]