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Last modified: December 9, 2021 ✓

Ex:  Newtonian, Lagrangian and Hamiltonian Mechanics
----------------------------------------------------

We will demonstrate the equivalency of Newtonian, 
Lagrangian and Hamiltonian mechanics by finding the 
equations of motion for the simple pendulum using 
all three methods.



Newtonian
---------

FT = -mgsinθ

Where,

T stands for tangential.

Now sinθ = θ for small θ.  

Therefore,

FT = -mgθ

Angular acceleration, α = dω/dt = d2θ/dt2

α = αT/r ∴ αT = rα = rd2θ/dt2

Where,

ω = angular velocity = dθ/dt.

Therefore, we can write:

-mgθ = mαT

-gθ = rd2θ/dt2

Or,

d2θ/dt2 + (g/r)θ = 0

Lagrangian
----------

Tangential velocity, v = rω = rdθ/dt 

KE = (1/2)mr2(dθ/dt)2  [ E = Fd]

U = -mgrcosθ   

L = KE - U = (1/2)mr2(dθ/dt)2 + mgrcosθ

Euler-Lagrange gives:
         .      .
πθ = ∂L/∂θ = mr2θ

∂L/∂θ = -mgrsinθ

dπ/dt - ∂L/∂θ = 0
  ..
mr2θ + mgrsinθ = 0

d2θ/dt2 + (g/r)sinθ = 0

Hamiltonian
-----------
      .
H = πθθ - L
             .
  = (1/2)mr2(θ)2 - mgrcosθ

  = KE + U

To get the equations of motion we need to get the KE
in terms of the angular momentum, L'.

KER = (1/2)Iω2

If we compare this to:

KET = (1/2)mv2

We get: 

L' ≡ p, I ≡ m and ω2 ≡ v2

KE = p2/2m => L'2/2I.  But I = mr2.  Therefore, 

KE = L'2/2mr2

H = KE + U

  = L'2/2mr2 - mgrcosθ

Hamilton's equations are:
         .
-∂H/∂q = p
        .
∂H/∂p = q

Or, in terms L' and θ:
.
L'/dt = -∂H/∂θ = -mgrsinθ
.
θ = ∂H/∂L' = L'/mr2

So,
..   .
 θ = L'/mr2

   = -mgrsinθ/mr2

   = (-g/r)sinθ

d2θ/dt2 + (g/r)sinθ = 0

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