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Units, Constants and Useful Formulas
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