Klassieke Mechanica 2

Klassieke Mechanica 2
Uitwerking huiswerk 1
Collegejaar 2013-2014
Periodically driven pendulum
The position of the suspension point is given by (x (t) , y (t)) = (a cos ωt, b sin ωt)
with a = 0 for case (a), b = 0 for case (b) and a = b for case (c). The position
of the mass of the pendulum is then given by
(x (t) , y (t)) = (a cos ωt + l sin θ (t) , b sin ωt − l cos θ (t)) .
The kinetic energy is then
2 2 1
1
2
2
˙
˙
T = m x˙ + y˙ = m −aω sin ωt + lθ cos θ + bω cos ωt + lθ sin θ
2
2
which can be rewritten as
1 ˙ (a sin ωt cos θ − b cos ωt sin θ)
T = m l2 θ˙2 + ω 2 a2 sin2 ωt + b2 cos2 ωt − 2lθω
2
The potential energy is given by
V = mg (b sin ωt − l cos θ)
This leads to the Lagrangian L θ, θ˙ = T −V . The first part of the EL equation
is
h
i
d ∂L
= ml lθ¨ + ω aθ˙ − bω sin ωt sin θ + ω bθ˙ − aω cos ωt cos θ .
dt ∂ θ˙
The second part is given by
−
∂L
˙ (a sin ωt sin θ + b cos ωt cos θ) + mgl sin θ.
= −mlθω
∂θ
Alltogether we obtain
d ∂L ∂L
−
= ml2 θ¨ − mlbω 2 sin ωt sin θ − mlaω 2 cos ωt cos θ + mg sin θ = 0
dt ∂ θ˙
∂θ
or shorter
ω2
g
θ¨ −
(a cos ωt cos θ + b sin ωt sin θ) + sin θ = 0.
l
l
The three cases follow then directly from this equation by iether setting a = 0,
b = 0 or a = b.
1
“Spontaneous symmetry breaking”
(a) The position of the bead is given by
(x (t) , y (t) , z (t)) = (R sin θ (t) cos ωt, R sin θ (t) sin ωt, −R cos θ (t)) .
This leads to the potential energy
V = −mgR cos θ
and kinetic energy
T =
1 2 ˙2
m R θ + R2 ω 2 sin2 θ .
2
This gives the Lagrangian
L=T −V =
1 2 ˙2
m R θ + R2 ω 2 sin2 θ + mgR cos θ.
2
From the Euler-Lagrange equation we find the equation of motion:
g
θ¨ − ω 2 sin θ cos θ + sin θ = 0.
R
(b) Equilibrium points are solutions with θ˙ = θ¨ = 0. This equation of motion
simplifies than to a condition on θ:
ω 2 sin θ cos θ =
g
sin θ.
R
This is always fulfilled for θ (t) ≡ 0 (“≡” means identical to, i.e. here the
function is constant and has the value 0). A second set of equilibrium positions
is characterized by the condition
cos θ =
g
.
Rω 2
This solution only exists if the hoop rotates fast enough such that g/Rω 2 > 0.
At that point there
is a bifurcation to two non-zero deflections, namely θ± =
± arccos g/Rω 2 .
(c) First look at the equilibrium position θ = 0. We assume a motion with
θ (t) 1. This allows us to simplify the equation of motion to
g
θ¨ +
− ω 2 θ = 0.
R
This is a harmonic oscillator with θ (t) = A cos (˜
ω t + φ) with angular frequency
r
g
− ω2 .
ω
˜=
R
For g/Rω 2 > 0 this position becomes instable. Instead new equilibrium positions form at θ± = ± arccos g/Rω 2 . Let consider small deviation ∆θ 1
2
from one of those positions. We find the equation of motion (let say around
θ+ ):
¨ − ω 2 sin (θ+ + ∆θ) cos (θ+ + ∆θ) + g sin (θ+ + ∆θ) = 0.
∆θ
R
We can Taylor expand sin (θ+ + ∆θ) ≈ sin θ+ + ∆θ cos θ+ and cos (θ+ + ∆θ) ≈
cos θ+ − ∆θ sin θ+ . Next insert this in the equation above. In leading order in
∆θ (i.e. we neglect terms of order ∆θ2 ) and using several that terms cancel
because of ω 2 sin θ+ cos θ+ = (g/R) sin θ+ one finds the equation of motion for
∆θ:
¨ + ω 2 sin2 θ+ ∆θ = 0.
∆θ
This is again the harmonic oscillator, namely θ (t) = θ+ + A cos (ˆ
ω t + φ), this
time with angular frequency
r
g 2
p
2
.
ω
ˆ = ω sin θ+ = ω 1 − cos θ+ = ω 1 −
Rω 2
3

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