Download Double Pendulum and Langrange Multipliers, Lecture Notes - Physics and more Study notes Mechanics in PDF only on Docsity!
Double Pendulum and Lagrange Multipliers
Adrian Down
September 20, 2005
1 Double pendulum
1.1 Motivation
This problem is a straightforward application of the method we have de-
veloped thus far. As with all mechanics problems, the art is choosing the
correct variables to parameterize the problem. It will also illustrate how to
take appropriate limits to simplify the equations of motion.
1.2 Setup
First mass hangs from a rod of length b at angle θ 1 from the vertical. The
second mass hangs from a rod of length a from the first mass at angle θ 2
from the vertical. Call the vector from the origin to the first mass ~r 1
, and
the vector from the origin to the second mass ~r 2
~r 1
= b sin θ 1
xˆ − b cos θ 1
yˆ
~r 2
= ~r 1
xˆ − a sin θ 2
yˆ)
= (b sin θ 1
)ˆx − (b cos θ 1
)ˆy
1.3 Compute the Lagrangian
1.3.1 Kinetic term
We need the squares of the time derivatives.
~r 1
= b
θ 1
cos θ 1
xˆ + b
θ 1
sin θ 1
yˆ
~r 2
~r 1
θ 2
cos θ 2
xˆ + a
θ 2
sin θ 2
yˆ)
= (b
θ 1 cos θ 1 + a
θ 2 cos θ 2 )ˆx + (b
θ 1 sin θ 1 + a
θ 2 sin θ 2 )ˆy
r ˙
2
1
= b
2 ˙ θ
2
1
cos
2
θ + b
2 ˙ θ
2
1
sin
2
θ 1
= b
2 ˙ θ
2
1
r ˙
2
2
= b
2 ˙ θ
2
1
cos
2
θ 1
θ 1
θ 2
cos θ 1
cos θ 2
2 ˙ θ
2
2
cos
2
θ 2
2 ˙ θ
2
1
sin
2
θ 1
θ 1
θ 2 sin θ 1 sin θ 2 + a
θ
2
2
sin
2
θ 2
= b
2 ˙ θ
2
1
2 ˙ θ
2
2
θ 1
θ 2
cos(θ 1
− θ 2
The last term of ˙r
2
2
is the cross term that mixes the motion of the two
masses and introduces nonlinear behavior.
T =
(m 1
)b
2 ˙ θ
2
1
m 2
a
2 ˙ θ
2
2
ab
θ 1
θ 2
cos(θ 1
− θ 2
1.3.2 Potential energy
Gravity is the only relevant potential.
U = m 1
gb(1 − cos θ 1
) + m 2
g (a(1 − cos θ 2
) + b(1 − cos θ 1
1.4 Euler-Lagrange equations
1.4.1 θ 1
∂L
θ 1
= (m 1
)b
θ 1
ab
θ 2
cos(θ 1
− θ 2
∂L
∂θ 1
= −m 2
ab
θ 1
θ 2
sin(θ 1
− θ 2
) − m 1
gb sin θ 1
− m 2
gb sin θ 1
1.6.2 θ 2
m 2
a
θ 2
θ 1
≈ −mgaθ 2
1.6.3 Combined equation
These are a set of coupled second order differential equations with constant
coefficients. Divide the first by
b
m 1 +m 2
to get
b
θ 1
m 2
m 1
a
θ 2
≈ −gθ 1
Divide the second by
a
m 2
to get
a
θ 2
θ 2
≈ −gθ 2
Using the reduced mass,
μ =
m 2
m 1
we can write these equations in a matrix,
b μa
b a
θ 1
θ 2
= −g
θ 1
θ 2
1.7 Comparison to regular pendulum
Consider a regular pendulum moving close to equilibrium. The equation of
motion is
θ +
g
l
sin θ = 0
This equation is nonlinear in its present form. We linearize with the small
angle approximation to get
θ + ω
2
0
θ ≈ 0
where
g
l
= ω
2
0
This type of differential equation should be familiar; the solutions are
sinusoids.
θ(t) = ηe
pt
θ(t) = A cos ω 0
t + B sin ω 0
t = F cos(ω 0
= De
ıω 0 t
−ıω 0 t
= Ge
ı(ω 0 t+θ 0 )
This motivates us to choose exponential solutions in the current case
θ 1
= η 1
e
ıωt
θ 1
= ıωη 1
e
ıωt
= ıωθ 1
θ 1
= −ω
2
θ 1
θ 1
= η 2
e
ıωt
θ 2 = ıωη 2 e
ıωt
= ıωθ 2
θ 2
= −ω
2
θ 2
1.8 Solving the differential equation
We go back to the matrix equation and replace the double time derivatives
−bω
2 −μaω
2
−bω
2 −aω
2
θ 1
θ 2
= −g
θ 1
θ 2
g − bω
2 −μaω
2
bω
2 g − aω
2
θ 1
θ 2
Solving the differential equation has now become a problem of finding the
eigenvalues of the above matrix. As with any eigenvalue problem, we take
the determinant and set it equal to 0.
(g − bω
2
)(g − aω
2
) − abμω
4
= 0
g
2
− aω
2
g − bω
2
g + abω
4
− abμω
4
= 0
ab(1 − μ)ω
4
− (a + b)gω
2
2
= 0
Solve for ω
2 using the quadratic formula. There are two solutions repre-
senting two modes of behavior for the pendulum system, each with a different
frequency.
ω
2
ω
2
−
This is what we have done in all previous cases.
The more general way is to introduce a function
f (x 1 , x 2 ) + λg(x 1 , x 2 )
λ is an undetermined constant, called a Lagrange multiplier. We find the
extremum of the new function with respect to both x 1
and x 2
and then
choose λ to satisfy the constraint.
2.3 Example
Suppose we want to minimize
f (x 1 , x 2 ) = 3x 1 + 4x 2
subject to the constraint
x
2
1
2
2
We look for extrema of
3 x 1
2
1
2
2
We take partial derivatives and set them equal to 0.
3 + 2λx 1 = 0 ⇒ x 1 = −
λ
4 + 2λx 2 = 0 ⇒ x 2 = −
λ
We plug these solutions for x 1
and x 2
into the constraint equation to
determine λ,
λ
2
λ
2
= 1 ⇒ λ
2
=
⇒ λ = ±
This gives the solutions,
x 1
x 2
x 1
x 2
The first is the minimum and the second is the maximum.
2.4 Justification
For small changes of dx 1
and dx 2
, the change in the given function is
∂f
∂x 1
dx 1
∂f
∂x 2
x 2
At the critical points, these partials are 0.
∂f
∂x 1
∂g
∂x 1
∂f
∂x 2
∂g
∂x 2
Adding the two,
∂x 1
(f + λg) = 0
∂x 2
(f + λg) = 0
We assume that λ is not a function of x 1
and x 2
2.5 Example: Asteroid
2.5.1 Situation
Consider an asteroid of mass m and volume v 0
. The mass density is
ρ =
m
v 0
We want to find the shape of the asteroid that gives the largest possible
vertical gravitational field.
The gravitational force experience by a test mass on the surface of a
sphere is as if all the mass of the sphere were at the center of the sphere.
The force goes as
1
r
2
, so moving closer to the center should Increase the force.
However, making the sphere flatter lessens the gravitational pull. We want
to flatten one edge of the sphere so that it remains symmetric about the
azimuth. Let r = 0 be the point of contact between the test mass and the
sphere. We want to find r(θ, φ).
