PHYS-3210: Classical Mechanics and Mathematical Methods II
Spring 2016
Dmitri Uzdensky, University of Colorado at Boulder
Chapter 7: Lagrangian Mechanics
Lecture 8, Jan. 29, 2016
Announcements:
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Homework # 2 is due today at 5 pm.
Solution for Homework # 1 is posted on D2L.
Homework # 3 is posted on D2L, due next Friday, Feb. 5, at 5 pm.
Lecture next Monday, Feb. 1: Tutorial.
New Office Hours: every Thursday 1:30 - 3:30 pm.
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Holonomic Constraints
One of the advantages of the Lagrangian equations is that they allow one to handle easily
mechanical systems with constraints. In the Newtonian approach, constraints come from
things like surfaces and ropes which provide forces of constraint (normal force and tension,
in these cases.) Quite often, we can write the constraints as algebraic equations which involve
only the coordinates and time,
f (~r1 , ~r2 , ..., ~rN , t) = 0.
(1)
Any constraint we can write in this form is called holonomic; if time doesn’t appear explicitly (∂f /∂t = 0), then we can further label it as normal. When time does show up
explicitly, the constraint is called forced (but it’s still holonomic!) Forced constraints tend
to crop up in systems being pushed around by external forces, as we’ll see.
————
Clicker Question:
Which of the following systems have holonomic constraints?
A. The double pendulum
B. The motion of a pool/billiard ball around a pool table (with no pockets)
C. A bead moving frictionlessly on a circular wire hoop, which is spinning at a constant
angular speed.
D. A and B
E. A and C
–2–
Correct Answer E:
The double pendulum is holonomic; the constraints relate the coordinates of the two bobs
to each other and to the ceiling. The bead on a rotating hoop is also holonomic; the
wire constraint can be written as a function of space and time, so it’s a forced holonomic
constraint. The pool ball, however, is not holonomic; the presence of the billiard table’s
walls limit, constrain the ball’s motion, but the constraints cannot be written as an equality,
only as an inequality.
————
In the problems we will consider in this class we will mostly be dealing with holonomic
constraints. The best way to recognize a holonomic constraint is actuallyto recognize cases
that are not holonomic. The two most common non-holonomic constraints to watch out for
are inequalities (like in the pool table considered above), or systems where the final state
depends not just on the coordinates but on the path taken.
Example: pendulum on a rotating support
Although I have told you that holonomic constraints and generalized coordinates are allowed
to depend explicitly on time, we haven’t seen any examples yet. So let us try one! Here we
consider a pendulum attached to a rotating support, rather than a fixed surface. Specifically,
we attach the end of our pendulum of length ` to a disc of radius a, spinning with angular
velocity ω. The plane of the pendulum’s swinging motion is assumed to coincide with the
plane of the rotating disk. The situation looks like this:
Fig. 1.— Sketch of the rotating-support pendulum problem.
All of the forces available are conservative; in fact, the only force acting is gravity (we don’t
count forces of constraint in the Lagrangian approach, remember.) As for the constraint
–3–
itself, it is holonomic, even though it depends on the angular speed ω; the key is that it
doesn’t depend on the speed of the pendulum bob itself, so we can write it as a function
only of the position of the bob and of time. (Remember, this is what we called a forced
constraint, as opposed to normal constraints which don’t depend on time.)
As always, we should start with a coordinate system that is inertial and fixed. We should
write down the expressions for the kinetic and potential energy of the system, and hence the
expression for the Lagrangian, in this coordinate system. Only after that we will consider
the constraints imposed on the system and will decide on the best choice for our generalized
coordinates.
So let us start with the inertial xy coordinate system I have drawn, with the origin at the
center of the spinning disc. The kinetic and potential energies of the pendulum’s bob in
these coordinates are nice and easy to write down:
1
m(ẋ2 + ẏ 2 ) ,
2
U = mgy ,
T =
(2)
(3)
and so the Lagrangian is
L=T −U =
1
m(ẋ2 + ẏ 2 ) − mgy .
2
(4)
Next, we want to change to our generalized coordinate system and this is a good moment
to think about the constraints and degrees of freedom. The pendulum bob’s motion takes
place in the 2D (xy) plane, but it is subject to the usual constraint that the pendulum has
a fixed length `, which can be written as
[x − x0 (t)]2 + [y − y0 (t)]2 = `2 ,
(5)
where x0 (t) and y0 (t) are the coordinates of the pendulum’s support (pivot) point, which
are explicit functions of time. Because of this explicit time dependence, this constraint is a
forced holonomic constraint. For definiteness, let us assume that at the initial moment of
time t = 0, this support point is located at it’s rightmost position, so that x(0) = a and
y(0) = 0.
Because of this constraint, the system has only one degree of freedom, and so required only
one generalized coordinate. As in the simple pendulum problem, it is convenient to choose
this generalized coordinate to be θ, the angle between the pendulum and the vertical (as
shown in the sketch).
————
Clicker Question:
What is the x coordinate of the bob in terms of θ?
–4–
A. x = a sin ωt + ` cos θ
B. x = a sin ωt + ` sin θ
C. x = a cos ωt + ` sin θ
D. x = a cos ωt + ` cos θ
Correct Answer C:
Some trigonometry practice! Since we know the initial condition for the support itself is
x = a at t = 0, the time-dependent piece has to go as cos ωt, eliminating A and B. With the
angle θ measured from the vertical as drawn, ` sin θ gives the x-component from the string,
which leads us to C as the correct answer.
————
With this choice, the original Cartesian coordinates of the pendulum bob can be expressed
in terms of θ as
x = a cos ωt + ` sin θ,
(6)
y = a sin ωt − ` cos θ.
(7)
Differentiating with respect to time yields
ẋ = − aω sin ωt + `θ̇ cos θ
(8)
ẏ = aω cos ωt + `θ̇ sin θ
(9)
Now we can write down our Lagrangian
1
1
L = T − U = mẋ2 + mẏ 2 − mgy
2
2
(10)
in terms of θ:
1 2 2
m a ω (sin2 ωt + cos2 ωt) + `2 θ̇2 (cos2 θ + sin2 θ)
2
+ 2`θ̇aω(sin θ cos ωt − cos θ sin ωt) − mg(a sin ωt − ` cos θ).
L =
(11)
This can be greatly simplified using the trigonometric identity sin2 θ + cos2 θ = 1 and also
sin θ cos ωt − cos θ sin ωt = sin(θ − ωt) .
(12)
(If you don’t have that one memorized - and I know I don’t! - you should at least recognize
that this looks like a trig identity situation, and be able to go look it up.) So now our
Lagrangian becomes neater and more compact:
1 L = m a2 ω 2 + `2 θ̇2 + 2`θ̇aω sin(θ − ωt) − mg(a sin ωt − ` cos θ).
2
(13)
–5–
The derivatives that we need are thus
∂L
= m`θ̇aω cos(θ − ωt) − mg` sin θ
∂θ
∂L
= m`2 θ̇ + m`aω sin(θ − ωt).
∂ θ̇
and finally the Euler-Lagrange equation is
m`θ̇aω cos(θ − ωt) − mg` sin θ − m`2 θ̈ + m`aω(θ̇ − ω) cos(θ − ωt) = 0.
(14)
(15)
Rearranging, we finally have
θ̈ =
ω2a
g
cos(θ − ωt) − sin θ.
`
`
(16)
We won’t try to solve this now, but notice that the limit in which the rotation vanishes
(ω = 0) gives us back the usual equation of motion for a pendulum, as it should.
As this example shows, the ability to include time dependence into generalized coordinates
lets us deal with rotating and accelerating systems in a relatively painless way. (Try this
one using F~ = m~a if you really want to appreciate how painless it is!) But again, be careful
to always start your work in an inertial, Cartesian frame, where you know what
the kinetic and potential energies look like, and then apply the coordinate change. Still,
for systems which are best described in terms of non-rectangular coordinates, or accelerating coordinates, or constraints relating different coordinates, dealing with the change of
coordinates in the Lagrangian approach is much, much easier.
End of Lecture 8