Revision: Previous Lecture Was About Applications of Euler-Lagrange Equations Euler-Lagrange Equations (Different Forms)
Revision: Previous Lecture Was About Applications of Euler-Lagrange Equations Euler-Lagrange Equations (Different Forms)
Revision
Previous lecture was about
Applications of Euler-Lagrange Equations
Euler-Lagrange Equations (Different Forms)
Some Examples related to Lagrangian and
Hamiltonian
FallingMass:
Consider a point mass m falling freely from rest. By
gravity a force F = mg is exerted on the mass
(assuming g constant during the motion). Take x to be the
coordinate, which is 0 at the starting point. The kinetic
energy is T = 1⁄2mv2and the potential energy is V = −mgx;
hence,
Then
which can be rewritten as , yielding the same result as
earlier.
Pendulum on a movable
support
Consider a pendulum of
mass m and length , which is
attached to a support with
mass M which can move along
a line in the x-direction.
Let x be the coordinate along
the line of the support, and let
us denote the position of the
pendulum by the angle from
the vertical.
The kinetic energy can then be shown to be
and the potential energy of the system is
The Lagrangian is therefore
Nowcarrying out the differentiations gives for the
support coordinate
therefore:
indicating the presence of a constant of motion.
Performing the same procedure for the variable yields:
Therefore
These equations may look quite complicated, but finding
them with Newton's laws would have required carefully
identifying all forces, which would have been much harder
and more prone to errors. By considering limit cases, the
correctness of this system can be verified: For example,
should give the equations of motion for a pendulum which
is at rest in some inertial frame, while should give the
equations for a pendulum in a constantly accelerating
system, etc. Furthermore, it is trivial to obtain the results
numerically, given suitable starting conditions and a chosen
time step, by stepping through the results iteratively.
Pendulum
The ideal, planar pendulum is a particle
of mass m in a constant gravitational
field, that is attached to a rigid, massless
rod of length L , as shown in Figure. The
canonical momentum of this system is
the angular momentum and the
potential energy is the gravitational
energy , where is the angle from the
vertical. The Hamiltonian is
This gives the equations of motion
While these equations are simple, their explicit solution requires
elliptic functions. However, the trajectories of the pendulum are
easy to visualize since the energy is conserved. When the
energy is below the angle cannot exceed and the
pendulum oscillates. Since the energy is conserved, the orbit
must be periodic.
For energies larger
than , the pendulum
rotates, and the
angle either
monotonically
grows with time (if
the angular
momentum is
positive) or
decreases
(negative ).
Springs
A harmonic spring has potential energy of the form
, where k is the spring's force coefficient (the force per unit
length of extension) or the spring constant, and x is the length
of the spring relative to its unstressed, natural length. Thus a
point particle of mass m connected to a harmonic spring with
natural length L that is attached to a fixed support at the
origin and allowed to move in one dimension has a
Hamiltonian of the form
and thus its equations of motion are
If the spring is hanging vertically in a constant gravitational
field, then the new equations are obtained by simply adding
the gravitational potential energy to .
A set of point masses that are coupled by springs has
potential energy given by the sum of the potential energies
of each spring in the system. For example suppose that
there are two masses connected to three springs as shown
in Figure.
The Hamiltonian is
One advantage of the Hamiltonian formulation of
mechanics is that the equations for arbitrarily
complicated arrays of springs and masses can be
obtained by simply finding the expression for the total
energy of the system (However, it is often easier to do
this using the Lagrangian formulation of mechanics which
does not require knowing the form of the canonical
momenta in advance).
Problem:
A torus of mass M and radius a rolls without slipping on a
horizontal plane. A pearl of mass m slides smoothly around
inside the torus. Describe the motion.
Solution:
The torus is rolling at angular speed . Consequently the linear
speed of the centre of mass of the hoop is , and the pearl
also shares this velocity. In addition, the pearl is sliding
relative to the torus at an angular speed and consequently
has a component to its velocity of tangential to the torus.
We are now ready to start.
The kinetic energy of the torus is the sum of its translational
and rotational kinetic
energies:
The kinetic energy of the pearl is
Therefore
+
The potential energy is
The Lagrange’s equations in and become
These, then, are two differential equations in the two
variables.
It is easy to eliminate and hence get a single
differential equation in :
Solving this gives as a function of the time. In the
meantime, we can get the as a function of . Thus, the
total energy is constant:
Here, the potential energy is being measured from the
centre of the circle. Also, if we assume that the initial
condition is that at time the kinetic energy was zero
and then
The End
