1. Physics 430: Lecture 17 Examples of Lagrange’s Equations
Dale E. Gary
NJIT Physics Department

2. 7.5 Examples Example 7.4: A Particle Confined to Move on a Cylinder
This section is a long one, where several examples are worked out in detail. In addition, there are many problems at the end of the chapter for you to work on. You should do as many as possible.
I will not be doing every example, but let’s do a couple of them:
Example 7.4: A Particle Confined to Move on a Cylinder
Statement of the problem:
Consider a particle of mass m constrained to move on a frictionless cylinder of radius R, given by the equation r = R in cylindrical polar coordinates (r, f, z). Besides the force of constraint (the normal force on the cylinder), the only force on the mass is a force F = -kr directed toward the origin. Using z and f as generalized coordinates, find the Lagrangian L. Write down and solve Lagrange’s equations and describe the motion.
Solution:
We are already told what the generalized coordinates are, so we can now proceed to find the Lagrangian L = T – U. To get the kinetic energy, we need the velocity in terms of the generalized coordinates. The z velocity is easy, just The f velocity (azimuthal) is just Thus, the kinetic energy is just
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3. Example 7.4, cont’d Solution, cont’d: z r
Now we need the potential energy (due to this central spring force), which is:
Notice that we could drop the R2 term, since it is constant, but it will not survive the Lagrange equation anyway.
Putting these together, the Lagrangian is
Now we form the Lagrange equations (one for z and one
for f) and solve. The z equation is
The f equation is
This just says that the angular momentum mR2w = constant. Thus, the particle executes simple harmonic motion in z, but rotates in f at a constant rate around the cylinder. r
r = R z
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4. Example 7.5: A Block Sliding on a Wedge
Let’s now do a non-trivial two-coordinate problem—the block sliding down a wedge that we showed in Phun last time.
We will take the generalized coordinates as shown in the figure. Notice that the first coordinate q1 is along the plane, so
that it has no component in the constraint force
direction. Also, q1 is measured relative to the
(moving) wedge.
The velocity of the wedge is just , but the
velocity of the block has components from both
q1 and q2:
Thus, the total kinetic energy is
The potential energy of the wedge can be taken as zero, and the block is just q2 q1 a
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5. Example 7.5, cont’d Putting these together, the Lagrangian is
There are two generalized coordinates, and so there are two Lagrange equations. The second is simplest:
Note that this is conservation of momentum in the x direction.
The first one is
This has two unknowns, but we can get another equation for the accelerations by differentiating the q2 relation: or
Finally, plug this into the above equation and solve
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6. Example 7.6: Bead on a Spinning Wire Hoop
Let’s do another one. How many coordinates? Since it is a wire, there is only one degree of freedom, angle q. What is the velocity to insert into the kinetic energy? This is a little tricky. The bead can slide up and down on the wire, which clearly has a velocity But the hoop is also rotating,
so there is another motion with velocity
The kinetic energy is then
The potential energy relative to its position at the bottom of
the hoop (when the hoop is not rotating and q = 0), is
The Lagrangian is
The sole Lagrange equation is then
Solving for the angular acceleration: w q R r
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7. Example 7.6, cont’d We cannot solve:
in terms of elementary functions, but we can look at the behavior to see if we can find an equilibrium and what the frequency of oscillation about any equilibria are.
This requires remembering all the way back to chapter 5! Think
physically about what will happen. For no rotation (w = 0), the
bead will act like a pendulum—equation
But for a rotating hoop, the bead will move up the wire and we
might expect a new equilibrium. Recall that at equilibrium the bead
will be at rest, i.e which occurs at either q = 0 (which we have
already examined) or at
Solving for the equilibrium position
We can find the oscillation frequency for small displacements q = qo + e. w q R r
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8. 7.6 Generalized Momenta and Ignorable Coordinates
Recall that we said the left-hand side of the Lagrange equation
is called the generalized force, while the right-hand side is the time derivative of the generalized momentum.
For normal coordinates like x, y, z, the generalized force is really the force, and the generalized momentum is really the momentum.
However, for angular coordinates we saw that the generalized force was the torque, and the generalized momentum was the angular momentum.
In any case, when the generalized force is zero (i.e. the Lagrangian is independent of the generalized coordinate qi, then the generalized momentum is conserved.
In such a case, we can say that the coordinate qi is ignorable. Another way to say it, which leads to an important idea in Physics, is to say that L is invariant under transformations (translations, rotations, etc.) in coordinate qi.
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9. 7.8 Hamiltonian
We will not study section 7.8, but I do want to introduce an idea of central importance to quantum mechanics—the Hamiltonian.
It comes about by considering the time derivative of the Lagrangian or
Now, if the Lagrangian does not depend on time, the last term on the right vanishes, so we have
Bringing the Lagrangian over to the right, the result is called the Hamiltonian:
When the Lagrangian does not depend explicitly on time, the Hamiltonian H is conserved.
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10. Hamiltonian-cont’d
The key point at the moment that I would like you to remember is that when the relationship between generalized coordinates and Cartesian coordinates is time independent (so that the Lagrangian is time independent), then the Hamiltonian is just the total energy H = T + U, and it is conserved.
The rest of section 7.8 goes on to show that when the Lagrangian is independent of time, then
Subtracting the Lagrangian (T – U) from this obviously gives T + U.
October 28, 2010