Transcript Examples of Lagrange`s Equations

Physics 430: Lecture 17
Examples of Lagrange’s
Equations
Dale E. Gary
NJIT Physics Department
7.5 Examples
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 vz  z. The f
velocity (azimuthal) is just vf  Rf. Thus, the kinetic energy is just
T  12 mz 2  12 mR 2f 2 .
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Example 7.4, cont’d

Solution, cont’d:

Now we need the potential energy (due to this central spring force), which is:
U  12 k ( R 2  z 2 ).

Notice that we could drop the R2 term, since it is constant, but it will not survive
the Lagrange equation anyway.
r=R
Putting these together, the Lagrangian is
L  12 mz 2  12 mR 2f 2 - 12 k ( R 2  z 2 ).



Now we form the Lagrange equations (one for z and one
for f) and solve. The z equation is
The f equation is
L d L

z dt z
L d L

f dt f
or
- kz  mz.
or 0 
d
mR 2f .
dt
z
r
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.
October 28, 2010
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
q1
that it has no component in the constraint force
direction. Also, q1 is measured relative to the
q2
(moving) wedge.
a
 The velocity of the wedge is just q2 , but the
velocity of the block has components from both
2
q1 and q2:
v2  vx2  vy2   q2  q1 cos a   q12 sin 2 a .
 Thus, the total kinetic energy is





T  12  M  m  q22  12 m q12  2q2 q1 cos a .

The potential energy of the wedge can be taken as zero, and the block is
just
U  -mgy  -mgq1 sin a .
October 28, 2010
Example 7.5, cont’d

Putting these together, the Lagrangian is
L  T -U 





1
2
 M  m  q22  12 m  q12  2q2q1 cos a   mgq1 sin a .
There are two generalized coordinates, and so there are two Lagrange
equations. The second is simplest:
L d L
L

or
 Mq2  m  q2  q1 cos a   const.
q2 dt q2
q2
Note that this is conservation of momentum in the x direction.
The first one is L d L

or mg sin a  m  q1  q2 cos a  .
q1 dt q1
This has two unknowns, but we can get another equation for the
accelerations by differentiating the q2 relation:
Mq2  m  q2  q1 cos a   0,
m
or q2  q1 cos a .
M m
g sin a
Finally, plug this into the above equation and solve q1 
.
m
2
1cos a
M m
October 28, 2010
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 Rq . But the hoop is also rotating,
w
so there is another motion with velocity rw  R sin q w .
 The kinetic energy is then



T  12 mv2  12 mR2 q 2  w 2 sin 2 q .

The potential energy relative to its position at the bottom of
the hoop (when the hoop is not rotating and q = 0), is
U  mgR(1 - cos q ).


The Lagrangian is L  T - U  12 mR2 q 2  w 2 sin 2 q - mgR(1 - cosq ).
 The sole Lagrange equation is then

L d L

q dt q

or mR 2w 2 sin q cos q - mgR sin q  mR 2q .
g
Solving for the angular acceleration: q   w 2 cos q -  sin q .
R

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q R
r
Example 7.6, cont’d





g

 sin q .
R

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.
w
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 q  -  g / R  sin q .
q R
But for a rotating hoop, the bead will move up the wire and we
r
might expect a new equilibrium. Recall that at equilibrium the bead
will be at rest, i.e. q  0, which occurs at either q = 0 (which we have
g
already examined) or at w 2 cos q -  0.
R
Solving for the equilibrium position
 g 
q o   arccos  2  .
w R 
We can find the oscillation frequency for small displacements q  qo  e.
We cannot solve: q   w 2 cos q -
October 28, 2010
7.6 Generalized Momenta and
Ignorable Coordinates





Recall that we said the left-hand side of the Lagrange equation
L d L

,
q1 dt q1
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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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
dL
L
L
L

qi  
qi  ,
dt
t
i qi
i qi
or
dL
d L
L
L

qi  
qi 
dt
t
i dt qi
i qi
   pi qi  pi qi  

i
L d
L
   pi qi  
t dt i
t
Now, if the Lagrangian does not depend on time, the last term on the right
vanishes, so we have
L    pi qi  ,

i
Bringing the Lagrangian over to the right, the result is called the
Hamiltonian:
H    pi qi  - L.
i
When the Lagrangian does not depend explicitly on time,
the Hamiltonian H is conserved.
October 28, 2010
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
  p q   2T .
i i
i

Subtracting the Lagrangian (T – U) from this obviously gives T + U.
October 28, 2010