PHYSICS 200A : CLASSICAL MECHANICS

SOLUTION SET #2
[1] [Jose and Saletan problem 3.11] Consider a three-dimensional one-particle system whose
potential energy in cylindrical polar coordinates {, , z} is of the form V (, k + z), where
k is a constant.
(a) Find a symmetry of the Lagrangian and use Noethers theorem to obtain the constant
of the motion associated with it.
(b) Write down at least one other constant of the motion.
(c) Obtain an explicit expression for the dynamical vector field (see JS eqn. 3.73) and
use it to verify that the functions found in (a) and (b) are indeed constants of the motion.
Solution :
(a) We have

L = 21 m 2 + 2 2 + z 2 V (, k + z) .

Consider now the one-parameter family of coordinate transformations,

()
+

z() z k .

Clearly

k + z = k + z ,

hence L does not vary with , and therefore


X L q 
Q=

q 

= m2 mkz

=0

is conserved: Q = 0.

(b) Since L
t = 0, the Hamiltonian H is conserved. And since the kinetic energy is homo z},
geneous of degree two in the generalized velocities {,
,
the Hamiltonian is simply the
total energy: H = T + U . Thus,


E = 12 m 2 + 2 2 + z 2 + V (, k + z)

is conserved: E = 0.

(c) The dynamical vector field is simply the total time derivative, expressed in terms of
derivatives with respect to coordinates and velocities:
=

= q
+ q
dt
q
q

+ z
+
+ z
+
+

z

z

The generalized accelerations follow from the equations of motion,

m
= m 2

d
V
V
= k
m2 =
dt

z
m
z=

V
,
z

which yield
=

1 V
m

k V
2
=

m2 z

z =

1 V
.
m z

Therefore, we have
Q = m2 mkz

= 2m + m2 + z (mk)
= 2m +




2
1 V
k V
2

m +
(mk)

m2 z

m z

=0.
We also have


E = 21 m 2 + 2 2 + z 2 + V (, k + z)

V
V
V
+
+ z
+ m 2 + m + m2 + mz z

z


V
V
V
1 V
2
2

=
+k
+ z
+ m + m

z
z
m




2
1 V
k V
2

+ mz
=0.
+ m
m2 z

m z

[2] [Jose and Saletan problem 3.24] Derive the equations of motion for the Lagrangian
i
h
L = et 21 mq2 21 kq 2 ,

where > 0. Compare with known systems. Rewrite the Lagrangian in terms of the new
variable Q q exp(t/2), and from this obtain a constant of the motion.

Solution : We have
p=

L
= mq et
q

F =

L
= kq et .
q

Mr. Newton then says

p = F

m q + m q = kq ,

which is the equation of a damped harmonic

oscillator. The phase curvespall collapse to the
p
origin, which is a stable spiral if < 2 k/m and a stable node if > 2 k/m.
In general, there is no reason for there to be a conserved quantity in a dissipative system like
this ... but ... consider the coordinate transformation Q q exp(t/2), which is inverted
trivially to yield q = Q exp(t/2). We have


q = Q 21 Q et/2
and therefore

L = 12 m Q 12 Q

2

21 kQ2

= 12 m Q 2 21 m Q Q

1
2

k 41 m 2 Q2 .

t) is independent of t, we have that H is conserved:

Since L(Q, Q,
H = Q

L
L
Q

= 21 m Q 2 +
=

2
1
2 m q

1
2

k 41 m 2 Q2 .

i
+ 21 m q q + 12 k q 2 et .

[3] A bead of mass m slides frictionlessly along a wire curve z = x2 /2b, where b > 0. The
wire rotates with angular frequency about the z axis.
(a) Find the Lagrangian of this system.
(b) Find the Hamiltonian.
(c) Find the effective potential Ueff (x).
(d) Show that the motion is unbounded for 2 > c2 and find the critical value c .
(e) Sketch the phase curves for this system for the cases 2 < c2 and 2 > c2 .
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(f) Find an expression for the period of the motion when 2 < c2 .
(g) Find the force of constraint which keeps the bead on the wire.
Solution : We will solve this problem for a general shape z(x). Since the curve is rotating,
we will use the radial coordinate instead of x, keeping in mind that the wire is a onedimensional object and not a two-dimensional surface. The coordinate then indicates the
direction along the wire but perpendicular to the z axis. Note that R may be positive
or negative.
(a) The Lagrangian is
L(, z, ,
z)
= 21 m 2 + 12 mz 2 + 21 m 2 2 mgz .
This is supplemented by the constraint
G(, z) = z z() = 0 .
Of course, we could eliminate z as an independent degree of freedom from the outset, and
write
h
i


L(, )
= 12 m 1 + [z ()]2 2 + 2 2 mgz() .
(b) The Hamiltonian is
H = p q L
= 21 m 2 + 12 mz 2 12 m 2 2 + mgz


= 21 m 1 + [z ()]2 2 + Ueff () .
(c) The effective potential is
Ueff () = mgz() 12 m 2 2
= 21 m (c2 2 ) 2 ,
where c

(). This is because

g/b. Note that we do not have m
= Ueff

p =

and thus
p =

L
= m 1 + [z ()]2 ,

2 
= 2 gz () z () z () 2 .
1 + [z ()
4

Figure 1: Level sets of the function C(u, v) = (1 + u2 ) v 2 + u2 superimposed on the phase

flow u = v, v = u (1 + v 2 )/(1 + u2 ). Note that the phase curves are bounded.

(d) Since L has no explicit time dependence, H is a constant of the moton:

H = 12 m 1 + [z ()]2 2 + Ueff ()


2 2 1
1
= 2 m 1 + 2 + 2 m(c2 2 ) 2 .
b
Note that if 2 > c2 that the level sets of H(, )
are unbounded. Hence the motion of the
system, which takes place along these level sets, is also unbounded.
(e) Let us define the dimensionless coordinate u /b and dimensionless time variable
s |c2 2 |1/2 t. Then conservation of H means that
is constant, where v =
dC
ds = 0, we obtain

du
ds

C = (1 + u2 ) v 2 u2

is the dimensionless velocity, and where sgn 2 c2 . Setting

du
dv
( v 2 ) u
=v
,
=
.
ds
ds
1 + u2
This phase flow has a single fixed point, at (u, v) = (0, 0), which is either a center ( 2 < c2 )
or a saddle point ( 2 > c2 ). A sketch of the phase flow for 2 < c2 is shown in Fig. 1; the
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Figure 2: Level sets of the function C(u, v) = (1 + u2 ) v 2 u2 superimposed on the phase

flow u = v, v = u (1 v 2 )/(1 + u2 ). Note that the phase curves are unbounded.
flow for 2 > c2 is shown in Fig. 2. The Mathematica plot in Fig. 1 was obtained from
the following commands:
<<GraphicsPlotField
G1 = ContourPlot[ (1+x^2) y^2 + x^2, {x,-4,4}, {y,-4,4}, PlotPoints -> 50,
Contours -> {0.1, 1, 4, 10, 20, 50, 100}, ContourShading -> False];
G2 = PlotVectorField[ {y, -(1+y^2) x / (1+x^2)}, {x,-4,4}, {y,-4,4},
PlotPoints -> 30, ColorFunction -> Hue, ScaleFactor -> 0.55];
Show[ {G1, G2} ]
It is worthwhile noting that other shapes z() may have fixed points for 6= 0. For example,
consider the shape
4
z() = 3 .
4b
If we define u = /b and c2 = g/b as before, but this time write s = c t, and define the
new dimensionless parameter 2 /c2 , we have that
C(u, v) = (1 + u6 ) v 2 + 41 u2 12 u2

is constant, and the dynamics is given by

du
=v
ds

dv
( u2 6 u4 v 2 ) u
=
.
ds
2 (1 + u6 )

This flow, shown in Fig. 3, exhibits a saddle point at (u, v) = (0, 0) and two centers at

(u, v) = ( , 0). The separatrix, which flows through (0, 0), has C = 0. All the phase
curves are bounded.

Figure 3: Level sets of the function C(u, v) = (1 + u6 ) v 2 + 14 u4 21 u2 superimposed on the

phase flow u = v, v = 12 u ( u2 6 u4 v 2 )/(1 + u6 ), for = 1. There are two centers, at
(1, 0), and a saddle at (0, 0). All phase curves are bounded.
(e) The equation of motion can be taken as H = 0, which yields


2 
+ z () z () 2 = 2 g z () .
1 + z ()

We can expand about an equilibrium solution gz ( ) = 2 , writing = + , in which

case
gz ( ) 2
= 2
,
2 =

2 .
1 + z ( )
Thus, the equilibrium at is stable if 2 < gz ( ) and unstable if 2 > gz ( ).

We can go even farther in this analysis, using the conservation of H, which allows us to
write the motion as a first order ODE,
q

2
1 + z ()
dt = q 
 d .
2
H

U
()
eff
m
Identifying the turning points as solutions to

H = Ueff ( ) ,
we have the period for motion T (H) is
+ (H)

T (H) =

m
2

(H)


2
1 + z ()
.
H Ueff ()

For the case z() = 2 /2b, we have

T (H) = p

4
c2 2

Z/2 s
d 1 +
0

2H sin2
.
mb2 (c2 2 )

(g) If we write G(, z) = z z() = 0 as a constraint, the equations of motion are

m
= m 2 z ()
m
z = mg + .

We now eliminate z = z(), in which case

z = z ()

z = z () + z () 2 .

We may now write

= mg + mz () + mz () 2
and, substituting this into the first of the equations of motion and collecting terms, we find

2 

= 2 gz () z () z () 2 .
1 + [z ()

As we have seen above, this result also follows from H = 0. We may now solve for in
terms of and :

m

2
2

=

2 g + z () + z () .
1 + z ()
The force of constraint supplied by the wire is
,
= (Q + Qz z)
Q = Qn
8

where

z () + z
=q
n

2
1 + z ()

is the unit vector locally orthogonal to the tangent to the curve. Thus,
Q=

q

2
1 + z ()

m g + z () 2 + 2 z ()
q
.
=

2

1 + z ()

We may further eliminate in favor of by invoking conservation of H, which says

2 =

2H
m

2gz() + 2 2
.
2
1 + [z ()

[4] A particle of mass m is embedded, a distance b from the center, in a uniformly dense
cylinder of mass M . (The mass of the cylinder plus the inclusion is thus M + m.) The
cylinder rolls without slipping along a plane inclined at an angle with respect to the
horizontal, under the influence of gravity. The axis of the cylinder remains horizontal
throughout the motion.
(a) Choose an appropriate generalized coordinate and find the Lagrangian.
(b) Find the equations of motion.
(c) Under what conditions does a stable equilibrium exist?
(d) Find the frequency of small oscillations about the equilibrium.

Figure 4: A cylinder of radius R with an inclusion rolls along an inclined plane.

Solution :
(a) Consulting the diagram in Fig. 4, let q be the distance of the cylinders point of contact
to the bottom of the wedge, and let be the angle the inclusion makes with respect to the
vertical, with = 0 pointing downward. The coordinates of the center of the cylinder, in
the plane perpendicular to its symmetry axis, are
xC = q cos R sin
yC = q sin + R cos .
Thus, the coordinates of the inclusion are
x = q cos R sin b sin

y = q sin + R cos b cos .

We may now write

x C = q cos
y C = q sin
x = q cos b cos
y = q sin + b sin ,
and thus the kinetic energy is
T = 12 M (x 2C + y C2 ) + 21 I 2 + 12 m (x 2 + y 2 )


= 12 M q2 + 21 I 2 + 12 m q2 + b2 2 2b q cos( + )


= 12 I + M R2 + mR2 + mb2 2mbR cos( + 2 ,

where in the last line we have implemented the holonomic constraint

q =

d
R( + ) = R .
dt

We can now write q = q0 + R, where q0 is a constant, in which case the potential is

U = M gyC + mgy
= (M + m)gR sin mgb cos + U0 ,
where U0 is a constant. The remaining generalized coordinate is , in which case


L = 12 I + M R2 + mR2 + mb2 2mbR cos( + ) 2 (M + m)gR sin + mgb cos ,

up to an irrelevant overall constant.

(b) The equations of motion, p = F , are found to be



I + M R2 + mR2 + mb2 2mbR cos( + ) + mbR sin( + ) 2 = U () ,
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where
U () = (M + m)gR sin mgb cos .
(c) Equilibrium requires U () = 0, which says


M R
sin .
sin = 1 +
m b
Note that sin < 0, which means that the inclusion must lie to the right of the vertical
midline shown in Fig. 4. In order for a solution to exist, we must have


M R
sin 1 .
1+
m b
Thus, no solution exists unless

M R sin
,
b R sin
and of course we must have b > R sin . In fact, there are two solutions:
"
#
#
"


M R
M R
1
1
sin
, 2 = 2 sin
1+
sin .
1 = + sin
1+
m b
m b
m

(d) Since U () = mgb cos , so the solution = 1 is unstable while = 2 is stable. The
equation for small oscillations is
= 2 ,
where = 2 + and
2 =

I+

M R2

mR2

mgb cos 2
.
+ mb2 2mbR cos(2 + )

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