PHYSICS 200A : CLASSICAL MECHANICSSOLUTION SET #2

[1] [Jose and Saletan problem 3.11] Consider a three-dimensional one-particle system whosepotential energy in cylindrical polar coordinates {, , z} is of the form V (, k+ z), wherek is a constant.

(a) Find a symmetry of the Lagrangian and use Noethers theorem to obtain the constantof 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) anduse it to verify that the functions found in (a) and (b) are indeed constants of the motion.

Solution :

(a) We haveL = 12m

(2 + 22 + z2

) V (, k + z) .Consider now the one-parameter family of coordinate transformations,

() + z() z k .

Clearlyk+ z = k+ z ,

hence L does not vary with , and therefore

Q =

L

q

q

=0

= m2mkz

is conserved: Q = 0.

(b) Since Lt = 0, the Hamiltonian H is conserved. And since the kinetic energy is homo-

geneous of degree two in the generalized velocities {, , z}, the Hamiltonian is simply thetotal energy: H = T + U . Thus,

E = 12m(2 + 22 + z2

)+ V (, k + z)

is conserved: E = 0.

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

=d

dt= q

q+ q

q

=

+

+ z

z+

+

+ z

z

1

The generalized accelerations follow from the equations of motion,

m = m 2 V

d

dt

(m2

)= V

= k V

z

mz = Vz

,

which yield

= 1m

V

, = k

m2V

z 2

, z = 1

m

V

z.

Therefore, we have

Q = (m2mkz)

= 2m+ m2 + z (mk)

= 2m +

( km2

V

z 2

)m2 +

( 1m

V

z

) (mk)

= 0 .

We also have

E = 12m(2 + 22 + z2

)+V (, k+ z)

= V

+

V

+ z

V

z+m 2 +m +m2 +mz z

= V

+ k V

z+ z

V

z+m 2 +m

( 1m

V

2

)

+m2 ( km2

V

z 2

)+mz

( 1m

V

z

)= 0 .

[2] [Jose and Saletan problem 3.24] Derive the equations of motion for the Lagrangian

L = et[12mq

2 12kq2],

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

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Solution : We have

p =L

q= mq et , F =

L

q= kq et .

Mr. Newton then says

p = F m q + m q = kq ,

which is the equation of a damped harmonic oscillator. The phase curves all collapse to theorigin, 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 likethis ... but ... consider the coordinate transformation Q q exp(t/2), which is invertedtrivially to yield q = Q exp(t/2). We have

q =(Q 12 Q

)et/2

and therefore

L = 12m(Q 12 Q

)2 12kQ2= 12mQ

2 12 mQQ 12(k 14m2

)Q2 .

Since L(Q, Q, t) is independent of t, we have that H is conserved:

H = QL

Q L

= 12mQ2 + 12

(k 14m2

)Q2 .

=[12m q

2 + 12 mq q +12k q

2]et .

[3] A bead of mass m slides frictionlessly along a wire curve z = x2/2b, where b > 0. Thewire 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 > 2c and find the critical value c.

(e) Sketch the phase curves for this system for the cases 2 < 2c and 2 > 2c .

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(f) Find an expression for the period of the motion when 2 < 2c .

(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 one-dimensional object and not a two-dimensional surface. The coordinate then indicates thedirection along the wire but perpendicular to the z axis. Note that R may be positiveor negative.

(a) The Lagrangian is

L(, z, , z) = 12m2 + 12mz

2 + 12m22 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, andwrite

L(, ) = 12m[(1 + [z()]2

)2 + 22

]mgz() .

(b) The Hamiltonian is

H = pq L

= 12m2 + 12mz

2 12m22 +mgz

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

)2 + Ueff() .

(c) The effective potential is

Ueff() = mgz() 12m22

= 12m (2c 2) 2 ,

where c g/b. Note that we do not have m = U eff(). This is because

p =L

= m

(1 + [z()]2

) ,

and thus

p =L

(1 + [z()

]2) = 2 gz() z() z() 2 .

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Figure 1: Level sets of the function C(u, v) = (1 + u2) v2 + u2 superimposed on the phaseflow u = v, v = u (1 + v2)/(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 = 12m(1 + [z()]2

)2 + Ueff()

= 12m

(1 +

2

b2

)2 + 12m(

2c 2) 2 .

Note that if 2 > 2c that the level sets of H(, ) are unbounded. Hence the motion of thesystem, which takes place along these level sets, is also unbounded.

(e) Let us define the dimensionless coordinate u /b and dimensionless time variables |2c 2|1/2 t. Then conservation of H means that

C = (1 + u2) v2 u2

is constant, where v = duds is the dimensionless velocity, and where sgn(22c

). Setting

dCds = 0, we obtain

du

ds= v ,

dv

ds=

( v2)u1 + u2

.

This phase flow has a single fixed point, at (u, v) = (0, 0), which is either a center (2 < 2c )or a saddle point (2 > 2c ). A sketch of the phase flow for

2 < 2c is shown in Fig. 1; the

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Figure 2: Level sets of the function C(u, v) = (1 + u2) v2 u2 superimposed on the phaseflow u = v, v = u (1 v2)/(1 + u2). Note that the phase curves are unbounded.

flow for 2 > 2c is shown in Fig. 2. The Mathematica plot in Fig. 1 was obtained fromthe following commands:

{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

z() =4

4 b3.

If we define u = /b and 2c = g/b as before, but this time write s = c t, and define thenew dimensionless parameter 2/2c , we have that

C(u, v) = (1 + u6) v2 + 14u2 12u2

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is constant, and the dynamics is given by

du

ds= v ,

dv

ds=

( u2 6u4 v2)u2 (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 phasecurves are bounded.

Figure 3: Level sets of the function C(u, v) = (1+u6) v2+ 14u4 12u2 superimposed on the

phase flow u = v, v = 12u ( u2 6u4 v2)/(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(1 +

[z()

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

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

= 2 , 2 = gz() 2

1 +[z()

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

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

dt =

1 +

[z()

]22m

[H Ueff()

] d .Identifying the turning points as solutions to

H = Ueff() ,

we have the period for motion T (H) is

T (H) =

m2

+(H)

(H)

d

1 +

[z()

]2H Ueff()

.

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

T (H) =4

2c 2

pi/20

d

1 +

2H sin2

mb2(2c 2).

(g) If we write G(, z) = z z() = 0 as a constraint, the equations of motion arem = m2 z()mz = 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(1 + [z()

]2) = 2 gz() z() z() 2 .

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

=m

1 +[z()

]2 (g + z() 2 + 2 z()) .

The force of constraint supplied by the wire is

Q = Q n= (Q +Qz z) ,

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where

n =z() + z1 +

[z()

]2is the unit vector locally orthogonal to the tangent to the curve. Thus,

Q = 1 +

[z()

]2

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

)1 +

[z()

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

2 =2Hm 2gz() + 22

1 + [z()]2 .

[4] A particle of mass m is embedded, a distance b from the center, in a uniformly densecylinder of mass M . (The mass of the cylinder plus the inclusion is thus M + m.) Thecylinder rolls without slipping along a plane inclined at an angle with respect to thehorizontal, under the influence of gravity. The axis of the cylinder remains horizontalthroughout 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.

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Solution :

(a) Consulting the diagram in Fig. 4, let q be the distance of the cylinders point of contactto the bottom of the wedge, and let be the angle the inclusion makes with respect to thevertical, with = 0 pointing downward. The coordinates of the center of the cylinder, inthe plane perpendicular to its symmetry axis, are

xC = q cosR sinyC = q sin+R cos .

Thus, the coordinates of the inclusion are

x = q cosR sin b siny = q sin+R cos b cos .

We may now write

xC = q cos

yC = q sin

x = q cos b cosy = q sin+ b sin ,

and thus the kinetic energy is

T = 12M(x2C+ y2

C) + 12I

2 + 12m (x2 + y2)

= 12Mq2 + 12I

2 + 12m(q2 + b22 2b q cos(+ ))

= 12

(I +MR2 +mR2 +mb2 2mbR cos(+

)2 ,

where in the last line we have implemented the holonomic constraint

q =d

dtR(+ ) = R .

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

U = MgyC +mgy

= (M +m)gR sinmgb cos + U0 ,

where U0 is a constant. The remaining generalized coordinate is , in which case

L = 12

(I +MR2 +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 +MR2 +mR2 +mb2 2mbR cos(+ )

)+mbR sin(+ ) 2 = U () ,

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whereU() = (M +m)gR sinmgb cos .

(c) Equilibrium requires U () = 0, which says

sin = (1 +

M

m

)R

bsin .

Note that sin < 0, which means that the inclusion must lie to the right of the verticalmidline shown in Fig. 4. In order for a solution to exist, we must have(

1 +M

m

)R

bsin 1 .

Thus, no solution exists unless

m MR sinbR sin ,

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

1 = pi + sin1

[(1 +

M

m

)R

bsin

], 2 = 2pi sin1

[(1 +

M

m

)R

bsin

].

(d) Since U () = mgb cos, so the solution = 1 is unstable while = 2 is stable. Theequation for small oscillations is

= 2 ,where = 2 + and

2 =mgb cos2

I +MR2 +mR2 +mb2 2mbR cos(2 + ) .

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