Mecflu - princ?pio de hamilton - lagrangiano e din?mica de hamilton

Mecflu - princ?pio de hamilton - lagrangiano e din?mica de hamilton

(Parte 1 de 2)

Hamilton’s Principle— Lagrangian and Hamiltonian Dynamics

7-1. Four coordinates are necessary to completely describe the disk. These are the x and y coordinates, the angle θ that measures the rolling, and the angle φ that describes the spinning (see figure).

x y Since the disk may only roll in one direction, we must have the following conditions:

cossindxdyRdφφθ+= (1)

These equations are not integrable, and because we cannot obtain an equation relating the coordinates, the constraints are nonholonomic. This means that although the constraints relate the infinitesimal displacements, they do not dictate the relations between the coordinates themselves, e.g. the values of x and y (position) in no way determine θ or φ (pitch and yaw), and vice versa.

7-2. Start with the Lagrangian

Now let us just compute dL d mv at m

According to Lagrange’s equations, (4) is equal to (5). This gives Equation (7.36)

To get Equation (7.41), start with Equation (7.40) cossinegaeθθηη−=− (7) A and use Equation (7.38)

to obtain, either through a trigonometric identity or a figure such as the one shown here, g a

ga+coeθ= 2 sin e

Inserting this into (7), we obtain

We know intuitively that the period of the pendulum cannot depend on whether the train is accelerating to the left or to the right, which implies that the sign of a cannot affect the frequency. From a Newtonian point of view, the pendulum will be in equilibrium when it is in line with the effective acceleration. Since the acceleration is sideways and gravity is down, and the period can only depend on the magnitude of the effective acceleration, the correct form is

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 183

7-3. θ

If we take angles θ and φ as our generalized coordinates, the kinetic energy and the potential energy of the system are

where m is the mass of the sphere and where U = 0 at the lowest position of the sphere. I is the moment of inertia of sphere with respect to any diameter. Since ()225Imρ=, the Lagrangian becomes

When the sphere is at its lowest position, the points A and B coincide. The condition A0 = B0 gives the equation of constraint:

fLd L dt

fLd L dt

After substituting (3) and fRθρ∂∂=− and fφρ∂∂=− into (5), we find

From (7) we find λ:

or, if we use (4), we have

Substituting (9) into (6), we find the equation of motion with respect to θ :

7-4. y r θ

If we choose (r,θ ) as the generalized coordinates, the kinetic energy of the particle is

Since the force is related to the potential by

U f

we find

where we let U(r = 0) = 0. Therefore, the Lagrangian becomes

Lagrange’s equation for the coordinate r leads to

Since is identified as the angular momentum, (6) implies that angular momentum is conserved. Now, if we use A, we can write (5) as 2mr θ = A

Multiplying (7) by , we have r

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 185 which is equivalent to d d A mr r

Therefore,

and the total energy is conserved.

7-5.

φ x y r m

Let us choose the coordinate system so that the x-y plane lies on the vertical plane in a gravitational field and let the gravitational potential be zero along the x axis. Then the kinetic energy and the potential energy are expressed in terms of the generalized coordinates (r,φ) as

sinA Urmgrαφα=+ (2) from which the Lagrangian is

Therefore, Lagrange’s equation for the coordinate r is

Lagrange’s equation for the coordinate φ is

Since 2mrφ is the angular momentum along the z axis, (5) shows that the angular momentum is not conserved. The reason, of course, is that the particle is subject to a torque due to the gravitational force.

7-6.

y M

S m α ξ x

Let us choose ξ,S as our generalized coordinates. The x,y coordinates of the center of the hoop are expressed by cos sin cos sin xS r yr S ξα α

1 cos sin

Tm x y I

In order to find the total kinetic energy, we need to add the kinetic energy of the translational motion of the plane along the x-axis which is

Therefore, the total kinetic energy becomes

The potential energy is

from which the Lagrange equations for ξ and S are easily found to be 2cossinmSmmgξαα0+− = (8)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 187 or, if we rewrite these equations in the form of uncoupled equations by substituting for ξ and , we have S cos 2s i

sin cos 2c os m Sg mg mM m ααξ n 0

Now, we can rewrite (9) as

where we can interpret ()mMξ+ as the x component of the linear momentum of the total system and cosmSα as the x component of the linear momentum of the hoop with respect to the plane. Therefore, (1) means that the x component of the total linear momentum is a constant of motion. This is the expected result because no external force is applied along the x-axis.

7-7.

x y y m y m x

cos

2121 cos cos sin sinxy 2 φ φφφ

2s in sin cos cos 2 m Tx y x y m φ φφ φ φ φ φ φ φ

A A=+ (3)

The potential energy is

A (5)

from which () sin 2 sin

2c os

sin sin cos

L m

L m

L m g

L m gφ φφ φ φφ

The Lagrange equations for 1φ and 2φ are

7-8.

U U y

Let us choose the x,y coordinates so that the two regions are divided by the y axis:

If we consider the potential energy as a function of x as above, the Lagrangian of the particle is

Therefore, Lagrange’s equations for the coordinates x and y are

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 189

x x dpdP dP Pdd x mx mx

(2) becomes

Integrating (5) from any point in the region 1 to any point in the region 2, we find

0xx dU xPd P dx dx

or, equivalently,

Now, from (3) we have

and is constant. Therefore, my

Adding (8) and (10), we have

From (9) we also have

Substituting (1) into (12), we find sin vU UvTθθ

This problem is the mechanical analog of the refraction of light upon passing from a medium of a certain optical density into a medium with a different optical density.

7-9.

O x y m φ

Using the generalized coordinates given in the figure, the Cartesian coordinates for the disk are (ξ cos α, –ξ sin α), and for the bob they are (A sin φ + ξ cos α, –A cos φ – ξ sin α). The kinetic energy is given by

T (1)

Substituting the coordinates for the bob, we obtain

The potential energy is given by

( )disk bob disk bob sin cosU U Mgy mgy M m g mgU ξ α= + = − + − A φ=+ (3)

Now let us use the relation ξ = Rθ to reduce the degrees of freedom to two, and in addition substitute 22IMR= for the disk. The Lagrangian becomes

The resulting equations of motion for our two generalized coordinates are

7-10.

y x x –yS

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 191

Let the length of the string be A so that

a) The Lagrangian of the system is

Therefore, Lagrange’s equation for y is

2 dL L My Mg

from which

Then, the general solution for y becomes

b) If the string has a mass m, we must consider its kinetic energy and potential energy. These are

2 string 2

Adding (8) and (9) to (3), the total Lagrangian becomes

Therefore, Lagrange’s equation for y now becomes

In order to solve (1), we arrange this equation into the form

dM d y

2y, (12) is equivalent to mgdM M y

which is solved to give

yA e Bm where

Then,

() ( )1c oshM yt tm

7-1. x x′ m y

The x,y coordinates of the particle are cos cos sin sin xR t R t yR t R t ωφ ω

sin sin cos cos xR t R t yR t R t ω ω φω φω

Since there is no external force, the potential energy is constant and can be set equal to zero. The Lagrangian becomes

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 193

Lm x y

R R osω φω ω φ ω φ=+

from which

dL d mR

Therefore, Lagrange’s equation for φ becomes which is also the equation of motion for a simple pendulum. To make the result appear reasonable, note that we may write the acceleration felt by the particle in the rotating frame as

where the primed unit vectors are as indicated in the figure. The part proportional to r′e does not affect the motion since it has no contribution to the torque, and the part proportional to i′ is constant and does not contribute to the torque in the same way a constant gravitational field provides a torque to the simple pendulum.

7-12.

r m θ Put the origin at the bottom of the plane

Lagrange’s equation for r gives 2sinmrmrmgtαα=−

The general solution is of the form phrrr=+ where is the general solution of the homogeneous equation and hr

So tthrAeBeαα−=+

For pr, try a solution of the form rCsinptα=. Then 2sinprCtαα=− . Substituting into (1) gives 22sin sin sinCt C t g tα αα α−− = − α

So

() 2 sin2 t g rtAeBetαααα−=++

We can determine A and B from the initial conditions:

Solving for A and B gives:

tg g

or

g rtttrtαααα=+−

7-13. a) θ a b

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 195 cos cos sin xa t b yb xa t b yb

1 2c os cos

Lm x y mgy ma t at b b mgbθ θθ =+ − dL L

2cos sin sinmat b mb mat b mgb

This gives the equation of motion b) To find the period for small oscillations, we must expand sin θ and cos θ about the equilibrium point 0θ. We find 0θ by setting 0θ= . For equilibrium,

00sincosgaθθ= or

0tan ag ag+

Using the first two terms in a Taylor series expansion for sin θ and cos θ gives

0tan sin

Thus

Substituting into the equation of motion gives g a ag g g a a

This reduces to

2 2 0 ga gabb

The solution to this inhomogeneous differential equation is

0cossinABθθωθωθ=++ where

Thus

θa b sin

cos sin ya t b xb ya t b

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 197

Lagrange’s equation for θ gives

2 sin cos sind

For small oscillations, sin θ θ

ag ππω==+ k mθ b = unextended length of spring

A = variable length of spring

bmgykbmgUkθ=−+=−−A

Taking Lagrange’s equations for A and θ gives

This reduces to k bg

gθθ θθ θ

A 7-16.

θ m b x = a sin ωt

For mass m: sin sin

cos cos cos sin xa t b yb xa t b yb

=+= Substitute into

Umgy= and the result is

Lagrange’s equation for θ gives

( cod )2s cos cos sin sinmab t mb mabw t mgbdt ωωθθθωθ+=−− θ

2sin sin cos 0 g a

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 199 7-17.

C y h mg

Using q and θ (= ω t since θ (0) = 0), the x,y coordinates of the particle are expressed as cos sin cos sin sin cos sin cos xh q h t q t t yh q h t q t t θ θω ω

from which sin cos sin cos sin cos xh t q t q t yh t q t q t ω ωω ω ω

Therefore, the kinetic energy of the particle is ( )

Tm x y

The potential energy is ()sincosUmgymghtqtωω==− (4)

Then, the Lagrangian for the particle is

=+ (5)

The complementary solution and the particular solution for (6) are written as ( ) ( ) cos

cos 2cP qt A i t g qt t so that the general solution is g qtAittωδω=+−ω (8)

Using the initial conditions, we have

0s in 0 g qA qi A δ ω

Therefore,

and

or,

In order to compute the Hamiltonian, we first find the canonical momentum of q. This is obtained by

Therefore, the Hamiltonian becomes

Hp q L mq m hq m h m q mq mgh t mgq t m qhω ωω ω ω=− =− − − − + − + ω so that

Solving (13) for and substituting gives q

H hp m q mgh t mgq tm ωωω−+−ω=+ (15)

The Hamiltonian is therefore different from the total energy, T + U. The energy is not conserved in this problem since the Hamiltonian contains time explicitly. (The particle gains energy from the gravitational field.)

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 201 7-18.

y S x –y x

From the figure, we have the following relation:

ACsRθ=−=−A (1) where θ is the generalized coordinate. In terms of θ, the x,y coordinates of the mass are cos sin cos sin cos sin cos sin xA C R R R yR AC R R θ θθ θ θ

(2) from which sin sin cos cosxRyR θθθ θ θ

Therefore, the kinetic energy becomes

The potential energy is

Now let us expand about some angle 0θ, and assume the deviations are small. Defining 0εθθ≡−, we obtain

The solution to this differential equation is cos sin

where A and δ are constants of integration and

is the frequency of small oscillations. It is clear from (9) that θ extends equally about 0θ when 02θπ=.

7-19. P m m m g m g

Because of the various constraints, only one generalized coordinate is needed to describe the system. We will use φ, the angle between a plane through P perpendicular to the direction of the gravitational force vector, and one of the extensionless strings, e.g., A, as our generalized coordinate. 2

The, the kinetic energy of the system is

The potential energy is given by

sin sin2

The Lagrangian equation for φ is

This is the equation which describes the motion in the plane . 12,,mmP

To find the frequency of small oscillations around the equilibrium position (defined by 0φφ=), we expand the potential energy U about 0φ:

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 203

cos tan

or, cos n

2c osmm m m si θφ θ+= ++ A A A A sin cos sin cot

cos sin cos cos2c os

2c os

Ug m m m gm m m m m m m gm m m m

A (9) Finally, from (6) and (9), we have

2c osmm m m g

which, using the relation, 2

can be written as

Notice that 2ω degenerates to the value gA appropriate for a simple pendulum when d → 0 (so that ). 12=A

7-20. The x-y plane is horizontal, and A, B, C are the fixed points lying in a plane above the hoop. The hoop rotates about the vertical through its center.

θ The kinetic energy of the system is given by

MR z

The potential energy is given by where we take U = 0 at z = –A.

Since the system has only one degree of freedom we can write z in terms of θ. When θ = 0, z=−A. When the hoop is rotated thorough an angle θ, then

R UM g

Mg θθ

From (1) and (6), the Lagrangian is

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 205 for small θ. The Lagrange equation for θ gives

which is the frequency of small rotational oscillations about the vertical through the center of the hoop and is the same as that for a simple pendulum of length A.

7-21. ω

From the figure, we can easily write down the Lagrangian for this system.

The equilibrium positions are found by finding the values of θ for which

Note first that 0 and π are equilibrium, and a third is defined by the condition

To investigate the stability of each of these, expand using 0εθθ=−

For 0θπ=, we have

which is stable if 2gRω< and unstable if 2gRω>. When stable, the frequency of small oscillations is 2gRω−. For the final candidate,

with a frequency of oscillations of ()22gRωω−, when it exists. Defining a critical frequency 2cgRω≡, we have a stable equilibrium at 0θ= when cωω<, and a stable equilibrium at

To construct the phase diagram, we need the Hamiltonian

which is not the total energy in this case. A convenient parameter that describes the trajectory for a particular value of H is

2cc cHmR θωK θ θωω ω

≡= (1)

so that we’l end up plotting

22c os sincc for a particular value of ω and for various values of K. The results for cωω< are shown in figure (b), and those for cωω> are shown in figure (c). Note how the origin turns from an attractor into a separatrix as ω increases through cω. As such, the system could exhibit chaotic behavior in the presence of damping.

HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 207

θ (b)

θ (c)

7-2. The potential energy U which gives the force

must satisfy the relation

we find

208 CHAPTER 7 Therefore, the Lagrangian is

The Hamiltonian is given by x L H px L x L

so that 2

2 txp k

The Hamiltonian is equal to the total energy, T + U, because the potential does not depend on velocity, but the total energy of the system is not conserved because H contains the time explicitly.

7-23. The Hamiltonian function can be written as [see Eq. (7.153)]

For a particle which moves freely in a conservative field with potential U, the Lagrangian in rectangular coordinates is

and the linear momentum components in rectangular coordinates are y z

(Parte 1 de 2)

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