Homework 8 6l5k5t

Classical Mechanics - Homework Assignment 8 Alejandro G´omez Espinosa∗ November 20, 2012

Goldstein, Ch.8, 14 The Lagrangian for a system can be written as L = ax˙ 2 + b

p y˙ + cx˙ y˙ + f y 2 x˙ z˙ + g y˙ 2 − k x2 + y 2 x

where a, b, c, f, g and k are constants. What is the Hamiltonian?. What quantities are conserved?. Let’s use the Hamiltonian formulation: H = q˙i pi − L,

pi =

∂L ∂ q˙i

(1)

Then, px = py = pz =

∂L = 2ax˙ + cy˙ + f y 2 z˙ ∂ x˙ b ∂L = + cx˙ + 2g y˙ ∂ y˙ x ∂L = f y 2 x˙ ∂ z˙

(2) (3) (4)

The Hamiltonian is given by: p y˙ − cx˙ y˙ − f y 2 x˙ z˙ − g y˙ + k x2 + y 2 x p b y˙ 2ax˙ 2 + cx˙ y˙ + f y 2 x˙ z˙ + y˙ + cx˙ y˙ + 2g y˙ 2 + f y 2 x˙ z˙ − ax˙ 2 − b − cx˙ y˙ − f y 2 x˙ z˙ − g y˙ 2 + k x2 + y 2 x x p 2 2 2 2 2 ax˙ + cx˙ y˙ + f y x˙ z˙ + g y˙ + k x + y p b z 1 2 2 2 2ax˙ + cx˙ y˙ + f y x˙ z˙ − ax˙ + p y − − 2 + k x2 + y 2 using (3) 2 x fy p 1 b z 2 xp ˙ x − ax˙ + p y − − 2 + k x2 + y 2 using (2) 2 x fy p ap2z 1 b z px pz − + p − − + k x2 + y 2 using (4) y f y2 f 2y4 2 x f y2

H = xp ˙ x + yp ˙ y + zp ˙ z − ax˙ 2 − b = = = = =

Finally, pz is conserved due to time. ∗

∂L ∂z

= 0 and the energy due to the Hamiltonian doesn’t depend on

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Goldstein, Ch.8, 15 A dynamical system has the Lagrangian L = q˙12 +

q˙22 + k1 q12 + k2 q˙1 q˙2 2 a + bq1

where a, b, k1 and k2 are constants. Find the equations of motion in the Hamiltonian formulation. For simplicity, let’s use the previous Lagrangian in the matrix notation: T q˙ q˙ L= 1 A 1 + k1 q12 , q˙2 q˙2

A=

Then, the momentum can be written as: ∂L q˙ p1 = 2A 1 = 2Aq˙ p= = q˙2 p2 ∂ q˙i

k2 2 1 a+bq12

1 k2 2

⇒

!

1 q˙ = A−1 p˙ 2

The Hamiltonian has the form: H = p · q˙ − L = p · q˙ − q˙ T Aq˙ − k1 q12 1 1 pA−1 p − pA−1 p + k1 q12 = 2 4 1 −1 = pA p + k1 q12 4 where: A−1 =

1 1 a+bq12

−

k12 4

1 a+bq12 − k22

using (5)

− k22

!

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Finally the equations of motion in the Hamiltonian formulation are:   2bq1 p21 1 − 2k q 1 1 2 2 2 k ∂H (a+bq1 ) 1  p˙ = − =  a+bq12 − 41 ∂qi 0 ∂H 1 q˙ = = A−1 p ∂pi 2

2

(5)

Goldstein, Ch.8, 26 A particle of mass m can move in one dimension under the influence of two springs connected to fixed points a distance a apart (see Figure 1). The springs obey Hooke’s law and have zero unstretched lenghts and force constants k1 and k2 , respectively.

Figure 1: Sketch of problem 26. A particle of mas m moving under the influence of two springs. (a) Using the position of the partice from one fixed point as the generalized coordinate, find the Lagrangian and the corresponding Hamiltonian. Is the energy conserved? Is the Hamiltonian conserved? For this system, the kinetic and potential energy are: T =

mx˙ 2 , 2

V =

k1 x2 k2 (a − x)2 + 2 2

and the Lagrangian is given by: L=

mx˙ 2 k1 x2 k2 (a − x)2 − − 2 2 2

(6)

Then, to find the Hamiltonian: px =

∂L = mx˙ ∂ x˙

⇒

x˙ =

px m

and the Hamiltonian can be written as: H = xp ˙ x−L p2 k1 x2 k2 (a − x)2 p2x − x + + = m 2m 2 2 p2x k1 x2 k2 (a − x)2 = + + 2m 2 2 As the Hamiltonian doesn’t depend explicitly on time, it is conserved. It is the same for the energy. (b) Introduce a new coordinate Q defined by: Q = q − b sin wt,

b=

k2 a k1 + k2

What is the Lagrangian in of Q? What is the corresponding Hamiltonian? Is the energy conserved? Is the Hamiltonian conserved? Considering the new coordinate: ⇒

q = Q + b sin wt 3

q˙ = Q˙ + bw cos wt

In the Lagrangian of equation (6), let’s replace x = q. Then: L=

2 k m˙ 1 Q + bw cos wt − (Q + b sin wt)2k2 (a − Q − b sin wt)2 2 2 2

And the momentum: P =

∂L = m Q˙ + bw cos wt ∂ Q˙

P Q˙ = − bw cos wt m

⇒

Finally, the Hamiltonian is given by: ˙ −L H = QP = =

2 P2 m P k1 − bwP cos wt − − bw cos wt + bw cos wt − (Q + b sin wt)2k2 (a − Q − b sin wt)2 m 2 m 2 2 2 k1 P − bwP cos wt − (Q + b sin wt)2k2 (a − Q − b sin wt)2 2m 2 2

In this case, the Hamiltonian does not depend explicitly on time and the potential does not depend on the velocity, then it is not conserved. In the case of the energy, as we do have any discipative force in this case, the energy must be conserved. Goldstein, Ch.8, 33 Two mass points, m1 and m2 , are connected by a string that acts as Hooke’s-law spring with force constant k. One particle is free to move without friction on a smooth horizontal plane surface, the other hangs vertically down from the string through a hole in the surface. Find the condition for steady motion in which the mass point on the plane rotates uniformly at constant distance from the hole. Investigate the small oscillations in the radial distance from the hole, and in the vertical height of the second particle. Let’s define the coordinates of the particle m1 in the plane as r, θ and the coordinate for m2 hanging from the plane as z. Then, the kinetic and potential energy are: T =

m1 r˙ 2 m1 r2 θ2 m2 z˙ 2 + + , 2 2 2

V =

k(r + z)2 − m2 gz 2

Using this expression, the Lagrangian is written as: L=

m1 r˙ 2 m1 r2 θ2 m2 z˙ 2 k(r + z)2 + + − + m2 gz 2 2 2 2

Calculate the momentum of each coordinate: pr = pθ = pz =

∂L = m1 r˙ ∂ r˙ ∂L = m1 r2 θ˙ ˙ ∂θ ∂L = m2 z˙ ∂ z˙

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⇒ r˙ =

pr m1

⇒ θ˙ = ⇒ z˙ =

pθ m1 r2

pz m2

The Hamiltonian is then given by: ˙ θ + zp H = rp ˙ r + θp ˙ z −L 2 2 p p2θ pr p2 k(r + z)2 p2 p2 = + θ2 + z − r − − z + − m2 gz 2 m1 m1 r m2 2m1 2m1 r 2m2 2 p2θ p2z kr2 p2r kz 2 + + + = + krz + − m2 gz 2m1 2m1 r2 2m2 2 2 Now, let’s use the Routh’s procedure. Find the expression of R using the previous Lagrangian and Hamiltonian: R = Hcycl (pi ) − Lnon−cycl (qi , q˙i ) =

p2θ m1 r˙ 2 m2 z˙ 2 kr2 kz 2 − − + + krz + + m2 gz 2m1 r2 2 2 2 2

Using this procedure, we have to solve first for the non-cyclic coordinates: d ∂R ∂R =0 − dt ∂ q˙i ∂qi Therefore: ∂R d = (−m1 r¨) = −m1 r¨ ∂ r˙ dt ∂R = kr + kz ∂r m2 r¨ + kr + kz = 0 d dt



∂R d = (−m1 z¨) = −m1 z¨ ∂ z˙ dt ∂R = kz + kr + m2 g ∂z m2 z¨ + kz + kr + m2 g = 0 d dt



Now, for the cyclic coordinates: ∂R = q˙i , ∂pi

∂R = p˙i ∂ q˙i

Thus, ∂R ∂pθ ∂R ∂ θ˙

=

pθ = θ˙ m1 r 2

= 0 = p˙θ

pθ = m1 r2 θ˙ = l = constant Then, the condition for steady motion is θ˙ =

l . m1 r 2

For the small oscillations, the kinetic energy is: T =

m1 r˙ 2 m2 z˙ 2 + = 2 2 5

m1 0 0 m2

and the potential energy: kr2 kz 2 l2 V = + krz + + m2 gz + = 2 2 m1 r 2

k+

6l2 m1 r4 k 2

k 2

!

k

including the conditions for equilibrium: 2l2 ∂L = kr + kz − =0 0 0 ∂r r=r0 m1 r03 ∂L = kz0 + kr0 − m2 g = 0 ∂z z=z0 2l2 = m2 g m1 r04 Finally, to find the frequency of small oscillations we compute the determinant: k + 6l2 − w2 m k 1 4 2 det |V − w2 T | = k m1 r0 =0 2 k − w m2 2

k2 3l2 2 k− − w m k − w2 m2 − =0 1 4 m1 r 4 Solving this equation we found the frequencies for small oscillations.

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