Exam: Analytical mechanics Lecturer: Dganit Meidan Tutor: Tomer Ygael

Exam: Analytical mechanics
Lecturer: Dganit Meidan
Tutor: Tomer Ygael
• Duration of the exam - 3 hours.
• No supporting material
• Answer the 3 questions below
Exercise 1: (35 points)
φ
M
R
m
θ
A bead of mass m slides on a thin ring of mass M in gravity, where the ring is held in place,
but free to rotate about the vertical axis.
1. Write down the Lagrangian for this system. Are there any cyclic coordinates? (8
points)
Solution: there are two DOF. We have a rigid body whose center of mass is fixed and
which is free to rotate only about one axis, leaving one degree of freedom. The bead
is free to move around the ring, but its distance from the center is fixed and its angle
in the horizontal plane is determined by the angle of the ring. Working in spherical
coordinates such that θ = 0 is the bottom for simplicity, we find that the Lagrangian is:
1 2 1 2
L =
IΩ + m ṙ + r2 θ̇2 + r2 sin2 θφ̇2 + mgr cos θ
(1)
2
2
1
1
=
M R2 φ̇2 + mR2 θ̇2 + sin2 θφ̇2 + mgR cos θ
(2)
4
2
φ is cyclic
1
2. Find the Hamiltonian of the system. (4 points)
Solution:
∂L
= (I + mR2 sin2 θ)φ̇
∂ φ̇
= mR2 θ̇
pφ =
(3)
pθ
(4)
where I = M R2 /2. The Hamiltonian is then:
H=
p2φ
p2θ
+
− mgR cos θ
2(I + mR2 sin2 θ) 2mR2
(5)
3. Check your answer to section 1 explicitly using Poisson brackets. (4 points)
Solution:
The equations of motion are given by
∂f
df
= [f, H] +
dt
∂t
If we want f to be a constant of motion it must obey
(6)
df
dt
= 0 giving us [f, H] = 0.
In our case this means [pφ , H] = 0 giving us
[pφ , H] =
∂H ∂pφ ∂H ∂pφ
∂H
=−
−
=0
∂pφ ∂φ
∂φ ∂pφ
∂φ
(7)
4. Find the Hamilton’s equations of motions for θ. (3 points)
Solution:
pθ
∂H
=
∂pθ
mR2
mR2 p2φ sin θ cos θ
∂H
= −
=
2 − mgR sin θ
∂θ
I + mR2 sin2 θ
θ̇ =
ṗθ
(8)
(9)
5. How many equilibrium points does the system have? Which are stable? (If you cannot
solve the equations, argue qualitatively). (12 points)
Solution:
The answer actually depends on the size of the conserved angular momentum. The
effective potential is:
Vef f =
p2φ
− mgR cos θ
2(I + mR2 sin2 θ)
(10)
such that
mR2 p2φ sin θ cos θ
∂Vef f
=−
2 + mgR sin θ
∂θ
I + mR2 sin2 θ
2
(11)
This vanishes for θ = 0, π. The second derivative gives:
mR2 p2φ
∂ 2 Vef f
|
=
−
± mgR
θ=0,π
∂θ2
I2
(12)
θ = π is unstable, while θ = 0 is stable for small pφ and unstable for large pφ , with the
critical value:
2 1/2
I g
c
pφ =
(13)
R
we note that for large pφ the two points θ = 0, π are both maximas. Hence there must
be a minimum for an intermediate value of θ inbetween.
6. Assuming a small angular velocity, determine the frequency of small oscillations around
equilibrium. (4 points)
Solution:
mgR p2φ
1 ∂ 2 Vef f
|θ=0 =
− 2
ω =
mR2 ∂θ2
mR2
I
2
(14)
Exercise 2: (30 points)
Approximate a (not very good) hockey player by a stationary solid disk of radius R. A
stream of hockey pucks of radius a approach it from the right and scatter elastically. The
hockey pucks are uniformly distributed over a length L. The total number of pucks is N 1.
A goal net is located a distance d away and has a width w. In the following you can assume
d R, d w and w a.
3
1. What is the relationship between the impact parameter b and the scattering angle θ of
the puck. (5 points)
Solution: for elastic collision the angle of incidence φ is equal to the angle of reflection
φ and χ = π − 2α. From geometry it follows that:
ρ
sin φ =
(15)
R+a
and therefore:
cos χ/2 = cos
π
2
− φ = sin φ =
ρ
R+a
(16)
2. Define a two dimensional scattering cross section so that NL σ(θ)dθ gives the number
of pucks scattered into an angle dθ. What is the differential cross-section σ(θ) in this
case? (5 points)
Solution: the number of particles in a range of impact parameters dρ that will scatter
into an angle dχ is given by:
N
N
dρ = σ(χ)dχ
L
L
(17)
dρ R + a χ σ(χ) = =
sin dχ
2
2
(18)
such that:
Note that this result is different from the 3D case.
3. Out of the N hockey pucks, how many will end up in the goal net? (10 points)
Solution:
Ngoal =
N
N R + a χ w
σ(χ)dχ =
sin L
L 2
2 d
(19)
4. Out of the N hockey pucks, how many will strike the player? (10 points)
Solution:
N
L
Z
π
N
σ(χ)dχ = (R + a)
L
−π
Z
π
sin
0
χ
N
= 2 (R + a)
2
L
(20)
The same result can be obtained from the fraction of particles with impact parameters
in the range −R − a < ρ < R + a
Exercise 3: (35 points)
A particle that has coordinate dependent mass M (q) = mq 2 is attached to a spring that has
spring constant k = mg. The particle’s Hamiltonian is:
p2
1
+ mgq 2
H=
2
2mq
2
4
1. Derive the Hamilton equations of motion for q and p. (5 points)
Solution:
∂H
p
=
∂p
mq 2
p2
∂H
= − 3 + mgq
−ṗ =
∂q
mq
q̇ =
(21)
2. Find the parameters α and β for which the coordinate transformation:
z = αq β
p
pz =
q
is a canonical transformation, and find its generating function (here α and β are parameters, that do not depend on q or p). (10 points)
Solution: We can show this in several ways. We can explicitly find the generating
function F (q, pz ):
∂F
∂q
∂F
∂pz
= p = qpz
(22)
1
= z = q 2 + g 0 (pz )
2
(23)
Integrating the first equation gives F = 21 pz q 2 + g(pz ). However, from the second equation we find that g 0 (pz ) = 0. Hence F (q, pz ) = 12 q 2 pz .
Another way is to show explicitly that Hamilton’s equations remain the same. For this
we write the Hamiltonian in terms of the new coordinates:
H0 =
p2z
+ mgz
2m
(24)
and verify that Hamilton’s equations are valid:
∂H 0
p
pz
=
=
∂pz
m
mq
∂H 0
= −
= −mg
∂z
q̇q = ż =
ṗ q̇p
− 2 = ṗz
q
q
(25)
(26)
which is satisfied from the equations of motion for q and p.
Last thing is to check that the preservation of poisson brackets:
[z, pz ] =
∂z ∂pz ∂z ∂pz
1
−
=q −0=1
∂q ∂p
∂p ∂q
q
(27)
3. Derive and solve the Hamilton equations of motion for z and pz . (5 points)
Solution: Hamilton’s equation are:
∂H 0
pz
=
∂pz
m
0
∂H
= −
= −mg
∂z
ż =
ṗz
5
(28)
(29)
The solution to the Hamilton equations is:
pz = pz0 − mgt
pz0
gt2
z = z0 +
t−
m
2
(30)
(31)
4. Let
P =
p2z
+ mgz.
2m
Find a coordinate Q such that Q and P make up a pair of canonical coordinates (Hint:
as a first step, find the generating function of the transformation). (12 points)
Solution: we look for a generating function of the form F (z, P ):
√
∂F
= pz = 2m(P − mgz)1/2
∂z √
2 2m
(P − mgz)3/2 + g(P )
F =−
3 mg
(32)
(33)
From here it follows that
√
∂F
2m
pz
Q=
=−
(P − mgz)1/2 + g 0 (P ) = −
+ g 0 (P )
∂P
mg
mg
(34)
where we can choose g 0 = 0. We may verify that the poisson brackets are conserved:
[Q, P ] = 1
(35)
5. Derive and solve the Hamilton equations of motion for Q and P . (3 points)
Solution
H = P
∂H
Q̇ =
=1
∂P
∂H
Ṗ = −
∂Q 0
with the solution Q = Q0 + t and P = P0
Useful formulas:
∂L
∂q
∂H
∂qk
=
d
dt
= −ṗk
∂L
∂ q̇
;
∂H
= q̇k
∂pk
6
(36)
(37)
(38)
df
dt
∂f
= [f, H] +
∂t
X ∂f ∂g
∂f ∂g
−
[g, f ] ≡
∂pi ∂qi ∂qi ∂pi
i
F1 (q, Q, t)
⇒
F2 (q, P, t)
⇒
F3 (p, Q, t)
⇒
∂F1
∂F1
∂F1
, P =−
, H0 = H +
∂q
∂Q
∂t
∂F2
∂F2
∂F2
, Q=
, H0 = H +
p=
∂q
∂P
∂t
∂F3
∂F3
∂F3
P =−
, q=
, H0 = H +
∂Q
∂p
∂t
p=
Good Luck!
7