Goldstein 9-39a (3rd ed. 9.39a)
(1) USING CARTESIAN COORDINATES
Using Cartesian canonical variables, the Hamiltonian for the Kepler problem
is
H=
k
p2
,
2m r
(1)
where
p2 = p2 + p2 + p2 ,
x
y
z
(2)
and
r=
p
x2 + y2 + z 2 .
(3)
mk
r.
r
(4)
The Laplace-Rung
Poisson Brackets for the Heavy Top
The rst thing to do is dene the canonical variables, i.e., the generalized
coordinates and momenta for the problem. The Euler angles , , and are
the generalized coordinates. The corresponding canonical momenta are found
Goldstein 9-37 (3rd ed. 9.37)
Let r be the position vector of the bob of mass m. In spherical coordinates
the components of r are
x = r sin cos ,
(1)
y = r sin sin ,
(2)
z = r cos ,
(3)
where r = |r| =constant, is the polar angle between r and the z axis,
Goldstein 9-33 (3rd ed. 9.35) plus an additional part (b)
(a) The Hamiltonian for this one dimensional system is
H=
p2
mk
+2.
2m
x
(1)
We want to nd y (t), where y is the quantity x2 , using the Poisson bracket
operator formalism. The formal solution is
y
Goldstein 9-28 (3rd ed. 9.30)
(a) Let H be a Hamiltonian for some dynamical system. If A and B are two
constants of the motion that explicitly depend on time, then it must be true
that
A
= [A, H ] = [H, A] ,
t
(1)
B
= [B, H ] = [H, B ] .
t
(2)
and
We want
(5)(a) The kinetic energy is the sum of Tcart and Tm, where Tcart
is the sum of the carts translational, i.e., center of mass
kinetic energy and the rotational kinetic energy of the wheels
(Trel) and Tm is the kinetic energy of the pendulum bob. In the
co
Goldstein 8-15 (3rd ed. 8.25)
This problem diers from 8-14 because the cylinder is now driven to rotate
at a constant angular speed , and its kinetic energy is no longer variable. Since
the cylinders angular motion is prescribed, we only need to consider
Goldstein 8-14 (3rd ed. 8.24)
This view is looking down on the cylinder from above. The z axis is coming up out of the page
through the center of the circular cylinder. The x axis is fixed in space. The x& axis is fixed in the
cylinder. The angle measures
GPS Problem 8.20 (2nd ed. 8.10) Modied Planar Double Pendulum
This problem is a variation of the planar double pendulum problem, illustratrated in Fig. 1.4. The solution to that problem, GPS 1.22 or 2nd ed. 1.20, will
be useful here. There are two major m
Goldstein 8-21 (3rd ed. 8.3)
We want to produce a o
n new Hamiltonian, or Gamiltonian G(q, p, t), in
which the quantities q, p are the natural independent variables. We can do
this with the following Legendre transform of the Lagrangian L(q, q, t),
G(q, p
Problem 8.1
(a) We can dene the Lagrangian L(qj , q j , t) as the Legendre transform of
the Hamiltonian H (qj , pj , t),
L=
X
j
pj q j H ,
(1)
where qj , qj , and pj are the j th generalized coordinate, velocity, and momentum,
respectively. To nd the EOM
Problem Goldstein 5-23 (3rd ed 5.10)
We will designate the body frame unit vectors as i, j, and k, and assume
they lie along the principal axes of the body. The space frame unit vectors
will be designated as x, y, and z. In order to use Eulers equations o
Heavy Top Angular Momentum Problem
(a)Long Way
Designate the body frame unit vectors as i, j, and k, and assume they lie
along the principal axes of the body. The space frame unit vectors will be
designated as x, y, and z. Since it is easiest to write L i
Problem Goldstein 5-24 (3 ed 5.11)
We will designate the body frame unit vectors as i, j, and k, and assume
they lie along the principal axes of the body. In this body frame the angular
momentum L is expressed as
L = 1 1 i + 2 2 j + 3 3 k
(1)
and the mag
Problem Kinetic Energy of a Suspended Sphere
(a) Because of the spheres symmetry, all o-diagonal components are zero,
and we have
0
0
0
Ixx = Iyy = Izz .
(1)
0
The easisest one to compute is Izz .
0
Izz
ZZZ
3M
=
(x + y )dxdydz =
4a3
5
3M
a4
=
2
.
3
4a
53
GPS 5-19 or Goldstein 5-13
This problem is distantly related to the Spring-Pendulum Problem from Ch.
1, but here, the spring is constrained to oscillate vertically and the mass point
is expanded into a rigid rod of mass and length 2 that swings in a verti
Problem: Inertia Tensor and Kinetic Energy of Two Cones Joined
at Their Tips
The two cones are joined at their tips. Each cone is an oblate ellipsoid with
1 = 2 3 . One cone is oriented with its principal symmetry axis along the
axis of a body frame, whi
Problem: Kinetic Energy of a Rod
(a) First we need to nd the principal moments of the rod. This can be
done with the origin either at the end of the rod (at the pivot) or at the center of
the rod (the COM). There is a conceptual advantage with the latter
Problem: Space and Body Cones for the torque free top
(a) One way to do this is as follows. Designate the body frame unit vectors
as i, j, and k, and assume they lie along the principal axes of the body. In the
body principal axis frame we can write the a
Problem Goldstein 5-19 (3rd ed 5.6)
(a) We will designate the body frame unit vectors as i, j, and k, and assume
they lie along the principal axes of the body. Now we want to show that the
angular momentum vector L rotates about the body symmetry axis k w
HW9 Problem 4(b)
Start with the fundamental equation governing the rate of change of angular
momentum from Chapter 1 or Chapter 5
L
= N
Next, take the dot product of Eq.(1) with L to get
L
= L N
Now expand the left dot product of Eq.(2) as
L
L
1 2
L
1 2
=
Problem GPS 5.3 (2nd ed. 5-7)
We have a collection of mass points rotating about a xed point, and the
positions ri and velocities vi of these points are measured in a space xed,
inertial frame whose origin is at the xed point. Start with the denition of T
Dropped Particle with Rebound Problem
(a) If we neglect the very small centrifugal force term in GPS Eq.(4.91),
the equation of motion for a particle of mass m in the rotating frame of the
earth is
dv
m
= mg + 2mv ,
(1)
dt
where v is the particle velocity
Problem GPS 4.22 (2nd ed. 4.23)
If we neglect the very small centrifugal force term in Eq.(4-129) [3rd ed.
Eq.(4.91)], the equation of motion for a particle of mass m in the rotating frame
of the earth is
dv
= mg + 2mv ,
(1)
m
dt
where v is the particle v
Problem GPS 4.21 (2nd ed. 4.22)
If we neglect the centrifugal force term in GPS Eq.(4.91), the equation of
motion for a particle of mass m in the rotating frame of the earth is
m
dv
= mg + 2mv ,
dt
(1)
where v is the particle velocity, is the earths angul
Problem GPS 4.20 (2nd ed. 4-9)
In general, we would need six generalized coordinates to describe the motion
of the sphere. Three of these would be for the center of mass (X , Y , Z ) and three
for the orientation of a body xed frame rotating with the sphe
Problem Goldstein 4-19 (3rd ed. 4.15)
The most direct way to do this problem is to take the body frame components
0 derived in class or in the text and convert them into space frame components
with the the help of the transposed Euler matrix
= AT 0 .
(1
Problem Goldstein 4-2 (3rd ed. 4.2)
Transposition
For two matrices, A and B, we want to prove
(AB)T = BT AT ,
(1)
where superscript T indicates a matrix transpose. Consider the ji element of
the left-hand-side
T
(AB)ji = (AB)ij .
(2)
By the rules of matri
Problem Goldstein 4-1 (3rd ed. 4.1)
Associativity
Say we have three matrices, P, Q, R, whose dimensions are such that the
matrix multiplications
PQ = S ,
(1)
QR = T ,
(2)
and
are dened. Then, to prove associativity, we need to show that
SR = PT .
(3)
Cons
Problem: Orbits with a 1-2 central potential
The equation of motion for the one-dimensional radial problem is
= () =
2
+ 3
(1)
where is the constant angular momentum, 2 , and is the reduced mass of
the two particle system. The 1-2 potential is dened as