Inverted Equilibrium of a Vertically Driven
Physical Pendulum
Woody Shew
Submitted to Dr. Don Jacobs
in partial fulfillment of Junior Independent Study
College of Wooster
4-24-97
Abstract: The suspension point of a physical pendulum with moment of inertia
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a
e
t
(l , 0,0)
10-2-1. A point charge was initially at the origin. Later after, it moves along the x-axis with a
constant acceleration of . Find the scalar and vector potentials at the moment of time and at
the position
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9-7-5. Using the definition of the electric dipole moment p and the continuity equation of electric
charge, prove that
dp(t )
= d 3 xj ( x , t )
dt
Solution:
Firstly
j ( x , t ) = j ( x , t ) (x ) = [ j ( x , t ) x ] x[
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5-3-2. There is a two-dimensional angular region with the boundaries being the two rays jointed
with an angle of / 3 . Suppose that the potential of the two boundaries have the potential 0 .
Find the equipotential surfac
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5-2-4. Under the external homogeneous electric field E0 , there is a homogeneous dielectric
sphere shell with dielectric constant , and the internal and outer radiuses of a and b
respectively. Find
(1) the electric field
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d
d > 11 + R2
R
R2
5-1-5. Two conduction cylinders have their axes to be parallel from each other, and are of radiuses
of and respectively. The distance between the two axes is with . The electric charges per unit
length
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12-1-1. An elliptic polarized plane electromagnetic wave
E = E1 cos t + E 2 cos t
with E1 E2 = 0 is scattered by a free electron. Ignoring the radiation damping force and the
relativistic effect, prove that the different
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10-4-1. An electron is moving with a velocity of v and an acceleration of a , and the angle
between the velocity and the acceleration is . Prove that in a plane spanned by v and a
,there is no radiation along the direct
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9-7-3. Suppose that there are two same electric dipoles oscillating with the same frequency and the
same amplitude p0 , the difference between the two oscillators is / 2 , and the angle between the
directions of the two
Chapter 2
Quantum mechanics in one
dimension
Following the rules of quantum mechanics, we have seen that the state of
a quantum particle, subject to a scalar potential V (r), is described by the
time-dependent Schrdinger equation,
o
i t (r, t) = H (r, t)
6.946 Driven Pendulum Project
Dennis V. Perepelitsa
18 November 2006
Gerald Jay Sussman, Jack Wisdom
Exercise 3.14: Periodically driven pendulum
We investigate several regions of parameter space of a driven pendulum system over which specic
phenomena occu
huwenhui 00.pdf
exercise 8.1 chain rule
a : 0 F (x, y ) = 2xy 3 ; 1 F (x, y ) = 3x2 y 2
b : 0 F (F, y ) = 4x3 y 9 ; 1 F (F, y ) = 9x4 y 8
4x3 y 9
9x4 y 8
c : 0 G(x, y ) and 1 G(x, y )
2xy 3
0
3x2 y 2
1
2ab3
d : DF (a, b) =
3a2 b2
750
0
DG(3, 5) =
675
1
Project 2.21:Rotation of Mercury
a:
Keplers rst law implies that the area enclosed by the orbit is ab
where a is the semi-major axis and b is the semi-minor axis of the ellipse.
Therefore the period
R(t)2 f = nab
(1)
b
= 1 e2
(2)
a
from equation (1) (2) w
Exercise 2.11:Kinetic energy of the top
the kinetic energy of rotation about the pivot is
1
m
2
(1)
where m is the constituent particle of the top is the vector from the
pivot to the small element
= center +
x
(2)
where center is the vector from the p
Exercise 3.1:Deriving Hamiltons equations
a:In this case the momentum p is
p = ml2
(1)
The Hamiltonian is
H (t, , p) =
p2
mglcos
2ml2
(2)
the Hamiltons equations are:
D(t) =
p
ml2
Dp(t) = mglsin
(3)
(4)
b:The Hamiltonian is
p2 + p2
x
y
+ (x2 + y 2 )/2 +
Exercise 3.8:Sleeping top :
from
p
p2
p2
H (t, , p ) =
+
+
tan2 + gM Rcos
2A 2C 2A
2
we can calculate that
(1)
p2
p2
+
tan2 + M gRcos H )2A
2C 2A
2
(2)
p =
and
dp
=
d
this is
(
p2 tan sec2 2AM gRsin
2
2
=0
2
2
p
p
tan2 M gRcos + H )2A
2 (
2C 2A
2
p2
cos4
Exercise 2.2: Steiners theorem :
Lets take the rst line as A the second line as B
If we separate the body into N parts then
N
i=1
2
mi li = I
where li is the distance from the small element i to A
And the distance Li = li + R
so
2
I = N mi L2 = N mi (li +
Exercise 1.26 :
Lagrangian equations for total time derivatives
if F=F(t,q);
F
F
+
q
G=
t
q
the Lagrangian equation f or G is :
2F
2F
2F
2F
+
+
q(
q) = 0
qt
q 2
tq
q 2
so the statement is proven
Exercise 1.27
Identifying total time derivatives
a:the i
problem 1.2
answer
(a): Every juggling pin has 6 degrees of freedom so 3 juggling pins have 18
degrees of freedom.
As for a juggling pin we can specify the conguration of it by giving the three
coordinates of the top point and the three coordinates of the
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6-5-6. In a superconductor where there is no resistivity the equation of motion of a conduction
electron with a mass of m and an electric charge of e is m& = eE . Assuming that the
x
number of conduction electrons per un
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6-9-1.
(1) Prove that the interaction energy of an electric dipole moment p with the external constant
electric field strength E is
W = p E
and derive from the above formula the force of the external electric field E act
C51
C51
unsigned char
0255
signed char
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unsigned int
065535
signed int
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unsigned long
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signed long
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float
1.175494E-383.402823E+38
*
13
bit
01
sfr
0255
sfr16
065535
sbit
01
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auto
break
case
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