Classical Mechanics Joel phần 1 pdf
Classical Mechanics Joel phần 1 pdf
... MoreGeneralMotion 10 0
4.4 Dynamics 10 7
4.4 .1 Euler’sEquations 10 7
4.4.2 Eulerangles 11 3
4.4.3 Thesymmetrictop 11 7
5 Small Oscillations 12 7
5 .1 Small oscillations about stable equilibrium 12 7
5 .1. 1 MolecularVibrations ... Externalandinternalforces 10
1. 3.2 Constraints 14
1. 3.3 Generalized Coordinates for Unconstrained Sys-
tems 17
1. 3.4 Kineticenergyingeneralizedco...
Classical Mechanics Joel phần 2 ppsx
... asym-
metry of the potential, and
the orbits become egg-shaped.
1
-1
x
p
1. 210 .8
0.6
0.40.2-0.2-0.4
0
0.3
0.2
0 .1
0
-0 .1
-0.2
-0.3
x
U
U(x)
Figure 1. 3. Motion in a cubic poten-
tial.
If the system has ... other than gravity. The Earth has a mass of 6.0 10
24
kg and
aradiusof6.4 × 10
6
m. Newton’s gravitational constant is 6.67 × 10
11
N ·
m
2
/kg
2
.
1. 2 In the discussion...
Classical Mechanics Joel phần 3 potx
... energy is
T =
1
2
m
1
˙r
2
1
+
1
2
m
2
˙r
2
2
=
1
2
m
1
˙
R −
m
2
M
˙
r
2
+
1
2
m
2
˙
R +
m
1
M
˙
r
2
=
1
2
(m
1
+ m
2
)
˙
R
2
+
1
2
m
1
m
2
M
˙
r
2
=
1
2
M
˙
R
2
+
1
2
µ
˙
r
2
,
where
µ ... invertible,
8
we also have ˙q = M
1
· P ,so
H = P
T
· ˙q −L
= P
T
· M
1
· P −
1
2
˙q
T
· M · ˙q −U(q)
= P
T
· M
1
· P −
1
2
P
T
· M
1
· M ·M...
Classical Mechanics Joel phần 4 ppt
... orthogonal and has determinant 1,
det
(A − 1I ) A
T
=det(1I−A
T
) = det(1I − A)
=det(A − 1I ) d e t ( A)=det(−(1I −A)) = ( 1)
3
det(1I − A)
= −det(1I −A),
so det(1I − A)=0and1I− A is a singular matrix. ... θ =0. Then
1
r
= A cos θ + B =
1
r
p
1 −
e
1+ e
(1 −cos θ)
=
1
r
p
1+ e cos θ
1+ e
where e = A/B.
What is this orbit? Clearly r
p
just sets the scale of the whole or...
Classical Mechanics Joel phần 5 potx
... I
1
= I
2
,sowehave
I
1
˙ω
1
=(I
1
−I
3
)ω
2
ω
3
,
I
1
˙ω
2
=(I
3
−I
1
)ω
1
ω
3
,
I
3
˙ω
3
=(I
1
−I
2
)ω
1
ω
2
=0.
We see that ω
3
is a constant. Let Ω = ω
3
(I
3
−I
1
)/I
1
.Thenweseethat
˙ω
1
= ... the gravitational field
15
.
T =
1
2
(ω
2
1
+ ω
2
2
)I
1
+
1
2
ω
2
3
I
3
=
1
2
˙
φ
2
sin
2
θ +
˙
θ
2
I
1
+
1
2
˙
φ cos θ +
˙
ψ
2
I
3
, (4.34)
U = Mg...
Classical Mechanics Joel phần 6 docx
... ˙η
=
1
2
˙η
T
·
O
1
1
·m ·O
1
· ˙η
=
1
2
˙η
T
·O
T
1
·m · (O
1
· ˙η)
=
1
2
(O
1
· ˙η)
T
· m · (O
1
· ˙η)
=
1
2
˙x
T
· m · ˙x.
Similarly the potential energy becomes U =
1
2
x
T
·O
1
· ... DYNAMICS 13 9
so
U(y
1
, ,y
i
, ,y
n
)
=
y
i
0
dy
i
τ
a
(2y
i
− y
i +1
−y
i 1
)+F (y
1
, ,y
i 1
,y
i +1
, ,y
n
)
=
τ
a
y
2
i
− (y
i +1
+ y
i 1
)y
i
+ F (y
1...
Classical Mechanics Joel phần 7 doc
... 2-forms
at a point, dx
1
∧ dx
2
, dx
1
∧ dx
3
,anddx
2
∧ dx
3
,wheredx
1
∧ dx
2
=
dx
1
⊗ dx
2
− dx
2
⊗ dx
1
, which means that, acting on u and v, dx
1
∧
dx
2
(u,v)=u
1
v
2
− u
2
v
1
. The product ... <i
k
ω
i
1
i
k
dx
i
1
∧···∧dx
i
k
over S by
S
ω
(k)
=
i
1
,i
2
, ,i
k
ω
i
1
i
k
(x(u))
k
=1
∂x
i
∂u
du
1
du
2
···du
k
.
Wehadbettergivesomeexa...
Classical Mechanics Joel phần 8 ppt
... this difference, d(ω
1
− ω
1
)=ω
2
− ω
2
=0,soω
1
− ω
1
is an closed 1-
form. Thus it is exact
11
, and there must be a function F on phase space
such that ω
1
− ω
1
= dF .WecallF the ... =tan
1
Q/P as
the new coordinate, and we might hope to have the radial coordinate
related to the new momentum, P = −∂F
1
/∂θ.AsP = ∂F
1
/∂Q is also
Q cot θ,wecantakeF
1
=
1
2...
Classical Mechanics Joel phần 9 doc
... symmetry point,
q = 0, to the next such crossing. The
-1
-0.5
0
0.5
1
-2 -1. 5 -1 -0.5 0.5 1 1.5
q
p
-1
-0.5
0
0.5
1
-2 -1. 5 -1 -0.5 0.5 1 1.5
q
p
Fig. 4. The trajectory in phase
space of the system ... a
circle.
0
10
20
30
-1. 5
-1
-0.5
0
0.5
1
1.5
-1. 5
-1
-0.5
0
0.5
1
1.5
Fig. 7. The surface Σ
1
for a harmonic
oscillator with a spring constant which
varies, for...