Сягло И.С. Теоретическая механика
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A ρ
A
V (t)
∆t
∆A = ∆t
Z
V (t)
∂ρ
A
∂t
dV + ∆ t
I
S
ρ
A
~v~n dS.
∆t ∆t
A
dA
dt
=
Z
V (t)
∂ρ
A
∂t
dV +
I
S
ρ
A
~v~n dS.
I
S
ρ
A
~v~n dS =
Z
V
∇(ρ
A
~v) dV.
dA
dt
=
Z
V (t)
(
∂ρ
A
∂t
+ ∇(ρ
A
~v)) dV =
Z
V (t)
(
dρ
A
dt
+ ρ
A
∇~v) dV.
ρ
A
= 1 A
dV
dt
=
Z
V (t)
∇~v dV.
∇~v = 0
A = m
m =
∂ρ
∂t
+ ∇(ρ~v) = 0,
dρ
dt
+ ρ∇~v = 0.
d~p
dt
=
d
dt
Z
V (t)
ρ~v dV =
Z
V (t)
(ρ
d~v
dt
+ ~v(
dρ
dt
+ ρ∇~v)) dV.
d~p
dt
=
Z
V (t)
ρ
d~v
dt
dV.
d
~
M
dt
=
Z
V (t)
ρ[~r ×
d~v
dt
] dV.
0
~a 0
~u
o
~u 0
~a
~a
0
= ~a + ~u −~u
o
.
0
~a
~u |~a|
~u
~u(~a) = ~u(0) +
∂~u
∂x
a
x
+
∂~u
∂y
a
y
+
∂~u
∂z
a
z
= ~u(0) +
3
X
j=1
∂~u
∂x
j
a
j
.
0
º
-
-
K
0
0
0
~u
o
~u
~a
~a
0
~a
0
= ~a +
3
X
j=1
∂~u
∂x
j
a
j
.
i
3
X
j=1
∂u
i
∂x
j
a
j
=
1
2
X
j
(
∂u
i
∂x
j
+
∂u
j
∂x
i
)a
j
+
1
2
X
j
(
∂u
i
∂x
j
−
∂u
j
∂x
i
)a
j
ε
ij
~χ
ε
ij
=
1
2
(
∂u
i
∂x
j
+
∂u
j
∂x
i
), ~χ =
1
2
[∇ ×~u].
~χ
~a
0
i
a
0
i
= a
i
+
3
X
j=1
ε
ij
a
j
+ [~χ ×~a]
i
.
~r
~a χ
~χ ~χ
ε
ij
~
b
0
~a
0
~
b
0
~a
0
~
b
0
= ~a
~
b + 2
X
i,j
ε
ij
a
i
b
j
.
~a
~
b
ox
l
l
02
= l
2
+ 2ε
11
l
2
, l
0
≈ l(1 + ε
11
),
l
0
− l
l
= ε
11
.
~a
~
b
ox oy ~a
~
b = 0
~a
0
~
b
0
= 2ε
12
.
cos(
π
2
− ϕ)
ϕ
ϕ
ε
12
=
1
2
~a
0
~
b
0
= a
0
b
0
cos(
π
2
− ϕ) ≈
ϕ
2
.
∆t ∆~u = ~v ∆t
~v
∆ε
ij
=
1
2
(
∂v
i
∂x
j
+
∂v
j
∂x
i
) ∆t, ∆~χ =
1
2
[∇ ×~v] ∆t.
∆t ∆t
d
ij
= lim
∆t→0
∆ε
ij
∆t
=
1
2
(
∂v
i
∂x
j
+
∂v
j
∂x
i
), ~ω = lim
∆t→0
∆~χ
∆t
=
1
2
[∇ ×~v].
~v ∆t ~u
~v(~a) = ~v(0) +
3
X
j=1
∂~v
∂x
j
a
j
,
v
i
= v
oi
+
3
X
j=1
d
ij
a
j
+ [~ω ×~a]
i
.
D
i
=
X
j
d
ij
a
j
.
~v = ~v
o
+ ~ω ×~a +
~
D.
~v
o
~ω ×~a
~
D
~
Φ =
∆
~
F
∆m
=
∆
~
F
ρ∆V
,
~
f =
∆
~
F
∆S
.