Яковенко Г.Н. Краткий курс теоретической механики
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m(t) = m
0
− m (t) + m (t), (23.1)
m
0
t
0
m (t) m (t)
[t
0
, t]
t
Q
∗
(t) t + ∆t
Q(t + ∆t) = Q
∗
(t) + ∆Q
∗
− ∆Q + ∆Q ,
∆Q
∗
t
∆Q ∆Q
dQ(t)
dt
= lim
∆t→0
Q(t + ∆t) − Q(t)
∆t
=
= lim
∆t→0
Q
∗
(t) + ∆Q
∗
− ∆Q + ∆Q −Q
∗
(t)
∆t
=
= lim
∆t→0
µ
∆Q
∗
∆t
−
∆Q
∆t
+
∆Q
∆t
¶
dQ
dt
= R + R (23.2)
˙
Q
∗
= R t
R
R = lim
∆t→0
µ
−
∆Q
∆t
+
∆Q
∆t
¶
(23.3)
dK
O
dt
= M
O
+ M
O
, (23.4)
M
O
M
O
M
O
= lim
∆t→0
µ
−
∆K
O
∆t
+
∆K
O
∆t
¶
. (23.5)
dT
dt
= N + N , (23.6)
N =
P
(F
i
, V
i
)
N
N = lim
∆t→0
µ
−
∆T
∆t
+
∆T
∆t
¶
. (23.7)
V(t)
n r
Q
m(t) = m
0
−
n
X
i=1
m
i
(t) +
r
X
l=1
m
l
(t),
Q =
(
m
0
−
n
X
i=1
m
i
(t) +
r
X
l=1
m
l
(t)
)
V,
dQ
dt
=
d(mV)
dt
=
dm
dt
V + m
dV
dt
=
=
Ã
−
n
X
i=1
dm
i
dt
+
r
X
l=1
dm
l
dt
!
V + m
dV
dt
.
(24.1)
∆t ∆m
i
∆m
i
∆Q =
n
X
i=1
∆m
i
C
i
, ∆Q =
r
X
l=1
∆m
l
C
l
, (24.2)
C
l
C
l
∆t
R = lim
∆t→0
µ
−
∆Q
∆t
+
∆Q
∆t
¶
= −
n
X
i=1
dm
i
dt
C
i
+
r
X
l=1
dm
l
dt
C
l
. (24.3)
m
dV
dt
= R −
n
X
i=1
dm
i
dt
u
i
+
r
X
l=1
dm
l
dt
u
l
, (24.4)
u
i
= C
i
−V u
i
= C
i
−V
m(t) = m
0
− m (t) ˙m(t) = − ˙m (t)
m
dV
dt
= −mg −
dm
dt
u. (24.5)
V (t) = V
0
− gt + u ln
m
0
m(t)
. (24.6)
g = 0
t m(t)
h =
R
V (t)dt
m(t) = m
0
− m (t)
∆t ∆m V − u
∆m = −∆m
N =
1
2
dm
dt
(V − u)
2
. (24.7)
dT
dt
=
d
dt
µ
mV
2
2
¶
= −mgV +
1
2
dm
dt
(V
2
− 2V u),
N + N + N ,
N = −mgV
N = −
dm
dt
u
2
2
.
z
I
z
(t)
dω
dt
= M
z
−
n
X
i=1
dm
i
dt
M
z
{u
i
} +
r
X
l=1
dm
l
dt
M
z
{u
l
}, (25.1)
I
z
(t) M
z
M
z
{·}
z
u
u = u
R
¡
I
0
− m (t)R
2
¢
dω
dt
=
dm
dt
Ru, (25.2)
I
0
t = 0 m (t) =
n
P
i=1
m
i
(t)
[0, t] M
z
{u
i
} = −Ru
ω(t) = ω
0
+
u
R
ln
I
0
I
0
− m (t)R
2
.
a
t
m(t) =
1
2
m
0
µ
1 + cos
πt
t
1
¶
, (25.3)
m
0
t
1
[0, t
1
]
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- - - - - -
a
O
x
x
C
=
Σm
i
x
i
M + m
0
Σm
i
x
i
= {M + m(t)}X(t) +
Z
t
0
x(τ)dm
}
(τ) = C
1
t + C
2
. (25.4)
X(t) m(t)
C
1
= V
0
/(M + m
0
) C
2
= X
0
/(M + m
0
)
dm
}
(τ)
τ ∈ [0, t] τ
X(τ) − a
˙
X(τ)
t
x(τ) = X(τ) − a +
˙
X(τ)(t − τ). (25.5)
dm (τ) τ
dm (τ) = ˙m (τ)dτ = − ˙m(τ )dτ, (25.6)
m(t) = m
0
− m (t) ˙m(t) = − ˙m (t)
X(t)
{M + m(t)}X(t) −
Z
t
0
n
X(τ) − a +
˙
X(τ)(t − τ)
o
˙m(τ )dτ = C
1
t + C
2
,
t
¨
X = −
a ¨m(t)
M + m(t)
.
m(t)
˙
X(0) = 0
V (t) =
˙
X(t) = a
π
t
1
(
π
t
1
t −
2M + m
0
p
M(M + m
0
)
arctg
Ã
r
M
M + m
0
tg
πt
2t
1
!)
.
t ∈ [0, t
1
] t
1
V = V (t
1
) = a
π
2
t
1
(
1 −
2M + m
0
2
p
M(M + m
0
)
)
.
M m
0
t V (t) > 0
˙
X(0) = 0
¨
X(0) > 0 t > t
1
V (t) < 0
t
O
O
r
0
(t) e
1
(t) e
2
(t) e
3
(t)
O
e
1
(t) e
2
(t) e
3
(t)
e
k
(t) i
1
i
2
i
3
e
k
=
3
X
l=1
a
k l
i
l
. (26.1)
A = ka
k l
k
a
k l
= cos
³
d
e
k
, i
l
´
δ
kj
= (e
k
, e
j
) =
3
X
l, s=1
a
k l
a
j s
(i
l
, i
s
) =
3
X
l, s=1
a
k l
a
j s
δ
ls
=
3
X
l=1
a
k l
a
j l
3
X
l=1
a
k l
a
j l
= δ
kj
AA
T
= E, (26.2)
A = ka
k l
k
A A
T
A
−
1
= A
T
i
l
i
l
=
3
X
j=1
a
j l
e
j
. (26.3)
e
j
(t) A(t) =
ka
k l
(t)k t t
3
X
l=1
˙a
k l
a
j l
= −
3
X
l=1
a
k l
˙a
j l
˙
AA
T
= −A
˙
A
T
. (26.4)
t
i
l
e
j
˙
e
k
=
3
X
l=1
˙a
k l
i
l
=
3
X
l=1
˙a
k l
a
j l
e
j
=
3
X
l=1
b
k j
e
j
. (26.5)
B = k b
k j
k = k˙a
k l
a
j l
k =
˙
AA
T
B
B
T
= (
˙
AA
T
)
T
= A
˙
A
T
(26.4)
= −
˙
AA
T
= −B,
B =
0 −ω
3
ω
2
ω
3
0 −ω
1
−ω
2
ω
1
0
.
ω
j
B ωωω
ωω
=
3
P
l=1
ω
j
e
j
˙
e
k
= [ωωω
ωω
, e
k
]
ωωω
ωω
a
k l
A = ka
k l
k
3
X
l=1
a
2
1 l
= 1,
3
X
l=1
a
2
2 l
= 1,
3
X
l=1
a
2
3 l
= 1,
3
X
l=1
a
1 l
a
2 l
= 0,
3
X
l=1
a
1 l
a
3 l
= 0,
3
X
l=1
a
2 l
a
3 l
= 0.
a
k l
i
1
i
2
i
3
e
1
e
2
e
3
O e
2
n n⊥i
3
n⊥e
3
θ =
[
i
3
, e
3
ψ =
d
i
1
, n
ϕ = [n, e
1
θ ψ ϕ
i
3
e
3
θ = 0 θ = π
ψ ϕ
q
y
j
i
1
i
2
i
3
n
e
3
e
1
O
y+j
i
1
i
2
i
3
e
3
e
1
*
*
O
θ ψ ϕ
i
3
ψ + ϕ
n θ
¤ e
∗
1
e
∗
3
ψ n
¥
ωωω
ωω
1
=
˙
ψi
3
ωωω
ωω
2
=
˙
θn