Манаков Н.Л., Некипелов А.А., Овсянников В.Д. Задачи по теоретической механике
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ϕ
ϕ(t
0
) = 0
t − t
0
=
s
l
2g
Z
ϕ
0
dϕ
√
cos ϕ − cos Φ
0
.
τ ϕ
−Φ
0
Φ
0
Φ
0
τ = 4
s
l
2g
Z
Φ
0
0
dϕ
√
cos ϕ − cos Φ
0
.
cos ϕ = 1 − 2 sin
2
(ϕ/2),
τ = 2
s
l
g
Z
Φ
0
0
dϕ
p
sin
2
(Φ
0
/2) − sin
2
(ϕ/2)
.
ξ
sin(ϕ/2) = sin(Φ
0
/2) sin ξ,
τ = 4
s
l
g
Z
π/2
0
dξ
p
1 − k
2
sin
2
ξ
,
k = sin(Φ
0
/2)
K(k) =
Z
π/2
0
dξ
p
1 − k
2
sin
2
ξ
τ = 4
s
l
g
K
µ
sin
Φ
0
2
¶
.
Φ
0
¿ 1, K(k)
k
2
sin
2
ξ
τ = 4
s
l
g
Z
π/2
0
µ
1 +
k
2
2
sin
2
ξ + . . .
¶
dξ = 2π
s
l
g
·
1 +
Φ
2
0
16
+ . . .
¸
.
m
1
, m
2
e
1
, e
2
E
m
1
¨
r
1
= e
1
E + F
12
m
2
¨
r
2
= e
2
E + F
21.
M = m
1
+m
2
r
c
= (m
1
r
1
+ m
2
r
2
)/(m
1
+ m
2
),
F
12
= −F
21
M
¨
r
c
= eE,
e = e
1
+ e
2
e
r
c
(t) =
et
2
2M
E +
˙
r
c0
t + r
c0
.
m v
1
x < 0
U
1
x > 0
U
2
-
x
6
y
*
ϑ
1
v
1
ϑ
2
v
2
mv
2
1
2
+ U
1
=
mv
2
2
2
+ U
2
v
2
= v
1
s
1 −
2∆U
mv
2
1
,
∆U = U
2
− U
1
.
v
1
sin ϑ
1
= v
2
sin ϑ
2
.
v
1
v
2
sin ϑ
1
sin ϑ
2
=
s
1 −
2∆U
mv
2
1
.
U
2
> U
1
∆U > 0 ϑ
2
> ϑ
1
sin ϑ
2
6 1
sin ϑ
1
/
s
1 −
2∆U
mv
2
1
> 1
U
2
U
1
N
U(r
1
, . . . , r
N
)
T =
1
2
V,
V =
N
X
k=1
µ
r
k
·
∂U
∂r
k
¶
f ≡ lim
τ→∞
1
τ
Z
τ
0
f(t) dt f(t).
T =
1
2
N
X
k=1
(
˙
r
k
· p
k
) =
1
2
d
dt
N
X
k=1
(r
k
· p
k
) −
1
2
N
X
k=1
(r
k
·
˙
p
k
).
˙
p
k
= −
∂U
∂r
k
,
T =
1
2
d
dt
N
X
k=1
(r
k
· p
k
) +
1
2
N
X
k=1
(r
k
·
∂U
∂r
k
) =
1
2
d
dt
N
X
k=1
(r
k
· p
k
) +
1
2
V.
T = lim
τ→∞
1
τ
Z
τ
0
(
1
2
d
dt
N
X
k=1
(r
k
· p
k
) +
1
2
V
)
dt =
=
1
2
lim
τ→∞
1
τ
(
N
X
k=1
(r
k
· p
k
)|
t=τ
−
N
X
k=1
(r
k
· p
k
)|
t=0
)
+
1
2
V.
τ
τ → ∞
l
ψ(r
1
, . . . , r
N
, v
1
, . . . , v
N
, t) = 0.
ψ
∂ψ
∂t
= 0
ψ
v
i
K K < 3N
K
ψ
α
(r
1
, . . . , r
N
, t) = 0, α = 1, 2, . . . , K.
F
i
R
i
R
i
R
i
=
K
X
α=1
R
αi
,
R
αi
i α
R
αi
ψ
a
r
i
R
αi
= λ
α
∇
i
ψ
α
(r
1
, . . . , r
N
, t) ≡ λ
α
∂ψ
α
∂r
i
.
λ
α
R
i
=
K
X
α=1
λ
α
∇
i
ψ
α
(r
1
, . . . , r
N
, t), i = 1, 2, . . . , N.
R
i
m
i
¨
r
i
= F
i
+
K
X
α=1
λ
α
∇
i
ψ
α
, i = 1, 2, . . . , N.
r
i
(t) K λ
α
3N + K
6N
r
i
(0) = r
0i
, v
i
(0) = v
0i
,
r
i
(t) λ
α
§ §§ §§ §§
α
-
x
6
z
½
½
½
½
½
½
½
sm
?
mg
α
z = x tg α
ψ(x, z) = x tg α − z = 0.
m
¨
r = mg + λ∇ψ
m¨x = λ tg α; m¨y = 0; m¨z = −mg − λ.
y
(
t
)
y(t) = y
0
+ ˙y
0
t.
z x x
λ = −mg cos
2
α.
R
x
= −
mg
2
sin 2α, R
y
= 0, R
z
= mg cos
2
α.
x(t) = −
gt
2
4
sin 2α + ˙x
0
t + x
0
;
z(t) = x(t) tg α = −
gt
2
2
sin
2
α + ( ˙x
0
t + x
0
) tg α .
y
2
= ax
y(0) = y
0
m
¨
r = mg + λ∇ψ,
ψ(x, y) = y
2
− ax = 0.
-
x
6
y
y
0
y
c
0
s
s
m¨x = −λa,
m¨y = −mg + 2λy.
x ¨y
m ˙y
2
+ y(2λy − mg) = −λa
2
/2.
y = y
c
λ = 0
gy
c
− ˙y
2
= 0.
˙y
m
2
( ˙x
2
+ ˙y
2
) + mgy = mgy
0
.
˙x = 2y ˙y/a
˙y
2
=
2g(y
0
− y)
1 + 4y
2
/a
2
.
y
c
y
3
c
+
3
4
a
2
y
c
−
a
2
2
y
0
= 0.
y = y(x)
x(0) = a v(0) = 0 τ
b
τ =
Z
b
a
s
1 + (dy/dx)
2
2g[y(a) − y(x)]
dx.
r
l > 2r
R
2
(R
1
+ R
2
+ mg) x z
-
x
6
z
O
ϕ
¶
¶
¶
¶
¶
¶
¶
¶
¶
¶7
A
?
q
mg
R
1
A
A
A
AK
R
2
O
0
q
B
O
0
O
0
B = l
Ã
7
8
−
r
1
64
+
2r
2
l
2
!
.
l
a < l
α β
ρ s
s =
(l − a) sin β
sin α + sin β
.
∆h = 0
¤
¤
¤
¤
¤
¤
¤
¤
β
J
J
J
J
J
J
J
J
d
d
d
d
d
d
d
d
d
s
ddddddddddddddddd
d
d
d
d
d
d
d
d
d
d
d
α β
m
a
v
0
α
R =
mv
2
0
cos
2
α
a
n n
a R
z
c
R = −
r
a
2
{3m(g · r) + 2E
0
}; z
c
=
2E
0
3mg
.
E
0
a b c F = −κr
κ = const
R = 2λ(
x
a
2
i +
y
b
2
j +
z
c
2
k); λ =
κ − m( ˙x
2
/a
2
+ ˙y
2
/b
2
+ ˙z
2
/c
2
)
2(x
2
/a
4
+ y
2
/b
4
+ z
2
/c
4
)
.
m
ω a z = a sin(ωt)
R = m(g − aω
2
sin ωt)k
aω
2
> g
E(t) = m( ˙x
2
0
+ ˙y
2
0
)/2 + ma sin ωt
£
g − (aω
2
/2) sin ωt
¤
l
v
0
z
0
R
z
v
min
v
max
R
max
= 10mg
R = m
£
v
2
0
+ g(2z
0
− 3z)
¤
/l, −l < z < l
v
min
=
p
g(3l − 2z
0
);
v
max
=
p
g(7l − 2z
0
).
ϑ
0
= const v
R ϑ
0
v(ϑ
0
) =
p
gl/ cos ϑ
0
sin ϑ
0
; R(ϑ
0
) = mg/ cos ϑ
0
.
r
i
R
i
R
i
q
i
, (i = 1, 2, . . . , s)
s
N K
s = 3N − K.
q
i
q = q
i
(t), i = 1, 2, . . . , s.
q
i
s
r
i
x
i
y
i
, z
i
q
i
t
r
i
= r
i
(q
1
, q
2
, . . . , q
s
), i = 1, 2, . . . , N.
0 ≡ 0
s = 3N
t
x
i
, y
i
, z
i
q
i
d
dt
∂L
∂ ˙q
i
−
∂L
∂q
i
= 0, i = 1, 2, . . . , s,
L
L = T − U,