Ein Körper der Masse m fällt entlang einer Kurve mit der Gleichung z = f(x) im Schwerefeld der Erde. Man bestimme (ohne Beachtung der Reibung) die Bewegungsgleichung des Körpers.
Es gilt:
z
=
f
(
x
)
{\displaystyle z=f(x)}
und wegen
z
˙
=
d
d
t
z
=
d
z
d
x
⋅
d
x
d
t
{\displaystyle {\dot {z}}={\frac {d}{dt}}z={\frac {dz}{dx}}\cdot {\frac {dx}{dt}}}
z
˙
=
f
′
(
x
)
⋅
x
˙
{\displaystyle {\dot {z}}=f'(x)\cdot {\dot {x}}}
1. Kinetische Energie:
T
=
1
2
m
⋅
(
(
x
˙
)
2
+
(
z
˙
)
2
)
{\displaystyle T={\frac {1}{2}}m\cdot (({\dot {x}})^{2}+({\dot {z}})^{2})}
T
=
1
2
m
⋅
(
(
x
˙
)
2
+
(
f
′
(
x
)
)
2
⋅
(
x
˙
)
2
)
{\displaystyle T={\frac {1}{2}}m\cdot (({\dot {x}})^{2}+(f'(x))^{2}\cdot ({\dot {x}})^{2})}
T
=
1
2
m
⋅
(
x
˙
)
2
[
1
+
(
f
′
(
x
)
)
2
]
{\displaystyle T={\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]}
Bezeichnet g die Fallbeschleunigung, dann gilt:
U
=
m
⋅
g
⋅
z
{\displaystyle U=m\cdot g\ \cdot z}
U
=
m
⋅
g
⋅
f
(
x
)
{\displaystyle U=m\cdot g\ \cdot f(x)}
L
=
T
−
U
{\displaystyle L=T-U}
L
=
1
2
m
⋅
(
x
˙
)
2
[
1
+
(
f
′
(
x
)
)
2
]
−
m
⋅
g
⋅
f
(
x
)
(
1
)
{\displaystyle L={\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]-m\cdot g\ \cdot f(x)\qquad (1)}
d
d
t
∂
L
∂
x
˙
=
∂
L
∂
x
{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {\partial L}{\partial x}}}
d
d
t
∂
L
∂
x
˙
{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}\,}
d
d
t
∂
L
∂
x
˙
=
d
d
t
∂
L
∂
x
˙
(
1
2
m
⋅
(
x
˙
)
2
[
1
+
(
f
′
(
x
)
)
2
]
)
(
E
i
n
s
e
t
z
e
n
a
u
s
(
1
)
u
n
d
∂
U
∂
x
˙
=
0
)
{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}({\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}])\qquad (Einsetzen\,aus\,(1)\,und\,{\frac {\partial U}{\partial {\dot {x}}}}=0)}
d
d
t
∂
L
∂
x
˙
=
d
d
t
(
m
⋅
(
x
˙
)
⋅
[
1
+
(
f
′
(
x
)
)
2
]
)
{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {d}{dt}}(m\cdot ({\dot {x}})\cdot [1+(f'(x))^{2}])}
d
d
t
∂
L
∂
x
˙
=
m
⋅
(
x
¨
)
⋅
[
1
+
(
f
′
(
x
)
)
2
]
+
2
⋅
m
⋅
(
x
˙
)
2
⋅
f
′
(
x
)
⋅
f
″
(
x
)
(
P
r
o
d
u
k
t
−
,
K
e
t
t
e
n
r
e
g
e
l
u
n
d
d
d
t
(
f
′
(
x
)
)
2
=
2
⋅
f
′
(
x
)
⋅
d
f
′
(
x
)
d
t
=
2
⋅
f
′
(
x
)
⋅
d
f
′
(
x
)
d
x
⋅
d
x
d
t
=
2
⋅
f
′
(
x
)
⋅
f
″
(
x
)
⋅
x
˙
)
{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}=m\cdot ({\ddot {x}})\cdot [1+(f'(x))^{2}]+2\cdot m\cdot ({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)\qquad \,(Produkt-,Kettenregel\,und\,{\frac {d}{dt}}(f'(x))^{2}=2\cdot f'(x)\cdot {\frac {df'(x)}{dt}}=2\cdot f'(x)\cdot {\frac {df'(x)}{dx}}\cdot {\frac {dx}{dt}}=2\cdot f'(x)\cdot f''(x)\cdot {\dot {x}})}
∂
L
∂
x
{\displaystyle {\frac {\partial L}{\partial x}}\,}
∂
L
∂
x
=
∂
∂
x
(
1
2
m
⋅
(
x
˙
)
2
[
1
+
(
f
′
(
x
)
)
2
]
−
m
⋅
g
⋅
f
(
x
)
)
{\displaystyle {\frac {\partial L}{\partial x}}={\frac {\partial }{\partial x}}({\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]-m\cdot g\ \cdot f(x))}
∂
L
∂
x
=
m
⋅
(
x
˙
2
)
⋅
f
′
(
x
)
⋅
f
″
(
x
)
−
m
⋅
g
⋅
f
′
(
x
)
{\displaystyle {\frac {\partial L}{\partial x}}=m\cdot ({\dot {x}}^{2})\cdot f'(x)\cdot f''(x)-m\cdot g\ \cdot f'(x)}
m
⋅
(
x
¨
)
⋅
[
1
+
(
f
′
(
x
)
)
2
]
+
2
⋅
m
⋅
(
x
˙
)
2
⋅
f
′
(
x
)
⋅
f
″
(
x
)
=
m
⋅
(
x
˙
2
)
⋅
f
′
(
x
)
⋅
f
″
(
x
)
−
m
⋅
g
⋅
f
′
(
x
)
{\displaystyle m\cdot ({\ddot {x}})\cdot [1+(f'(x))^{2}]+2\cdot m\cdot ({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)=m\cdot ({\dot {x}}^{2})\cdot f'(x)\cdot f''(x)-m\cdot g\ \cdot f'(x)}
(
x
¨
)
⋅
[
1
+
(
f
′
(
x
)
)
2
]
+
(
x
˙
)
2
⋅
f
′
(
x
)
⋅
f
″
(
x
)
+
g
⋅
f
′
(
x
)
=
0
(
D
i
v
i
s
i
o
n
d
u
r
c
h
m
u
n
d
Z
u
s
a
m
m
e
n
f
a
s
s
e
n
g
l
e
i
c
h
a
r
t
i
g
e
r
T
e
r
m
e
)
{\displaystyle ({\ddot {x}})\cdot [1+(f'(x))^{2}]+({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)+g\ \cdot f'(x)=0\qquad (Division\,durch\,m\,und\,Zusammenfassen\,gleichartiger\,Terme)}
x
¨
+
(
x
˙
)
2
⋅
f
′
(
x
)
⋅
f
″
(
x
)
1
+
(
f
′
(
x
)
)
2
+
g
⋅
f
′
(
x
)
1
+
(
f
′
(
x
)
)
2
=
0
(
D
i
v
i
s
i
o
n
d
u
r
c
h
1
+
(
f
′
(
x
)
)
2
)
{\displaystyle {\ddot {x}}+{\frac {({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)}{1+(f'(x))^{2}}}+{\frac {g\ \cdot f'(x)}{1+(f'(x))^{2}}}=0\qquad (Division\,durch\,1+(f'(x))^{2})}
x
¨
+
(
x
˙
)
2
2
⋅
d
d
x
(
l
n
(
1
+
f
′
(
x
)
2
)
+
g
⋅
f
′
(
x
)
1
+
f
′
(
x
)
2
=
0
{\displaystyle {\ddot {x}}+{\frac {({\dot {x}})^{2}}{2}}\cdot {\frac {d}{dx}}(ln(1+f'(x)^{2})+{\frac {g\cdot f'(x)}{1+f'(x)^{2}}}=0}
![{\displaystyle T={\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f525d316a74fb67f8cd7ae9ad21b7dbc2fb315e)
![{\displaystyle L={\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]-m\cdot g\ \cdot f(x)\qquad (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5d091318f55fb1f91566e8d0c5d97d27784f80f)
![{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}({\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}])\qquad (Einsetzen\,aus\,(1)\,und\,{\frac {\partial U}{\partial {\dot {x}}}}=0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00d82c383fe2ecd42216d3cc6d31c655163fe1a4)
![{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}={\frac {d}{dt}}(m\cdot ({\dot {x}})\cdot [1+(f'(x))^{2}])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b32d621fa1762fbf90391323bb4079f05875d68b)
![{\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {x}}}}=m\cdot ({\ddot {x}})\cdot [1+(f'(x))^{2}]+2\cdot m\cdot ({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)\qquad \,(Produkt-,Kettenregel\,und\,{\frac {d}{dt}}(f'(x))^{2}=2\cdot f'(x)\cdot {\frac {df'(x)}{dt}}=2\cdot f'(x)\cdot {\frac {df'(x)}{dx}}\cdot {\frac {dx}{dt}}=2\cdot f'(x)\cdot f''(x)\cdot {\dot {x}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87db8503a935ef820b574a84cefdde5d6413d684)
![{\displaystyle {\frac {\partial L}{\partial x}}={\frac {\partial }{\partial x}}({\frac {1}{2}}m\cdot ({\dot {x}})^{2}[1+(f'(x))^{2}]-m\cdot g\ \cdot f(x))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe3b3563b4b0a2fa611f75a792d7404a88daa8a0)
![{\displaystyle m\cdot ({\ddot {x}})\cdot [1+(f'(x))^{2}]+2\cdot m\cdot ({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)=m\cdot ({\dot {x}}^{2})\cdot f'(x)\cdot f''(x)-m\cdot g\ \cdot f'(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fccdf6b642f2cdc0ee4ea19303fd257d0783c939)
![{\displaystyle ({\ddot {x}})\cdot [1+(f'(x))^{2}]+({\dot {x}})^{2}\cdot f'(x)\cdot f''(x)+g\ \cdot f'(x)=0\qquad (Division\,durch\,m\,und\,Zusammenfassen\,gleichartiger\,Terme)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a83f66e1d9e776c32b6ba1529252ad8849e4277)
