Lösung 7 - TUM

M l
m v0
hmax θmax
v0
hmax θmax vo = 36 m
s m = 22 g M = 350 g l = 1 m
FR = − (m + M) g sin (θ) = l (m + M) ¨ θ s (t) = lθ ˙s (t) = v (t) = l ˙ θ
¨s (t) = a (t) = l ¨ θ vApfel (t = 0) = vAnf
θ (t) sin (θ) ≈ θ
hmax θmax
=⇒
pvorher = pnachher
mv0 = (m + M) vAnf
vAnf = m
m+M v0

1
2 (m + M) v2 Anf = (m + M) ghmax
hmax = v2
Anf
2g =
m
m+M
2 v 2
0
2g
cos(θ) = h′
l
= l−h
l
= 1 − h
l
→ θmax = arccos 1 − hmax
l
hmax =
θmax = arccos 1 −
2 22 g
22 g+350 g · 362 m2
s2 2·9.81 m
s2 0.23 m
1 m
m
= arccos 1 − m+M
= 39.65 ◦
= 0.23 m
2 2
v0 2gl
m1 = 1 kg m2 = 0.4 kg
2.8 m
v1 =
−3.0 s v2
−3.5 m
=
−3.0 s
vsp vrel
p s 1 ps 2
vsp = m1
m1+m2 v1 + m2
m1+m2 v2 =
= 1
2.8 m 0.4 −3.5 m
1.4 −3.0 s + 1.4 −3.0 s =
1.0
−3.0
6.3 m
vrel = v1 − v2 =
0.0 s
m
s
v s 1/2
v s
1.8 m
1 = v1 − vsp =
0 s =⇒ ps 1 = m1v s
1.8 kg·m
1 =
0 s
v s
−4.5 m
2 = v2 − vsp =
0 s =⇒ ps 2 = m2v s
−1.8 kg·m
2 =
0 s = −ps 1
E Lab
kin
= 1
2 m1 |v1| 2 + 1
2 m2 |v2| 2 = 12.67 J

E sp 1
kin = 2m1 |v s 1| 2 + 1
2m2 |v s 2| 2 = 5.67 J
∆Ekin = 1
2 (m1 + m2) |vsp| 2 = 1
2
(1.0 kg + 0.4 kg) · 10 m2
s 2 = 7.00 J
p s 1 + p s 2 = p ′ s
1 + p ′ s
2 = 0 =⇒ p s 2 = −p s 1 p′ s
2 = −p ′ s
1
|ps
1 |2
2m1
+ |ps
2 |2
| p s 1 | 2
2m2 = |p′ s
1 |2
2m1 + |p′ s
2 |
2m2
1 1 + 2m1 2m2
=| p ′ s
1 | 2
1 1 + | p 2m1 2m2
s 1 | 2 =| p ′ s
1 | 2 p s 1 = −p ′ s
1
p s 1 = p ′ s
1 p ′ s
2
→ v ′ s
1 = −v s
−1.8 m
1 =
0.0 s v′ s
2 = −v s
4.5 m
2 =
0.0 s
→ v ′
1 = v ′
s
−0.8
1 + vsp =
v
−3.0
′
2 = v ′
s
5.5
2 + vsp =
−3.0
v ′
2
= 8.42 J − 4.82 J = 3.6 J
△E = E1,kin − E ′ 1
1,kin = 2m1 |v1| 2 − 1
2m1
1
z
ρ
(x, y, z) r(x, y, z)
ρ · r (x, y, z) 2
dx dy dz
r (x, y, z) = x 2 + y 2 ≤ a · √ 2
J = ´ r2dm ρ = m
V → m = ρV → dm = ρdV
→ J = ´ r2dm = ρ ´ r2dV á á á
2 2 = ρ x + y dxdydz = ρ
= ρ
0 0 0
á
0
1
3 a3 y + 1
3ay3 a
0
á
dz = ρ
0
dV =dxdydz
= ρ ´ ´ ´ r2dxdydz = ρ ´ ´ ´ x2 + y22dxdydz = ρ ´ ´ ´ x2 + y2 dxdydz
3x3 + y2x á á
a
1
dydz = ρ
0 3a3 + ay2 dydz
á á
1
0 0
1
3 a4 + 1
3 a4 dz = ρ
0 0
á
2
3a4dz = ρ 2
3a5 = m 2
V 3a5 = m
a3 2
3a5 = 2
3ma2 0