fisica1-youn-e-freedman-exercicios-resolvidos

9.79: a) See Exercise 9.50.
K =
π
T
2
24
6
Ι 2π
(0.3308)(5.97 × 10 kg)(6.38×
10 m)
=
2
(86,164 s)
2
2
2 29
= 2.14×
10
2
2
2
24
11 2
1 ⎛ 2πR
⎞ 2π
(5.97 × 10 kg)(1.50 × 10 m)
33
b) M ⎜ ⎟ =
= 2.66×
10 J.
7 2
2 ⎝ T ⎠
(3.156×
10 s)
c) Since the Earth’s moment on inertia is less than that of a uniform sphere, more of
the Earth’s mass must be concentrated near its center.
J.
9.80: Using energy considerations, the system gains as kinetic energy the lost potential
energy, mgR. The kinetic energy is
1 2 1 2 1 2 1
2 1
2 2
K = Ιω + mv = Ιω + m(
ωR)
= ( Ι + mR ) ω ⋅
2 2 2 2 2
1 2
Using Ι = mR and solving for ω,
2
2 4 g
ω = , and ω =
3 R
4 g
3 R
.