fisica1-youn-e-freedman-exercicios-resolvidos

9.90: a) In the case that no energy is lost, the rebound height h′ is related to the speed
2
υ
v by h ′
h
= , and with the form for h given in Example 9.9, ′ = . b) Considering
2g
h
1+
M 2m
the system as a whole, some of the initial potential energy of the mass went into the
kinetic energy of the cylinder. Considering the mass alone, the tension in the string did
work on the mass, so its total energy is not conserved.
9.91: We can use Κ ( cylinder) = 250 J to find ω for the cylinder and v for the mass.
Ι
= MR
1 2 1
2
2
= ( 10.0 kg)(0.150 m) = 0.1125 kg ⋅ m
2
2
2
K = 1 Iω
so ω = 2K
I =
2
66.67 rad s
v = Rω
=10.0 m s
Use conservation of energy K
1
+ U1
= K
2
+ U
2.
Take y = 0 at lowest point of the
mass, so y = and y = , the distance the mass descends. K = U = so U = .
2
0
1
h
1 2 1 2
=
2
mv +
2
I , where m =
2
MR and
2
1 2 1 2
= mv Mv
2
4
mgh ω
12.0 kg
1 2 1 2
For the cylinder, I = 1 ω = v R,
so Iω
= Mv .
mgh +
2
v ⎛ M ⎞
h = ⎜1
+ ⎟ = 7.23 m
2g
⎝ 2 m ⎠
2
4
1 2
0
1
K2