fisica1-youn-e-freedman-exercicios-resolvidos

1 2 2
9.45: From Eq. (9.17), K − K = I(
ω − ), and solving for I,
2 2 2
ω1
I
= 2
( ω
( K − K )
2
2
2
1
2
1
− ω
= 2
((520 rev min)
)
2
= 0.600 kg ⋅ m .
2
( −500 J)
− (650 rev
min)
2
)
π rad s
( )
30 rev min
2
9.46: The work done on the cylinder is PL, where L is the length of the rope. Combining
Equations (9.17), (9.13) and the expression for I from Table (9.2(g)),
PL =
1
2
ω
v
g
2
,
or
2
1 ω v
P =
2 g L
2
(40.0 N)(6.00 m s)
=
= 14.7 N.
2
2(9.80 m s )(5.00 m)
a
1 2
2 rad
9.47: Expressing ω in terms of α
rad, ω = . Combining with I = MR , Eq. (9.17)
R
2
2
1 1 (70.0 kg)(1.20 m)(3500 m s)
4
becomes K = MRarad =
= 7.35×
10 J.
2 2
4
2
9.48: a) With I = MR , with expression for v is
v =
2gh
.
1+
M m
b) This expression is smaller than that for the solid cylinder; more of the cylinder’s
mass is concentrated at its edge, so for a given speed, the kinetic energy of the cylinder is
larger. A larger fraction of the potential energy is converted to the kinetic energy of the
cylinder, and so less is available for the falling mass.