fisica1-youn-e-freedman-exercicios-resolvidos

9.40: a) In the expression of Eq. (9.16), each term will have the mass multiplied by
3
f and the distance multiplied by f , and so the moment of inertia is multiplied by
3 2 5
5
8
f ( f ) = f . b) (2.5)(48) = 6.37 × 10 .
9.41: Each of the eight spokes may be treated as a slender rod about an axis through an
end, so the moment of inertia of the combination is
2 ⎛ mspoke
⎞ 2
I = mrimR
+ 8 ⎜ ⎟ R
⎝ 3 ⎠
⎡ 8 ⎤
2
=
⎢
(1.40 kg) + (0.20 kg) (0.300 m)
⎣ 3 ⎥
⎦
= 0.193 kg ⋅ m
2
9.42: a) From Eq. (9.17), with I from Table (9.2(a)),
1 2 1
2 rev 2π
rad rev
K = mL ω = (117 kg)(2.08 m) (2400 × )
2 12 24
min 60 s min
b) From mgy = K,
1 2 2
6
y
K
= mg
6
( 1 .3×
10 J) 3
= 1.16 × 10 m = 1.16 km.
=
(117 kg)(9.80 m s
2
)
= 1.3×
10
J.
2
9.43: a) The units of moment of inertia are [kg][m ]
and the units of ω are equivalent
−1
2
to [s ] and so the product 1 Iω has units equivalent to −2
2
]
2
[kg ⋅ m ⋅ s ] = [kg ⋅ (m s) ,
which are the units of Joules. A radian is a ratio of distances and is therefore unitless.
b) K = π
2 Iω 2 1800 , when is in rev min.
9.44: Solving Eq. (9.17) for I,
2K
2(0.025J)
−3
2
I = =
= 2.25×
10 kg ⋅ m .
2 2π
rad s 2
ω (45 rev min × )
60 rev min