fisica1-youn-e-freedman-exercicios-resolvidos

2
1
9.56: a) Ι =
1 Μa b) Ι = Μb
12
12
2
=
M
. . , Ιcm
L and d L h ,
12
2
9.57: In Eq ( 919) ( 2 ) so
2
⎡ 1 L ⎤
2 ⎛ ⎞
Ιp = Μ ⎢ L + ⎜ − h⎟
⎥
⎢⎣
12 ⎝ 2 ⎠ ⎥⎦
⎡ 1 2 1 2
= Μ
⎢
L + L − Lh + h
⎣12
4
⎡1
2
2⎤
= Μ
⎢
L − Lh + h
⎥
,
⎣3
⎦
which is the same as found in Example 9.12.
=
−
⎤
⎥
⎦
2
9.58: The analysis is identical to that of Example 9.13, with the lower limit in the integral
2
being zero and the upper limit being R, and the mass Μ = πLρR . The result is
2
Ι = 1
ΜR , as given in Table ( 9 .2(f)).
2
9.59: With dm = dx
L
M
L
3
2 M M x M
Ι = ∫ x dx = =
L L 3 3
0
L
0
2
L .