fisica1-youn-e-freedman-exercicios-resolvidos

9.60: For this case, dm = γ dx.
a)
L
2
x
Μ = ∫ dm = ∫ γx dx = γ
2
L
0
L
0
2
yL
=
2
4
x
4
2
γL
b) Ι = ∫ x ( γx)
dx = γ = =
4
4
0
L
0
Μ
2
2
L .
This is larger than the moment of inertia of a uniform rod of the same mass and length,
since the mass density is greater further away from the axis than nearer the axis.
L
2
c) Ι = ∫(
L − x)
γxdx
0
L
2
= γ∫
( L x − 2Lx
0
⎛
= γ
⎜ L
⎝
2
4
L
= γ
12
M 2
= L .
6
2
3
+ x ) dx
2
3 4
x x x ⎞
− 2L
+
2 3 4
⎟
⎠
This is a third of the result of part (b), reflecting the fact that more of the mass is
concentrated at the right end.
L
0
9.61: a) For a clockwise rotation, ω r will be out of the page. b ) The upward direction
crossed into the radial direction is, by the right-hand rule, counterclockwise. ω r and r are
r r
r
perpendicular, so the magnitude of ω × is ωr = v.
c) Geometrically, ω is
perpendicular to v r r r
, and so ω × v has magnitude ωv = arad,
and from the right-hand
rule, the upward direction crossed into the counterclockwise direction is inward, the
direction of a r
rad.
Algebraically,
r r r r r r
a
rad
= ω × v = ω × ( ω ×
)
r r r r r r
= ω ( ω ⋅
) −
( ω ⋅ ω )
2 r
= − ω
,
where the fact that ω r and r are perpendicular has been used to eliminate their dot
product.