Problem: KineticEnergyo
faRod
(a)
First we need to
f
nd the principal moments of the rod.
This can be
done with the origin either at the end of the rod (at the pivot) or at the center of
the rod (the COM). There is a conceptual advantage with the latter approach,
so we’ll take that route. Start with the general expression for the
αβ
component
of the inertia tensor (
α
,
β
=
x
,
y
,
z
)
I
αβ
=
Z
ρ
(
r
)(
r
2
δ
αβ
−
x
α
x
β
)
d
3
r.
(1)
Note that
r
is measured from the origin of the body frame coordinate system,
r
2
=
x
2
+
y
2
+
z
2
,and
x
x
=
x
,
x
y
=
y
, etc. For this rod, the linear mass density
λ
is constant, and its mass
M
is
M
=2
lλ,
(2)
where 2
l
is the length of the rod.
Given the geometry of the object, it is
straightforward to use cartesian coordinates to do the required integrations.
Let the rod lie along the
z
axis, so that
x
=
y
=0
along the length of the
rod. The density can then be written as
ρ
(
r
)=
λδ
(
x
)
δ
(
y
)
,u
s
ingD
i
racde
l
ta
functions to constrain the
x
and
y
values. The easiest moment to evaluate is
I
zz
,
I
=
λ
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 Fall '10
 Wilemski
 mechanics, Energy, Kinetic Energy, Mass, Moment Of Inertia, inertia tensor

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