Tensor exercise  Solutions
Problem 1:
Although
d
ˇ
I
dt
6
= 0 in general, show that
ω
·
d
ˇ
I
dt
·
ω
= 0
.
Solution:
Since
ˇ
I
=
∑
3
i
=1
I
i
e
i
e
i
, and
˙
I
i
= 0, we have
d
ˇ
I
dt
=
X
i
(
˙
I
i
e
i
e
i
+
I
i
˙
e
i
e
i
+
I
i
e
i
˙
e
i
)
=
X
i
(
I
i
˙
e
i
e
i
+
I
i
e
i
˙
e
i
)
=
X
i
I
i
(
ω
×
e
i
)
e
i
+
I
i
e
i
(
ω
×
e
i
)
.
Now contract the left leg of each term with
ω
:
ω
·
d
ˇ
I
dt
=
X
i
I
i
[
ω
·
(
ω
×
e
i
)]
e
i
+
I
i
(
ω
·
e
i
)(
ω
×
e
i
)
.
Each term in the equation is now just a vector. The first term on the right hand
side vanishes [because
ω
·
(
ω
×
e
i
) = 0], but the second term does NOT vanish in
general. (In the second term, we are taking the dot product of
ω
with
e
i
,
not with
ω
×
e
i
!) So
ω
·
d
ˇ
I
dt
=
X
i
I
i
(
ω
·
e
i
)(
ω
×
e
i
)
.
Since (
ω
·
e
i
) is just a scalar, and (
ω
×
e
i
)
·
ω
= 0, each term on the right hand side
of the above equation is a vector perpendicular to
ω
, so the whole expression
ω
·
d
ˇ
I
dt
is a vector perpendicular to
ω
. Therefore
ω
·
d
ˇ
I
dt
·
ω
= 0.
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 Spring '11
 Tan
 mechanics, Angular Momentum, Moment Of Inertia, Rigid Body, Rotation, ω · ei, ijk ωk ej

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