Environmental Engineering Reference
In-Depth Information
10.5 The moment of inertia tensor is already diagonal in
x
3
so we have the first
eigenvector
5
3
Ma
2
.
The other two eigenvectors
γ
=
e
3
with eigenvalue
I
γ
=
satisfy
2
a
1
a
2
−
I
−
3
=
0
.
−
38
−
I
Hence
(
1
,
√
2
Ma
2
6
3
√
2
)
α
=
−
1
,
0
)
α
=
(
5
−
and
√
2
3
√
2
).
Ma
2
6
β
=
(
1
,
−
−
1
,
0
)
β
=
(
5
+
10.6 Use cylindrical co-ordinates
(r,θ,z)
with the
z
-axis as the symmetry axis.
Then any pair of axes
x
and
y
perpendicular to the
z
axis will serve as
principal axes. Calculate
ρ
h/
2
−
2
π
R
πρR
4
h
2
r
3
d
r
d
θ
d
z
I
z
=
=
h/
2
0
0
and
1
3
R
2
.
ρ
h/
2
−
2
π
R
πρR
4
h
4
h
2
(r
2
sin
2
θ
z
2
)r
d
r
d
θ
d
z
=
=
+
=
+
I
x
I
y
h/
2
0
0
The barrel of the gun ensures that the angle between the symmetry axis and
the angular momentum is small. From the geometry,
I
3
/I
=
6
/
19 so the
symmetry axis precesses about 15
.
8 times per second.
10.7 As with Feynman's plate we can take the
e
3
principal axis to be normal to
the plate, and
I
1
=
I
2
=
I,
so that
I
3
=
2
I.
Euler's equations then give us
I
ω
1
+
˙
Iω
2
ω
3
=−
kω
1
,
I
ω
2
+
˙
Iω
3
ω
1
=−
kω
2
,
ω
3
=
˙
0
.
As with the free top,
ω
3
is constant. Multiply the second equation by
i
and
add to the first equation, then write in terms of
η
to obtain:
I
η
˙
−
iIω
3
η
+
kη
=
0
,
which has a general solution:
=
−
+
+
η
A
exp[
kt/I
i(ω
3
t
φ)
]
.
