Game Development Reference
In-Depth Information
1cos2
θ
2
(14.67)
sin
θ
=
2
(see Appendix B, Section B.4), we can evaluate the remaining integral:
2
π
2
π
(
)
2
sin
θdθ
=−
11
22
cos 2
θ dθ
0
0
]
2
π
=−
=
1
θ
1
sin 2
θ
2
4
0
π
.
(14.68)
The moment of inertia about the x and y axes is therefore given by
4
3
2
 
==
1
πρhR πρhR
mR
+
1
11
22
4
12
=
2
+
mh
2
.
(14.69)
1
1
4
12
The product of inertia 1 is equal to the integral
h π R
22
 
3
=−
xyρdV ρ
=−
r θ dr dθdz
sin
cos
.
(14.70)
12
V
h
20 0
Since
2
π
sin
θ dθ
cos
=
0
,
(14.71)
0
it is the case that
  . It can also be shown that all of the other products
of inertia are equal to zero, so the inertia tensor of a cylinder has the form
==
0
12
21
2
2
1
mR
+
1
mh
0
0
4
12
2
2
=
0
1
mR
+
1
mh
0
,
(14.72)
4
12
2
0
0
1
2
mR
where m ρV
=
is the mass of the cylinder.
Nonzero products of inertia arise when we consider a solid box that rotates
about an axis passing through one of its vertices. The significance of an inertia
tensor that is not diagonal is discussed in Section 14.2.4.
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