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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