Page 220 - Dynamics of Mechanical Systems

0593_C07_fm Page 201 Monday, May 6, 2002 2:42 PM

Inertia, Second Moment Vectors, Moments and Products of Inertia, Inertia Dyadics 201

n a
n a
n b d a
P(m) P(m)

p p

O O
FIGURE 7.3.1 FIGURE 7.3.2
A particle P, reference point O, and unit vectors Distance from particle P to line through O and
n a and n b . parallel to n a .
Observe that by substituting for I PO from Eq. (7.2.1) that I P O and I PO may be expressed
a aa ab
in the form:

p n =
×
×
×
I PO = m p ×( n × ) ⋅ m( p n )⋅( p n ) = m( p n ) 2 (7.3.3)
aa a a a a a
and
×
p n =
×
I PO = m p ×( n × ) ⋅ m( p n )⋅( p n ) (7.3.4)
ab a b a b
Observe further that (p × n ) may be identified with the square of the distance d from
2
a
a
P to a line passing through O and parallel to n (see Figure 7.3.2). This distance is often
a
referred to as the radius of gyration of P relative to O for the direction n .
a
Observe also for the product of inertia of Eq. (7.3.3) that the unit vectors n and n may
a
b
be interchanged. That is,
×
×
×
I PO = m( p n )⋅( p n ) = m( p n ) = I PO (7.3.5)
ab a b a ba
Note that no restrictions are placed upon the unit vectors n and n . If, however, n and
a
b
a
n are perpendicular, or, more generally, if we have three mutually perpendicular unit
b
vectors, we can obtain additional geometric interpretations of moments and products of
inertia. Specifically, consider a particle P with mass m located in a Cartesian reference
frame R as in Figure 7.3.3. Let (x, y, z) be the coordinates of P in R. Then, from Eq. (7.2.1)
the second moment vectors of P relative to origin O for the directions n , n , and n are:
x
z
y
I PO = m p ×( n × p)
x x
z) [
x (
x (
= mx n + y n + z n × n × x n + y n + z n z)]
y
x
y
or
my +
I PO = ( [ 2 z 2 n ) − xy n − xz n z] (7.3.6)
x x y