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0593_C09_fm Page 284 Monday, May 6, 2002 2:50 PM

284 Dynamics of Mechanical Systems

where the middle two terms in the last line are zero because G is the mass center of B (see
Section 6.8). The first term may be recognized as the moment of the linear momentum of
a particle with mass M at G about Q. That is,

×
×
MQG v = QG × Mv G = D QG L = D A (9.4.8)
G
G
GQ
The last term of Eq. (9.4.7) may be expressed in terms of the central inertia dyadic (see
Sections 7.4 and 7.6) as follows: Let n be a unit vector with the same direction as ωω ωω. Then,
ω
ωω ω ω may be expressed as:
ωω= ωn ω (9.4.9)

where ω is the magnitude of ωω ωω. Hence, we have:

N
N
i ∑
∑ m r ×(ωω × r ) = N m r ×(ω n × r i) = ω ∑ m r ×( n × r i) (9.4.10)
ω
ω
ii
i i
i i
i=1 i=1 i=1
From Eqs. (7.2.2) and (7.4.11), this last expression in turn may be expressed in terms of
the inertia dyadic as:
N
∑ m r ×( BG BG
ω ii n × r i) = ωI ω = ωI n ⋅ ω
ω
i=1 (9.4.11)
= I BG ⋅( ωn ) = I BG ⋅ωω
ω
We can obtain yet a different form of the last term of Eq. (9.4.7) by recognizing that the
/
parenthetical term (ωω ωω × r ) may be identified as v PG or as v − v G (see Eq. (4.9.4)). Hence,
i
P i
i
we have:
N
∑ m r ×(ωω × r ) = N m r ×( v − )
i ∑
G
v
P i
ii
i i
i=1 i=1
N
N
= ∑ m r × v − ∑ m r × v G (9.4.12)
P i
ii
ii
i=1 i=1
= A − 0
BG
where the last equality is obtained from Eq. (9.4.4) and by recalling that G is the mass
center of B. Then, by comparing the results in Eqs. (9.4.11) and (9.4.12), we have:
N
∑ m r ×(ωω × r i) = I BG ⋅ωω = A BG (9.4.13)
ii
i=1