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0593_C10_fm Page 329 Monday, May 6, 2002 2:57 PM

Introduction to Energy Methods 329

where M is the mass of B (the total mass of all the particles composing B), and the middle
term is zero because G is the mass center of B.
Consider the last term of Eq. (10.5.5). By expanding and rearranging the terms we have:

N
N
N
i [
i (
i (
1 ∑ m ωω × ) = 1 ∑ m ωω × )⋅( ωω × r i) = 1 ∑ m ωω⋅ r ×( ωω × r i)]
2
r
r
2 i 2 i 2 i
i= 1 i= 1 i= 1
N
N
i[
×
= 1 ωω⋅ ∑ m ×( ωω × r i) = 1 ωω⋅ ∑ m r i ( nω × r i)]
r
2 ii 2 ω
i= 1 i= 1 (10.5.6)
= 1 ωω⋅I BG ω ω = 1 ωω ⋅I BG ⋅n ω
2 ω 2 ω
= 1 ωω ⋅I BG ⋅ωω
2
where ω is the magnitude of ωω ωω, n is a unit vector parallel to ωω ωω, I BG is the second moment
ω
ω
vector of B for G for the direction of n , and I B/G is the central inertia dyadic of B (see
ω
Sections 7.4 and 7.6.).
By substituting from Eq. (10.5.6) into (10.5.5), the kinetic energy of B becomes:
1
K = 1 v 2 + ωω ⋅I BG ⋅ωω (10.5.7)
2 G 2
Finally, suppose that n , n , and n are mutually perpendicular principal unit vectors of
3
2
1
B for G (see Section 7.7), then K may be expressed as:
1
K = 1 Mv 2 + ( 11 2 + I ω 2 + I ω 2 ) (10.5.8)
I ω
2 G 2 1 22 2 33 3
where I , I , and I are the corresponding principal moments of inertia of B for B, and
22
33
11
the ω (i = 1, 2, 3) are the n components of ωω ωω. Observe that in all the expressions for kinetic
i
i
energy no accelerations appear, only velocities.
10.6 Work–Energy Principles

We can obtain a relationship between work and kinetic energy by integrating equations
obtained from Newton’s laws. To this end, consider first a particle P with mass m subjected
to a force F as in Figure 10.6.1. Then, from Newton’s laws (see Section 8.2), we have:

F = m a (10.6.1)

where a is the acceleration of P in an inertial frame R. If we project the terms of Eq. (10.6.1)
along the velocity v of P in R we have:

⋅
Fv⋅= m av (10.6.2)