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0593_C10_fm Page 331 Monday, May 6, 2002 2:57 PM

Introduction to Energy Methods 331

P (m ) B B
2 2
P (m ) F i
1 1
P (m ) G
i i
F G
2 T
F F
1
P (m )
N N
R R
FIGURE 10.6.3 FIGURE 10.6.4
A rigid body B subjected to forces. Equivalent force system on rigid body B.

By adding together the terms of similar equations for each of the particles, we have:

dW dK
= (10.6.9)
dt dt

where W is the work on S from all the forces, and K is the kinetic energy of S. By integrating
Eq. (10.6.9), we have:

W = K − K = ∆ K (10.6.10)
2 1

where, as in Eq. (10.6.5), K and K represent the kinetic energy of S at the beginning and
1
2
end of the time interval that the forces are applied to S.
Finally, consider a rigid body B acted upon by a system of forces as in Figure (10.6.3).
As before, consider B to be composed of fixed particles P with masses m as shown. Let
i
i
the force system applied to B be replaced by an equivalent force system (see Section 6.5)
consisting of a single force F passing through the mass center G of B together with a couple
with torque T as in Figure 10.6.4.
From d’Alembert’s principle (see Section 8.3) we can represent the inertia force system
*
*
on B by a force F passing through G together with a couple with torque T where F and
*
T are given by (see Eqs. (8.6.5) and (8.6.6)):
*
α
*
*
I
F =−M a and T =− ⋅ −αωω × I⋅ ( ωω ) (10.6.11)
G
where, as before, M is the mass of B, I is the central inertia dyadic of B, a is the acceleration
G
of G in inertia frame R, and ωω ωω and αα αα are the angular velocity and angular acceleration of
B in R. Then, from Newton’s laws and d’Alembert’s principle, we have:
F = M a (10.6.12)
G
and

T =⋅ +αωω × I⋅ ( ωω ) (10.6.13)
α
I