Page 349 - Dynamics of Mechanical Systems

0593_C10_fm Page 330 Monday, May 6, 2002 2:57 PM

330 Dynamics of Mechanical Systems

P (m ) S
2 2
. .
.
P (m ) P (m )
1 1
P(m) i i
F 2
F . .
N
.
F 1
P (m )
R N N
F R
FIGURE 10.6.1 FIGURE 10.6.2
A force applied to a particle. A set of particles subjected to forces.

The left side of Eq. (10.6.2) may be recognized as the power of F (see Eq. (10.4.2)). The
right side of Eq. (10.6.2) may be expressed in terms of a derivative of the square of the
velocity of P. That is,

mav⋅= ( 2 dt = d  1 mv 2  (10.6.3)
md v ) 2
dt  2 

From Eqs. (10.4.1) and (10.4.2) we recognize the left side of Eq. (10.6.2) as the derivative
of the work W of F. In like manner, from Eq. (10.5.1), we recognize the right side of Eq.
(10.6.3) as the derivative of the kinetic energy K of P. Hence, we have:

dW dK
= (10.6.4)
dt dt
or

W = K − K = ∆ K (10.6.5)
2 1
where K and K represent the kinetic energy of P at the beginning and end of the time
2
1
interval during which F is applied to P.
Next, consider a set S of particles P (i = 1,…, N) subjected to forces F as in Figure 10.6.2.
i
i
Then, for a typical particle P , Newton’s laws become:
i
F = m a (no sum on i) (10.6.6)
i i i
where a is the acceleration of P in inertia frame R. By multiplying the terms of Eq. (10.6.6)
i
i
by v , the velocity of P in R, we obtain:
i
i
Fv = m a v = d  1 m v 2  = dK i (10.6.7)
⋅
⋅
i i i i i i i 
dt  2 dt
where K is the kinetic energy of P . i
i
We can recognize the left side of Eq. (10.6.7) as the derivative of the work W of F . Hence,
i
i
Eq. (10.6.7) becomes:
dW dK
i = i (10.6.8)
dt dt