Page 308 - Dynamics of Mechanical Systems

0593_C09_fm Page 289 Monday, May 6, 2002 2:50 PM

Principles of Impulse and Momentum 289

By setting moments about Q equal to zero we have:

×
×+
×
P
*
QP F QP F = or QP F = QP × md v dt (9.6.3)
0
Consider the final term in Eq. (9.6.3). By the product rule for differentiation we have:
P
P
QP × md v dt = ( d QP × md v ) dt −(d QP ) × m v P
dt
= d A dt − v PQ × m v P
Q
(9.6.4)
−( P Q P
= d A dt v − ) × m v
v
Q
Q
= d A dt + v × m v P
Q
Observe that QP × F is the moment of F about Q. Hence, by substituting from Eq. (9.6.4)
into (9.6.3) we obtain:
M = d A dt + v × m v P (9.6.5)
Q
Q Q
If the velocity of Q is zero (that is, if Q is fixed in R), then we have:

M = d A dt (9.6.6)
Q Q

By integrating over the time interval in which F is applied, we have:

t 2 t 2
∫
∫ M dt = ( d A Q ) A t () − A t () (9.6.7)
dt dt =
1
Q
2
Q
Q
t 1 t 1
or
J =∆ A (9.6.8)
Q Q
where J is the angular impulse of F about Q during the time interval (t , t ). That is, the
Q
1
2
angular impulse about a point Q fixed in an inertial reference frame is equal to the change
in angular momentum about Q. This verbal statement of Eq. (9.6.8) is called the principle
of angular momentum. Note, however, it is valid only if Q is fixed in the inertial reference
frame.
Consider next a set S of N particles P with masses m (i = 1,…, N) moving in an inertial
i
i
reference frame R as in Figure 9.6.3. Let the particles be acted upon by forces F (i = 1,…,
i
N) as shown. Consider a free-body diagram of a typical particle P as in Figure 9.6.4 where
i
F * is the inertia force on P given by:
i i
F =− m a =− md v P i dt (no sum on i) (9.6.9)
*
P i
i i