Page 317 - Dynamics of Mechanical Systems

0593_C09_fm Page 298 Monday, May 6, 2002 2:50 PM

298 Dynamics of Mechanical Systems

and

t t
∫ O dt + ∫ Pdt = mv (9.8.12)
y
0 0
Similarly, the angular impulse–momentum principle about G for n leads to:
z
t t

 ∫
P x −  l dt − ∫ l O dt = 1 ml ω
2
y
0  2 0 2 12
or
t t
 x − ∫ dt − l ∫ O dt = 1 ml ω 2
 l
2
 2 P 2 y 12 (9.8.13)
0 0
t
∫
By eliminating Pdt between Eqs. (9.8.12) and (9.8.13) we have:
0
 t  t
 ∫
 x −  l mv − ∫ O dt − l O dt = 1 ml ω
2
 2    0 y  2 0 y 12 (9.8.14)
Because B is pinned at O, G moves in a circle about O with radius /2. Then, v and ω are
related by:

v = ( ) 2 ω (9.8.15)
l
t
∫
By solving Eq. (9.8.14) for O y dt and by using Eq. (9.8.15) we have:
0
t
∫ Odt =   23l + − x  mv (9.8.16)
x 
y
 l
0
Observe that if B is struck at a point such that x is 2 /3 the pin reaction is zero. That
is, if the impact force is applied at a point 2/3 along the bar length, no pin reaction is
generated. This means that even if the bar is not pinned it will initially move such that
its end away from the impact has zero velocity. This point of application of the impact
force is called the center of percussion.

Example 9.8.4: A Pinned Double Bar Struck at One End
As an extension of the foregoing example consider two identical, pin-connected bars each
having length and mass m and resting on a smooth horizontal surface as depicted in
Figure 9.8.7. Let the end O of one of the bars be struck with the striking force being
perpendicular to the bars and having a magnitude P as shown. The objective is to deter-
mine the motion of the bars just after the impact.
Consider a free-body diagram of the bars and also a free-body diagram of the left, or
unstruck, bar as in Figures 9.8.8 and 9.8.9, where as in the foregoing example the inertia