Page 319 - Dynamics of Mechanical Systems

0593_C09_fm Page 300 Monday, May 6, 2002 2:50 PM

300 Dynamics of Mechanical Systems

where the third and sixth equations are obtained by evaluating moments about Q and G ,
1
2
and where I is the central moment of inertia: (1/12)m . Observe from the kinematics of
the bars that we can obtain the expressions:
v = v , v = v −(l 2)ω , v = v − lω , v = v −(l 2)ω (9.8.19)
1 x 2 x 2 y 0 2 Q 0 2 1 y Q 1

Hence, from the first expressions of Eqs. (9.8.17), (9.8.18), and (9.8.19), we find:

v = v = 0, Q = 0 (9.8.20)
1 x 2 x x

t t
Also, by eliminating the linear impulses ∫ Pdt and ∫ Qdt from Eqs. (9.8.17) and (9.8.18),
y
respectively, we obtain: 0 0

lmv + lmv = ( l 2) mv +( 1 12) m l ω −( l 2) mv +( 1 12) m l ω (9.8.21)
2
2
1 y 2 y 2 y 2 1 y 1
and
( l 2)mv 1y = ( 1 12)m l ω 1 (9.8.22)
2

Then, by substituting for v and v from Eqs. (9.8.19), we have:
1y
2y
v − lω −(l 2)ω + v −(l 2)ω = ( 1 2)[ v −(l 2)ω 2]
0 2 1 0 2 0
+( 112)lω 2 ( 1 2)[ v − lω 2 (l 2)ω 1] ( 112)lω 1 (9.8.23)
+
−
−
0
and
( 12)[ v 0 − lω 2 −(l 2)ω 1] ( 1 12)lω 1 (9.8.24)
=

or, after simplification:

v = ( 512)lω +( 11 12)lω
0 1 2 (9.8.25)

and

v = ( 23)lω + lω
0 1 2 (9.8.26)

Finally, by eliminating v , we obtain:
0
ω − 3 ω or ω = −( 1 3) ω (9.8.27)
2 1 1 2

Observe that the bars rotate in opposite directions immediately after the impact and that
the left bar rotates clockwise.