Page 161 - Dynamics of Mechanical Systems

0593_C05_fm Page 142 Monday, May 6, 2002 2:15 PM

142 Dynamics of Mechanical Systems

Because Q also moves in a circle about R, we have:

a = αα 3 × RQ + ωω 3 ×(ωω 3 × RQ)
Q
= αα 3 n × − ( 495 λλ 3 ) +− ( 1 81. n ) ×− ( [ 181 n ) ×− ( 4 95 λλ 3 )]
.
.
.
z
z
z
=−495. α 33 + 1621. λλ 3
νν
)
+
−
.
=−495. αα 3 ( [ 2/ n ) 2 x ( 2/ n ) 2 y] + 1621 [ ( 2/ n ) 2 x ( 2 2/ n y] (5.5.26)
n
=−3 5αα n − 3 5. αα n + 11 46. n − 11 46. n
.
3 x 3 y x y
−
= (11 46 3 5αα 3 n ) x ( − . αα 3 n ) y
+−11 46 3 5.
.
.
Comparing Eqs. (5.5.25) and (5.5.26), we have:
−23 467 1 5α = 11 46 3 5α (5.5.27)
−
−
.
.
.
.
2 3
and
−
−58 93 2 6α = −11 46 3 5α (5.5.28)
+
.
.
.
.
2 3
Solving for α and α we obtain:
2
3
α = 306 rad sec and α = 1128 rad sec 2 (5.5.29)
2
.
.
2 3
Hence, the acceleration of Q is:
Q
.
a =−28 02 n − 50 94 n ft sec 2 (5.5.30)
.
x y
Observe how much more effort is required to obtain accelerations than velocities.
5.6 Chains of Bodies
Consider next a chain of identical pin-connected bars moving in a vertical plane and
supported at one end as represented in Figure 5.6.1. Let the chain have N bars, and let
their orientations be measured by N angles θ (i = 1,…, N) that the bars make with the
i
vertical Z-axis as in Figure 5.6.1. Because N angles are required to define the configuration
and positioning of the system, the system has N degrees of freedom. A chain may be
considered to be a finite-segment model of a cable; hence, an analysis of the system of
Figure 5.6.1 can provide insight into the behavior of cable and tether systems.
A kinematical analysis of a chain generally involves determining the velocities and
accelerations of the connecting joints and the centers of the bars and also the angular
velocities and angular accelerations of the bars. To determine these quantities, it is easier
to use the absolute orientation angles of Figure 5.6.1 than the relative orientation angles