Page 285 - Dynamics of Mechanical Systems

0593_C08_fm Page 266 Monday, May 6, 2002 2:45 PM

266 Dynamics of Mechanical Systems

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where a is the n component of a and T is the n component of T ; a may be obtained
2
2
2
3
3
directly from Eq. (8.12.8), and, because n , n , and n are parallel to principal inertia
3
1
2
directions of B for G, T is given by Eq. (8.6.12) as:
3
1 (
T =−α I + ω ω I − ) (8.12.13)
I
3 3 33 2 11 22
where α , ω , and ω can be obtained directly from Eqs. (8.12.5) and (8.12.6) and where
2
1
3
from the inertia properties of a slender rod, I , I , and I are:
22
11
33
I = 0, I = I = ml 2 12 (8.12.14)
11 22 33
Finally, by substituting from Eqs. (8.12.5), (8.12.6), (8.12.10), (8.12.13), and (8.12.14) into
(8.12.12), we obtain:
[
)
˙˙
˙˙
θ
− ( ) 2 sinθ − ( ) ( ) −( ) 2 Ω 2 sin cosθ ] (ml 12 θ
l 2 θ
−
2
mg l
l
m l 2
)
+(ml 12 Ω 2 sin cosθ = 0
θ
2
or
θ
˙˙ 2l ) sinθ − Ω 2 sin cosθ = 0 (8.12.15)
θ +(3g
˙
Observe that Ω does not appear in this governing equation. This means that the angular
speed change of S does not affect the movement of B, except indirectly through Ω. Observe
further that, if Ω is constant, B has two equilibrium positions, that is, positions of constant θ:
θ = 0 and θ = cos −1 (3g 2lΩ 2 ) (8.12.16)
This last result shows that the inertia torque is important even when a body is moving at
a uniform rate. Indeed, if the body is not moving with uniform planar motion, its inertia
torque is not generally zero. Consider again the expression for T of Eq. (8.12.13):
3
1 (
T =−α I + ω ω I − ) (8.12.17)
I
3 3 33 2 11 22
Observe that even if α is zero, T is not zero unless the second term is also zero.
3
3
The erroneous neglect of the inertia torque can of course produce incorrect results. For
example, suppose that in the current problem, where we are seeking the equilibrium
position with Ω constant, we construct an incorrect free-body diagram, as in Figure 8.12.4.
By setting moments about O equal to zero, we would obtain:
(
2
m l 2) Ω sin cosθ − ( = 0 (8.12.18)
θ
mg l 2)sinθ
2
or
cosθ = 2g lΩ 2 or θ = cos (2g lΩ 2 ) (8.12.19)
−1
The incorrectness of this result is seen by comparison with Eq. (8.12.16).