Page 516 - Dynamics of Mechanical Systems

0593_C14_fm Page 497 Tuesday, May 7, 2002 6:56 AM

Stability 497

D
D
R
where the α and the ω (i = 1, 2, 3) are the n components of αα αα and ωω ωω and where I ,
R
i
11
i
i
I , and I are:
33
22
I = I = Mr 4 , I = Mr 2 (14.6.23)
2
2
11 33 22
Consider next the free-body diagram itself in Figure 14.6.2. By setting moments of the
forces, about C, equal to zero we have:
rn ×( W F ) + rn + rd ) ×( + Q) + T =
G (
+
*
*
*
3 3 3 w F D 0 (14.6.24)
Then, by substituting from Eqs. (14.6.13) through (14.6.18), we obtain the scalar equations:
(
Mgrsinθ + Mra + mr + cosψ) a Q
G
1
2 2
(14.6.25)
+ T + mgr + cosψ) sinθ = 0
(1
1
Q
−Mra G − mr (1 + cosψ )a Q + mra sinψ
1 1 3
(14.6.26)
ψ
+T + mgrsin cosθ = 0
2
and
−mra sinψ + T − mgrsin sinψ = 0 (14.6.27)
θ
Q
2 3
R Q
R G
where the a G and the a Q are the n (i = 1, 2, 3) components of a and a . Then, by
i i i
substituting from Eqs. (14.6.12), (14.6.13), (14.6.25), (14.6.26), and (14.6.27), we have:
(
+
˙ ˙
Mgrsinθ + Mr − + ˙ 2 sin cosθ ψφ cosθ)
θ φ
˙˙
θ
2
+
θψφ
− I ( ˙˙ − ˙ ˙ I − )( ˙ ψ φ ˙ sinθ φ ) ˙ cosθ
I
11 cosθ) +( 22 33
)(
+
−
˙ ˙
θ
+ mr (1 + cosψ) ( [ 1 + cosψ ψφ cosθ φ ˙ 2 sin cosθ θ) (14.6.28)
˙˙
2
˙
+ φ ˙˙ sin cosθ φψ ψ θ θψ ψ]
+
ψ
˙ ˙ cos cosθ + 2 ˙ sin
+mgr (1 + cos ψ)sin θ = 0
(ψφ
˙ ˙
˙˙
+
˙ ˙
−Mr ˙˙ + sinθ + 2θφ cosθ ) − ( ˙˙ ψ θφ cosθ
2
I
22
+ sinθ ) +(I − )θφ cosθ − mr 2 (1 + cosψ ) ( [ ψ
φ
˙˙
˙ ˙
˙˙
I
33 11
+ sinφ ˙˙ θ + 2φθ cosθ ) + cosψ ) −( ˙ ψ 2 + )sinφ 2 ˙ ψ
˙ ˙
(1
]
−2 ˙ sin sinψ ++ mr sinψ − ( [ φ 2 ˙ sin θ (14.6.29)
θ
φψ
˙
2
2
ψφ
ψ
− ) + cosψ
ψφ
− ˙ ˙ sinθ θ 2 ˙ ( 1 ) − ˙ ˙ cos sinθ
˙˙
ψ
− ˙˙ sinψ φ
θ
− ˙ cosψ ψ − sin sinψ ]
2
+mgrsin cosθ = 0
ψ