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0593_C14_fm Page 491 Tuesday, May 7, 2002 6:56 AM

Stability 491

and

φ ˙˙* cosθ + 2 ψ θ = 0 (14.5.21)
˙ ˙ *
0 0
In view of Eq. (14.5.15), the sum of the first three terms of Eq. (14.5.19) is zero, thus they
may be neglected. Also, Eqs. (14.5.20) and (14.5.21) may be integrated as:

˙
ψ +
3 ˙ * 3φ ˙ * sinθ + 5φ θ * cosθ = c * (14.5.22)
0 0 0 1
and

φ ˙ * cosθ + 2 ˙ ψ θ = c * (14.5.23)
*
0 0 2
where c * and c * are constants. Solving Eq. (14.5.23) for φ ˙ * we have:
1 2
˙ ψ θ + )
*
φ =− ( 2 * c cosθ (14.5.24)
˙ *
0 2 0
Then, by substituting into Eq. (14.5.22), we have:
ψ +− ( 6 ˙ ψ θ + ) ˙ *
*
3 ˙ * * 3c tanθ + 5φ θ * cosθ = c (14.5.25)
0 2 0 0 0 1
ψ
ψ
Finally, by solving for ˙ * and by substituting for φ ˙ * and ˙ * into Eq. (14.5.19) (without
the first three terms) we have:

˙˙ θλθ = κ * (14.5.26)
+
*
*
where λ and κ are defined as:
ψ )
˙
D
λ =(12 5 ˙ 2 +(14 5 φ ψ ˙ sin θ + φ 2
0 ) 0 0 0 0
(14.5.27)
−(4gr )cos θ 0
5
and

κ * =( ) ψ ˙ c * + (2 φ ) 5 ˙ c * cos θ + (45 ˙ c * sin θ (14.5.28)
D
65
02 0 1 0 φ ) 0 2 0
*
We recall from our previous analyses that stability will occur if the coefficient λ of θ in
Eq. (14.5.26) is positive; as a corollary, instability will occur if λ is negative. (λ = 0 represents
a neutral condition, bordering between stability and instability.) Recall also from Eq.
(14.5.15) and by inspection of Figure 14.5.1, that if ˙ ψ 0 and φ ˙ 0 are positive, then θ must
0
be negative. Hence, from Eq. (14.5.27) we see that λ is positive; thus, stability occurs, if:

( 12 5)ψ 2 0 + φ 0 2 ˙ > 4 ( gr 0 ( 14 5)φ ˙ ˙ sinψ 0 θ 0 (14.5.29)
5 )cosθ
−
˙
0