Page 114 - Dynamics of Mechanical Systems

0593_C04*_fm Page 95 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 95

B
R
Then α is:
d ωω
R αα = R R B + R
B
dt = ˙ ω n
11 w dn 1 dt + ˙ ω n 2
1
2
+ω R dn dt + ˙ ω n + ω R dn dt
2 2 3 3 3 3
(4.8.10)
= ˙ ω n + ˙ ω n + ˙ ω n
1 1 2 2 3 3
+ω R dn dt + ω R dn dt + ω R dn dt
1 1 2 2 3 3
Observe that the last three terms each involve derivatives of unit vectors fixed in B. From
Eq. (4.5.2) these derivatives may be expressed as:
R dt= ωω B × i ( )
R
dn n i = 12 3 (4.8.11)
,,
i
Hence, the last three terms of Eq. (4.8.10) may be expressed as:
R
R
R
ω dn dt + ω dn dt + ω dn dt
1 1 2 2 3 3
R
R
R
= ω ωω B × n + ω ωω B × n + ω ωω B × n
1 1 2 2 3 3
R
R
R
= ωω B × ω n + ωω B × ω n + ωω B × ω n (4.8.12)
1 1 2 2 3 3
R
= ωω B ×( ω n + ω n + ω n )
1 1 2 2 3 3
R
R
= ωω B × ωω B = 0
B
Therefore, αα αα becomes:
R
R B
αα= ˙ ω n + ˙ ω n + ˙ ω n (4.8.13)
1 1 2 2 3 3
Example 4.8.1: A Simple Gyro
See Example 4.7.1. Consider again the simple gyro of Figure 4.7.5 and as shown in Figure
4.8.1. Recall from Eq. (4.7.13) that the angular velocity of the gyro D in the fixed frame R
may be expressed as:
ωω= ( Ω + s ˙ ψ φ Y ) N 1 + c ˙ ψ N Y2 + φN Y3 (4.8.14)
R D ˙
φ
θ
˙
where as before Ω (the gyro spin) is ; φ and ψ are orientation angles as shown in Figure
4.8.1; and N , N , and N are unit vectors fixed in the yoke Y. Determine the angular
Y2
Y1
Y3
acceleration of D in R.
Solution: By differentiating in Eq. (4.4.14) we obtain:
αα= ( Ω ψ s + ˙ ˙ c φ Y ) N 1 +( Ω ψ φ dN Y1 dt
ψφ
+ ˙
+ ˙˙
R D ˙ s ) R
φ
+( ψ c − ˙ ˙ s φ Y ) N 2 + ˙ ψ c dN Y2 dt (4.8.15)
ψφ
˙
R
φ
φ
˙˙
+φN Y3 + φ R dN Y3 dt