Page 105 - Dynamics of Mechanical Systems

0593_C04*_fm Page 86 Monday, May 6, 2002 2:06 PM

86 Dynamics of Mechanical Systems

To see this, let c be expressed in terms of the unit vectors n , n , and n . That is, let
1
3
2
c = c n + c n + c n (4.5.3)
1 1 2 2 3 3
Because c is fixed in B, the scalar components c (i = 1, 2, 3) are constants; hence, the
i
derivative of c in R is:
=
+
+
c
d dt c dn dt c n dt c n dt (4.5.4)
1 1 2 2 3 3
R
B
Next, consider the product ω × n :
1
  dn   dn   dn  


R B 2 3 1

ωω× n =   ⋅nn +  ⋅nn +  ⋅nn  × n
1  3  1  1  2  2  3 1
 dt dt dt 
(4.5.5)
 dn   dn 


=− 3 ⋅nn +  1 ⋅nn
 dt 1  3  dt 2  2
Recall that:
⋅
⋅
nn = 1 and nn = 0 (4.5.6)
1 1 1 3
By differentiating these expressions we have:
dn d n d n
1 ⋅ n = 0 and 1 ⋅ n = − n ⋅ 3 (4.5.7)
dt 1 dt 3 1 dt
Hence, Eq. (4.5.5) may be written as:
 dn   dn   dn 


R B =  1 +  1 +  1

ωω× n ⋅nn ⋅nn ⋅nn (4.5.8)
1  dt 1  1  dt 2  2  dt 3  3
Because any vector V may be expressed as (V · n )n + (V · n )n + (V · n )n , we see that
1
3
2
3
2
1
the right side of Eq. (4.5.8) is an identity for dn /dt. Thus, we have the result:
1
R B =
ωω× n dn dt (4.5.9)
1 1
Similarly, we have the companion results:
R B = R B =
ωω × n dn dt and ωω × n dn dt (4.5.10)
2 2 3 3
Finally, by using these results in Eq. (4.5.4) we have:
ddt = c ωω B × n + c ωω B × n + c ωω B × n
R
R
R
c
1 1 2 2 3 3
R
= ωω B c n + c n + c n ) (4.5.11)
2 2 3 3
×( 1 1
R
= ωω B × c