Page 245 - Dynamics of Mechanical Systems

0593_C07_fm Page 226 Monday, May 6, 2002 2:42 PM

226 Dynamics of Mechanical Systems

where, as before, the a and b are the n components of n and n . Then, because n and
a
b
i
i
i
a
n are unit vectors, the a and b must satisfy the equations:
b
i
i
1− aa = 0 and 1− b b = 0 (7.10.11)
ii i i
If n and n are also perpendicular to each other, we have:
a
b
ab = 0 (7.10.12)
ii
Equations (7.10.11) and (7.10.12) form three constraint equations to be satisfied while
maximum values of I of Eq. (7.10.10) are sought. Then, by generalizing the foregoing
ab
procedure, we form a function h(a , b , λ, µ, ν) defined as:
i
i
i (
ha b,, , ,ν) = a b I + ( 1 a a ) + ( 1 b b ) + ν a b (7.10.13)
µ −
λ −
λµ
ii ij
ii
i i
ii
i
where λ, µ, and ν are Lagrange multipliers. By setting the derivative of h with respect to
a and b equal to zero, we obtain:
k
k
ν
∂∂ha = I b − 2λ a + b = 0 (7.10.14)
k kj j k k
and
ν
∂∂hb k = I a − 2µ b k + a k = 0 (7.10.15)
ik i
If we multiply these equations by a and b , respectively (and sum over k), we obtain:
k
k
aI b − 2λ = 0 (7.10.16)
kkj j
ν
bI b += 0 (7.10.17)
kkj j
ν
aI a += 0 (7.10.18)
kik i
bI a − 2µ = 0 (7.10.19)
kih i
where we have used Eqs. (7.10.11) and (7.10.12). By comparing Eqs. (7.10.16) and (7.10.19)
and by recalling that I is symmetric, we have:
ik
λ = µ (7.10.20)
Then, Eqs. (7.10.14) and (7.10.15) may be rewritten in the forms:

Ib = 2λ a − ν b (7.10.21)
kj j k k
and

Ia = 2λ b − ν a (7.10.22)
kj j k k