be a right handed system of unit basis vectors. Thus k
(t)
= i
(t)
×j
(t)
and each vector has unit length. This represents a moving coordinate system. We assume
that i
(t)
,j
(t)
,k
(t)
are each continuous having continuous derivatives, as many as
needed for the following manipulations for t in some open interval. The various rules of
differentiation of vector valued functions will be used to show the existence of an angular
velocity vector.
Lemma 17.4.1The following hold. Whenever r
(t)
,s
(t)
are two vectors from
{i(t),j(t),k (t)}
,
r (t)⋅s′(t) = − r′(t)⋅s (t)
In particular, the case wherer = s,implies r^{′}
(t)
⋅ r
(t)
= 0.
Proof: By assumption, r
(t)
⋅ s
(t)
is either 0 for all t or 1 in case r = s. Therefore,
from the product rule,
r(t)⋅s′(t)+ r′(t)⋅s(t) = 0
which yields the desired result. ■
Then the fundamental result is the following major theorem which gives the existence
and uniqueness of the angular velocity vector.
Theorem 17.4.2Let
(i(t) ,j(t),k(t))
be a right handed orthogonal system ofunit vectors as explained above. Then there exists a unique vector Ω
(t)
, the angularvelocity vector,such that for r
(t)
any of the
{i(t),j(t),k(t)}
,
r′(t) = Ω(t)× r(t)
Proof:First I will show that if this angular velocity vector Ω
(t)
exists, then it
must be of a certain form. This will prove uniqueness. After showing this, I will
verify that it does what it needs to do by simply checking that it does so. In all
considerations, recall that in the box product, the × and ⋅ can be switched. I
will use this fact with no comment in what follows. So suppose that such an
angular velocity vector exists. Then i^{′}
(t)
= Ω
(t)
× i
(t)
with a similar formula
holding for the other vectors. Also note that since this is a right handed system,
i
(t)
×j
(t)
= k
(t)
,j
(t)
×k
(t)
= i
(t)
, and k
(t)
×i
(t)
= j
(t)
as earlier. In addition, if you
want the component of a vector v with respect to some r
(t)
, it is v ⋅ r
(t)
= v_{r}
(t)
.
Thus
v =vii(t)+ vjj(t)+ vkk(t), vr = v⋅r(t) for each r(t) ∈ {i(t),j(t),k(t)}
is the angular velocity vector and there is only one. Of course it might
have different descriptions but there can only be one and it is the vector just described.
■
This implies the following simple corollary.
Corollary 17.4.3Let u
(t)
be a vector such that its components with respectto the basis vectors i
(t)
,j
(t)
,k
(t)
are constant. Then u^{′}
(t)
= Ω
(t)
× u
(t)
.
Proof:Say u
(t)
= u_{i}i
(t)
+ u_{j}j
(t)
+ u_{k}k
(t)
. Then
u′(t) = uii′(t)+ ujj′(t)+ ukk′(t)
= uΩ (t) ×i(t)+ u Ω(t)× j(t)
i j
+ukΩ (t)× k (t)