Rate of change of a vector

Question

I'm studying the book "Classical Mechanics" by Goldstein together with a coursebook my professor provided.

I'm having trouble grasping how to intuitively determine what the rate of change of a vector is affected by.

Earlier on in the book, I saw a body with 3 frames of reference:
$Oxyz$ which is just an arbitrary coordinate system.
$O'x'y'z'$ which has a fixed origin in a point of the body and fixed axes that rotate corresponding to the body.(Body fixed?)
$O'xyz$ which has a fixed origin in a point of the body and fixed axes that are parallel with the original coordinate system.(Space fixed?)

Now, in my professors book it says the following:

It is clear that there are 2 sources of time-dependence in the Carthesian components of a general vector in the (body-fixed) coordinate system:
On one hand: The intrinsic time-dependence of the vector.
On the other hand: The time dependence of the basisvectors $\vec n'_i$ of the moving coordinate system, with respect to which the Carthesian components (orthogonal projection of the vector onto $\vec n'_i$) will be determined.

I'm confused as to what exactly is meant by this paragraph.
What exactly is the intrinsic time-dependence of the vector? To my knowledge, in the body-fixed coordinate system every vector stays constant, so there is no rate of change.
I understand that the $\vec n'_i$ vectors move through space, but during a translation they wouldn't change since they are their orientation stays the same. During a rotation however I understand they would change.

What am I getting wrong here? I have a feeling I'm mixing these frames of reference up since it's not so clear in my book. I think that I'm misunderstanding how these frames of reference behave as the body moves.

Answer 1

1. Consider a changing vector $\vec{A}(t)$ on a fixed coordinate system. The rate of change of the vector components are $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t}= \dot{\vec{A}}(t) $$
2. Consider a fixed vector $\vec{A}$ on a rotating coordinate system, with angular velocity $\vec\Omega(t)$. The rate of change of the vector components with respect to the attached frame are zero, but with respect to an inertial frame the rate of change is $$\frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \vec{\Omega}(t) \times \vec{A} $$
3. Combined the changing $\vec{A}(t)$ vector on a rotating frame $\vec\Omega(t)$ is $$\boxed{ \frac{{\rm d}\vec{A}(t)}{{\rm d}t} = \frac{\partial \vec{A}(t)}{\partial t} + \vec{\Omega}(t) \times \vec{A}(t)}$$

The first part is the intrinsic change along the rotating frame, and the second part the change due to the rotation of the coordinates.

Example

Two bodies are connected with a slider joint, defined by an axis $\vec{e}$ fixed to the first body which is rotating by $\Omega(t)$. If the joint distance is $\chi(t)$ then the position of the 2nd body is defined relative to the first body as $$\vec{r}_2(t) = \vec{r}_1(t) + \vec{e}\, \chi(t)$$

The position vectors are differentiated to derive the velocity kinematics as:

$$ \vec{v}_1 = \frac{{\rm d}\vec{r}_1(t)}{{\rm d}t} \\ \vec{v}_2 = \frac{{\rm d}\vec{r}_2(t)}{{\rm d}t} $$

$$\begin{align} \vec{v}_2 & = \vec{v}_1 + \frac{{\rm d}(\vec{e}\,\chi(t))}{{\rm d}t} \\ & = \vec{v}_1 + \frac{\partial \vec{e} \chi(t)}{\partial t} + \vec{\Omega}(t) \times \vec{e} \,\chi(t) \\ & = \vec{v}_1 + \vec{e} \dot\chi (t) + \vec{\Omega}(t) \times \left( \vec{r}_2(t) - \vec{r}_1(t) \right) \end{align}$$

_{See related answer: Derivation of Euler's equations for rigid body rotation}