Did I Miss Something in Disputing a Popular Book's Solution on Relative Motion?

[guv]

Homework Statement

200 More challening physics problems question #2

Ann is sitting on the edge of a carousel that has a radius of 6 m and is
rotating steadily. Bob is standing still on the ground at a point that is 12 m from the
centre of the carousel. At a particular instant, Bob observes Ann moving directly
towards him with a speed of 1 m s−1. With what speed does Ann observe Bob to
be moving at that same moment?

The hint from the book: Be careful, the transformation principle due to Galileo Galilei applies
only to inertial reference frames. The idea that Ann simply observes Bob moving
towards her with a speed of 1 m s−1 is false.

Solution: some complex calculation involving using the center of the carousel. Bob's velocity relative to the center is decomposed into tangetial and radial components, the tangential components $\sqrt{3}$ m/s is used as solution.

Relevant Equations

$$v_{a/b} = - v_{b/a}$$

I do not agree, this is bullocks. We can simply set up position vector of $\vec A(t)$ and $\vec B(t)$ with respect to the fixed center of the carousel, their relative velocity is simply $\frac{d (A-B)}{dt}$ or $\frac{d (B-A)}{dt}$

Since this is a pretty popular book, I am wondering if I overlooked any detail in disputing the book's solution.

 

[etotheipi]

The book is right! The answer is that in general, for frames that are permitted to rotate, the derivatives of vectors depend on the angular velocity of the frame! As a theorem, if frame $\mathcal{B}$ rotates at $\boldsymbol{\omega}$ relative to frame $\mathcal{A}$, then the derivative of an arbitrary vector $\mathbf{u}$ with respect to these two frames are related by$$\left( \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \right)_{\mathcal{A}} = \left( \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \right)_{\mathcal{B}} + \boldsymbol{\omega} \times \mathbf{u}$$It is because of how we define the derivatives of vectors with respect to $\mathcal{A}$ and $\mathcal{B}$. If $\{\mathbf{e}_i \}$ is the basis of $\mathcal{A}$, for instance, and $\{\tilde{\mathbf{e}}_i \}$ a basis for $\mathcal{B}$ then you have $\mathbf{u} = u_i \mathbf{e}_i = \tilde{u}_i \tilde{\mathbf{e}}_i$, and we define:$$\left( \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \right)_{\mathcal{A}} := \sum_i \frac{\mathrm{d}u_i}{\mathrm{d}t} \mathbf{e}_i$$i.e. treating the $\mathcal{A}$ basis as constant, whilst$$\left( \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \right)_{\mathcal{B}} := \sum_i \frac{\mathrm{d}\tilde{u}_i}{\mathrm{d}t} \tilde{\mathbf{e}}_i$$i.e. treating the $\mathcal{B}$ basis as constant. You can see how they are related by writing $\tilde{\mathbf{e}_i} = R_{ij}(t) \mathbf{e}_j$ where $R_{ij}(t)$ is a time-dependent rotation matrix.

So even though the relative position vector $\mathbf{x}$ is indeed invariant (as all vectors are), its time-derivatives with respect to both frames are different. For more info consult a classical mechanics text e.g. Douglas Gregory.

 

[guv]

Very good, that was the detail I overlooked. Thanks.