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

AI Thread Summary
The discussion centers on the disagreement with a popular book's solution regarding relative motion in rotating frames. It emphasizes that the derivatives of vectors in rotating frames depend on the angular velocity of the frame, which is a key detail often overlooked. The relationship between the derivatives of vectors in different frames is established through a mathematical theorem, highlighting the importance of the frame's basis. The invariant nature of the relative position vector is acknowledged, but its time-derivatives differ across frames. This clarification resolves the initial confusion about the book's explanation.
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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.
 
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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.
 
Very good, that was the detail I overlooked. Thanks.
 
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