General question about a derivation

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The discussion centers on the derivation of the equation v_{B}= \omega_{B|A} r_{A}, which relates linear velocity (v) to angular velocity (\omega) and the radius (r) in the context of a wheel's rotation. The user seeks clarification on how this formula is justified, particularly when point "A" is at the center of the wheel and point "B" is at the edge. The relationship v= \omega r is established, but the user expresses confusion over the absence of a clear derivation in their reference material. The discussion concludes that the book simplifies the explanation by omitting certain details to streamline the mathematical process.

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wolfmanzak
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I just have a question of "why/how?" I know that for instance \mathbf v=\omega \hat k \times \mathbf r where \mathbf v is my vector for velocity, \omega is my angular velocity and \mathbf r is my position vector from a point on the axis of rotation of a wheel to a point on the outer edge of the wheel. I also know that v= \omega r

But I'd like to understand how it's possible to derive/justify the following from what I have above or if there is another means by which this justification is made. I'm just trying to understand a formula.

v_{B}= \omega_{B|A} r_{A}

This question came up because I saw the final formula at the bottom used in part to solve for the angular velocity of a wheel rotating about a fixed axis where point "A" was at the center of the wheel and point "B" was along the wheel's edge. I guess I'm just trying to figure out why this equation was used, as I don't see any derivation or reasoning for it in text that I'm using and I wouldn't necessarily have thought to use it like shown if I were solving a similar problem. Any explanation as to why/how or what prompted the book to use the equation in this way would really help my understanding of the topic. Thanks in advance.
 
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Change your vector equation to one that deals only in magnitudes and you'll get the one at the bottom.
The book is not making a definition and leaves some details implicit in order to make the math easier.
 

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