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Tinkering around with rotational movement I ended up with this equation:

diagram of the situation

$$\omega=\frac{v_t}{r}=\frac{v}{r}\sin\theta = \frac{\sin \theta}{r} \frac{dr}{dt} = \frac{\sin \theta}{r} \frac{dr}{d\phi}\frac{d\phi}{dt} = \frac{\sin \theta}{r} \frac{dr}{d\phi} \omega$$

$$r = \frac{dr}{d\phi} \sin \theta$$

I can't understand exactly what this means. Is it correct? If so, what is the intuition behind it?

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  • $\begingroup$ v is not #\frac{dr}{dt}#. You can see that is not right if you consider uniform circular motion. r is constant so all its derivativs are zero so by tour formulas you will have r and omega equal to zero, $\endgroup$
    – nasu
    Commented Feb 11, 2022 at 13:59
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    $\begingroup$ The magnitude of r is not a function of $\phi$ thus $\dfrac{dr}{d\phi }=0$ $\endgroup$
    – Eli
    Commented Feb 11, 2022 at 15:46

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v is not $\frac{dr}{dt}$. You can see that is not right if you consider uniform circular motion. r is constant so all its derivativs are zero so by your formulas you will have r and omega equal to zero,

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