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What is the relation between linear and angular velocity? How is $v = r\omega$ derived?

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closed as off-topic by garyp, Yashas, Kyle Kanos, Jon Custer, David Hammen Mar 9 '17 at 18:06

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    $\begingroup$ I'm voting to close this question as off-topic because there is no evidence of prior effort. $\endgroup$ – garyp Mar 9 '17 at 11:02
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Mar 9 '17 at 17:17
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Hint- $v={ds}/{dt}$. What does this $s$ represent(i.e. which 'path' is your object on)? We know angular velocity is the rate of change of angular displacement, i.e. $\omega=d\theta/dt$.What is the relation between $s$ and $\theta$? Are they proportional? Plug in the known data in the first equation.

I believe this should be sufficient.

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  • $\begingroup$ Yes it suffices the need $\endgroup$ – Aaron John Sabu Mar 9 '17 at 12:03
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Thanks to @GRrocks for guiding me through. This is how $v = r\omega$ is derived:

  1. $velocity = {displacement\over time}$
  2. $v = {ds\over dt}$
  3. Since our displacement/distance is the length of the arc which is made by the angle $d\theta$: $s = rd\theta$
  4. $v = {rd\theta\over dt}$
  5. Since $r$ is constant: $v = r{d\theta\over dt}$
  6. ${d\theta\over dt} = \omega$
  7. Therefore! $v = r\omega$
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  • $\begingroup$ Prefer infinitesimally small terms - namely dr, d(theta), dt. $\endgroup$ – Aaron John Sabu Mar 9 '17 at 12:04
  • $\begingroup$ @AaronJohnSabu Thanks, I have edited my answer to reflect this. $\endgroup$ – Roshan Tabriaz Mar 9 '17 at 12:58
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Hint

$l=r\theta$

Differentiating with respect to time.

$\dfrac{dl}{dt}=r\dfrac{d\theta}{dt}$

V=r$\omega $

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  • $\begingroup$ Ask me in comments if you don't understand anything $\endgroup$ – Fawad Mar 9 '17 at 11:17
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It's simple since velocity = $\frac{dx}{dt}$ and displacement is the arc length so $x = r\theta$ and only theta varies not the radius so $\frac{dx}{dt}=r\frac{d\theta}{dt}$ which is $v= r\omega $.

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