# What is the relation between linear and angular velocity? How is v = rω derived? [closed]

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

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

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• I'm voting to close this question as off-topic because there is no evidence of prior effort. – garyp Mar 9 '17 at 11:02
• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Mar 9 '17 at 17:17

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.

• Yes it suffices the need – Aaron John Sabu Mar 9 '17 at 12:03

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! $v = r\omega$
• Prefer infinitesimally small terms - namely dr, d(theta), dt. – Aaron John Sabu Mar 9 '17 at 12:04
• @AaronJohnSabu Thanks, I have edited my answer to reflect this. – Roshan Tabriaz Mar 9 '17 at 12:58

Hint

$l=r\theta$

Differentiating with respect to time.

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

V=r$\omega$

• Ask me in comments if you don't understand anything – Fawad Mar 9 '17 at 11:17

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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