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circular motion - how to prove/derive v=rw?

title says it all :smile:
Reply 1
Avatar for -G-a-v-
-G-a-v-
16
You're moving in a circle of radius r, so the distance you travel in one full trip round the circle is 2.pi.r and you're doing it in one time period, T

speed, v = distance/time = 2.pi.r/T

w = angular speed, and you complete 2 pi radians in time T, so w = 2pi/T

so v=rw
Thankyou! :smile:

you can have a little dribble of green tomorrow! (alredy repped today :s-smilie:) :biggrin::p:
Reply 3
Avatar for -G-a-v-
-G-a-v-
16
Heh, thankyou muchly :smile: I shall look forward to it :p:
or you could use calculus to prove it:

S=rθ→dSdt=d(rθ)dt=rdθdt+θdrdt S = r \theta \to \frac{d S}{dt} = \frac{d (r \theta)}{dt} = r \frac{d \theta}{dt} + \theta \frac{d r}{dt}

If your moving on a circle, your radius is not changing with time so drdt=0 \frac{dr}{dt} = 0

so v=dSdt=rdθdt=rω v = \frac{dS}{dt} = r \frac{d \theta}{dt} = r \omega
0 div curl F
or you could use calculus to prove it:

S=rθ→dSdt=d(rθ)dt=rdθdt+θdrdt S = r \theta \to \frac{d S}{dt} = \frac{d (r \theta)}{dt} = r \frac{d \theta}{dt} + \theta \frac{d r}{dt}

If your moving on a circle, your radius is not changing with time so drdt=0 \frac{dr}{dt} = 0

so v=dSdt=rdθdt=rω v = \frac{dS}{dt} = r \frac{d \theta}{dt} = r \omega


ahhh clever!

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