
#1
Mar1113, 01:53 PM

P: 333

Could someone please explain why PQ in the diagram below is rΔθ? Isn't rΔθ arc length?
The best reason I can think of is that it's only an approximation for when the angle is very small, so PQ≈arclength=rΔθ. Not 100% sure though. The diagram is from the first volume of the Feynman lectures in 183, in the section where he talks about rotation of rigid bodies. 



#2
Mar1113, 02:08 PM

Mentor
P: 10,767

It is an approximation for small ##\Delta \theta##, right. I would expect that it is used in a differential somewhere, where the approximation gets exact.




#3
Mar1113, 02:14 PM

P: 333

Thanks mfb




#4
Mar1113, 05:35 PM

HW Helper
P: 6,925

Is this arc length?
Start with the actual distance:
PQ = sqrt(Δr^{2} + (r sin(Δθ))^{2}) If this is circular motion, then Δr = 0, and as Δθ > 0, then sin(Δθ) > Δθ, and you end up with lim Δθ > 0 of sqrt((r sin(Δθ))^{2}) > sqrt((r Δθ)^{2}) > r Δθ. If r is some function of θ, then as long as Δr approaches zero more rapidly than r sin(Δθ), then lim Δθ > 0 of f(Δr, r sin(Δθ)) > f(0, r sin(Δθ)). 


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