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Is this arc length?

by autodidude
Tags: length
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autodidude
#1
Mar11-13, 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 18-3, in the section where he talks about rotation of rigid bodies.
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mfb
#2
Mar11-13, 02:08 PM
Mentor
P: 11,580
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.
autodidude
#3
Mar11-13, 02:14 PM
P: 333
Thanks mfb

rcgldr
#4
Mar11-13, 05:35 PM
HW Helper
P: 7,033
Is this arc length?

Start with the actual distance:

PQ = sqrt(Δr2 + (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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