Feynman Lectures on Physics - Rotation in 2 dimensions

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Homework Help Overview

The discussion revolves around understanding concepts presented in section 18-2 of Volume 1 of The Feynman Lectures on Physics, specifically related to rotation in two dimensions. The original poster raises questions about the geometric relationships and mathematical expressions involved in the reasoning presented by Feynman.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the validity of a right angle in a specific geometric configuration and seeks clarification on the relationship between arc length and angular displacement. Participants explore the implications of small angle approximations and Taylor series in this context.

Discussion Status

The discussion is active, with participants providing insights into the mathematical reasoning behind the original poster's questions. Some guidance has been offered regarding the small angle approximations, but the inquiry about the right angle remains open.

Contextual Notes

The original poster references a specific section of Feynman's lectures and has checked for errata, indicating a desire for clarity on the material presented. There is an emphasis on understanding rather than solving a specific problem.

ZirconiumAce
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This isn't really a 'problem', I'm just trying to follow Feynman's reasoning in section 18-2 of Volume 1 The Feynman Lectures on Physics. I've attached a png of the paragraph in question.

I have 2 issues with this:

1. If the length OQ = OP, how can there be a right angle at P(x,y)?

2. I don't understand why QP = r \Delta\theta, rather than r * tan (\Delta\theta)

Am I missing something? I've checked the latest errata and this isn't mentioned.

Thanks for your help.
 

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Welcome to PF!

Hi ZirconiumAce! Welcome to PF! :wink:
ZirconiumAce said:
1. If the length OQ = OP, how can there be a right angle at P(x,y)?

2. I don't understand why QP = r \Delta\theta, rather than r * tan (\Delta\theta)

Because that's the way these ∆s work …

we ignore anything that's small compared with a ∆ (in the same way that we ignore dh2, compared with dh, when we do calculus proofs)

So ∆θ = sin∆θ = tan∆θ. :wink:

(sin∆θ = ∆θ - (∆θ)3/3! + …)
 
Ah yes a Taylor series. Mathematics back in the days of the slide rule I suppose. Now we don't bother with simplifications as much. Cheers! How about the right angle?
 
I've got it.
 

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