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I'm trying to follow the Feynman lectures and I'm stuck on a particular piece. Let me frame it for you before I ask my question. Nobody does it better than the author himself,

we must find out how a particular particle moves when the angular velocity is such and such. To do this, let us take a certain particle which is located at a distance $r$ from the axis and say it is in a certain location $P(x,y)$ at a given instant, in the usual manner. If at a moment $\Delta t$ later the angle of the whole object has turned through $\Delta \theta$, then this particle is carried with it. It is at the same radius away from $O$ as it was before, but is carried to $Q$. The first thing we would like to know is how much the distance $x$ changes and how much the distance $y$ changes. If $OP$ is called $r$, then the length $PQ$ is $r \Delta \theta$, because of the way angles are defined. The change in $x$, then, is simply the projection of $r \Delta \theta$ in the $x$-direction: $$\Delta x = -PQ \sin \theta = -r \Delta \theta \dot (y/r) = -y \Delta \theta$$ Similarly, $$\Delta y = +x \Delta \theta$$

Angular velocity problem

Source: Feynman Lecturs on Physics

Now I will try to reproduce this in a way I find logically consistent: we can imagine an isosceles triangle formed by $ABC$ where the angle $ACB$ is $\gamma$ and the line segment $AC$ is $b$, the line segment $CB$ is $a$ and the line segment of $AB$ is $c$. The law of cosines says that, $c^2 = a^2 + b^2 - 2ab$. In the case of an isosceles triangle we can simplify this by saying $a=b=r$ ($r$ because that is what we have in our example), therefore our law simplifies to: $c^2 = r^2 + r^2 - 2r^2 \cos \gamma = 2r^2(1 - \cos \Delta \theta)$.

If we suppose that $c$ is the length of the line segment $PQ$ then it should be: $$c = \sqrt{ 2r^2(1 - cos \Delta \theta)}$$ Instead, Feynman says that this should be $r\Delta \theta$. Okay, maybe it is only an implicit approximation for infinitesimal angles, so we can look and see that $\cos \Delta \theta \approx 1$ for very small angles which immediately implies that $c \approx 0$. So why and how does Feynman find $c=r \Delta \theta$? Where did I go wrong?

Bernhard
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sciencenewbie
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1 Answers1

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Can you get any further when you know the Taylor expansion of $cos(\Delta \theta)$

$$ cos(x) \approx 1 - \frac{1}{2}x^2 + \mathcal{O}(x^4)$$

Bernhard
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  • Well, that appears to change every thing... $c= \sqrt{ 2 r^2 (1- (1 - \frac{1}{2}x^2 + \mathcal{O}(x^4) ) )}$ but I assume $\mathcal{O}$ is the big-O notation I've heard so much about and for a very small $x$, $x^4$ small enough to be neglected. So $c=\sqrt{2r^2 (1-1+\frac{1}{2}\Delta\theta^2)} = \sqrt{r^2 \Delta\theta^2} = c \Delta \theta$. I feel so stupid for not thinking about using a Taylor series. Thank you! I would upvote you if I could. It was not obvious to me that a Taylor series would give me a different result, do you just recognize this by experience? – sciencenewbie Dec 10 '12 at 09:48
  • You already used a Taylor expansion, when assuming $cos(x)\approx1$. Only, if you find out that everything cancels, you just take one more term :) – Bernhard Dec 10 '12 at 10:43