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In the book "Fundamental Laws of Mechanics" by I.E Irodov, on page 15, he writes as a footnote,

note the in general case $ |\mathrm{d}\textbf{r}| \neq \mathrm{d}r$ where $r$ is the modulus of the radius vector $\textbf{r}$, and $ v \neq \frac{\mathrm{d}r}{\mathrm{d}t}$. For example, when $\textbf{r}$ changes only in direction, that is the point moves in a circle, then $ r= \textit{const}$; $\mathrm{d}r=0$ but $ |\mathrm{d} \textbf{r}| \neq 0$.

The reason this doesn't make sense to me, is suppose we have a position vector modelling circular motion, suppose,

$$\vec{ r(t)} = \vec{r_o} + ( \cos \theta \vec{i} + \sin \theta \vec{j})$$

Where, $ \theta$ is a function of time (with $ \frac{d \theta}{dt} =\omega$ and $r_o$ is a constant vector, then,

$ \mathrm{d(\vec{r})} = \omega ( -\sin \theta \vec{i} + \cos \theta \vec{j}) dt $

now,

$ |\mathrm{d(\vec{r})}| = \omega dt$

Now, neither of these are zero, so I'm looking for a simpler explanation of what Irodov is saying

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    $\begingroup$ Why are you using a different scenario other than what is described in the quote? It says "For example, when $\mathbf r$ changes only in direction", and then you say "this doesn't make sense because think of this scenario where $\mathbf r$ is changing in both magnitude and direction." What Irodov is saying has no application to the scenario you have made up. It might help to explain why you think they are the same thing. $\endgroup$ Jul 31, 2020 at 12:51
  • $\begingroup$ doesn't the equation I have written model motion of a particle around a circle centered at $r_o$ $\endgroup$ Jul 31, 2020 at 13:12
  • $\begingroup$ Yes, that is what your equations are. But since you have moved the origin $\mathbf r$ is no longer just changing direction. $\mathbf r$ depends on where your origin is, since it is a vector pointing from the origin to some other point in space. Is this the issue then? $\endgroup$ Jul 31, 2020 at 13:13
  • $\begingroup$ I am starting to realize that now, that I've read the answer of zero the hero as wel $\endgroup$ Jul 31, 2020 at 13:14

2 Answers 2

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In your example $\mathbf{r}$ does not change in direction only. There is no reason to believe that (for $\theta=0$) $\vec r_0+\hat i$ has the same magnitude as (for $\theta=\pi/2$) $\vec r_0+\hat j$.

In the statement “the point moves in a circle”, the center of the circle is assumed to be at the origin.

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  • $\begingroup$ Ok, that works so he meant a circle cantered at origin $\endgroup$ Jul 31, 2020 at 13:13
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I think the author is not refering to the same situation you are thinking on: if the origin does not coincide with the center of the circle, $|dr|$ is never zero, because $|\vec{r}|$ varies between $r_0+R$ and $r_0-R$. The affirmation the author makes is only valid if $r_0=0$. In that case, $\vec{r}$ changes its direction, but not its modulus.

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