0
$\begingroup$

The relationship between angular and linear displacement is given by: $$\vec{s}=\vec{\theta} \times \vec{r}$$

If we want to calculate the differential, it is:

$$d\vec{s}=d\vec{\theta} \times \vec{r} + d\vec{r} \times \vec{\theta}$$

In all the derivations I've came across the second term is zero, see here, right under Figure 10.39, because $d\vec{r}$ is zero for some reason. Why is that the case? It's a position vector of our body (say, a single particle), so if it's rotating around origin, its position vector $\vec{r}$ is constantly changing. Then how can $d\vec{r}$ be zero? If the particle is rotating around the origin, even though its length stays constant, the vector itself is changing.

Is this because we're consindering the system in a rotating frame of reference attached to the rotating body?

$\endgroup$
7
  • 1
    $\begingroup$ It is $0$ when walking in a circle because $dr=0$ (no change in radius). It is not true for cases not in a circle $\endgroup$ May 10, 2020 at 15:16
  • $\begingroup$ Ok, but $r$ is a vector, not just a single number (radius). $\endgroup$ May 10, 2020 at 15:25
  • $\begingroup$ In this case, they mean radius. This formula isn't correct if $dr\neq 0$, for example, for a straight line going out from the origin, $d\theta = 0$ but $ds\neq 0$ $\endgroup$ May 10, 2020 at 15:46
  • $\begingroup$ In the linked article they wrote: 'Note that $d\overset{\to }{r}$ is zero because $\overset{\to }{r}$ is fixed on the rigid body from the origin O to point P'. I'm assuming this is wrong then? It should say simply $r$. $\endgroup$ May 10, 2020 at 15:57
  • $\begingroup$ I mean, the radius vector $r$ stays constant if the distance from the centre of rotation doesn't change, even though the particle is rotating and changing its position. Is that how I should understand it? $\endgroup$ May 10, 2020 at 16:06

1 Answer 1

3
$\begingroup$

I don't know what your first equation, $\vec {s}=\vec {\theta} \times \vec {r}.$ means, if the quantities are finite vectors. Just try representing the equation as a diagram! What I think you need is $$\vec {ds}=\vec {d\theta} \times \vec {r}.$$

Here $\vec {ds}$ is the (tangential) displacement of a particle that has displacement $\vec {r}$ from a point on the axis of rotation, that turns through angle $\vec {d\theta}$ about the axis of rotation.

The equation is more usually presented as a relation between a particle's tangential velocity and its angular velocity $\vec \omega$ about the axis: $$\vec v=\vec {\omega} \times \vec {r}.$$

Your worry is therefore unfounded.

$\endgroup$
6
  • $\begingroup$ Well, I took it from here - check for the section below figure 10.39. It states that $\overset{\to }{s}=\overset{\to }{\theta }\,×\,\overset{\to }{r}$, then an attempt is made to derive $\vec {ds}$ from that. Is the article wrong then? $\endgroup$ May 10, 2020 at 18:59
  • $\begingroup$ I think that the article is suspect. I've found the equation s ⃗=𝜃⃗×𝑟⃗ (without the $\times$). It refers the reader to the figure, but this figure doesn't show $s$ or $\theta$. A little later it claims that $\vec{dr}$ is zero for a rotating rigid body. This is not the case. It is $dr$ that is zero. There may well be other mistakes. $\endgroup$ May 10, 2020 at 19:45
  • $\begingroup$ Ok, so $d\vec{r}$ is not zero because the vector IS changing, because the particle IS changing its position, but its magnitude ($r$) is not, therefore $dr = 0$, is that right? $\endgroup$ May 10, 2020 at 19:47
  • $\begingroup$ Yes indeed..... $\endgroup$ May 10, 2020 at 19:50
  • $\begingroup$ This article is apparently the main source of my confusion then. Thanks for clarification! I just assumed I didn't understand that, but I felt there was something wrong with it as well. $\endgroup$ May 10, 2020 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.