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?

  • 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


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.

  • $\begingroup$ 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

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