Displacement vector for rotational motion

Question

When I try to find out how to compute work for rotational motion. I found an equation from a book online with a figure and equation as follows:

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

$Thus,$

$d\vec{s}\ =d\left(\vec{\theta}\ \times\vec{r}\ \right)=d\vec{\theta}\ \times\vec{r}\ +d\vec{r}\ \times\vec{\theta}\ =d\vec{\theta}\ \times\vec{r}\ $

$Note\ that\ d\vec{r}\ is\ zero\ because\ \vec{r}\ is\ fixed\ on\ the\ rigid\ body\ from\ the\ origin\ O\ to\ point\ P. $

What I don't understand is that it says $ \vec{s}\ =\vec{\theta}\ \times\vec{r}$, it means the direction of $ \vec{\theta}\ $ is perpendicular to the direction of $ \vec{s}\ $. However, $ d\vec{s}\ $ and $\hat{\theta}$ are in the same direction in the figure.

In addition, from what I learned in my other physics class: $ d\vec{s}\ = r·d\theta·\hat{\theta} $ and it makes sense to me since $ d\vec{s}\ $ is the infinitesimal arc length in $\hat{\theta}$ direction(⊥ to $ \hat{r} $).

Therefore, I am not comfortable with $ \vec{s}\ =\vec{\theta}\ \times\vec{r} $ which is claiming that $ \vec{\theta}\ $ is perpendicular to $ \vec{s}\ $. I was wondering if someone could please give me some insights on this?

Reference Book: Physics book

Answer 1

This seems to be a case of the book not clarifying what notation is being used. The vector in the figure, $\hat{\theta}$, is the unit vector which points in the direction of increase of the angle $\theta$.

The vector $\vec{\theta}$, however, is a different vector defined to be the vector out of the page corresponding to the rotation via the right hand rule (see Equation 10.2.2 from the same book here). In this case, $\vec{\theta}$ will be a vector out of the page from $O$. The cross product now makes sense, with $\vec{s}$ being perpendicular to both $\vec{\theta}$ and $\vec{r}$.