I have the following exercise in my notes and I need some explanations.
We consider a fluid which rotates around the $z-$ axis with constant angular speed $w$
$$\overrightarrow{r}=x \hat{\imath}+y \hat{\jmath}$$
$$\overrightarrow{\theta}=-y \hat{\imath}+x \hat{\jmath}$$
$$\hat{\theta}=\frac{1}{r}(-y \hat{\imath}+x \hat{\jmath})$$
$$\overrightarrow{v}=wr(\frac{-y \hat{\imath}+x \hat{\jmath}}{r})$$
$$\overrightarrow{v}=w(-y \hat{\imath}+x \hat{\jmath})$$
Which is the general formula of $v$ that we have used?
Isn't it $wr$ as it is shown at the picture?
Let $(x,y)$ denote the rectangular coordinates of a point and $(r,\theta)$ the corresponding polar coordinates where $x=r\cos(\theta),y=r\sin(\theta). $ The position vector of a fluid particle located at $(x(t),y(t))$ at time $t$ is
$$\overrightarrow{r}(t)= x(t)\overrightarrow{i}+y(t)\overrightarrow{j}=r(t)[\cos(\theta(t))\overrightarrow{i}+\sin(\theta)(t)\overrightarrow{j}].$$
The velocity of this particle is
$$\overrightarrow{v}(t)=\frac{d}{dt}\overrightarrow{r}(t)= r'(t)[\cos(\theta(t))\overrightarrow{i}+\sin(\theta(t))\overrightarrow{j}]+r(t)\theta'(t)[-\sin(\theta(t))\overrightarrow{i}+\cos(\theta(t))\overrightarrow{j}]$$
In a rigid body rotation with angular speed $w$, we have
$$r'(t)=0, \\\ \theta'(t) = w,$$
so
$$r(t)=r=const, \\\ \theta(t) = wt +\theta_0,$$
Hence,
$$\overrightarrow{v}(t)= rw[-\sin(\theta(t))\overrightarrow{i}+\cos(\theta(t))\overrightarrow{j}]= w[-y(t)\overrightarrow{i}+x(t)\overrightarrow{j}]$$
The magnitude of the velocity (speed) is therefore $rw$.