Differece between torque $\tau$ and angular velocity $\omega$ The title is a weird question, i know. But hear me out: Take the equation that relates the tangential velocity $\vec v$ and the angular velocity $\vec \omega$ of an object moving on a circle with radius $r$.
$\vec v = \vec \omega × \vec r$
If we want to know the magnitude of the axial quantity $\omega$, we need to divide the tangential quantity $v$ by the radius $r$.
$v = \omega r \Leftrightarrow \omega = \frac{v}{r}$
Now let's take a look at the case of a torque.
$\vec \tau = \vec r × \vec F$
Here, we can just multiply the radius $r$ by the tangential quantity $F$ to get the axial quantity $\tau$.
$\tau = r F$
Why is it that we multiply in one case and divide in the other, if we want to know the magnitude of the axial quantity, even though the overall structure of these three vectors is the same in both cases. I mean, you can more or less make an ignorant one to one correspondence beteween $\vec \tau$ and $\vec \omega$, $\vec F$ and $\vec v$ and well, $\vec r$ and $ \vec r$, if you comapre both scenarios. This ignorant comparison is depicted below. I mean you can literaly draw these two scenarios in the exact same diagram. But we still end up with two different equations. I find that a bit odd.

I reckon it has something to do with the fact that in the case of the velocity we have a differntial relationship, which is not present in the case of the torque. I. e. the fact that $\vec v = \frac{d\vec r}{dt}$ but $ \vec \tau ≠ \frac{d \vec F}{dt}$. But i can't really pin down a complete explanation.
 A: The magnitude of $ \,\vec c= \vec a\times \vec b $ is
$$|c|=|a||b|\sin(\phi)$$
Where $\phi$ is the angle between vector a and vector b
A: The overall structure of the two equations aren't actually the same despite both using the cross product. The first equation $v = \omega \times r$  is multiplying an axial vector with a vector, and the second equation $\tau = r \times F$ is multiplying a vector with a vector. In three dimensions both of them can be expressed with the cross product, but this isn't true in other dimensions.
The "correct" equations, in the language of geometric algebra, and the correct generalization beyond three dimensions, are actually $v = r \cdot \omega$, where $\cdot$ is the inner product between a vector and a bivector, and $\tau = r \wedge F$ where $r \wedge F$ is the exterior product between a vector and a vector. (Note that geometric algebra generalizes the inner product between a vector and a bivector in a way that doesn't mean you just dot product the vector $r$ and the axial vector $\omega$.)
A: Here's one way to think about why they transform seemingly oppositely. Force and velocity form a pair in the sense that when multiplied you get power:
$$W = \int F \cdot v\,dt.$$
The same is true for torque and angular velocity:
$$W = \int \tau \cdot \omega\,dt.$$
So if to go from force to torque we multiply by $r$, we need to compensate by dividing velocity by $r$.
A: Velocity is the moment of rotation, just as torque is the moment of force, as I explained in this answer.
$$\begin{aligned} 
\vec{v} & = \vec{r} \times \vec{\omega}  \\
\vec{\tau} & = \vec{r} \times \vec{F}  \\
\end{aligned}$$
where $\vec{r}$ points from the measuring/reference point (the origin) to the axis (rotation axis, or line of action axis).
If you assume the perpendicular distance to the axis is $d$ then
$$\begin{aligned} 
\|\vec{v}\| & = d\, \|\vec{\omega}\|  \\
\|\vec{\tau}\| & = d\, \|\vec{F}\|  \\
\end{aligned}$$
so your question is odd to me. In both cases you multiply by the moment arm $d$ to get the magnitude of the moment-of quantity. Just in your post you re-wrote the second expression as $\| \vec{F} \| = \frac{ \| \vec{\tau} \|}{d}$, which means you are interested in the axial vector and not the moment-of vector. You can do the same with motion and write $\| \vec{\omega} \| = \frac{ \| \vec{v} \|}{d}$ if you are interested in the axial vector of motion.
A: Generally torque is just force (angular force that's what I call it) and angular velocity is moment of rotation( as other have described)
Torque occurs in rotational body if some kind of force was occurred which we denoted as $\vec F$. If you remember your study than you may remember that we write $F=dp/dt$. So force is actually changes in momentum with respect to time. The same thing happens with torque. We denote angular momentum $\vec L$. The equation of angular momentum is $\vec L=m\vec r \times \vec v$. You can write torque something just like this : $\tau=\dfrac{dL}{dt}$
