Why is Torque defined the way it is? According to my professor, Torque is like the connection of force in the angular world. But, we know that most quantities such as angular acceleration, angular velocity, etc. are defined as something divided by $r$. So then why is torque defined as the product of $r$ and $F$, when intuitively it should be $F/r$?
 A: Torque, or moment, is when the application of a force causes, or potentially causes, rotation. The radius of the rotation is the moment arm. The greater the moment arm the greater the torque for a given force, which is why torque is the product of the force and radius.
A simple example is a wrench, where the force you apply is perpendicular to the end of the handle of the wrench in order to tighten or free a bolt. The longer the wrench handle (radius of rotation) the less force you need to apply (the easier it is to tighten or free the bolt.)
Hope this helps
A: Intuitively, the radius can be somewhat compared to a lever. The longer the lever the greater the torque that the force applies. Similarly the farther out on the radius, the greater the torque. This is why r and F are multiplied.
A: Tl;dr: how much spin an object around a point when a force is acted, depends on where the body is relative to that point and how much force exerted.
For example, imagine you are on a circular race track, then the amount of 'effort' you have to do to increase your speed tangential to the race track increases as you increase the radius of the circle which you are moving around in.
In another way, the amount of 'effort' you need to increase your spin speed around some axis depends on the perpendicular distance from that axis to you.

For example, consider the work doing in moving a body around a small infinitesiam circular arc with the arc's center at some origin $O$. Say a force acts on it for some arc distance $ds$ then(*),
$$ \mathrm{dW} = \vec{F}\cdot \vec{ds}$$
Now since it's a circular path,
$$ \vec{ds} = |r| d \theta \vec{ \theta}$$
Hence,
$$ \mathrm{dW} = |r||F_{\perp}| d \theta$$
Or,
$$ \frac{dW}{d \theta} = |r|F_{\perp}|$$
So, it's pretty easy to see that the rate of change of work as the angle changes depends on how the $r$ and $F$ this quantity on the RHS. This quantity we can define as torque with:
$$ W_{angular} = \int \vec{\tau} d \theta$$

More generally speaking, we can say that the $\vec{ds}$ has to components, the arc component and a radial component:
$$ \vec{ds} = r d \theta \hat{\theta} + dr \vec{r}$$
So, $$ F \cdot ds = rF_{tngt} d \theta + F_{radial} dr $$
The second component is the radial force component while the first is the angular / tangential force component, so the total work:
$$ W = \int F_{rdl} dr + \int rF_{tngnt} d \theta$$

Note:
(*): When you dot two vectors only the components which they don't have in common goes to zero:
Eg:
Imagine a vector $q$ having components perpendicular and parallel to vector $p$
$$ (q_{ \perp} + q_{\parallel} ) \cdot (  p) =  q_{ \parallel} \cdot p$$
So, dotting the force with the circular arc, extracts out the component which is acting along the tangential direction immediately.
Refer here
