newtonian mechanics - Why is the Torque divided by the radius but other rotational analogs multiplied?

I'm having trouble building intuition for circular motion. I understand that torque is the rotational analog of force. Why do we multiply the tangential force by the radius while we multiply the rotational analogs of distance, velocity, and acceleration by the radius?

I understand that the moment of inertia is responsible for the r term for a point mass, but is there any intuitive ways of relating it?

Answer

Torque is not the equivalent of force for rotation. Torque is a measure of distance for a force. The definition of torque is exactly the moment of force

$$\mathbf{T} = \mathbf{r} \times \mathbf{F}$$

Conversely, velocity isn't the equivalent of rotational velocity. Velocity is a measure of distance for rotational motion. The definition of velocity is exactly the moment of rotation

$$ \mathbf{v} = \mathbf{r} \times \boldsymbol{\omega}$$

Additionally, angular momentum is the moment of momentum

$$\mathbf{L} = \mathbf{r} \times \mathbf{p}$$

_{Above $\times$ is the vector cross product}

The fundamental relationship in mechanics, is newton's 2^nd law, and Euler's law of rotation as commonly expressed at the center of mass C:

$$\begin{aligned} \mathbf{F} & = \frac{{\rm d}}{{\rm d}t}( m \mathbf{v}_C) \\ \mathbf{T}_C & = \frac{{\rm d}}{{\rm d}t}( \mathrm{I}_C \boldsymbol{\omega}) \\ \end{aligned} $$

But to understand them fundamentally consider the following tweak:

$$\begin{aligned} \mathbf{F} & = m \frac{{\rm d}}{{\rm d}t}( \mathbf{r}_{\rm rot} \times \boldsymbol{\omega}) \\ \mathbf{r}_{\rm force} \times \mathbf{F} & = \frac{{\rm d}}{{\rm d}t}( \mathrm{I}_C \boldsymbol{\omega}) \\ \end{aligned} $$

You can see now how the geometry (relative location vectors $\mathbf{r}_{\rm rot}$ and $\mathbf{r}_{\rm force}$) enter at different parts of the equations. And that is why you end up with an r multiplied for torque and divide for velocity.