Angular velocity of rotating rod Consider the following system:

Newton's second law for rotational motion:
\begin{equation}\tau=I\alpha \Leftrightarrow rF=\frac{1}{3}mr^{2}\alpha \Leftrightarrow \frac{d\omega}{dt}=\frac{3F}{mr}\end{equation}
Considering RHS constant, we get $\omega=\frac{3F}{mr}t.$
I'm not sure if the angular velocity whould be inverse proportional to the radius (from natural experience I know that pushing farther requiers lower force).
Also what happens if the bar is not fixed and the two opposite forces are acting at the ends of the bar.
Since their sum is $\vec{0}$ there is translational equilibrum and so the axis of rotation is at the $C.M.$ but will the action of the two forces change the angular velocity from the previous situation?
 A: 
I'm not sure if the angular velocity whould be inverse proportional to the radius (from natural experience I know that pushing farther requiers lower force).

You are forgetting that you are also using $r$ as the length of the rod. So if you increase $r$ you are increasing $\tau$ that scales linearly with $r$, but then you also increase $I$ which scales quadratically with $r$. So it makes sense that you end up with a $1/r$ for $\alpha$. 

I know that pushing farther requiers lower force).
  Also what happens if the bar is not fixed and the two opposite forces are acting at the ends of the bar.
  Since their sum is $\vec 0$  there is translational equilibrum and so the axis of rotation is at the C.M. but will the action of the two forces change the angular velocity from the previous situation?

I would suggest just doing your previous analysis. Determine the net torque acting on the rod about the C.M., and then you can determine the angular acceleration about the C.M. using your rotational analog of Newton's second law.
A: Consider the FBD of the situation, where the reaction force $R$ below is drawn in a positive sense. Also shown is the center of mass (blue dot), located exactly at $\tfrac{r}{2}$ from the pivot.



*

*Case A, pivoted arm means the linear acceleration of the center of mass is coupled with the angular acceleration $$ \ddot{y}_{\rm cm} = \tfrac{r}{2} \ddot{\theta} $$ and the two (linear + angular) equations of motion being $$ \left. \begin{aligned} R + F & = m\, \ddot{y}_{\rm cm} \\ \tfrac{r}{2} F - \tfrac{r}{2} R & = I \ddot{\theta} \end{aligned} \;\right\}\; \begin{aligned} R & = \tfrac{F}{2} \\ \ddot{\theta} & = \tfrac{3 F}{m r} \end{aligned} $$ when $I=\tfrac{m}{12} r^2$. So yes, rot. acceleration is inversley proportional to $r$ since despite the moment arm being less, the mass moment of inertia is reduced by $r^2$.

*Case B, free arm with equal and opposite $R=-F$ applied on the end of the rod. Same equations of motion, but now $R$ is known, and linear acceleration is not known. The result is $$\begin{aligned} \ddot{y}_{\rm cm} & =0 \\ \ddot{\theta}  = \tfrac{r F}{I} & = \tfrac{12 F}{m r} \end{aligned} $$ Again rot. acceleration is inversley proportional to $r$ for the same reason, but the quanity differs from Case A.

NOTE: Just by a cursory look you see in the first case the center of mass accelerating and in the second it isn't. So the two situations differ significantly. 
EXTRA CREDIT: If the force $F$ is not applied at the end, but at some other point $d<r$ from the pivot, then the pivot reaction $R$ will be different. Is there a value of $d$ that makes $R$ equal to zero?
