The direction of friction in rolling without slipping I know there are multiple SE questions talking about this subject but I still don't understand:
If I have three balls as seen in the picture, what is the direction of friction on each ball if:

*

*No forces are applied on the first ball except gravity and N in the Y axis (I don't about friction, that's why I'm asking)


*Some force passing through the COM is applied on the ball


*Some force not passing through the COM is applied on the ball.
In the 3rd case, does it matter where the force is applied or how big it is?

 A: yes, It matters


*

*where the force is applied. 
for eg, if force is applied above z>r then the torque(analogue of newton second law) will be in opposite direction to the torque if the force applied above the height r (radius of the object).

*And also how big it is
let's suppose that the torque by force f (assuming the force is applied at z>r) is bigger than the torque applied by friction then it will try to rotate the object in a clockwise direction. It also decides the moving direction of object. 
and if the condition for rolling is not satisfied by object (given in answer by user777777) then object is slipping at that moment. 




A: In the first scenario, assuming perfectly rigid bodies, the ball will continue to roll indefinitely. There is no friction and the net force acting on the ball is zero. The equation $v = r \omega$ is satisfied (or else there will be friction).
In the second and third scenarios, the ball will have a net acceleration. Newton's Second Law still applies: $F - F_f = ma$, where $F$ is the applied force and $F_f$ is the friction. 
Additionally, the rotational analog of Newton's Second Law reads $F_f r + Fs = I_{CM} \alpha$, where $r$ is the radius, $s$ is the perpendicular distance of $F$ from the center, and $\alpha$ is its angular acceleration. (In the second scenario, $s = 0$.)
If the ball rolls without slipping, then the condition $a = r \alpha$ also holds.
A: Let's start with case 3

Some force not passing through the COM is applied on the ball.

Let the mass of the ball be $M$, the radius of the ball be $R$, and let the external force $F_E$ be applied at position $r$ where $r=R$ is the top of the ball, $r=0$ is the center of the ball, and $r=-R$ is the bottom of the ball where it touches the ground. The external force $F_E$ points to the right and the friction force $F_F$ points to the left, the ball has a linear acceleration $a$ to the right and an angular acceleration $\alpha$ clockwise.
By Newton's 2nd law we have $F_E-F_F=M a$. Then since the moment of inertia about the center of the ball is $I=\frac{2}{5}MR^2$ we have $F_E r + F_F R = \frac{2}{5}MR^2 \alpha$. And the no-slip condition gives $a=R \alpha$. We have three equations so we can slove for $F_F$ and $a$ while eliminating $\alpha$. Doing so we get: $$F_F=\frac{F_E}{7R}(2R-5r)$$ $$a=\frac{5 F_E}{7MR}(R+r)$$
The one that is interesting for you is the first equation. If we plot it we get

where positive numbers are forces to the left. Note that at $r>0.4 R$ the friction force is negative so actually points to the right (forward)! So for 3 the direction of the frictional force does matter where the external force is applied, although the magnitude of the external force only affects the magnitude of the friction force, not its direction.
Now we can look at case 2

Some force passing through the COM is applied on the ball

Substituting $r=0$ we get $F_F=\frac{2}{7}F_E>0$ so the frictional force points to the left.
Finally, for case 1

No forces are applied on the first ball

Substituting $F_E=0$ we get $F_F=0$ so unsurprisingly there is no frictional force in case 1.
