frictioninertiarotational-dynamicsrotational-kinematics
I was under the impression that for a ball/cylinder to roll without slip, there must be a static friction force that opposes the direction of motion. Why, in this case, does the friction act in the same direction of motion? I've posted the question and answer. Can someone help me understand this?
I have modified your diagram a little
The torque does depend on the frictional force $f$ even when calculated about the axis through $P$ because the applied horizontal force $F$ and the frictional force are related.
Assuming the no slipping condition $a_{\rm G} = r \alpha$
The horizontal translational equation of motion for an applied force is given by $F-f = Ma_{\rm G}$ where $M$ is the mass of the cylinder and this equation holds irrespective of the axis that is considered for rotation.
For rotational motion about the centre of mass $fr=I_{\rm G}\alpha$.
Using the point of contact as the axis you get $Fr = I_{\rm P} \alpha$
However note that $F$ and $f$ are linked $F = f + Ma_{\rm G} = f+Mr\alpha$
So $Fr = (f+Mr\alpha)r = fr + Mr^2\alpha = I_{\rm P}\alpha = (I_{\rm G} +Mr^2)\alpha \Rightarrow fr = I_{\rm G}\alpha$
Why does this frictional force go the same direction as the force that's accelerating the wheel?
Because it is the force that's accelerating the wheel. There are no other forces pushing forward (to the right). Without static friction the bike would not move forward at all.
Try to lift the wheel off of the ground - you can spin the pedals all you want, but you won't move forward. Or try cycling on slippery ice with no friction. You won't be able to move anywhere.
This is Newton's 3rd law. You apply a force and a responding force (or a reaction force) acts on yourself the opposite way. This is what happens for a rolling wheel at the contact point.
And this is how a bicycle is able to push itself forward by exerting a backwards push in the ground.
Best Answer
Assume that your external force $F$ was applied at some distance $r$ from the centre of mass of the spool.
To satisfy the rolling condition the linear acceleration of the centre of mass $a$ must equal the radius $R$ of the spool times its angular acceleration about the centre of mass $\alpha$.
$a = R \alpha$
Just suppose that there is no friction and only the external force $F$ is acting on the spool.
The linear acceleration of the centre of mass of the spool is $a = \dfrac F m$ and the angular acceleration is $\alpha = \dfrac {2Fr}{MR^2}$ and so $R\alpha = \dfrac {2Fr}{MR}$.
Comparing $a = \dfrac F m$ and $R\alpha = \dfrac {2Fr}{MR}$ we note that if $r = \dfrac R 2$ the rolling condition is satisfied and hence in the context of the origial question the frictional force would be zero.
Now make $r$ larger than $\dfrac R 2$.
The linear acceleration of the centre of mass stays the same but the angular acceleration increases.
To get to the rolling condition the linear acceleration needs to increase (larger force to the right) and/or the angular acceleration needs to decrease (smaller clockwise torque about the centre of mass.
That is precisely what the frictional force $f$ in the diagram does.
If on the other hand $r$ is less than $\dfrac R 2$ then the frictional force will be to the left to decrease the linear acceleration and/or increase the angular acceleration and so the rolling condition is again satisfied.