First things first, Torque is always relative to a given point. You can calculate the torque about the centre of mass or about the point of contact.
Next, Any object, moving howsoever, can be analysed in one definite way, by considering the pure translational motion of the Centre of mass and the pure rotation of the body about the centre of mass. (This is true because in the COM reference frame, the pseudo-forces due to the rotating create torques which are balanced about the axis passing through centre of mass.) Therefore, you can treat the rolling wheel as a translation of the centre of mass +the rotation about the centre of mass. There is another way to treat a moving body. It can be treated as a pure rotational motion about an axis called Instantaneous Axis Of Rotation(IAOR). In your case of the rolling wheel, the IAOR passes through the point of contact of the wheel with the incline. Moreover, the $\omega$ for rotation about IAOR is the same as the rotation about COM in your case.
Let us use the first method to analyse your case. The forces acting are:-
1)$F_g$ which acts through the centre of mass and hence can provide no torque for the rotation.
2)$N$ or the normal reaction which again passes through COM and hence produces no torque.
3)The only force that can provide any torque is the friction $f$.
Now, all of these can accelerate the centre of mass. But in the direction perpendicular to the incline, all forces are balanced and there is no acceleration. Along the incline, the translational acceleration is $a=\frac{F_g\sin\theta-f}{m}$. To find the friction, remember, friction tries to prevent relative motion. It can do so by making the rolling a pure rolling. Here the point of contact is (momentarily) stationary and hence no relative motion. for pure rolling, $v=\omega R$ and $a=\alpha R$ must hold true. From this $\alpha=\frac{fR}{I}$, where $I$ is the moment of inertia about COM axis, $R$ the radius and $\alpha$ the angular acceleration. This determines all the variables with $I$.
Using IAOR in your analysis, the torques change but the physical quantities $a$ and $speed$ remain the same. You can try that out for yourself.
I understand now, that the static friction opposes the tendency of a surface to slip over another.
Exactly! Let's remember this point in the next sentences.
When a wheel rotates, static friction pushes the wheel forward to keep the contact surface stationary with respect to the ground.
Not necessarily forward. It can also be backwards. It pulls in whatever direction necessary to keep the contact point stationary. And only if necessary (e.g. there is no static friction when a wheel rolls at constant speed over a horizontal surface. There are no forces at all present, so nothing for the static friction to oppose. The rolling motion just continues effortless until stopped.)
When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too.
No. It pulls uphill.
Think for a moment of a star rolling down. It's legs touches the ground one at a time. While a leg touches the ground, it must not slide. We can think of it as a stationary object in that moment while it is touching. Downwards forces (gravity) must be opposed by static friction upwards, just like for a stationary object resting on the hill.
Then the next leg takes over, and the same thing is the case. Static friction must hold back upwards to avoid sliding of the leg.
Now add more legs to the star. Many, many more. Soon you almost have a round circle, where the legs are the "points" of the circle that touch the ground for only a split-second. Nevertheless, the same is the case; while touching, static friction holds on to the touching point upwards to avoid it from sliding because of gravity.
[...] for the car to go up the hill, where the static friction is again pushing the car forward, countering the slipping of the tires that would result due to the rotation of the tires.
Yes. Again static friction pulls uphill to prevent sliding because gravity pulls down. The direction of static friction does not depend on the rolling direction; it doesn't care if you roll up or down the hill.
If the wheel accelerates at the same time, the static friction might point differently. This is again regardless of the rolling direction but only the acceleration direction comes into play.
How can you get a constant velocity when going down hill, when the static friction and gravity are both creating a net force down the hill (if you ignore rolling friction)?
You are exactly on point here. It can't have constant speed, if all forces pull the same way. Such thinking will lead to the understanding that static friction in fact must point the other way.
A question asks me to find the steepest hill a car can descend at constant speed given the static friction coefficient, but I think what is needed is the rolling friction coefficient. I don't think its possible to answer the question without rolling friction.
As mentioned in a comment, most questions would assume ideal world-models. No deformation of surfaces e.g. So rolling friction will be assumed 0 most often, unless you drive on a clearly non-ideal surface, like a sandy beach or a flexible trampoline.
The idea to solve this question is to remember the formula for maximum static friction:
$$f_s\leq \mu_s n$$
If you have a certain friction coefficient $\mu_s$, you can do your Newtons 2nd law calculations on the car and put this formula in with an $=$ instead of $\leq$, because you are looking for the maximum limit. Then solve it for the slope angle (the angle will be a part of the force components).
Best Answer
"Rolling friction" is a misnomer. Never be confused by it. Its not friction. Its ROLLING RESISTANCE-the correct name. It comes into play because objects are not perfect rigid in real life. It comes into play because of the deformation of shape of objects when in contact. When an object rolls on another WITHOUT slipping, there is a surface contact between them.
It is the normal reaction that is responsible look at the diagram. A body rolling like this (in the diagram) is deformed and the normal forces on the front parts of the surface are always more (whether road is hard or soft) and results in a net backward force that will gradually stop the rolling body. The deformation depends on the nature of the two bodies, depending on their rigidity. Rolling resistance is usually less than static friction. Both are expressed in the same form where the coefficients are called coefficient of rolling resistance and coefficient of static friction. Experimentally, coefficient of rolling resistance is found to be smaller than coefficient of static friction for bodies with same mass. Coefficients are determined experimentally.