Let's assume a wheel is rolling without slipping across a horizontal surface.

If it moves with a constant velocity, then the angular velocity, $\omega$, is also constant and therefore both the net force and the net torque are zero. The force of friction on the wheel is therefore zero.

Only when the wheel is accelerating will there be a force of friction acting on the wheel. If, as assumed, the wheel is rolling without slipping, then by that definition, the force is a static frictional force. Where the wheel makes contact with the ground, it is motionless with respect to the ground (just like your shoe when you plant your foot to walk forward).

Figuring out the direction of the static frictional force on the wheel is not trivial. If the wheel is accelerating to the right, then the angular velocity is clockwise and the angular acceleration is also clockwise (the wheel's rotation is speeding up). If friction is the only force causing a torque on the wheel, then the force of static friction must point left, to cause the CW angular acceleration of the wheel.

The description of the previous paragraph is puzzling, though: how can the wheel accelerate forward if the only horizontal force acting is pointing backward? Obviously it cannot be true. I can think of two ways to rectify this:

1. If the wheel described above is the front wheel of a bicycle, then the bicycle can accelerate because of a forward frictional force on the rear wheel. This is possible on the rear wheel because the chain can provide a larger CW torque on the wheel, such that the net torque (and angular acceleration) from the chain torque and the frictional torque is CW. Of course, for the bicycle to accelerate forward, the (forward) frictional force on the rear wheel must be larger than the (backward) frictional force on the front wheel.
2. If this is a unicycle, then there is an additional torque on the wheel due to the pedals. Then, just like the rear wheel on the bicycle, the pedaling can provide a large CW torque on the wheel, letting the frictional force point forward (with CCW torque). In that way there can be a net forward force on the unicycle (provided by friction) causing it to accelerate, while the net torque is CW, letting $\alpha$ be in the same direction as $\omega$.

Let's assume a wheel is rolling without slipping across a horizontal surface.

If it moves with a constant velocity, then the angular velocity, $\omega$, is also constant and therefore both the net force and the net torque are zero. The force of friction on the wheel is therefore zero.

Only when the wheel is accelerating will there be a force of friction acting on the wheel. If, as assumed, the wheel is rolling without slipping, then by that definition, the force is a static frictional force. Where the wheel makes contact with the ground, it is motionless with respect to the ground (just like your shoe when you plant your foot to walk forward).

Figuring out the direction of the static frictional force on the wheel is not trivial. If the wheel is accelerating to the right, then the angular velocity is clockwise and the angular acceleration is also clockwise (the wheel's rotation is speeding up). If friction is the only force causing a torque on the wheel, then the force of static friction must point left, to cause the CW angular acceleration of the wheel.

The description of the previous paragraph is puzzling, though: how can the wheel accelerate forward if the only horizontal force acting is pointing backward? Obviously it cannot be true. I can think of a few ways to rectify this:

1. There could be a forward pushing force, larger than the backwards frictional force, and acting at the center of mass (causing no torque).
2. If the wheel described above is the front wheel of a bicycle, then the bicycle can accelerate because of a forward frictional force on the rear wheel. This is possible on the rear wheel because the chain can provide a larger CW torque on the wheel, such that the net torque (and angular acceleration) from the chain torque and the frictional torque is CW. Of course, for the bicycle to accelerate forward, the (forward) frictional force on the rear wheel must be larger than the (backward) frictional force on the front wheel.
3. If this is a unicycle, then there is an additional torque on the wheel due to the pedals. Then, just like the rear wheel on the bicycle, the pedaling can provide a large CW torque on the wheel, letting the frictional force point forward (with CCW torque). In that way there can be a net forward force on the unicycle (provided by friction) causing it to accelerate, while the net torque is CW, letting $\alpha$ be in the same direction as $\omega$.