Wheel moving without sliding Suppose we have a wheel moving on an horizontal surface, with constant velocity $v$, without sliding. This latter condition implies that the wheel rotates around its centre with angular speed $\omega = v / r$, $r$ being the radius. The rotation of the wheel is caused by the torque of the (static) friction force in the contact point between the wheel and the surface. Because static friction only exists when an opposing force is applied on the object, this means that in order to maintain a constant velocity, an external force must exist. My question is, what happens when this force is no longer applied? I imagine that the wheel will slow down until it stops, but from the equations of kinetic energy, I do not see how to deduce this behaviour. The potential energy is constant because we move horizontally, and the kinetic energy is: $E_k = E_{k_r} + E_{k_t} = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$ which equals the total mechanical energy. If the external force stops, then both $v$ and $\omega$ decrease, but why? After all, the mechanical work of the friction force is zero, so why does the wheel slow down?
Thanks in advance!
 A: The problem with your argument lies with this:

The rotation of the wheel is caused by the torque of the (static) friction force in the contact point between the wheel and the surface.

That isn't quite correct. When the wheel is spinning up, then yes, static friction is what accelerates it, but once it is rotating, no torque is required to keep the rotation going. So once the wheel is moving at some constant velocity, there is no static friction, and in fact no external force at all, thus there are no changes in energy.
In reality, every wheel has some amount of dissipative force acting on it, usually due to deformation. The collective effect of the dissipative forces acting on a rolling wheel is sometimes called "rolling friction," and it acts on the wheel in much the same way as normal friction: it removes energy. Given the constraint that $v = \omega r$, the energy of a wheel is equal to
$$E = \frac{1}{2}(I + mr^2)\omega^2 = \frac{1}{2}\biggl(\frac{I}{r^2} + m\biggr)v^2$$
so when energy is removed, both the angular and linear speeds will decrease. This is what causes the wheel to slow down and eventually stop. Note that in this case, there will be some nonzero amount of static friction as required to prevent slipping.
A: 
The rotation of the wheel is caused by the torque of the (static) friction force in the contact point 

That's just not true.  Newtons's laws of motion dictate that a body in motion continues to stay in motion in the absence of forces acting on it.  Take away both the ground and gravity in your mental picture of the wheel rolling.  It continues to both move forward and spin, as angular momentum is conserved by the same law that conserves momentum.
There may be some confusions as to where the friction of a rolling wheel comes from.  There is certainly friction with the ground but it's hard to pin it down as static of kinetic.  In the case of a car, for instance, the deformation of the elastic tire dissipates energy and creates friction.  I would hesitate to classify that as either static or kinetic friction as typically defined in physics classes.
I'll offer a specific case.  Consider a wheel rolling with some fiction with the ground that is balanced by a force that acts at the center of the wheel in the direction of motion.  Friction decreases both the linear momentum and rotational momentum.  It has to, the observation of a wheel rolling to a stop wouldn't make any sense otherwise.  The exact direction of the force is something I had to struggle with myself a little bit.  The normal force is generally below the axle of the wheel, but when being slowed down by friction the force from the ground acts slightly in front of the axle.  This causes a small amount of torque opposite the direction of the spinning, which is expected since it must be slowing down the spinning.
Imagine a wheel with many spokes of springs around it.  When pushed in they don't spring back to their original position right away - this is its means of dissipating energy.  That means that the springs are longest right before they make contact with the ground, which places the normal force slightly in front of directly below the wheel axle.  As they leave the ground they are not quite as long and do not exactly balance the front force.  This is how wheel friction contributes to retarding both the forward motion and the forward spin.
