newtonian mechanics - Rigid bodies - the wheel

As I've been taught lately in my mechanics course:

the wheel has a unique property: at every moment of motion, the touching point between the wheel and the ground is not in movement and therefore no work is done by the friction force.

Now, many of those problems are solved by using the 2nd Newton law and its rotational analog.

For instance consider having a wheel with a mass $m$ and a radius $R$ rolling on a slope that creates an angle of $\theta$ and we want to calculate its acceleration then we can start by writing: $$ma=mg\sinθ−F_f$$

and the analog equation for torque: $$F_f R=I\alpha.$$

where $F_f$ is the frictional force. Now, the first equation is the 2nd Newton law applied on the centre of mass of the wheel, and as we see, one of the forces is the external frictional force. Now, though the touching point is not in movement at the moment, the center of mass is, and in the equation we assume there is a friction force on the center of mass and therefore work is done indeed. Now, after thinking about this for a while, I've come to the conclusion that this makes sense, cause if we see the wheel as point of mass located in the center, then energy is not preserved because some of it is transfered to the spin and that's why we have the second equation.

The question I'm having trouble with is whether the "work" of the friction force on the center of mass is equal to the energy transfered to the spin of the wheel?

Answer

The best treatment that I've seen of these types of questions comes from Sherwood and Chabay, in Matter and Interactions.

If you look at the wheel as a particle (the "point particle" system), then it cannot rotate, because particles have no physical extent. That means that the distance in the definition of work is the distance that the center of mass travels. That also means that the particle-wheel can only have translational kinetic energy. Let the displacement of the center of mass be $\Delta x$, which is a distance $d$ down the plane.

$$W_{net,PP} = m\vec{g}\cdot\Delta x+\vec{F}_s\cdot\Delta x = mgd-F_sd= \Delta E = \frac{1}{2}mv_f^2$$

If, however, the wheel is modeled as a physical object (the "real" system), then the point of application of each force is its real contact point, which isn't moving for the friction force, but is moving for the weight (because it's the CM). However, it can now have rotational kinetic energy.

$$W_{net,R} = m\vec{g}\cdot\Delta x+{F}_s\cdot0 = mgd = \Delta E = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2$$

Combining the expressions shows that:

$$F_sd = \frac{1}{2}I\omega_f^2$$

You could also modify the system in either case to include the Earth in the system, which would convert that positive work done by gravity on the LHS to a loss in $U_g$ on the RHS.