Work using angle vs. displacement [closed]

Question

I am trying to calculate the work done on this wheel as it undergoes one full revolution, and is rolling without slipping.

I am aware that work can be calculated either using the integral of force with respect to displacement, or the integral of torque with respect to angle.

Since the speed of the contact point B is zero, the friction does not do any work. The weight and normal force are perpendicular to the displacement so neither do they. There is only F. The point A moves at twice the speed as point C, thus undergoes the displacement $4\pi r$. This means the work is

$$ \int_0^{4\pi r}Fdr = 4\pi r F. $$

But if I use the integral with torque instead, I get

$$ \int_0^{2\pi}Fr d\theta = 2 \pi r F. $$

These aren't equal so I must be missing something. Isn't the angle rotated $2\pi$? Or should I not be considering rotation around C but rather B? These are the only explanations I can find but I don't know what is correct.

Answer 1

For a rigid body like a wheel rolling without slipping, but accelerating under a force, we may consider the force as acting at the center of mass, or at point C in the diagram. This is the only force that dos work on the wheel, whose center travels a distance of $s=2\pi R$ in one complete revolution, so that he work done is according to: $$W=Fs=F2\pi R.$$ TO do the problem with angular variables you must know the amount of torque $\tau$, applied to the wheel during an angular displacement $\theta$, of $2\pi$ for a full revolution so that the work is according to: $$W=\tau \theta=2\pi\tau.$$