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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?

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Can you clarify your question? In your first line you say (correctly) that wheel-surface friction does no work, but at the end you mention "work" of the friction force. If this helps: that the friction force does no work doesn't mean you can ignore it - without it there would be no rotation and the wheel would simply slide down. – Kvothe Jan 17 '14 at 16:17
Have you tried to solve this problem? Write down the expression for the "power" exerted by friction (as defined in your question statement). Then write down the expression for the power done by the torque. See if they are equal. – NowIGetToLearnWhatAHeadIs Jan 17 '14 at 16:38
I've solve the problem. But still I am having trouble because by looking at the 2nd Newton equation there is a force that works on a moving object (the center of mass). So looking at that equation alone shows that in the linear world there is energy loss (and it makes sence because it goes to the spin that is not expressed in the newton law but only in hte other equation). My question is if that "energy loss" from the linear world to the spin equals to the "work" that the friction does on the center of mass. formaly I am asking weather Fdx=dE while F=friction, x=c.m, dE=loss of linear energy – Itamar Vigi Jan 18 '14 at 12:28

2 Answers 2

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The magnitude of the work done by friction in linear motion is equal to the work done by the torque of friction only if the wheel is smooth rolling.

In smooth rolling,we have

$$v_{cm}=\omega R$$(This is equivalent to the fact that the point of contact is at rest)

or equivalently:

$$dx=d\theta R$$

Now the work done by friction in linear motion:

$$dW_f=-F_f \cdot dx$$

and the work done by the torque of friction is:

$$dW_f=\tau \cdot d \theta=F_fRd \theta=F_fR \frac{dx}{R}=F_fdx$$.

So the the "two works" are equal in magnitude and no energy is dissipated.

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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.

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