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## Summary:

- A derivation is provided to support the statement that work is not done by static friction when accelerating a car.

## Main Question or Discussion Point

A recent thread posed the question whether work is done by static friction in the case of an accelerating car. Before I had a chance to reply, the thread was closed on the grounds that the subject was "beaten to death". Undaunted, I am determined to deliver the

We assume rolling without slipping. Let

##I_{cm}=qmR^2~~~(0 \leq q\leq 1)## = the moment of inertia of the wheel, radius ##R## and mass ##m##, about its CM.

##F## = the force acting on the wheel at the CM; it is the torque on the wheel divided by the radius ##R##.

##f##= the force of static friction on the wheel.

From the FBD shown above, we get

##F-f=ma_{cm}##

##fR=I_{cm}\alpha=qmR^2(a_{cm}/R)##

These equations give

##a_{cm}=\dfrac{F}{(q+1)m}~\rightarrow~\alpha=\dfrac{F}{(q+1)mR};~~~~f=\dfrac{qF}{q+1}##

We invoke the SUVAT equation

$$\Delta K_{trans}=\frac{1}{2}m(2a_{cm}\Delta s_{cm})=\frac{1}{2}m\left[2\frac{F}{(q+1)m}\Delta s_{cm}\right]=\frac{F\Delta s_{cm}}{(q+1)}$$

$$\Delta K_{rot}=\frac{1}{2}I_{cm}(2\alpha\Delta \theta)=\frac{1}{2}qmR^2\left[2\frac{F}{(q+1)mR}\frac{\Delta s_{cm}}{R}\right]=\frac{qF\Delta s_{cm}}{(q+1)}=f\Delta s_{cm}$$

The input work crossing the system boundary is ##F\Delta s_{cm}## and is equal to ##\Delta K=\Delta K_{trans}+\Delta K_{rot}## in agreement with the work-energy theorem. Static friction does no work on the system but partitions the input work between two internal degrees of freedom, translational and rotational, according to the size of parameter ##q##. Specifically, ##f\Delta s_{cm}## is the amount of input work diverted into change in rotational energy. When ##q=0## (point mass) all the input work goes into translational internal energy; when ##q=1## (a ring with all the mass at ##R##) we have equipartition of energy. R.I.P.

Note: It often helps to view energy transformations in terms of the first law of thermodynamics. This idea is presented among others in a trilogy of insight contributions currently under preparation.

*coup de grâce*here with a simple derivation.We assume rolling without slipping. Let

##I_{cm}=qmR^2~~~(0 \leq q\leq 1)## = the moment of inertia of the wheel, radius ##R## and mass ##m##, about its CM.

##F## = the force acting on the wheel at the CM; it is the torque on the wheel divided by the radius ##R##.

##f##= the force of static friction on the wheel.

From the FBD shown above, we get

##F-f=ma_{cm}##

##fR=I_{cm}\alpha=qmR^2(a_{cm}/R)##

These equations give

##a_{cm}=\dfrac{F}{(q+1)m}~\rightarrow~\alpha=\dfrac{F}{(q+1)mR};~~~~f=\dfrac{qF}{q+1}##

We invoke the SUVAT equation

*sans*time in the linear and rotational forms to find the changes in translational and rotational kinetic energy, after the CM of the wheel has advanced by ##\Delta s_{cm}##.$$\Delta K_{trans}=\frac{1}{2}m(2a_{cm}\Delta s_{cm})=\frac{1}{2}m\left[2\frac{F}{(q+1)m}\Delta s_{cm}\right]=\frac{F\Delta s_{cm}}{(q+1)}$$

$$\Delta K_{rot}=\frac{1}{2}I_{cm}(2\alpha\Delta \theta)=\frac{1}{2}qmR^2\left[2\frac{F}{(q+1)mR}\frac{\Delta s_{cm}}{R}\right]=\frac{qF\Delta s_{cm}}{(q+1)}=f\Delta s_{cm}$$

**Interpretation**The input work crossing the system boundary is ##F\Delta s_{cm}## and is equal to ##\Delta K=\Delta K_{trans}+\Delta K_{rot}## in agreement with the work-energy theorem. Static friction does no work on the system but partitions the input work between two internal degrees of freedom, translational and rotational, according to the size of parameter ##q##. Specifically, ##f\Delta s_{cm}## is the amount of input work diverted into change in rotational energy. When ##q=0## (point mass) all the input work goes into translational internal energy; when ##q=1## (a ring with all the mass at ##R##) we have equipartition of energy. R.I.P.

Note: It often helps to view energy transformations in terms of the first law of thermodynamics. This idea is presented among others in a trilogy of insight contributions currently under preparation.