Work and energy for an accelerating car

Someone wrote to me with questions about work vs pseudowork (extended system vs point-particle system). At one point he asked me to analyze a subsystem of an accelerating car, the car minus the wheels, and I learned something in the process that’s worth sharing. It’s yet another example of the need to be careful about the fact that in calculating work it’s critically important to pay close attention to the individual displacements of the points of applications of each force, not simply use the displacement of the center of mass.

A car moves with constant acceleration a in the +x direction, with no tire slippage. Simple model: there’s a forward static friction force f applied by the road to the bottom of each tire, where the instantaneous speed is zero, and therefore this force does no work on the bottom of the wheel. Total mass of car is M, mass of each wheel is m, each wheel is like a bicycle wheel, with all mass at the rim of radius R so the moment of inertia of the wheel is mR^2. From the Momentum Principle (Newton’s 2nd law) we have Ma = 4f. No work is done on the car, so \Delta E_{\text {car}} = 0, where \Delta E_{\text {car}} = \Delta K_{\text {trans}} + \Delta E_{\text {int}}, where \Delta K_{\text {trans}} = \frac{1}{2}mv_{\text {cm}}^2 is the translational kinetic energy of the car and \Delta E_{\text {int}} includes kinetic energy of wheels, pistons, camshaft, chemical energy of gasoline, thermal energy of engine block, etc. For a point-particle system subjected to the same forces, \Delta K_{\text {trans}} = 4fd, where d is the distance the car moves, and this is also the translational kinetic energy of the actual car.

Now consider a system I’ll call “sys” consisting of the car minus the wheels, with mass M-4m. In this simple model, suppose the engine pushes on the top of each wheel. When I was a kid there were little electric motors you could mount on the top of the front wheel of a bike. The motor turned a small wheel in contact with the big wheel, to drive the big wheel. Or you could imagine an arm or arms continually pushing the top of the wheel, then being lifted and retracted. What are the energetics of “sys”?

Start by analyzing the wheel alone, which is acted upon by the force f of the road, the force of the car axle on the inside of the hub of the wheel, and the force of the engine along the top of the wheel. By determining these forces we get by reciprocity of electric forces the forces the wheel exerts on “sys”. In the +x direction the engine exerts a force +f_2 and the axle exerts a force -f_3 (the engine force pushes the hub of the wheel against the car’s axle). Remember that Ma = 4f, so f = Ma/4.

Momentum Principle: ma = f + f_2 - f_3

Angular Momentum Principle (-z direction): I\alpha = Rf_2 - Rf = mR^2(a/R) = Rma, or ma = f_2 - f, and f_2 = ma + Ma/4 = (m+M/4)a

From the Momentum Principle we have ma = Ma/4 + (m+M/4)a - f_3 , so f_3 = Ma/2 , which is twice as large as f (which is a bit surprising)

Summary for wheel: The wheel is acted upon by the road, f = Ma/4 in the +x direction, the axle, f_3 = Ma/2 = 2f in the -x direction, and the engine, f_2 = (m+M/4)a in the +x direction. The engine force is only slightly larger than f, by the amount ma which is small compared to f = Ma/4, since the wheel mass m is very small compared to the large mass M of the car. It’s interesting that the force of the axle on the wheel is so large, twice as big as f. Of course if the mass of the wheel is distributed differently the values of f_2 and f_3 will be different, related to I not being simply mR^2.

Now we can look at “sys”, the system consisting of the car minus the wheels, with mass M-4m. The forces in the +x direction acting on “sys” are the forces due to the hubs of the wheels, +4\times 2f = 8Ma/4 = 2Ma, and the forces due to the tops of the wheels on the engine, -4\times f2 = -4(m+M/4)a.

Momentum Principle: (M-4m)a = f_3 - f_2 = 2Ma - 4(m+M/4)a = Ma-4ma = (M-4m)a, which checks.

What about the Energy Principle for “sys”? Here’s the element in the analysis that I found particularly interesting. The forces of the hubs of the wheels are applied to the axle, which in a small time dt moves through a small displacement vdt, where v is the instantaneous speed of the car. But the forces of the tops of the wheels on the engine act through a distance 2vdt ! The instantaneous speed of the top of the wheel is 2v (speed of hub is v, speed of bottom of wheel is zero).

Energy Principle: \Delta E_{\text {sys}} = W = 4f_3\times vdt - 4f_2\times 2vdt = 4(Ma/2 - 2(m+M/4)a)\times vdt = -8ma\times vdt

As usual, real work must be calculated by integrating EACH force through ITS point of application, THEN you add up all the contributions to the total net work. Here, as in all cases of deformation or rotation, there is the possibility that different forces act through different distances. The case here is particular striking, because the force at the top of the wheel acts through twice the distance of the force at the hub.

It may at first glance seem odd that the work done by the wheels on “sys” is negative. However, note that “sys” increases the wheels’ translational and rotational motions, thereby increasing the energy of the wheels, so there must be a small decrease in the energy of “sys” associated with giving energy to the wheels. There is of course a much larger decrease in chemical energy associated with accelerating the mass of the “sys” system.

Compare with car: \Delta E_{\text {car}} = \Delta E_{\text {sys}} + \Delta E_{\text {wheels}} = 0, so \Delta E_{\text {sys}} = -\Delta E_{\text {wheels}}, which is negative, with \Delta E_{\text {wheels}} = 8ma\times vdt

The energy of one wheel is E = K_{\text {trans}} + K_{\text {rot}} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(v/R)^2 = mv^2

The rate of energy change is dE/dt = 2mv\times dv/dt = 2mva, and for 4 wheels we have dE/dt = 8mva. In a time dt the amount of work done on the wheels is 8ma\times vdt, which is indeed what we found above.

Bruce Sherwood

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One Response to Work and energy for an accelerating car

  1. josé leme says:


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