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In which I find the solution by going back in time

October 12, 2010

I knew there were was something about this wheel problem that felt familiar. Turns out I’d wrestled with the same problem 10 years ago. I know this because I looked it up in Swartz and Miner’s excellent book, Teaching Introductory Physics, a source book, which does a great job of going into deeper detail on a number of interesting problems in intro physics. For my benefit, I’m going to try to reproduce the essentials of the argument here, but I’ve also included photos of the pages in case you’re interested in checking my work.

This time, maybe by putting it on the web, I’ll be able to find it easier on my iBrain in 2020.

So here we go:

  • Big idea 1: Wheels deform. This has the effect of making the normal force of the road not point directly through the axle, which means the normal force will exert an external torque on the wheel.
  • An image of the deformation of wheels, from Schwartz and Miner. The size of the offset depends on the acceleration of the wheel—more on this later

  • Corollary: The asymmetry of the wheel causes a retarding frictional force opposite the translational motion of the wheel. The pressure on the tire in front of the vertical through the axle is slightly higher, and this leads to a backward force on the tire at the intersection with the road.
  • From Schwartz and Miner. The asymmetry of the wheel leads to slightly larger forces in the front of the wheel than the rear, which cause a backward reaction force (retarding friction).

  • Big idea 3: This retarding frictional force is independent of speed or acceleration. So say Schwartz and Miner. I can believe this, it’s hard for the wheel to change how it pushes on the ground, so the ground can’t change how it pushes on the wheel.
  • The combination of the normal and retarding force of the road create an external torque on the wheel. Let’s take our axis of rotation as the axle. 3 cases:
    • constant velocity: In this case, offset must be such that the force of the road points directly through the axle so that the external torque is zero. Let’s call this the equilibrium offset.
    • speeding up: Here you need a clockwise torque to increase the angular velocity of the wheel, so the offset decreases, and the line of the force of the road passes behind the axis of rotation, creating a clockwise torque. This makes some sense, because we can think of the axle getting a big ahead of the wheel, which would decrease the offset.
    • slowing down: Here we need a counterclockwise torque, and so the offset increases, making the line of force pass in front of the axle, and creating a CCW torque.

This in essence, is what Shwartz and Miner call rolling friction—they aren’t needing to talk about all the internal sources of friction inside the engine, axle, etc. Now we need to account for the Traction force. Here’s the quote from Schwarz and Miner:

The driving torque \tau transforms the force couple F and F acting through the axle and rim. These behave as if they were external forces acting on the wheel.

(you know, it would be much easier to understand all this if Swartz and Miner followed my convention about labeling their forces).

But the basic gist seems to be the torque causes the tire to push on the road backward, which causes a forward force of the road on the tire (traction). We all knew this.

When the car is going at constant velocity, the net torque and force on the wheel must be zero. this means that the traction force (F in the diagram) must equal the retarding force (f) in the diagram. The road can exert 2 forces on the tire since the the tire isn’t contacting the road in a single point, there is an area of contact (and this helps to understand some of Noah’s question about how the placement of a banana under one’s foot might cause you to fall forward or backward). The key idea is these are point objects, or points of contact, so you can have multiple forces at a single surface. Here’s a diagram from Swartz and Miner:

a wheel driven by an internal torque at constant velocity. From Swartz and Miner

You also need to have a net torque of zero. Since the resultant of the retarding force and normal force pass through the axle, this must mean that the torque of the traction force must cancel out the torque of the engine on the axle. Again, I can believe this. But this still feels fuzzy to me, since the engine really is an internal torque, and it shouldn’t be be able to change the angular momentum of the wheel, right?

And finally, towed wheels (like the rear wheels of a front wheel drive car) don’t have traction forces. They only feel retarding forces.

But even though I’ve gone through all this, and thought about this problem for more than an hour when I should be writing comments, I still have some questions I can’t resolve:

  • If a car is experiencing drag, how can the traction force equal the retarding force? It would seem to me that drag would exert a backward force on the car, which would cause a backward force on the axle, which would exert a backward force on the wheel. Thus for the wheel to be moving at constant velocity, the traction force would need to be greaterthan the retarding force.
  • If we consider the car (assumed to be front wheel drive) as a whole, we can see that there are 2 traction forces in the forward direction, 4 backward retarding forces on each wheel, and a drag force. Thus the traction force must be as big as all of these backward forces to maintain constant velocity.
  • I’m still a bit unclear on the internal torque of the engine. Can this really change the angular momentum of the wheel? I think it can, since if the wheel is our system, the axle is actually providing an external torque.

One last big insight from Swartz and Miner that I’ll just retype:

Note, incidentally, that the reaction force of the road during braking or acceleration produces a torque that is in the wrong direction for the expected angular acceleration of the wheel. If the wheel accelerates forward, the traction reaction from the road is forward, but that force by itself would produce a torque that would slow the wheel. Remember, however, that the traction force arises as a force pair to one part of the force pair representing the driving torque. In this representation, the two forces at the rim cancel each other out (both act as external forces on the tire), leaving the other half of the torque component at the axis to exert the forward force. The accelerating torque arises from the shift of the offset ebetween the vertical weight through the center of gravity and the normal force of the road.

This doesn’t make complete sense to me. If the traction and retarding force cancel, this would seem to leave the normal force to exert a torque on the wheel that would tend to slow the angular velocity of the wheel, which isn’t right.

Anyway, I think this brings this problem a bit closer to solution. In case you want to read Swartz and Miner in all its detail, I’ve put the relevant pages below (click to embiggen).

Here’s a link to a paper in the Physics Teacher that does a pretty good job of describing this as well (you’ll need a TPT account to access):

Ok, time to write some comments.

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