Quiz Question

This question never appeared on a quiz. Motorcycle racers are going to get this one fairly quickly. But first the story behind it.

Back in my college days I knew these two racers, Mike Bonnell and John Horn, both college students. They went up to Meadowdale, Illinois, to a professional race. The tales they brought back were mind boggling. For one, there was this turn that was called the wall. I will explain. See the image below.

HighTraction

These bikers are cranked over to the max executing a tight turn. Look at the angle the wheels make with the ground. See the following image.

Traction-01

Diagram 1

This shows a schematic of a motorcycle tire when the bike is making a tight turn. If you know the coefficient of friction between the tire and the pavement, you can calculate the maximum lean angle, which translates to how tight you can turn. When the track surface is banked into the turn you can turn even tighter, lean even further, for the same coefficient of friction.

Traction-02

Diagram 2

At Meadowdale “The Wall” was so named because the pavement was absolutely vertical.

Traction-03

Diagram 3

Yes, you can ride the wall, provided you have enough traction. But what if the surface is inclined beyond the vertical, as in the next image.

Traction-04

Diagram 4

If you have enough traction, is it possible, in principle, for the tire to cling to the surface and support the motorcycle when there is nothing below the motorcycle but air? That is the quiz question. Post your answer as a comment below. Provide some justification for your answer, some analysis. I will post the correct answer next week, which could be as early as Sunday.

Solution

Prasad has answered the question with a “yes,” which is the correct answer.

Yes. This is in a turn only and with a sufficient speed of the bike. The centrifugal force compensates over the gravitational force.

However, he has not provided an adequate explanation. Here is a diagram that explains the forces acting on the motorcycle.

ResolutionForces-01

Diagram 5

The pavement curves to the right in the above diagram. This causes it to exert a rightward force on the motorcycle, diagrammed here as a tire cross-section only. Gravity acts down on the center of gravity. The frictional force acts parallel to the area of contact with the pavement. The acting forces are resolved this way.

  1. Combined gravity and friction forces produce a net CW torque on the motorcycle. The centripetal force produces a CCW torque. The two torque components cancel out, so the motorcycle does not rotate in the plane of view.
  2. The friction force contains a vertical component that exactly matches gravity. This keeps the motorcycle stable in the up-down direction. The motorcycle does not fall away from the pavement surface.

Update

Prasad has added comments too complex to be analyzed in the comments section. I will address his comments here. First Prasad’s comments:

September 15, 2015 at 23:59 Edit

Sorry, I didn’t have the time to tackle the write-up.

However, your diagram needs a little more clarity, for lack of better words.

Acceleration is not a force. The mass of the motorcycle plus the motorcyclist does exert a force on the road surface when it accelerates but the beauty is that the motorcycle doesn’t need to be accelerating; it is just an additional benefit if you will. It does need sufficient velocity entering the turn and to counter the little friction which slows down the bike.

THE most important thing here is ‘angular momentum.’

The centrifugal force compensates for the gravity component. That is the most important thing. This does not happen in linear motion — ONLY when there is angular momentum.

The components of the forces are for students. It’s not too difficult to draw but I don’t want to draw them on a computer for lack of time.

I will tackle the various parts one at a time:

Yes, acceleration is not a force.

The motorcycle does need to be accelerating. The arrow labeled “Acceleration” in the Diagram 5 shows the direction of acceleration. Acceleration is to the right, away from the local surface. The acceleration is toward the center of the turn radius. That’s what keeps the motorcycle traveling in a circle while in the turn.

This problem has nothing to do with angular momentum. The wheels have angular momentum with respect to their turning about their axles. The motorcycle has angular momentum about it’s turn radius. Neither of these enter into the problem.

No “centrifugal force” compensates for the gravity component. The only thing that compensates for the gravity component is the vertical component of the Friction Force so labeled in Diagram 5.

Additional explanation on request. Keep reading, and stay involved.

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6 thoughts on “Quiz Question

  1. Pingback: Spooky Physics and Lunar Thinking | Skeptical Analysis

  2. Sorry, I didn’t have the time to tackle the write-up.

    However, your diagram needs a little more clarity, for lack of better words.

    Acceleration is not a force. The mass of the motorcycle plus the motorcyclist does exert a force on the road surface when it accelerates but the beauty is that the motorcycle doesn’t need to be accelerating; it is just an additional benefit if you will. It does need sufficient velocity entering the turn and to counter the little friction which slows down the bike.

    THE most important thing here is ‘angular momentum.’

    The centrifugal force compensates for the gravity component. That is the most important thing. This does not happen in linear motion — ONLY when there is angular momentum.

    The components of the forces are for students. It’s not too difficult to draw but I don’t want to draw them on a computer for lack of time.

  3. Pingback: Quiz Question | Skeptical Analysis

  4. Pingback: Program For September | North Texas Skeptics

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