# Can a car drive around a frictionless mountain?

1. Mar 9, 2016

### invariant99

1. The problem statement, all variables and given/known data

This question was posed on a recent test I had to write. Our professor does not let us keep our graded tests nor does she share solutions, so all I have is the problem statement. I've made an effort to recreate the solution I had, but I got it wrong in any case and I'd like to understand the problem better.

A car drives around a mountain on a frictionless path that is bank outwards. Assume the car's wheels still have traction. The car's route traces out a horizontal circle. Is this possible? Explain your reasoning.

Here is the image that was provided to us:

2. Relevant equations

Let theta be the angle of the banked path from horizontal axis. I used:
• F_{Earth on car} = mg
• F_{road on car} = mg * cos(theta)
• F_{traction} = u_{kinetic} * F_{road on car} = u_{kinetic} * mg * cos(theta) = 0
In addition, I used the fact that an object in moving around in a circle has its net force pointing towards the centre of the circle.

3. The attempt at a solution

I started by creating a free-body diagram of the forces acting on the car. I drew F_{traction} as a dashed line because I'm not sure if it belongs there. I'm confused about how the path can be frictionless, but the wheels can still have traction. That makes no sense to me. I listed it as F_{traction} = 0 above, but I would like verification and clarification on this matter.

From here, I considered the forces along the horizontal and vertical.

F_{net, x} = -F_{road on car} * cos(theta) + F_{traction} * cos(theta)
= -mg * cos(theta) * cos(theta) + 0
= -mg * cos(theta) * cos(theta)

F_{net, y} = F_{road on car} * sin(theta) - F_{Earth on car}
= mg * cos(theta) * sin(theta) - mg
= 0

At this point I got stuck, but I don't think that continuing with these calculations is going to help me. Nor am I convinced that these calculations are correct. Any advice?

2. Mar 9, 2016

### andrewkirk

This is the way I see it.

The car will not be able to drive around the mountain. It will slide down. The forces on the car are:
1. normal force of mountain on car: perpendicular to the mountain surface, which means it points out and up.
2. gravity, which points straight down.
3. the propulsion force of the car, which is perpendicular to the above two.

The net of 1 and 2 is a force directly down the mountain, perpendicular to the direction in which the car is pointing. Since the track is frictionless, and 3 is perpendicular to that force, that net force is unopposed, and the car slides down.

One slightly weird, but theoretically possible, way for the car to get forwards traction but no sideways grip would be if the wheels were cogs and the mountain surface was made of splines going straight down the mountain. Then the cogs would provide forward propulsion without any need for friction, but the car would at the same time be able to slide downwards down the splines, and that's what it would do.

By the way, your drawing doesn't seem to match your description. You have drawn the wheels as circles, which makes it look like the car is trying to drive up the mountain, not around it. If it's driving around it - ie towards the viewer - the wheels should have a rectangular profile.

3. Mar 9, 2016

### Simon Bridge

... so am I.

I suspect the idea is to avoid the answer that if there is no friction the car cannot drive because the wheels will slide. i.e. there is forward traction but no up/down traction.
To go around in the circle, there has to be a net centrpetal force... the sum of the forces has to be horizontal and to the center.
You can answer just from personal experience - in real life, the roads are tilted at corners. Do they usually tilt towards the middle or away from the middle?
Compare with the mountain.

4. Mar 9, 2016

### Staff: Mentor

The answer would be quite different and the picture less illogical were this stated "that is banked on the outside".

Hence, I'm inclined to believe it probably didn't say the path was banked outwards, so that could be why your answer scored poorly.

5. Mar 9, 2016

### billy_joule

I'd tend to agree. Although in that case the mountain is a total red herring and only serves to confuse as it's just a circular banked track that just happens to encircle a mountain.

6. Mar 9, 2016

### invariant99

@andrewkirk Sorry about the drawing. You are correct that the tires should be rectangular. Thank you for the explanation, especially about the analogy about cogs.

@NascentOxygen I copied the question verbatim from the test. She allowed me to do that and that was the exact phrase that she used "banked outwards". I'm not sure what the difference between "banked outwards" and "banked on the outside" is, though.

Anyway, thank you everybody for the feedback. I think have a clearer understanding of what was being asked and how to approach the problem.

7. Mar 9, 2016

### haruspex

It is a poorly specified question. As far as I am aware, there is no standard terminology for describing the banking as inwards or outwards. The only other reference I found in a quick search uses it the other way around: http://dynref.engr.illinois.edu/avb.html.

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