I’ve recently had a look at the 2010 New Zealand Scholarship Physics exam, for the first time. (This is the exam taken by the top final year school students in physics – the best performers get rewarded with scholarships that will help them financially at university).

The scholarship exams are hard. There’s no denying that. For physics, this has often been seen in the way that several concepts can be mixed up in one question. To do well, the students have to be good at identifying what is going on, physically, in a particular situation, and pulling together their knowledge over different areas of physics to make sense of it. The famous ‘equation-spotting’ method (see what information you are given, see what you are told to find, and find an equation from the list given that contains all those quantities in it) does not work at scholarship level. There’s an infamous example from a few years back (the ‘phugoid oscillator’ – go look up what those are if you’re interested) that needed five different equations just to answer one part of the question.

However, in the 2010 scholarship exam I think there was a subtle shift in the type of question asked. Instead of having questions where the student had to identify and pull together the concepts, the questions seemed more focused on individual concepts, but having the student drill down to the core of that concept to see how well they really understood it.

At this point, a big disclaimer is in order: I DO NOT SET SCHOLARSHIP EXAMS; I DO NOT MARK THEM, AND HAVE NOTHING WHATSOEVER TO DO WITH THEIR IMPLEMENTATION; I HAVE NO INSIDER-KNOWLEDGE TO GIVE YOU; what I write here is MY interpretation. I might be completely mistaken.

So, here’s an example, straight off the exam (which you can access from www.nzqa.govt.nz – just do a search on ‘scholarship physics’).

"A television safety advertisement features a car taking corners at dangerously high speeds. The danger is symbolised by land-mines appearing scattered around the corners. As the vehicle approaches a corner, the voice-over says ‘There is more force taking you off the road and less force keeping you on it.’ The car skids across the road and rolls over an embankment. Discuss the accuracy of the voice-over statement, with reference to centripetal force and friction."

First of all, what a lousy presentation of physics. Why didn’t I blog about that when the advert was showing? Anyway, in terms of the question, it gets to the heart of what happens with circular motion. The student really needs to show they have grasped what goes on when a car travels round a curve, that there is centripetal force on the car towards the centre of the circle, and that this force is provided by friction between the tyres and the road. A common misconception is that centripetal force is somehow generated by an object moving in a circle – that it is an ‘extra’ force in addition to all the other forces acting on it. It isn’t an extra force – it is simply the resultant of summing all the forces acting on the car. The car is accelerating towards the centre of the circle, therefore the net force on it must be towards the centre.

So…the statement is wrong on both counts. First of all, there is no force ‘taking you off the road’. Secondly the force keeping you on it actually goes up at high speed – it’s the frictional force between the tyres and road. But take the corner at too high a speed and the frictional force is no longer adequate to keep you on it – and you go sliding.

If you don’t really UNDERSTAND centripetal force and circular motion, you don’t have the slightest chance of answering this question. By understand, I don’t mean being able to say F = m v^2 / r, but I mean knowing physically what causes something to move in a circle. Our centripetal force equation is easy to write, but not so easy to grasp.

## Paula says:

Just curious – why does the frictional force between the tyres and the road increase with high speed?

Thanks,

Paula

## Marcus Wilson says:

Hi Paula. Sorry for the slow response to this.

In this instance, we are talking about the lateral (sideways) frictional force on the tyres. That’s different from the frictional force that goes against the direction of travel of the car. For circular motion, the centripetal force is mv^2/r, where m is the mass of the object moving in the circle (the car), v is the speed of the car, and r the radius of the circle it is turning. The centripetal force in this case comes from the lateral frictional force on the tyres. At higher speeds (higher v), the force goes up.

However, it can’t go up indefinitely as v increases. At some point, the tyre/road interface is unable to provide the required force and the tyre slips on the road, and the car skids. Just how big that limit is will depend on the properties of the tyre (e.g. how well inflated it is, does it have any tread left on it) and road (e.g. is it dry or wet), and weight of the car.

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