Oxford PAT 2012, Question 17

You obviously don’t need to write out the answer in such detail in a multiple-choice test, but it is worth doing it so that you can see exactly why the answer is (A).  This question requires that you understand circular motion and are able to write down the forces and apply the equation F = ma, without getting confused by doubts about ‘centrifugal force’.


People often find circular motion hard to understand, and there are two confusions that crop up again and again.  So it pays to remember these two principles:

  1. Centripetal acceleration is a geometric property of circular motion.  That is, even in a world where there were no such thing as mass, or force, a point moving in a circle would still be accelerating towards the centre of the circle.  It’s just because of the geometry of the situation.  It is worth working out for yourself the centripetal acceleration of a point moving in a circle, without any mention of mass, just to ensure you remember this.
  2. If you find yourself thinking about ‘causality’, stop.  In mechanics there is no such thing as causality: there are masses, forces and accelerations, and a law linking them together in an equation.  So statements like “the object has inertia, which makes it want to go straight on and this causes it to exert a centrifugal force on the track” are a recipe for disaster.

So do what we have done in the question above: find the kinetic energy (and hence the velocity) of the puck in terms of its loss in potential energy, write down the forces on the puck, and the acceleration that the puck is undergoing, find a suitable direction to resolve the force and acceleration vectors, and apply F = ma.

Once you have done this, there is a trick that comes up very often in mechanics.  For an object to stick to another object, there is a condition that they are pushing against each other with a force F, where F must be greater than or equal to zero.  If the pushing force is less than zero, the objects won’t remain touching.


13 thoughts on “Oxford PAT 2012, Question 17

  1. Hi,

    Could you not just reason in terms of conservation of energy? If the object is dropped from point S, and there are no resistive forces, then it will reach T with zero Kinetic energy exactly, as it has the same PE as S. But in order for it to stay in contact with the top of the circle the instant before T, its speed would have to be above a threshold speed, thus it does not reach T.

    Thank you

    1. Yes that’s absolutely right, and it’s obviously the kind of reasoning you would use when thinking about the situation in the test. But, as mentioned below, it’s a bit unsatisfactory to use a reductio ad absurdum argument.

  2. Hi,

    Could you explain how you determined the value of ‘d’? Also is your reference point for the gravitational potential energy from the top of the circle?


    1. Look at the small triangle in the diagram. The acute angle in the triangle I have labelled θ, and this vertex of the triangle is at the centre of the circle, so the hypotenuse of the triangle has length r; the vertical side of the triangle therefore has length r sinθ. But the bottom of the vertical side of the triangle is a distance r below T (because the bottom of the triangle is horizontal and the other end goes through the centre of the circle). Therefore the distance d from the top of the triangle to the hotizontal line going through T, is r – r sinθ, which equals r (1- sinθ).

      Yes my reference point for gravitational pe is the top of the circle.

  3. I understand your solution mathematically, but I don’t think this quite explains the situation in its very essence, simply as it is without any long calculations. Could you please describe the distribution of energy qualitatively?

    1. Hi Rishi — the mg sin θ force is the component of the gravitational force parallel to the line from the mass to the centre of the circle; so the total centripetal force is F (the force with which the track presses on the mass plus mg sin θ. Since mg sin θ is a component of a force, it has to be less than or equal to the force itself — notice sin θ is less than or equal to 1, whereas 1/sin θ is greater than or equal to 1, so from this analysis you should be able to spot that the component of the gravitational force can’t be mg / sin θ

      1. if mg is the ‘resultant’ weight couldn’t it be made op of components that are bigger that the force? for example in the right angled triangle in the diagram mg would be made up of a vector pointing to the centre of the circle and a vector pointing to the right perpendicular to mg?

  4. Your explanation here is very well written, thanks for the effort!
    As I was reading through it, however, I thought about approaching the problem from the conservation of energy perspective. At the bottom of the track, all potential energy has been converted into kinetic energy. As the slider begins to go up the loop, some kinetic energy is converted into potential energy, but some is also converted into rotational kinetic energy. Since some potential energy is lost to the rotational, the slider cannot reach the same height as before.
    Is my explanation correct? If you could give some feedback, I would appreciate that very much.
    This a lot of work for two marks, but they’re also marks that I lost on my first try!

    1. Hi. I think your instincts are good: sometimes you can solve a problem much more simply by considering energy rather than forces. But with this particular argument I think you’re on rather dodgy ground. Suppose the slider were a point mass: in that case it would have no moment of inertia and therefore no kinetic energy of rotation, but it would still not reach point T.

      So where is the initial potential energy going? Well, when it goes past (under!) point T, the mass is moving, so it has got kinetic energy. If you want to come up with an argument based on conservation of energy, it is this kinetic energy that you need to focus on.

      Here is an argument based on conservation of energy that I think does work. Let’s assume the slider were at the point T. The downwards force on it would have to be at least mg, and so its acceleration downwards (towards the centre of the circular track) would have to be at least g. But if the slider were at T, it would have been moving in a circle all the way round the track, and would still be moving in a circle, so it would have to have a horizontal speed of at least sqrt(rg). But by conservation of energy, it can’t have any speed at all, so our assumption must be false: the slider can’t be at point T.

      I think this argument is correct, but it’s a tad unsatisfactory to me because we don’t normally argue by reductio ad absurdum in mechanics. Maybe you or someone else can come up with a more satisfactory argument based on conservation of energy.

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