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:
- 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.
- 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.
Oxford PAT 
If we consider the point S with 0 potential then we can use conservation of energy directly till T. delta(KE) + delta(GPE) = 0 as delta(GPE) = 0 we get 0.5mv(at T)^2 = 0 do V(at T)=0 which means it cannot get to T in the first place. Does this make sense?
This is all totally true but we should be very careful: it would be entirely plausible from the above argument that the truck makes it up to T at the precise moment it reaches zero velocity (and then falls straight down). So, I do not think this is a good argument.
Hi there,
I understand everything besides why you equated 0.5mv^2 to mgd, why is d used and not 2r in this equation?
Thank you in advance!
Hi Andreea,
I’m measuring the zero point of height from the height where the mass is released from. In case this weirds you out, it’s equivalent to adding/subtracting a constant from the energy which is allowed. Consider it this way – when it falls the distance d, it gains energy mgd, and so has velocity given by 1/2 mv^2 = mgd.
Ah I see, thank you very much!
Hi, this website is so helpful thank you very much. Very de-stressing, good job.
I have a question and I think the answer is yes, but I just want to ask it and then feel gratified as if I have understood the concept.
At the point where the slider loses contact with the inside of the circular track, will it fly off following an upside down parabolic path, with a maximum turning point in line with ST before it begins to descend again? As the remaining KE converts to GPE? So it will, at some point re-reach its initial height at S.
Unfortunately that’s not quite the case Joey! You’re right that it will fly off, but the maximum point will not be at S, it will end up flying out to the side. Think of it this way: to reach S, all the kinetic energy would have to be in the ‘up’ direction. When it flies off, it will have some sideways momentum so cannot have enough energy to reach the top.
Hi, I did not understand why at top the sin(theta) is more than 2/3 and hence Force=0
Read carefully the algebra – in order to stick to the track, there must be a positive centrifugal force. This gives us the necessary condition in terms of the angle for the force to be positive. When sin(theta)>= 2/3, this condition no longer holds. At the top, sin(theta)>2/3 because, well, that’s how angles work. If theta > ~42 degrees, then sin(theta)>2/3.
This may have been mentioned in earlier comments, but it is valid to look at resistive forces in the horizontal plane? If the slider travels over half way from the bottom to the top of the loop (as it must) then after that there is no resistive horizontal force to reduce the horizontal velocity to zero before it reaches T. In which case the slider cannot reach T and maintain conservation of energy.
So, if there were friction in the problem, you would be right in giving a reason why it can’t reach the top.
However, the mass doesn’t reach T even without friction. Maybe think of it this way: in order to maintain circular motion, it needs to travel at a certain velocity. To reach T it has to be undergoing circular motion, hence must have a sideways component to the velocity, and hence some kinetic energy. However, there is no energy in the system at T, therefore the ball can never reach it.
Maybe my original comment wasn’t clear enough, but it was based on there being no friction. If there is no friction then there is no resistive force to reduce the horizontal component of the velocity to zero. Therefore the slider cannot get to T and satisfy conservation of energy.
I’m sorry, but shouldn’t the total force acting on the slider towards the centre be F + mgcosθ?
Hi Ben,
I see where you’ve got confused, but no. The angle between F and mg is 90-theta (draw a picture!), so the resolution of mg along F ends up as mg cos(90-theta) = mg sin(theta).
hello, instead of using forces I looked at it from an energy conservation point of view, where the car would have to fall through 2R height (R being the radius of the circular track) to get to the very bottom, gaining mgh=2mgR worth of kinetic energy. using 1/2mv^2, the maximum velocity at the bottom of the track is v^2=4gR. as centripetal force is mv^2/R, this velocity would require 4mgN of centripetal force.
as the force towards the centre at the top of the track is mg, the centripetal force required to keep in contact is larger than the force provided. hence, the car doe not reach the track.
is there a flaw in my workings or can I approach future questions this way as well?
thanks a ton!
I don’t think the second paragraph is a convincing argument. But there are some OK arguments presented purely in terms of energy in the discussions below, so have a look at them. It’s also worth really understanding the technique in the method I use above because it’s applicable to lots of problems.
gotcha, thanks! so your method involves assuming the maximum height it can reach is d from the top, and working out how high d is (in terms on its angle) to find out if it can reach (or exceed) T, right?
To be exact, my method is to find the maximum height it can reach before it loses contact with the track. After losing contact with the track it is still going up, but just not going up as fast as the track is. And if it loses contact with the track below T, it’s never going to reach T.
Why does it come off the track if F is smaller than 0? Because the component of the weight still acts towards the centre, and as long as there is a resultant force towards the centre there is circular motion, so surely F just has to be larger than -mgsinθ?
Because F is the force with which the track is pushing the mass and therefore equivalent in size to the force with which the mass is pushing the track, They point where the mass and the track stop pushing against each other (i.e. when F is zero) is the point where they come apart.
Hi, another way of doing this question is to simply calculate the height the slider would have to be in order to reach the top. Then you have in terms of energy: mgh = mg(2r) + 0.5mv^2 –>
gh = 2gr + 0.5v^2 (r is the radius of the circle); we are able to calculate the necessary velocity through centripetal Force which has to be equal or greater than gravity: mg=mv^2/r -> v^2 = gr. If we put this in the first equation we get: gh = 2gr + 0.5(gr) –> h = 2.5r . In conclusion, for the slider to reach the top it has to start at a height of 2.5r which is higher than 2r like stated in the question -> it won’t reach the Top. I think it’s a nice way of doing this question because you don’t have to start thinking about trigonometry.
Hi — yes you are basically right. Have a look down at the bottom of the comments where people have discussed similar approaches. Since the question only asks for a multiple choice answer you wouldn’t do it the way I have done in an exam, but it’s a useful opportunity to go through the general technique for solving this kind of problem.
Hi,
I do not understand the middle part (“Resolving along F” and the following two lines). The only parts I do recognise are the formula for centripetal force and maybe the formula for downhill-slope force. Would you mind further explaining the middle part about forces?
Kind thanks
F is the force exerted on the slider by the circular track (and the point where F reaches 0 is the one where the slider loses contact with the track). If we resolve parallel to F we get the total force pushing the slider towards the centre (which is F + mgsinθ). We know that because the slider is travelling in a circle the centripetal acceleration must be v²/r. So, because force is equal to mass times acceleration we conclude that F + mgsinθ = mv²/r. But earlier in the question we worked out a formula for v², which is 2gr(1-sinθ), and therefore F + mgsinθ = 2mg(1-sinθ).
Thanks for your reply!
Although I understand most of it now, I have to admit that I am still quite unsure about the “Resolving parallel to F” and F + mgsinθ part.
My current guess is that F is the net force and mgsinθ is the part pushing towards the centre. Also, I assume that we only consider the component of gravitational force that acts towards the centre of the circle in this calculation.
I probably wrongly thought of F as an individual force opposing mgsinθ resulting in F < mgsinθ instead of
F < 0 as a requirement for leaving the track.
Is this correct or am I following an erroneous path?
My current guess is that F is the net force and mgsinθ is the part pushing towards the centre.
That’s not really correct but some of the other things you say are exactly right — so let me try to state clearly what the forces are. The weight of the slider (i.e. the force of gravity acting downwards on the slider) is mg acting directly downwards: so mgsinθ is the component of the weight of the slider that acts towards the centre. F is the force with which the track is pushing the slider towards the centre (and notice that since the track is frictionless F is also the total force exerted by the track on the slider).
Then there are two key steps: (1) applying F=ma to equate the total force acting on the slider towards the centre (which is F + mgsinθ) to the mass of the slider times its acceleration towards the centre; (2) applying the circular motion formula (centripetal acceleration = v²/r) to equate the slider’s acceleration towards the centre with v²/r (which is 2g(1-sinθ)).
You are absolutely right that if F becomes less than zero then the slider leaves the track, so we are looking for the point where F is zero, which is the last point where the slider is in contact with the track. By putting together steps (1) and (2) above we derive that when F is zero, mgsinθ = 2mg(1-sinθ), or more simply mg(2-3sinθ) = 0.
All right, I think I got it now. F and mgsinθ are pushing in the same direction (centre of circle). You did indicate that in your drawing and formula (addition in F + mgsinθ), I just couldn’t wrap my head around it somehow.
Thanks a lot for your detailed explanation!
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
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.
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?
Thanks!
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.
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?
Check out the comment and replay at the bottom of this page, which I think is a bit closer to what you are looking for.
Hi could you please explain why the centripetal force is F + mgsintheta?
Shouldn’t it be F + mg/sintheta?
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
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?
Hi can you please explain why d = r(1-sintheta)
It’s alright, I figured it out 🙂
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!
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.