Model
Symbols I will use in my calculations:
 r = radius of the loop
 k = constant coefficient of r for specific loop where kr = minimum height required
 R = Normal reaction force between track and particle
 m = mass of object
 a = acceleration towards the centre of the circle
 θ = angle that the ramp makes with the horizontal
 g = acceleration due to gravity

Ek = Kinetic energy

Ep = Gravitational potential energy
 v = velocity at a given point
Analysis
Using the conservation of energy I can say that the total energy at A equals the total energy at B which also equals the total energy at C.
Model to “Just” reach the top of the loop:
Ep at A = Ep at C
mgkr = mg2r divide both sides by mg
kr = 2r divide both sides by r
k = 2
Therefore for the object to “just” reach the top of the loop h = 2r
From this I can calculate the velocity at the bottom of the loop to “just” reach the top:
Ep at A = Ek at B
mg2r = ½ mv2 Divide both sides by m and x by 2
4gr = v2 square root both sides
v = 2 gr
The height and velocity I have just calculated are a minimum bound because to be completely certain that the roller coaster completes a loop it must still be in contact at point C at the top of the loop to ensure it does not fall off. A real life roller coaster runs on wheels which keeps it on the track. Therefore would only need to just get past the point C to get complete a loop and would not fall off.
Diagram to show the forces at point C when the object is still in contact:
From the above diagram you can see that there are two forces acting on it. Because it is still in contact with the surface then there must be a reaction force, therefore R>0.
We know that the object is in circular motion therefore there must be a resultant force and it is equal to (mv2)/r , the resultant force is R + mg. We can use this to show:
R + mg = (mv2)/r  mg from both sides
R = (mv2)/r – mg we know that R>0 therefore
(mv2)/r – mg > 0 + mg to both sides and x by r
mv2 > mgr divide both sides by m
v2 > gr
Now I will create a model to look at the height when the object is still in contact at point C and will complete a loop:
Using the conservation of energy:
Ep at A = Ep at C + Ek at C
mgkr = mg2r + ½ mv2 divide by throughout by m and x by 2
2gkr = 4gr + v2  4gr from both sides
v2 = 2gkr – 4gr but v2 > gr therefore
2gkr – 4gr > gr + 4gr and divide through by g
2kr > 5r divide both sides by 2r
k > 2 ½
Therefore h > 2 ½ r
This is only just higher that the value of h to “just” reach the top so it looks a bit small to be correct. I will do an experiment to validate my model for getting all the way around this and compare my results with my model.
Experiment
For the experiment I will use a marble and a loop the loop made from plastic curtain rail. I started by dropping the marble from the height h = 2r and measured the height it reached, the distance travelled until It completed the loop. I think that the height required will be more that 2.5r even thought that is what the model predicted because I have ignored friction. Mass of marble = 4.39g.
As you can see the experiment has a curved ramp unlike the model.
Results:
I found my minimum release height to complete a loop was 25 cm. These results were not what my model predicted. I think this is mainly due to leaving friction out of the calculations. The marble needed to be dropped from a bigger height that 2.5r because there was work done against friction. The extra height required and therefore the extra potential energy put into the system will be the work done against the frictional force between the marble and track. Also my result will be affected by the fact that my experimental model had a curved ramp rather than a straight ramp I used in my model.
Improving My Model to Include Friction
I know that the work done against friction is the extra potential energy put into the system. Therefore using work done = force x distance I can find an average force for the friction for the marble and the track in my experiment:
Force = Work done/ distance
Force = (mg(3.572.5)r)/55
Force = 5.87N
You can see that the distance travelled while doing work against friction is dependant on the angle of the ramp and the height released. Therefore the distance can be generalised and then be used to find a term for work done against friction:
Distance = kr/sinθ + πr
Wok done against friction = F((kr/sinθ) + πr)
Model to complete a loop including friction:
Using the conservation of energy:
Ep at A = Ep at C + Ek at C + Work done against friction
Using the improved model I will find a new theoretical value for k:
F = 5.87N
m = 4.39g
r = 7cm
I worked out θ by knowing what the distance travelled down the ramp was and the vertical height :
θ = arcsin (25/557π)
θ = 49
I found a theoretical value for k of 3.57 which is exactly equal to what my experimental value of k was found to be.
Interpretation
After recalculating the model to include friction and comparing theoretical answers against the experimental answers I am very pleased with how close my improved model predicts the minimum height.
I think that my original model was inaccurate mainly because I had ignored friction after finding that a roller coaster minimises it by using bearings and lubrication. The friction in my experiment was higher that I had thought. Also my experiment differed from my model because I used a curved track for the slope where my model had a straight piece. This would result in different values for the distances travelled while working against friction which would affect the accuracy of my model. I also think that when the marble travelled around the track it would have done a small amount out work in deforming and bending the track slightly which mean that my theoretical value for h is slightly smaller that reality.
It would be interesting to try and model the roller coaster as a long vehicle rather that a particle and experiment with similar objects because I think if you did that then the value of h would be different because of increased friction mainly. If it was a long carriage then it would have a greater tendency to get “stuck” at the top of the loop if you did not drop it from a high enough point.
If I had more time I would experiment more with a different track and objects with different values of friction, this would allow me to test my model even more to see if it is accurate. Maybe I could investigate a track that had a noncircular loop on it.
This would be difficult because it Is undergoing circular motion with a changing radius and maybe my model would involve differential equations.
mgkr = mg2r + ½ mv2 + F((kr/sinθ) + πr) rearrange and x2
2[mgkr  mg2r  F((kr/sinθ) + πr)] = mv2 divide throughout by m
(2/m)[mgkr  mg2r  F((kr/sinθ) + πr)] = v2 but we know v2 > gr therefore
(2/m)[mgkr  mg2r  F((kr/sinθ) + πr)] > gr rearrange to get
mgkr  mg2r  F((kr/sinθ) + πr) > ½ mgr rearrange and expand
mgkr  F(kr/sinθ)  Fπr > 2½ mgr + Fπr to both sides
mgkr  F(kr/sinθ) > 2½ mgr + Fπr factorise the left side
k ( mgr + Fr/sinθ) > 2½ mgr + Fπr divide by (mgr + Fr/sinθ)
k > (2½ mgr + Fπr)/( mgr + Fr/sinθ) simplify fraction by factor r
k > (2½ mg + Fπr/( mg + F/sinθ)