# Minimum safe height for a roller coaster

1. Dec 17, 2016

### jmm5180

1. The problem statement, all variables and given/known data
Based on the height of a first hill (114.5m), mathematically determine the minimum height (circular motion and radius) of the next hill based on the speed generated on the first hill.

At the bottom of the first hill, the velocity of the roller coaster is 47.38m/s. There is no friction force, so the velocity is the same at the bottom of the second hill.
There is a flat section of the track between the two hills that is 93.81m long. At the top of the second hill, when it is 112m tall, the velocity is 7.03m/s, and the roller coaster is able to go over and down the second hill safely.

2. Relevant equations
KE = mgh
Fnet= N-W= -ma
a= v^2/r

3. The attempt at a solution

When the hill isn't tall enough, the roller coaster launches off the top of the hill. When the hill is too tall, the coaster doesn't reach the top and slides back down.

I tried to calculate the maximum height,
KE (at bottom of the second hill) = mgh
0.5mv^2 = mgh
mass cancels
0.5v^2 = gh
H= (0.5v^2)/9.8
H= 114.5 m, which is the same height as the first hill so the coaster will be able to reach the top without sliding back down.

When I tried to get the minimum safe height, I had issues.
I tried summation of forces.
Y direction:
Fnet= -N-W = -ma (since acceleration is directed downward toward the middle of the hill)
The coaster flies off when the normal force = 0.
-W=-ma
Mg=ma
G=a
g= v^2/r
r= v^2/g
I plugged in the velocity at the bottom of the hill for v, and got 229.06 m, which exceeds the max safe height.

If it helps, it took 11.72s to reach the bottom of the second hill, then 1.98s to travel the flat section of the track, then 6.11s to reach the top of the second hill.

Any help is appreciated. Thanks.

Last edited: Dec 17, 2016
2. Dec 18, 2016

### haruspex

That does not make sense. Given the height, you can determine the minimum radius; given the radius, you can determine the minimum height.

3. Dec 18, 2016

### Cutter Ketch

You have the right idea. You say the train leaves the track when the normal force falls to zero. You constructed the right formula: g=v^2/r. You just applied it in the wrong place.

They could certainly introduce curvature that the train would fly off of at any time. An unspoken premise of the problem is that they are asking about the top of the second hill. The second hill won't be a semicircle right from the bottom. It will have a hill shape and will have its minimum approximately circular curvature right at the top. Assume you safely reach the top of the second hill. Find a relation between the height of the hill and the radius of curvature at the top of the hill. i.e. having worked out g=v^2/r can you replace v with a function of the height of the hill?

4. Dec 18, 2016

### jmm5180

I'm only in AP Physics 1, so we haven't done any functions yet, but the hill can only be as high as the kinetic energy from the first hill will allow it to be...
0.5mv^2 = mgh
V^2 = 2gh

r = 2gh/g or 2h, so double the height of the last hill, but that can't be right.

5. Dec 18, 2016

### Staff: Mentor

That is the upper limit, and you found it already.

Can you calculate the speed at the top of the hill?

If the roller coaster is not supposed to take off, what is the minimal curvature radius of the track with that speed?

6. Dec 18, 2016

### jmm5180

The radius is half the height, so 56m at that speed of 7.03m/s (this was an online simulation and that speed was measured in the simulation).

7. Dec 18, 2016

### Staff: Mentor

No it is not. The rollercoaster is not a circle, and your approach would have nothing to do with the idea of not leaving the tracks.

8. Dec 18, 2016

### jmm5180

I am not sure how to deal with curvature except in circles; we haven't covered it yet.

9. Dec 18, 2016

### Staff: Mentor

You can assume that the top part is like a circle, but the second hill is not a giant circle starting at the bottom: you don't get its radius from the geometry of the rollercoaster.

Let's start from scratch because I don't think the current discussion is making progress:

What is the condition for the rollercoaster to stay on top of the track? Can you find an inequality where the rollercoaster leaves the track when this inequality is violated?

10. Dec 18, 2016

### Cutter Ketch

Still with us jmm? Circles will be fine. In reality the hill would be something closer to a parabola. The curvature will be the tightest right at the apex and less everywhere else. But that's ok. Just understand that the question is about that tightest curvature right at the top of the hill. You can treat it as a circle. You can say what radius circle would be a problem right at the apex where the coaster is at its slowest. Just don't try and figure out what radius is acceptable anywhere else where the velocity is higher. Assume that the designers already worked that out.

11. Mar 16, 2017

### Clive Bellamy

roller coasters have wheels that hold the coaster on the tracks so it will not fly off at the top of hill two. all SLC coasters have this function to hold the cars in place. So the physics can and should only be applied if you accept conditions that are not real world situations. In other words you don't have enough information to make an accurate calculation to start with. include the force of the coaster with negative g and the friction on the wheels while in neg g. As usual assumption is the cause of confusion.