# Rollercoasters and Work-Energy

1. Jan 19, 2014

### aerospacedout

1. The problem statement, all variables and given/known data
A rollercoaster has an initial hill that leads to a circular loop of radius R. (a) Show that the top of the hill must be at least 1/2 R higher than the highest part of the loop. (b) Discuss additional factors that must be considered in the design of an actual rollercoaster of this type.

2. Relevant equations
PE = mgy
KE= 1/2mv^2
Ac=V^2/r
W=Fd= change in KE
2piR=C
Work done by friction(?)= Fnu(c)

3. The attempt at a solution

A) I started by realizing that the loop must be lower than the hill due to conservation of energy, because PEi=KEf(assuming no friction). What I don't understand is if there only needs to be enough energy to get the rollercoaster to the top of the loop and a little past it, why is the question telling me that it has to be at least 1/2 R greater? Why not 1/5 R? If there is friction, I don't know the coefficient so I have no idea whether or not this extra 1/2R of distance is needed to overcome friction. I'd appreciate some guidance as to where to go from here, thanks in advance.

2. Jan 19, 2014

### voko

If the initial hill is as high as the loop, then at the highest point of the loop, the velocity is zero. What happens to the train next?

3. Jan 19, 2014

### NasuSama

If you place an object on the hill at some height $h$, then it only has a potential energy, which is a total energy. Let it go, so without friction, the moving object, which makes back at a height $h$, has that amount of potential energy and reaches zero velocity. However, zero velocity at a same height $h$ doesn't happen for all cases; for instance, the loop de loop example you have; the moving object (with decreasing velocity) will not stay at some height on the loop. It will then fall off from the loop. Do you know why this happens, in physic term? It's not amount of energy. It's... (You determine this part by yourself.)

4. Jan 19, 2014

### aerospacedout

If the v at the top of the loop=0, then the rollercoaster car will just stop there, so the initial hill has to be higher. I understand that, but why is a specific number such as 1/2R needed? Even if the initial hill was a very small amount higher, wouldn't the car be able to move enough to continue the loop?

5. Jan 19, 2014

### voko

Stop there? What happens with an object stuck immobile high in the air?

So instead of "stopping", the car must be on a circular path. What is required for that?

6. Jan 19, 2014

### aerospacedout

I assumed it would stop because the car is conncected to the track. But if it was not connected to the track it would fall straight down.

To stay in a circle it requires centripetal acceleration, v^2/r

V^2/r= ac, v^2/r(1/2)=2ac? Or v^2/1/2r=mgh Am i on the right track? No pun intended

Im not sure how to relate centripetal acceleration to the fact that energy is conserved

Oh, at the top, v=0, so centripetal acceleration at the top would be zero, so it would just be gravity

7. Jan 19, 2014

### voko

If v = 0 at the top, then the car falls down. You do not want that.

What is the acceleration of the car at the top? What is the speed of circular motion that corresponds to that acceleration?

8. Jan 19, 2014

### aerospacedout

Ok so Ac at the top is v^2/r
Solving for velocity mgh=1/2mv^2 you get (2gh)^(1/2)
Plug that into v^2/r you get 2gh/r=ac

Solve for r you get r=2gh/ac... did i do something backwards? I can make it r/2=ghac But i feel like thats wrong, unless you take out the gh because theyre constants in it... then youd have ac=r/2

9. Jan 19, 2014

### voko

Again, what is the acceleration at the top? What forces are responsible for it?

10. Jan 20, 2014

### aerospacedout

The acceleration would be g i believe, or g plus the normal force of the rollercoaster/track system

I GOT IT OMG

V^2/R=g
V=(gr)^(1/2)
Mgh=1/2mv^2
Gh=1/2(gr)
H=1/2R

Thank you everyone. I believe this is the correct solution.

11. Jan 20, 2014

### voko

Very well.

Just a word of caution. At some point you sounded like "because the thing moves in a circle, it has centripetal acceleration". That is a common mistake. It moves in a circle because it is acted upon by a force, or forces. The centripetal acceleration, like any other acceleration, is always a consequence of force. Very frequently it is a force of reaction alone, but it can be gravity or an electric force, or any combination.

12. Jan 20, 2014

### aerospacedout

Thank you very much, I will keep this in mind.