Roller Coaster Hills and Loops

In summary, based on the information given, the height at which the passenger experiences a 1G force into his seat is equal to the height of the initial hill plus 1.5 meters.
  • #1
JebsCripedìas
4
0

Homework Statement



A roller coaster begins with a steep downhill descent followed by a vertical loop. How much higher does the initial hill have to be to than the height of the loop so that the riders experience a force of 1G keeping them in their seats at the top of the loop? Assume that the rolling friction involved is negligible.

Homework Equations



Fc = mV2/r
Fc = FT - Fg (at the top of the loop)
ac = V2/r
Ep = mgh
Ek = mV2/2
ET1 = ET2

The Attempt at a Solution



For starters, I imagine there's going to be a lot of algebra involved here as no information is given. I've tried working under the assumption that m=1 and r=1. I'm fairly certain I can do the same thing with V, but I'm not going to risk sabotaging my work any more than I might have already :redface: Also, I figure Fc must be 9.8 N to account for the 1G, although I'm not convinced I'm right there. I've tried determining the velocity through centripetal force and I've tried applying the law of conservation of energy. Unfortunately, it's been about 4 months since my last physics course, so a lot of this stuff has left me already.

Thanks :smile:
 
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  • #2
Welcome to PF!

Hi JebsCripedìas! Welcome to PF! :smile:
JebsCripedìas said:
A roller coaster begins with a steep downhill descent followed by a vertical loop. How much higher does the initial hill have to be to than the height of the loop so that the riders experience a force of 1G keeping them in their seats at the top of the loop? Assume that the rolling friction involved is negligible.

Come on … think! …

How much force (they mean acceleration, of course) are you experiencing keeping you in your seat at the moment? :wink:
 
  • #3
Honestly, I'm ashamed of myself. I know this is an easy question. Mass is canceled (at least when applyign the LCE), that much is obvious. The problem for me is that my h1 always ends up being a fraction of h2. I can't wrap my head around this and it's really getting on my nerves. I've somehow ended up with different answers, too. I've got h1 = 1/20, 1/2, and 1/r. The latter seems like the most likely to me, but I just don't see it.
 
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  • #4
JebsCripedìas said:
Honestly, I'm ashamed of myself. I know this is an easy question. Mass is canceled (at least when applyign the LCE), that much is obvious. The problem for me is that my h1 always ends up being a fraction of h2. I can't wrap my head around this and it's really getting on my nerves. I've somehow ended up with different answers, too. I've got h1 = 1/20, 1/2, and 1/r. The latter seems like the most likely to me, but I just don't see it.

I think you are thrashing about a bit. Draw a simple force diagram for the moment at which they want the passenger to be experiencing a 1g force into his seat. He's upside down at the top of the loop isn't he? His upward force needs to be g into the seat and there is also gravity working against that, so the radial acceleration must be ... what?

Now look at Kinetic energy he would be carrying at that point. From the height of the drop how much would that be?

Don't you now have a pretty good idea of how to express the height he started above that point in terms of r?
 
  • #5
Well here's what I'm doing, and I'm always getting h1 = h2/4. It's what I had done when I said one of my results for h1 was 1/2

Fc = FT - Fg = mV2/r
FT = m(V2/r + g)

Since FT - Fg has to yield 1g, or 9.8, I know FT = 19.6

19.6 = m(V2/r + 9.8)

I'm operating under the assumption that m = 1

19.6 = V2/r + 9.8
9.8 = V2/r

Now if r = 1 (in which case h2 = 2)

9.8 = V2 and 3.1 = V

So now I apply the law of conservation of energy.

Ep = Ek
mgh1 = mV2/2
gh1 = V2/2
9.8h1 = 9.8/2
h1 = 1/2 = 0.5m

Now I have h1 = 0.5m and h2 = 2m. I have to be missing something when I do the LCE. Actually, I think I see it now.

Ep1 = Ek + Ep2
mgh1 = mV2/2 + mgh2
gh1 = V2/2 + gh2
9.8h1 = 4.9 + 19.6
9.8h1 = 24.5
h1 = 2.5m

So now I have h1 = 5h2/4. It looks good to me. Tell me this is right please :smile:
 
  • #6
JebsCripedìas said:
… So now I apply the law of conservation of energy.

Ep = Ek
mgh1 = mV2/2
gh1 = V2/2
9.8h1 = 9.8/2
h1 = 1/2 = 0.5m
…
So now I have h1 = 5h2/4. It looks good to me. Tell me this is right please :smile:

HI JebsCripedìas! :smile:

Yes, that's right …

but it would be a lot simpler if you left g as "g", and r as "r" rather than 1:

mgh = mv2/2, and v2/r = g, so h = … :wink:
 
  • #7
Well I want to write h1 as a function of h2, so I don't see any other way to say it. I guess I must not be trying hard enough.
 

1. What causes the feeling of weightlessness on a roller coaster?

The feeling of weightlessness on a roller coaster is caused by the laws of physics. When the roller coaster goes over a hill or loop, it accelerates downward, creating a sensation of falling. Since both the rider and the coaster are falling at the same rate, there is no force pushing against the rider, resulting in a feeling of weightlessness.

2. How are roller coaster hills and loops designed?

Roller coaster hills and loops are designed using mathematical principles, such as calculus and physics equations. Engineers use these principles to determine the appropriate height, angle, and speed of the hills and loops to create a thrilling and safe ride.

3. Why do some roller coasters have more hills than loops?

The number of hills and loops on a roller coaster is determined by the overall design of the ride. Some roller coasters may have more hills to create a longer ride or to provide a different type of thrill. Additionally, loops require a lot of energy to keep the train moving, so roller coasters with more loops may have fewer hills to conserve energy.

4. How do roller coaster designers ensure the safety of riders?

Roller coaster designers use advanced computer simulations and physical testing to ensure the safety of riders. They also follow strict design standards and regulations set by organizations such as the International Association of Amusement Parks and Attractions (IAAPA) to ensure that the roller coaster is structurally sound and can withstand the forces of the ride.

5. Can roller coaster hills and loops be changed or modified?

Yes, roller coaster hills and loops can be changed or modified. Roller coaster designers often make adjustments to the hills and loops during the design process to fine-tune the ride experience. Additionally, some roller coasters may undergo renovations or upgrades over time, which may include changes to the hills and loops.

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