Roller Coaster, Cosine 'ramp', Exit Velocity?

In summary, a roller coaster has a cosine shape, starts at the top with near 0 velocity, has no friction, and goes down a cosine shaped ramp at a constant velocity. The velocity at the minimum height is found to be the same as if the coaster was simply dropped off the edge of the track.
  • #1
jim.nastics
14
0

Homework Statement


The teacher screwed up and gave us a problem that he could not solve.
I am trying to solve it for fun.

A roller coaster has a cosine shape.
The coaster starts at the top (0 pi) with near 0 velocity
Friction is 0, and there is no air resistance.


Homework Equations


Want a hint... Not a full solution


The Attempt at a Solution


Tried to integrate vdv = sin(theta)d(theta) from 0 to pi/2
Answer = 0 (doh!) Should have seen that before I started!

Teacher conjectures that the result will be the same as if the coaster went down a straight ramp at 45 degrees, but can't prove it. It is bugging him too.

Thanks in advance,
Jim
 
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  • #2
Sorry, but what is the question again?
 
  • #3
Hi Robb,

After the coaster goes down the cosine shaped ramp, what is the exit velocity?

Also, I typoed...
Meant to enter: dv = sin(theta)d(theta) from 0 to pi/2

Regards,
Rudy
 
  • #4
So a coaster starts at a known height, the amplitude of the cos function, and is released from rest. Find the speed at the minimum height? i.e. at the position corresponding to the minimum of the cos function. No friction either. Is this it?
 
  • #5
(edit: Reformatted using LaTex, now I know how to use yet another language...)
Yes, that is the question. I am struggling with this, and decided to approach it differently. I keep coming up short needing t, so I decided to look at t, and consider a simpler model, the inclined plane.

Let r be the height of the inclined plane
Let L be the length of the plane
Let Φ be the angle of the plane from horizontal
Let m be the mass on the plane
Let g be acceleration due to gravity
Let w be the downward vector due to gravity
The plane is frictionless

Starting with:
[tex]a = \frac{f}{m}[/tex]


[tex]f = w sin\phi[/tex]

But
[tex]w = mg[/tex]

So:
[tex]a = \frac{mg sin\phi}{m} = g sin\phi [/tex]; how fast the mass moves

How far does it go?
[tex]L = \frac{r}{sin\phi} = \frac{at^2}{2}[/tex]

Substituting...
[tex]t = \sqrt{\frac{2L}{a}} = \frac{1}{sin\phi} \sqrt{\frac{2r}{g}}[/tex]

So that is a ramp...
What if the ramp is a quarter circle?
Then:
[tex] L = \frac{\pi r}{2}[/tex]

I can conclude(can I?) that:
[tex] t = \sqrt{\frac{\pi r}{gsin\phi}}[/tex]

But Φ is constantly changing, so t is a function of Φ

If I am OK to here, then I can do this:
[tex]\frac{dt}{d\phi} = -\sqrt{\frac{\pi r}{4g}} \frac{cos\phi}{(\sqrt{sin\phi})^3}[/tex]

[tex] dt = -\sqrt{\frac{\pi r}{4g}} \frac{cos\phi}{(\sqrt{sin\phi})^3}d\phi[/tex]

let [tex]u = sin\phi[/tex], [tex]du = cos\phi d\phi[/tex]

Wow, this can be integrated from 0 to π/2.

Did I make it this far?

Regards,
Jim
 
Last edited:
  • #6
I had a strange nightmare last night... It relates to this problem.

I was in my car rolling down a hill, thinking about this problem (don’t know why the engine wasn’t running). I guess I wasn’t paying attention to the road, and ran into someone :bugeye: (I think it was my physics teacher...). I almost woke up. Almost, but not quite.

The collision sent him over the edge of the road, perpendicular to my path, at the same velocity as I was traveling at the time, with only a horizontal component. I know his velocity was the same as mine, as I could see him in my side mirror, and he was always at an angle of 45 degrees on my left.

As he went over the edge, his path became parabolic in the z direction, just as a roller coaster would actually be built to provide that freefall feeling. He was neglecting air resistance (wouldn’t you at such a time?).

I continued down the slope, and followed it to the left, going up and down several times, at varying angles, finally becoming parallel to the path of the teacher, at the bottom of the hill.

Now he was on a roller coaster (told you this was a strange dream), and he was out ahead of me (glad he is OK, I was worried).

What I saw woke me up.

He was out ahead of me because he reached the bottom before I did. We had the same velocity, he was not getting further away from me, neither was I catching up to him.

So, my intuition tells me that time is in some way proportional to the angle of descent, but only the starting altitude affects final velocity. More importantly, final velocity is completely independent of time and angle of descent.

The physics tools I have at this time are kinematics, circular motion, and momentum. Can I prove this conjecture? If I can, then the velocity at the bottom of the track is the same as if the coaster was simply dropped off the edge of the track (of course I mean magnitude of velocity, not direction).

Regards,
Jim
 
  • #7
Do you know about energy conservation?
 
  • #8
No, I don't. Looking at the course syllabus, I will know about it in two weeks (after spring break). I am, of course, willing to read ahead. Since you are asking, I assume it would help me out.

Oh, and my proposed integration in post 5? Looks like it would take forever to get moving (division by 0, limit from the right approached infinity), and then get the rest of the way down the curve pretty quick.
Regards,
Rudy
 
  • #9
jim.nastics said:

Homework Statement


The teacher screwed up and gave us a problem that he could not solve.
I am trying to solve it for fun.

A roller coaster has a cosine shape.
The coaster starts at the top (0 pi) with near 0 velocity
Friction is 0, and there is no air resistance.


Homework Equations


Want a hint... Not a full solution


The Attempt at a Solution


Tried to integrate vdv = sin(theta)d(theta) from 0 to pi/2
Answer = 0 (doh!) Should have seen that before I started!

Teacher conjectures that the result will be the same as if the coaster went down a straight ramp at 45 degrees, but can't prove it. It is bugging him too.

Thanks in advance,
Jim

It is trivial to solve using energy considerations, but a nightmare using F=ma.
What are the forces? There is gravity and a normal force...But in addition, there must be a normal force "inward" to keep the roller coaster from "flying off". (In practice, the wheels of the roller coaster are "locked" between some rails). This all makes it pretty much hopeless with F=ma. But with energy, again, it's trivial (assuming no friction)
 
  • #10
Thanks, robb and nrged.

I'll go read ahead in the book. I need to find out for myself if my intuition is correct. Thanks for providing guidance, and for not posting spoilers.

I do agree that there must be a better way than starting with f=ma, especially if getting an answer means tossing out the infinity part and keeping the difference (even if that difference looks correct).

I'll post again when I have a question (or a solution).

Regards,
Rudy

p.s. this is a great forum!
 
  • #11
jim.nastics said:
(edit: Reformatted using LaTex, now I know how to use yet another language...)
Yes, that is the question. I am struggling with this, and decided to approach it differently. I keep coming up short needing t, so I decided to look at t, and consider a simpler model, the inclined plane.

Let r be the height of the inclined plane
Let L be the length of the plane
Let Φ be the angle of the plane from horizontal
Let m be the mass on the plane
Let g be acceleration due to gravity
Let w be the downward vector due to gravity
The plane is frictionless

Starting with:
[tex]a = \frac{f}{m}[/tex]


[tex]f = w sin\phi[/tex]

But
[tex]w = mg[/tex]

So:
[tex]a = \frac{mg sin\phi}{m} = g sin\phi [/tex]; how fast the mass moves

How far does it go?
[tex]L = \frac{r}{sin\phi} = \frac{at^2}{2}[/tex]

Substituting...
[tex]t = \sqrt{\frac{2L}{a}} = \frac{1}{sin\phi} \sqrt{\frac{2r}{g}}[/tex]

So that is a ramp...
What if the ramp is a quarter circle?
Then:
[tex] L = \frac{\pi r}{2}[/tex]

I can conclude(can I?) that:
[tex] t = \sqrt{\frac{\pi r}{gsin\phi}}[/tex]
No, you can't conclude that (I am guessing that you mean by this "t" the time for the object to slide down? No, there is absolutely no way to get directy an equation for this time. One has to start from F=ma and get from that a differential equation for the angle as a function of time.
 

1. What is a roller coaster?

A roller coaster is an amusement ride that consists of a track with steep drops, sharp turns, and high-speed loops. Passengers are seated in train-like cars and experience a thrilling and exhilarating ride as the train moves along the track.

2. What is a cosine 'ramp'?

A cosine 'ramp' is a mathematical term used to describe the shape of a roller coaster track. It refers to a smooth, curved section of the track that resembles the shape of a cosine curve.

3. How does the cosine 'ramp' affect the ride experience?

The cosine 'ramp' is designed to gradually change the direction of the roller coaster train, creating a smooth transition between different sections of the track. This helps to reduce the G-forces experienced by passengers and makes the ride more comfortable.

4. What is exit velocity in a roller coaster?

Exit velocity is the speed at which the roller coaster train leaves a particular section of the track. It is typically measured in miles per hour (mph) or kilometers per hour (km/h). The exit velocity can vary depending on factors such as the height of the drop and the design of the track.

5. How is exit velocity calculated?

Exit velocity is calculated by using the principles of physics and the laws of motion. Factors such as the height of the drop, angle of descent, and friction of the track are taken into consideration to determine the final velocity of the roller coaster train as it exits a particular section of the track.

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