Friction between roller coaster and track

  • #1

Homework Statement


A roller coaster is going down on a track (see image below, the track itself is highlighted in red) without initial velocity. What is the coefficient of friction between the roller coaster and the track?

height at which the cart starts (h) = 12m
weight of the car (m) = 620kg
angle of the track (α) = 40°
radius of the circle (r) = 8m
normal force acting on the cart in the moment it reaches the lowest point (N) = 22,900N

Homework Equations


I managed to calculate the energy lost during the process, but I have no idea what to do next.

The Attempt at a Solution


I calculated the energy of the cart at the highest and lowest points (hopefully my thought process is visible on the image) and the difference of the two is the energy lost due to friction.
 

Answers and Replies

  • #2
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  • #3
PeroK
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Okay, so you've got the energy loss. Where did that energy go?
 
  • #4
All of that is lost through friction. I just don't know how to figure out the coefficient, since N is changing throughout the movement
 
  • #5
PeroK
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All of that is lost through friction. I just don't know how to figure out the coefficient, since N is changing throughout the movement

Okay, I was wondering what you thought about that! Do you think you are supposed to make some assumption? If this is "introductory" physics, it looks a little hard. If not, then you'll have to integrate round the curved bit of the track, which isn't so easy. Do you know the maths to do that?
 
  • #6
My high school teacher gave this problem so I thought it's introductory. I don't know the maths for that, but I think it has something to do with finding the area below a curve. I can look it up if that's what it takes to solve this. Could you give me some hints on how to go on from here?
 
  • #7
PeroK
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My high school teacher gave this problem so I thought it's introductory. I don't know the maths for that, but I think it has something to do with finding the area below a curve. I can look it up if that's what it takes to solve this. Could you give me some hints on how to go on from here?

First, you need to do a bit of geometry. You're diagram is not bad, but note that if ##r = 8m## and ##h = 12m## then the circle is much bigger if drawn to scale. You need to work out how far down the slope the point of intersection is. Hint: first find the angle subtended at the centre of the circle.

Then, you'll need an expression for the normal force at each point of the curve and you'll have to integrate that round the curve.

If there's a quick way, I don't see it.
 
  • #8
PeroK
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Here's an idea: redraw your diagram so that the straight slope finishes just at the ground, where the circular track begins. Don't have the straight track meet the circular track above the ground. That makes it easier and I think that's the problem that was intended.
 
  • #9
I could calculate that the circle and the straight part of the track meet at a height of 1.9m
The function for the normal force is:
f(x) = gravitational force*cos(x)+centripetal force or in terms of numbers f(x) = 6200*cos(x)+16740
x is between 0°(at the lowest point of the circle) and 40°(where the circle becomes part of the track). I tried to understand integrals, but it seems like a much harder topic than I could cover in one day so I'm stuck at this point
 
  • #10
PeroK
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I could calculate that the circle and the straight part of the track meet at a height of 1.9m
The function for the normal force is:
f(x) = gravitational force*cos(x)+centripetal force or in terms of numbers f(x) = 6200*cos(x)+16740
x is between 0°(at the lowest point of the circle) and 40°(where the circle becomes part of the track). I tried to understand integrals, but it seems like a much harder topic than I could cover in one day so I'm stuck at this point

It's definitely too hard to learn in one day. See my suggestion above. I think the circular track is supposed to start at the bottom and curve behind the initial slope. Like a real rollercoaster!
 
  • #11
Oh, I think there is a misunderstanding. The car doesn't travel all the way around the circle. The circle is just there to show the curve of the track after the straight part. After it reaches the lowest point it doesn't go up, it continues straight, but that's not part of the problem anyways. So the whole movement is just going down the slope and the small curve of the circle where the track becomes flat. Sorry if I phrased it wrong, English isn't my primary language
 
  • #12
PeroK
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Oh, I think there is a misunderstanding. The car doesn't travel all the way around the circle. The circle is just there to show the curve of the track after the straight part. After it reaches the lowest point it doesn't go up, it continues straight, but that's not part of the problem anyways. So the whole movement is just going down the slope and the small curve of the circle where the track becomes flat. Sorry if I phrased it wrong, English isn't my primary language

If it doesn't go up and start a loop, then there would be no centripetal acceleration.
 
  • #13
How should I approach the problem then?
 
  • #14
PeroK
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How should I approach the problem then?

Just assume the slope goes (almost) all the way to the bottom. This minimises the length of the circular track before it gets to the bottom and means that the length of curved track with variable normal force can be neglected. Effectively assume the track is straight all the way to the ground, where the circular upward loop begins.
 
  • #15
haruspex
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Seems to me your teacher has set a far more difficult problem than intended. For the curved section, I get an equation of the form ##yy'=A\sin(x)-\mu A \cos(x)-\mu y^2## where A=g/r. I very much doubt that can be solved.
PeroK's approximation in post #14 looks as good as anything.
 
  • #16
haruspex
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Seems to me your teacher has set a far more difficult problem than intended. For the curved section, I get an equation of the form ##yy'=A\sin(x)-\mu A \cos(x)-\mu y^2## where A=g/r. I very much doubt that can be solved.
PeroK's approximation in post #14 looks as good as anything.
Correction: the differential equation can be solved, but the resulting equation for mu involves a mixture of functions which can only be solved by numerical approximation.
 

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