Finding lower and upper limit for time related to roller coaster

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  • #1
songoku
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Homework Statement:
The data below shows the height and speed of a roller coaster ride when it moves through the circular part of a track having diameter of 160 m.

The height is measured in meter and speed in meter per second. The bottom of the circular track is 20 meter from ground.

Calculate the lower and upper limit for the time for the roller coaster to travel from a height of 100 meter to a height of 20 meter. State the assumption and additional information (if needed) to be able to calculate the time
Relevant Equations:
Hyperbolic Function

Riemann Sum
height from groundspeed
10040
8048
6060
4072
2080

I tried to plot the points (speed on x-axis and height on y axis) and I got more or less like a straight line but I am not sure whether the graph would help to calculate the upper and lower limit of the time.

I also tried to use Riemann sum but just stuck because as far as I know, Riemann sum is used to calculate area under curve while for this question the time is not the area under the curve.

I am also not sure what assumption I need to make. Since this is math question, I don't think the assumption will be physics-related, such as frictionless track (I even used conservation of energy for this question and got nowhere). And I don't really understand about "additional information" part, basically because I have no idea how to calculate the time so I don't know whether the data given is sufficient or not

Please give me hint to start. Thanks
 

Answers and Replies

  • #2
BvU
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Relevant Equations:: Hyperbolic Function
Huh ? why ?
Please give me hint to start. Thanks
You should know by now that 'don't know how to start' isn't good enough for PF HW forums !

I tried to plot the points (speed on x-axis and height on y axis) and I got more or less like a straight line
If there is no friction a physicist expects mechanical energy conservation, so a plot of ##v^2## versus ##h## might yield a straight line -- but it doesn't. Wind drag spoils that.

[I also tried to use Riemann sum
That is your other relevant equation, and I suggest it's the one to work with.

You have speed and position, so you can calculate path length on the arc and time !

##\ ##
 
  • #3
Office_Shredder
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Given the speed on each section, you can compute how long it would take to traverse each section. You actually have two speeds, one at each endpoint of each section. One gives a lower bound of the time to traverse the segment, and one gives an upper bound.

One thing to watch out for: the roller coaster is not falling straight down! You'll have to do some geometry.
 
  • #4
BvU
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PF rules also forbid handing over the answer -- especially if the OP hasn't posted any own work yet !

##\ ##
 
  • #5
BvU
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a physicist expects mechanical energy conservation

BTW does anyone spot a big flaw in this exercise :wink: ?

##\ ##
 
  • #6
songoku
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You should know by now that 'don't know how to start' isn't good enough for PF HW forums !
PF rules also forbid handing over the answer -- especially if the OP hasn't posted any own work yet !
I don't ask for answer, only a hint for the correct direction to approach this question. I thought I have provided what I did, such as I drew a graph of height against speed, tried to use Riemann sum but failed because I didn't know to apply it for this question since this question is not about finding the area, tried conservation of energy.

The only reason I didn't post my graph because I think that's not the correct approach for this question and it won't result in any meaningful discussion about graph of height vs speed for this question. My #1 post is already my best knowledge for this question

If the OP is inappropriate and not in line with PF rules, I apologize

Huh ? why ?
That's the latest chapter (with Riemann sum) that I learn and since I don't know how to approach this question, I just wrote it down. Actually, the reason I drew graph height vs speed was to compare whether the graph would be similar to any hyperbolic functions

You have speed and position, so you can calculate path length on the arc and time !
Given the speed on each section, you can compute how long it would take to traverse each section. You actually have two speeds, one at each endpoint of each section. One gives a lower bound of the time to traverse the segment, and one gives an upper bound.

One thing to watch out for: the roller coaster is not falling straight down! You'll have to do some geometry.
This is what I come up after trying to understand this hint

Let say the roller coaster starts from the leftmost part of the circular track (since the radius is 80 m and the bottom part of the track is 20 m above ground so 100 m will be at the middle of the circular track).

It moves downward to height of 80 m and the area swept by it will be in the shape of sector of circle. I try to find the angle of the sector, ##\theta## , by using trigonometry:
$$\sin \theta = \frac{20}{80}$$

Then I use the formula of arc of circle to find the path length: ##\text{arc}=\theta.r##

Next, I find the time taken to travel each arc, using ##t=\frac{\text{arc}}{\text{speed}}##

I repeat this working for each part of the arc traveled by the roller coaster (there are 4 arcs in total) and add all the time to get the time needed by roller coaster to travel from height of 100 m to 20 m

Is this the correct approach?

That is your other relevant equation, and I suggest it's the one to work with.
If my working is correct, I think I don't use Riemann sum at all?

Thanks
 
  • #7
Office_Shredder
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I'm not sure about ##\sin(\theta)=20/80##, what triangle did you draw to get that?

I think what you're doing here is basically just Riemann sums. Let's consider a simpler example.

Suppose you had the speed at various times and you were asked to compute an upper and lower bound for distance traveled. You could multiply the speed times the time difference between each measurement to get the distance between each pair of measurements, and add them up. No Riemann sums required?

But that's exactly what a Riemann sum is! Same thing here. You have some function t(h) where time is a function of height. You compute how much t(h) changes between consecutive measurements, which is really an estimate of t'(h). Integrating t'(h) yields t(0)-t(100), which is the same thing as taking the amounts that t changed between consecutive measurements, and adding them to appropriately.


I think the really point here is that Riemann sums aren't some scary weird thing to do, they're just kind of the obvious thing you do.
 
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  • #8
songoku
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I'm not sure about ##\sin(\theta)=20/80##, what triangle did you draw to get that?
I am sorry I don't have my phone right now to take the picture of my drawing. I will send the picture as soon as I have my phone

I think what you're doing here is basically just Riemann sums. Let's consider a simpler example.

Suppose you had the speed at various times and you were asked to compute an upper and lower bound for distance traveled. You could multiply the speed times the time difference between each measurement to get the distance between each pair of measurements, and add them up. No Riemann sums required?

But that's exactly what a Riemann sum is! Same thing here. You have some function t(h) where time is a function of height. You compute how much t(h) changes between consecutive measurements, which is really an estimate of t'(h). Integrating t'(h) yields t(0)-t(100), which is the same thing as taking the amounts that t changed between consecutive measurements, and adding them to appropriately.


I think the really point here is that Riemann sums aren't some scary weird thing to do, they're just kind of the obvious thing you do.
Your explanation makes so much sense to me. I need to re-read my note and other sources to gain more understanding about Riemann sum.

Thanks
 
  • #9
songoku
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1619231487376.png
 
  • #10
Office_Shredder
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That looks good to me. Sorry, I had drawn my own picture but I did it in a more complicated way and mislabeled something and got the wrong answer myself.
 
  • #11
songoku
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Thank you very much for the help and explanation BvU and Office_Shredder
 
  • #12
BvU
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BTW does anyone spot a big flaw in this exercise :wink: ?

For the record: 40 m/s is 144 km/h. And with $$\Delta {1\over 2} mv^2 = mg \Delta h$$ a drop of 80 m would increase ##v## to 56 ms (203 km/h).

Which is too fast to be reasonable (but it seems to happen -- see below). And the 80 m/s (288 km/h) in the exercise is idiotic. The patients in the cars would suffer a vertical acceleration of ##\text {g} + v^2/r = 9\;\text {g}## and some would die.

https://en.wikipedia.org/wiki/G-force
Wiki said:
A typical person can handle about 5 g0 (49 m/s2) (meaning some people might pass out when riding a higher-g roller coaster, which in some cases exceeds this point) before losing consciousness, but through the combination of special g-suits and efforts to strain muscles—both of which act to force blood back into the brain—modern pilots can typically handle a sustained 9 g0 (88 m/s2)​


##\ ##
 
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