Rollercoasters- find the g forces


by studenthelp10
Tags: forces, rollercoasters
studenthelp10
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#1
Jun26-12, 11:51 PM
P: 26
Hi
Im a bit confused about where to go with this problem - could anyone possibly help please ?

I have to determine the g forces for the kingda ka rollercoaster
according to research the kingda ka rollercoaster is 139m tall
it shoots up on a 90 degree angle to the right- reaches the peak of the hill then spirals downward on 270 degree angle

according to : http://www.sixflags.com/greatadventu.../kingdaka.aspx

I went from here to calc the g force when it drops using the equation (height lost= speed gained) : 1/2mv^2=mgh (independant of mass) so cancel (m's)... -> 1/2v^2=gh -> v^2=2gh -> v=sqrt(2gh)

h=139m
g=10
calc (speed)= 52.73ms^-1

Equation for g force at bottom of loop: (v^2/r)+g/g

*note* i calc the heights of other points and the radius of the ride by making it proportional to the rest of the ride e.g. 139/7.3=x/1.5 -> x= 28.5

anyways , I then calc the g force using equation from above and got a g force of 10!!! this must be wrong or something??? because it should be 4 or less

The only explanation i can think of is maybe friction or banking???? but i dont know how to get the angle for banking - with only the given vertical 270 degree spiral downward- i think the angle is always changing ? could anyone explain SIMPLY please - Im new to this type of physics and any help would be appreciated :)
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rcgldr
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#2
Jun27-12, 01:35 AM
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Quote Quote by studenthelp10 View Post
Equation for g force at bottom of loop ...
There isn't any loop, just curves that transition the coaster from horizontal to vertical (for the climb) and later from vertical to horizonta (after descent). The shape of these curves is not stated. You'd need to know the radius of curvature versus the expected speed in order to determine the g forces. For roller coasters with loops, usually something similar to a clothoid loop is used to reduce peak g forces:

wiki_vertical_loop_physics.htm

clothoid_loop.htm
haruspex
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#3
Jun27-12, 01:54 AM
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There's no great g force generated during the 'spiral' (actually, helical) descent because the axis of rotation is in line with the velocity. The 'g' force is a cross product of linear velocity with angular velocity, so is maximised when they're at right angles.
The 'g' forces will kick in, as rcgldr says, when it bottoms out. The curve there looks fairly gentle.
I'm not sure whether your 4g figure includes the normal 1g or is in addition to it. One way makes the radius about 70m, the other about 90m.

studenthelp10
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#4
Jun27-12, 06:29 PM
P: 26

Rollercoasters- find the g forces


Thank you for the replies :0
so how would you then calculate the g forces when it swings by the bottom ? also how did you get the radius measures?

I have no idea what you mean by "axis of rotation is in line with the velocity. The 'g' force is a cross product of linear velocity with angular velocity, so is maximised when they're at right angles." Im sure g force was defined as how many times being larger than gravity or smaller .e.g. gravity acceleration is 10ms^-2 /10 =1g, however 20ms^-2/10=2gs but that is all I know as well as a few other equations + energy conversion
rcgldr
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#5
Jun27-12, 09:25 PM
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Quote Quote by studenthelp10 View Post
so how would you then calculate the g forces when it swings by the bottom?
You would need to know the speed versus radius of curvature of the two sections of track where the coaster transitions from horizontal to vertical and later from vertical back to horizontal. This data was used during the design of the roller coaster, but I'm not sure where you could obtain the actual data for a specific roller coaster.
haruspex
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#6
Jun28-12, 02:43 AM
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Quote Quote by studenthelp10 View Post
how did you get the radius measures?

I have no idea what you mean by "axis of rotation is in line with the velocity. The 'g' force is a cross product of linear velocity with angular velocity, so is maximised when they're at right angles."
I took the claim that the max force was 4g, your figure for the speed, and deduced the radius according to acceleration = v2/r. As I said, I wasn't sure whether the 4g included gravity or was in addition.
On the long descent, before it starts to level out, the track screws around in a tight helix. Although the radius is small, most of the velocity is straight down, along the axis of the helix. Since this component of the velocity remains straight down, unaffected by the twisting, it does not contribute to any forces. Only the relatively small horizontal component of the velocity gets screwed around in a circle.
studenthelp10
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#7
Jun28-12, 06:37 PM
P: 26
thanks, for the explanation :). I think i get it a bit more now
It hard to find any sites showing the radius of the track so I think i will use the equations to try and find the radius like a=v^2/r
I have to log off now, but ill will post back later
Thanks :)


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