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Rollercoasters find the g forces 
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#1
Jun2612, 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 :) 


#2
Jun2712, 01:35 AM

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wiki_vertical_loop_physics.htm clothoid_loop.htm 


#3
Jun2712, 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. 


#4
Jun2712, 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 


#5
Jun2712, 09:25 PM

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#6
Jun2812, 02:43 AM

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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. 


#7
Jun2812, 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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