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# Roller Coaster Physics

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## Brendan Donnelly

on 13 November 2012

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#### Transcript of Roller Coaster Physics

Roller Coaster Physics A2 Physics - Circular Motion THE STEEL CYCLONE The Station This is the reference point for all calculations so: v=0, h=0, Ek=0, Egp =0 On top of the hill: Some key data: mass of car, m=800kg
friction is neglgible H=120m

Egp = mgh
Egp = 800 x 9.81 x 120
= 941760J And at the bottom: Ke = 1/2mv^2 = 941760J r=15 Going over the hill: r = 22m Loop the Loop: G-Force Safety Regulations v = sqrt(Ke/ 1/2m) v = sqrt(941760 / 1/2x800) v = sqrt(2354.4) = 48.5 ms^-1 (3s.f.) mg-S=mv^2 / r, S=0 so: mg=mv^2 / r When the support force is 0 the roller coaster will lift from the track. v = sqrt(mgr / m) So 12.1 ms-1 is the maximum velocity at which gravity will keep the train in contact with the track. The Steel Cyclone however travels more than 4x than this.

This is safe as the Coaster is attatched to the track, meaning that the coaster will stay maintain contact with the track but riders will be lifted out of their seats, making it a super awesome ride. v = sqrt(gr) v = sqrt(gr) =14.69m/s v = sqrt(9.81 x 22) v = sqrt(215.82m/s) v = sqrt(800x9.81x15 / 800) v = sqrt(147.15) v = 12.1 ms-1 All of the Egp gained going uphill is converted to Ek: v = 48.5 ms-1 Safety Features Headrests are specifically designed to work with Harnesses. In most roller coasters, the head moves and twists on the neck. The headrest is the safety device that prevents the head from snapping backwards, support for when G-force is experienced and prevents whiplash. A Key safety requirement on a Roller coaster are the Harnesses that come down and keep you in your seat and as still as possible so that during the rapid acceleration you are held still and cannot injure yourself. In order for a roller coaster to be classed as 'non high-g' and therefore safe for most riders it must not exert a force of more than 3.5g on the rider. However, higher G-forces are acceptable as long as safety precautions are put in place. Safety a = v^2 / r a = 30^2 / 15
a = 60ms-2

g-force = a / g
g-force = 60/9.81
g-force = 6.12g This would put our roller coaster up with the highest-g thrill riders in the world. In order for a roller coaster carriage to make it through the loop must exceed : As the speed of our roller coaster far exceeds this it will have no problem completing the loop. Braking Rollercoasters employ a wide range of braking systems, from pnuematic clamps to magnets.

It is important to control the speed as the levels of friction are often so low that friction is not sufficient. Welcome To The STEEL CYCLONE QUEUE HERE Making it the 4th fastest Roller Coaster in the world.
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