The Science of

Gravity (g) Gravity is a force of attraction between objects.

The more massive the

object, the greater the pull. However, the object has to be really

massive, like Earth, for the pull to be obvious.

NASA at the Amusement Park

g forces

Earth’s gravity = 1 g Provides a force of

acceleration known as free fall (9.8 m/s2).

High g’s Any acceleration greater than free fall.> 1 g> 9.8 m/s2

Low g’s Any acceleration

less than free fall.< 1 g< 9.8 m/s2

NASA at the Amusement Park

G forces

Stealth Facts

• Height Height 62 metres62 metres• Launch timeLaunch time 2.3 seconds2.3 seconds• Train weightTrain weight 8 tonnes (unladen)8 tonnes (unladen)

10 tonnes (laden)10 tonnes (laden)• Launch SpeedLaunch Speed 38 metres per second38 metres per second• Dive SpeedDive Speed 35 metres per second35 metres per second

Stealth acceleration

Acceleration = v – u (m/s) t (s) Acceleration = 38 – 0 2.3

Acceleration = 16.5m/s2

16.5 = 1.68 x gravity 9.8(v = final speed, u = initial speed, t =

time)

How far to reach 80mph?

S = ut + 1/2 at2

S = (0 x 2.3) + (16.5 x 2.3 x 2.3) 2

S = 0 + 43.6

S = 43.6m

S = distance, u = initial velocity, a = acceleration, t = time

Force to launch Stealth

Force = mass (kg) x acceleration (m/s2)

F = m x aF = 10000kg x 16.5m/s2

F = 165000kgm/s2

165 000N (Newtons)

10 tonnes = 10000Kg

Work done in launching Stealth

Work = Force (N) x Distance (m)Work =165 000 x 43.6Work = 719400Joules

Work = 7 194 KJWork = 7.19 MJ

The Power of Stealth

Power = work (J) time (s)Power = 7194000

2.3Power = 3127826Watts

Power = 3.1MW

Will you make it? –

Potential energy needed= mghPE = 10000 x 9.8 x 62

PE = 6 076 000J Kinetic energy = I/2mv2

KE = I/2 x 10000 x 382

KE = 7220000J

As long as KE is greater than PE ………

………Stealth will make it over the top

On the way up!Curve radius = 35m

Centripetal force = mv2

r Centripetal force = 100 x 38 x 38

on 100Kg rider 35

Centripetal force = 4125NForce of 1g on a 100Kg person = 1000N

4125N = 4.1gTotal force on rider = 4.1g + 1.0g =

5.1g

On top of the world!

PE = 6 480 000J KE = 7 220 000J

KE – PE = 740 000JSurplus KE = 740 000J

KE = I/2mv2

v2 = 2KE = 2x740000 = 148 m 10000

v = 12+ m/s

On top of the world

Outside curve radius 8mCentripetal force = mv2

r Centripetal force = 100 x 12 x 12

8 Centripetal force = 1800N

Force of 1g on a 100Kg person = 1000N

Resultant force 800N = 0.8g(almost weightless!)

On the way down!

Curve radius 40mCentripetal force = mv2

r Centripetal force = 100 x 35 x 35

40 Centripetal force = 3062N

Force of 1g on a 1000Kg person = 1000N Resultant force on rider = 3.1g + 1.0g =

4.1g

Nitrogen accumulators

Hydraulic fluid

The motors - Medusa

Cylinder block

Winch drum

The catch car

Wire cable and return drum

Brakes!• A metal plate moves

through a permanent magnetic field.

• Eddy currents in the field produce a force to oppose the motion.

• The higher the speed – the greater the force.

• Magnetic brakes never stop you completely. Air brakes finish the job

Brakes!

Brakes!

Keeping on track

Vortex• Manufactured by

KMG Europe/Chance Rides 2001

• Height 20m• Max 120 degrees

above vertical• 4.5g max.• Ride capacity 32• 500 per hour• Pendulum 9.2 rpm• Carousel 7.5 rpm• Duration 3

minutes

Pendulum

Vortex - Pendulum

Length of arm 8.5m

Period T = 2π√l /g = 2x 3.14√8.5/9.8

= 5.82s

Frequency = 10.3 rpmActual = 9.2 rpm

Vortex - Pendulum

Data: 9.2 rpm Length, l = 8m Displacement, x =

8m

Frequency of oscillation, f =9.2/60 = 0.15/s(Hz)

Acceleration, a = (2Πf)2x = (6.28x1/6.5)2x8 = 7.46ms-2 (0.76g)

Motion in a circle

Vortex - CarouselData: diameter, d= 8m, radius, r = 4m,Frequency, f = 7.5/60 = 0.125 /s (Hz)

Angular velocity ω = 2Πf = 6.38 x 0.125 = 0.785 radians

Linear speed, v = ωr = 0.785 x 4 = 3.14m/s (7mph)

Vortex - Carousel

Centripetal accelerationa = v2/r

= 3.142/4 = 2.46ms-2 (0.25g)

6. On one type of theme park ride, a boat swings freely along a circular path from successively higher starting positions. As the boat moves through the lowest point on its swing, the riders are traveling at high speeds, and feel quite big forces on them.With some rides, such as Rush at Thorpe Park, the highest starting point is with the supporting arm horizontal as shown.

Rush boat at starting point

18m

Boat

The length of the supporting arm of Rush is 18m. The mass of a typical rider is 70 kg.(a) Calculate the maximum "g-force" felt by a typical rider on Rush.

“g-force" = force from seatweight of rider

Salters Horners Jan 2008 A2 paper

Questions?