The Science of. Gravity (g) Gravity is a force of attraction between objects. The more massive the...
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![Page 1: 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.](https://reader033.fdocuments.in/reader033/viewer/2022042822/56649eb25503460f94bb99dd/html5/thumbnails/1.jpg)
The Science of
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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
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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
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G forces
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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
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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)
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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
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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
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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
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The Power of Stealth
Power = work (J) time (s)Power = 7194000
2.3Power = 3127826Watts
Power = 3.1MW
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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 ………
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………Stealth will make it over the top
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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
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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
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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!)
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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
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Nitrogen accumulators
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Hydraulic fluid
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The motors - Medusa
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Cylinder block
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Winch drum
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The catch car
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Wire cable and return drum
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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
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Brakes!
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Brakes!
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Keeping on track
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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
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Pendulum
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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
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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)
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Motion in a circle
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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)
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Vortex - Carousel
Centripetal accelerationa = v2/r
= 3.142/4 = 2.46ms-2 (0.25g)
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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
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Questions?