Transcript The Science of Gravity (g)

The Science of
Gravity is a force of
Gravity
(g)
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.
>1g
> 9.8 m/s2
Low g’s
Any acceleration less
than free fall.
<1g
< 9.8 m/s2
NASA at the Amusement Park
G forces
Stealth Facts
• Height
• Launch time
• Train weight
• Launch Speed
• Dive Speed
62 metres
2.3 seconds
8 tonnes (unladen)
10 tonnes (laden)
38 metres per second
35 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=mxa
F = 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 43.6
Work = 719400Joules
Work = 7 194 KJ
Work = 7.19 MJ
The Power of Stealth
Power = work (J)
time (s)
Power = 7194000
2.3
Power = 3127826Watts
Power = 3.1MW
Will you make it? –
Potential energy needed= mgh
PE = 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 = 4125N
Force of 1g on a 100Kg person = 1000N
4125N = 4.1g
Total 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 000J
Surplus KE = 740 000J
KE = I/2mv2
v2 = 2KE = 2x740000 = 148
m
10000
v = 12+ m/s
On top of the world
Outside curve radius 8m
Centripetal 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 40m
Centripetal 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 rpm
Actual
= 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 - Carousel
Data:
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 acceleration
a = v2/r
= 3.142/4 = 2.46ms-2 (0.25g)
Salters Horners Jan 2008 A2
paper
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 seat
weight of rider
Questions?