Transcript PhysRollerCoastersx - University of San Diego Home Pages

UPWARD BOUND 2016
THE PHYSICS OF
ROLLER COASTERS
UNIVERSITY OF SAN DIEGO
TOM SKELTON
CONNER HOUGHTBY
WHAT DO YOU HOPE TO LEARN?
MY TWO QUESTIONS:
1. WHAT PULLS THE TRAIN?
2. WHY NOT CIRCLES FOR THE LOOPS?
COURSE GOAL: BASICS OF
ROLLER COASTERS
COURSE LEVEL:
MOST OF PHYSICS
SOME OF THE MATH
OUTLINE:
DEFINITIONS
NEWTON’S LAWS
APPLICATIONS (!)
DEFINITIONS WE WILL NEED
VECTOR: HAS MAGNITUDE AND DIRECTION
SPEED: HOW FAST IT’S MOVING
VELOCITY: SPEED AND DIRECTION
ACCELERATION: HOW FAST VELOCITY
IS CHANGING - A VECTOR
THIS MEANS THAT A CAR IS ACCELERATING IF IT IS
SPEEDING UP, SLOWING DOWN, OR ROUNDING A CURVE
YOU NEED TO REMEMBER THESE
NEWTON’S LAWS
1. An object at rest tends to remain at rest, and an object in motion
tends to remain in motion in a line, unless acted on by a NET force.
2. Net Force equals mass x acceleration
3. If Object A exerts a force on Object B, then Object B exerts
a force on Object A which is equal and opposite.




The second law is written F NET = m a or a = F NET / m
The NET FORCE is the VECTOR SUM of all forces
Examples of forces
Fg = Gravitational force – weight, pointed downwards
FN = Normal Force – support force from tabletop, rails, etc.
“normal” means the direction is perpendicular.
Ff = Friction force – direction is along surface, rails, etc.
FD = Drag force – opposite direction as velocity
FT = Tension force – pull by a string, rope, etc.
A FREE BODY DIAGRAM HELPS ANALYSIS
What force or
forces act on
the box (car)?
A FREE BODY DIAGRAM HELPS ANALYSIS

F g is fixed; and

F NET



= F g + FN must point in the direction as a
MOTION IN A CIRCLE AT CONSTANT SPEED:
Is it accelerating?
MOTION IN A CIRCLE AT CONSTANT SPEED:
Is it accelerating?
YES!! THE
VELOCITY
IS CHANGING!
MOTION IN A CIRCLE AT CONSTANT SPEED:
Is it accelerating?
THE ACCELERATION IS TOWARDS THE
CENTER OF THE CIRCLE
MOTION IN A CIRCLE AT CONSTANT SPEED:

Is it accelerating? YES; | a | = v2/R

| a | = v2/R
v = speed
R = radius
THE ACCELERATION IS TOWARDS THE
CENTER OF THE CIRCLE
APPLY WHAT WE HAVE LEARNED TO DRIVING IN CIRCLE
ON A BANKED ROAD
SKETCH
TOP VIEW.
TRACK IS
BANKED.
SKETCH
HORIZONTAL
VIEW.
CAR COMING
AT YOU.
NO FRICTION.
FREE-BODY DIAGRAM
FN
a
d
F NET
Fg
IN NORMAL DRIVING, FRICTION WOULD MEAN
THAT SPEED v DOESN’T HAVE TO BE EXACT.
APPLY WHAT WE HAVE SO FAR TO TOP OF LOOP:
SKETCH
FREE-BODY DIAGRAM
FN
a
F NET
Fg
What you feel is FN ,
the force on your body
F NET = m a
F NET = m v2/ R
F NET = Fg + FN
Fg + FN =d m v2/ R
FN = m v2/ R - Fg
FN MIGHT BE NEAR ZERO; IF NEGATIVE …
APPLY WHAT WE HAVE SO FAR TO BOTTOM OF LOOP:
SKETCH
FREE-BODY DIAGRAM
FN
a
F NET
Fg
F NET = m a
F NET = m v2/ R
F NET = FN - Fg
FN - Fg d= m v2/ R
FN = m v2/ R + Fg
FN WILL BE MUCH GREATER THAN Fg; FURTHERMORE…
…
ENERGY IN A ROLLER COASTER
GENERAL REMARKS ON ENERGY:
ALWAYS CONSERVED – ONLY CONVERTED FROM
ONE FORM TO ANOTHER
SEVERAL FORMS:
KINETIC – ASSOCIATED WITH MOTION
GRAVITATIONAL POTENTIAL – HEIGHT
THERMAL – BASICALLY SAME AS HEAT
ELECTRIC
CHEMICAL – SUCH AS A BATTERY
OTHERS
ENERGY IN A ROLLER COASTER
THE TOTAL ENERGY STAYS
CONSTANT. POTENTIAL ENERGY
CHANGES FORM TO KINETIC ENERGY
P
K
K
K
P
P
P
K
ENERGY IN A ROLLER COASTER
K = ½ mv2
P = mgh
ETOT = K + P = constant
Since there is really some
friction, some energy is
converted to heat. Ignore…
P
K
K
K
P
P
P
K
K = ½ m v2
m = mass; often confused with weight, since they are similar in
everyday life. Mass is the total amount of material. Weight is
the gravitational pull on it, Fg.
Fg = m g
g is the acceleration due to gravity; same for any object.
v is the speed, as before.
P = mgh
h is height above baseline.
ETOT = K + P = constant (ignoring loss to heat)
ENERGY IN A ROLLER COASTER
K = ½ mv2
P = mgh
ETOT = K + P = constant
Since there is really some
friction, some energy is
converted to heat. Ignore…
P
K
K
K
P
P
P
K
A CAR IS TO BE RELEASED AT POINT “A.”
WHICH TRACK DESIGN WILL HAVE IT GOING THE FASTEST AT “B?”
A
B
A CAR IS TO BE RELEASED AT POINT “A.”
WHICH TRACK DESIGN WILL HAVE IT GOING THE FASTEST AT “B?”
A
B
ALL THE SAME! SINCE THE CHANGE IN HEIGHT IS THE SAME, THE
CHANGE IN POTENTIAL ENERGY IS THE SAME, AND SO IS THE
CHANGE IN KINETIC ENERGY
B
A
WILL THE CAR MAKE IT OVER HILL B, WHICH IS HIGHER THAN HILL A?
K
K
K
B
P
A
P
P
WILL THE CAR MAKE IT OVER HILL B, WHICH IS HIGHER THAN HILL A?
YES; IT HAS PLENTY OF TOTAL ENERGY.
GO BACK TO THE TOP OF LOOP:
SKETCH
FREE-BODY DIAGRAM
FN
a
F NET
Fg
What you feel is FN ,
the force on your body
F NET = m a
F NET = m v2/ R
F NET = Fg + FN
Fg + FN =d m v2/ R
FN = m v2/ R - Fg
If FN = 0, then Fg = m v2top / R. A little more algebra and v2top = Rg
GO BACK TO BOTTOM OF LOOP:
K
K
P
A LITTLE ALGEBRA, AND FN = 6 Fg AT THE BOTTOM
BIOLOGICAL CONSIDERATIONS
Increased blood pressure
Can rupture weak spots in arteries, etc
Can damage arteries
Brain Damage
6 g’s is the limit for many people; too much for some
ENGINEERING CONSIDERATIONS
There is friction – and it varies with wear, lubrication
There is air drag – and it varies with wind
Safety in keeping car on the track
Structural strength of the support
Speed adjustment – compensate for variations
MY TWO QUESTIONS:
1. WHAT PULLS THE TRAIN?
2. Nothing, after the original lift … Energy
2. WHY NOT CIRCLES FOR THE LOOPS?
Too many g’s …