UNIVERSITY OF SAN DIEGO 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 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.

forces act on the box (car)? 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 d FREE-BODY DIAGRAM SKETCH SKETCH HORIZONTAL VIEW. CAR COMING AT YOU. NO FRICTION. FN TOP VIEW. TRACK IS BANKED. a 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: FREE-BODY DIAGRAM d F NET = m a F NET = m v2/ R F NET = Fg + FN Fg + FN = m v2/ R FN = m v2/ R - Fg SKETCH FN Fg a F NET What you feel is FN , the force on your body FN MIGHT BE NEAR ZERO; IF NEGATIVE …

APPLY WHAT WE HAVE SO FAR TO BOTTOM OF LOOP: FREE-BODY DIAGRAM d F NET = m a F NET = m v2/ R F NET = FN - Fg FN - Fg = m v2/ R FN = m v2/ R + Fg SKETCH FN a F NET 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 K P K K K P P P

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… K P K K K P P P

g is the acceleration due to gravity; same for any object. 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 = m g h 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… K P K K K P P P

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: FREE-BODY DIAGRAM d F NET = m a F NET = m v2/ R F NET = Fg + FN Fg + FN = m v2/ R FN = m v2/ R - Fg SKETCH FN Fg a F NET What you feel is FN , the force on your body If FN = 0, then Fg = m v2top / R. A little more algebra and v2top = Rg

GO BACK TO BOTTOM OF LOOP: 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: WHAT PULLS THE TRAIN? Nothing, after the original lift … Energy 2. WHY NOT CIRCLES FOR THE LOOPS? Too many g’s …