Presentation on theme: "Athlete or Machine? A STEM Resource www.raeng.org.uk/athleteormachine."— Presentation transcript:
Athlete or Machine? A STEM Resource
Bob Skeleton 1500m track 150 m vertical drop 80 mph + 33 – 43 kg sled Amy Williams - Olympic gold 2010 Great context A STEM activity
Investigate the big question: athlete or machine? Practical activities and testing Mathematics activities Science activities Engineer/athlete video Student led Independent investigation Higher level thinking Scheme of work for STEM day or STEM club
Make a 1:5 bob skeleton sled 90 minute make Cheap materials Basic tools and equipment
Make a launcher
Make some timing gates (if you have the time) Investigate launch pressure consistency
Factors Weight The athletes shape The athletes position Aerodynamic lift Steering Clothing and equipment Starting Corners Ergonomics (how the body fits a product) Track incline (the slope down the length of the track) Friction on the ice Aerodynamic drag (air resistance) Tuning the characteristics of the skeleton Material choice Sled runners
Energy transfer Potential Energy (PE) = m x g x h Change in PE for our athlete and sled = Joules (J) Kinetic Energy (KE) = ½ x m x v x 97 kg x (40.23 x 40.23) = J Why isnt the all of the athlete and sleds potential energy transferred into kinetic energy? Amy Williams max speed Max speed if all PE transferred into KE Mass (m) of athlete and sled = 97kg Vertical drop of track (h) = 152m 1450m (diagram not to scale) Gravity (g) = 9.81 m/s 2
Which two forces resist the forward movement of the athlete and sled down the track? friction aerodynamic drag (air resistance)
Friction force Friction is a force that resists the movement of two surfaces against each other. Which combinations provide a lot or a little friction? A lot of frictionA little friction rubber / rubber (1.16) rubber / concrete (1.02) steel / wood ( ) felt / wood (0.22) steel / ice (0.03) rubber / concrete felt / wood rubber / rubber steel / ice steel / wood
Calculating friction force Friction is a force that resists the movement of two surfaces against each other. We can investigate the affect friction has on the model bob skeleton sled using the following equation. Force is measured in Newtons (N). F f = x m x g =Mu, the coefficient of friction. m =Mass (kg). g =The acceleration due to the gravity, which is 9.81 m/s 2.
Calculating friction force Ff = x m x g What is the friction force acting on the runners of a bob skeleton sled and athlete with the combined mass of 110 kg (athlete = 75 kg, sled = 35 kg)? Ff = 0.03 x 110 x 9.81 = N =Mu, the coefficient of friction (steel on ice = 0.03). m =Mass (kg). g =The acceleration due to the gravity, which is 9.81 m/s 2.
Calculating friction force Ff = x m x g Amy WilliamsKristan Bromley Athlete mass63 kg72 kg (+15%) Sled mass29 kg29 kg Total mass92 kg101 kg (+10%) What effect does a 15% increase in athlete mass have on friction? =Mu, the coefficient of friction (steel on ice = 0.03).m =Mass (kg). g =The acceleration due to the gravity, which is 9.81 m/s 2.
What effect does a 15% increase in athlete mass have on friction? Friction force(Ff) x m x g Amy Williams 0.03 x 92 x 9.81 = 27 N Kristan Bromley 0.03 x 101 x 9.81 = 29 N (+7%) A 15% increase in athlete mass doesnt result in a 15%increase in friction. This might be significant for the engineer. =Mu, the coefficient of friction (steel on ice = 0.03).m =Mass (kg). g =The acceleration due to the gravity, which is 9.81 m/s 2.
Aerodynamic drag force The resistance provided by the air passing over a shape is a force called aerodynamic drag. Which shapes have a higher or lower coefficient of drag? Higher C D Lower C D C D = 1.05 C D = 0.5 C D = 0.47C D = 0.42
Calculating drag force The resistance provided by air passing over the sled is a force called aerodynamic drag. F DRAG = ½ x x C D x A f x V 2 =Density C D =Drag coefficient A f =Frontal area V 2 =Velocity = 1000 kg/m 3 = kg/m 3 C D = 1.05 C D = 0.47 A f = m 2
Calculating drag force What is the drag force acting on the athlete and sled as they travel down the track at 5 m/s? F DRAG = ½ x x C D x A f x V 2 F DRAG = 0.5 x 1.2 x 0.45 x x 25 = 0.94 N =1.2 kg/m 3 (density of air) C D =0.45(drag coefficient of athlete and sled) A f =0.139 m 2 (frontal area of athlete and sled) V 2 =5 m/s(velocity - 5 m/s = mph)
Calculating drag force F DRAG = ½ x x C D x A f x V 2 F DRAG = 0.5 x 1.2 x 0.45 x x 25 (1) = 0.94 N What happens to drag force if you increase frontal area by 15 %? F DRAG = 0.5 x 1.2 x 0.45 x x 25 = What happens to drag force when the velocity increases by 15 %? F DRAG = 0.5 x 1.2 x 0.45 x x 33 (2) = What happens when frontal area and velocity increase? F DRAG = 0.5 x 1.2 x 0.45 x x 33 (2) = = density of airC D = drag coefficient of athlete and sledA f = frontal area of athlete and sled V 2 = velocity - 5 m/s = mph (1) 5 m/s(2) 5.75 m/s 1.08 N (+15%) 1.24 N (+31 %) 1.43 N (+52 %)
TASKS 1. In your groups complete the activities, tasks and questions in the booklets.(10 min) 2. In your groups discuss the questions: Athlete or Machine? Which is more important in the bob skeleton event? What could be done to reduce friction and drag? (Make sure you can justify your answers) 3. Choose a spokesperson who will communicate your groups answer to the rest of the class. 4. Check your answers against Kristan Bromleys (2008 World Skeleton Champion and engineer).