Athlete or Machine? A STEM Resource www.raeng.org.uk/athleteormachine

Bob Skeleton 1500m track 150 m vertical drop 80 mph + 33 – 43 kg sled Amy Williams - Olympic gold 2010 Great context A STEM activity www.youtube.com

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 athlete’s shape The athlete’s 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

Potential Energy (PE) = m x g x h Kinetic Energy (KE) = ½ x m x v2 Energy transfer Potential Energy (PE) = m x g x h Change in PE for our athlete and sled = 144 639 Joules (J) Kinetic Energy (KE) = ½ x m x v2 0.5 x 97 kg x (40.23 x 40.23) = 78495 J Why isn’t the all of the athlete and sled’s potential energy transferred into kinetic energy? Mass (m) of athlete and sled = 97kg Vertical drop of track (h) = 152m 1450m (diagram not to scale) Gravity (g) = 9.81 m/s2 Amy Williams max speed Max speed if all PE transferred into KE The line graph above shows that if all the potential energy (PE) were to be transformed into kinetic energy (KE) then the athlete and sled would need to travel at 55 m/s (122 miles per hour) to reach a KE figure of 144 639 J. However, the 2010 bob skeleton Olympic champion, Amy Williams, is known to travel at a maximum speed of 90 mph (40.23 m/s). Our simple analysis of the energy transfer over estimates the maximum speed of the athlete and sled by 15 m/s or 37% because it neglects the affects of aerodynamic drag and friction.

aerodynamic drag (air resistance) Which two forces resist the forward movement of the athlete and sled down the track? friction aerodynamic drag (air resistance)

Friction force A little friction A lot of friction rubber / concrete Friction is a force that resists the movement of two surfaces against each other. Which combinations provide a lot or a little friction? rubber / concrete felt / wood rubber / rubber steel / ice steel / wood A little friction A lot of friction rubber / rubber (1.16) steel / ice (0.03) rubber / concrete (1.02) steel / wood (0.2 - 0.6) felt / wood (0.22)

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). Ff =  x m x g = Mu, the coefficient of friction. m = Mass (kg). g = The acceleration due to the gravity, which is 9.81 m/s2.

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 = 32.37 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/s2.

Calculating friction force Ff =  x m x g Amy Williams Kristan Bromley Athlete mass 63 kg 72 kg (+15%) Sled mass 29 kg 29 kg Total mass 92 kg 101 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/s2. Friction force Amy Williams 0.03 x 92 x 9.81 = 27 N Kristan Bromley 0.03 x 101 x 9.81 = 29 N Increase in friction = new amount - original amount x 100 original amount Difference in friction force between Amy Williams and Kristan Bromley is 7%

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 doesn’t 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/s2. Friction force Amy Williams 0.03 x 92 x 9.81 = 27 N Kristan Bromley 0.03 x 101 x 9.81 = 29 N Increase in friction = new amount - original amount x 100 original amount Difference in friction force between Amy Williams and Kristan Bromley is 7%

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? Lower CD Higher CD CD = 0.42 CD = 1.05 CD = 0.47 CD = 0.5

Calculating drag force The resistance provided by air passing over the sled is a force called aerodynamic drag. FDRAG = ½ x  x CD x Af x V2 = Density CD = Drag coefficient Af = Frontal area V2 = Velocity = 1000 kg/m3 = 19 300 kg/m3 CD = 0.47 Af = 0.139 m2 CD = 1.05

Calculating drag force What is the drag force acting on the athlete and sled as they travel down the track at 5 m/s? FDRAG = ½ x  x CD x Af x V2 FDRAG = 0.5 x 1.2 x 0.45 x 0.139 x 25 = 0.94 N = 1.2 kg/m3 (density of air) CD = 0.45 (drag coefficient of athlete and sled) Af = 0.139 m2 (frontal area of athlete and sled) V2 = 5 m/s (velocity - 5 m/s = 11.18 mph)

Calculating drag force FDRAG = ½ x  x CD x Af x V2 FDRAG = 0.5 x 1.2 x 0.45 x 0.139 x 25 (1) = 0.94 N What happens to drag force if you increase frontal area by 15 %? FDRAG = 0.5 x 1.2 x 0.45 x 0.160 x 25 = What happens to drag force when the velocity increases by 15 %? FDRAG = 0.5 x 1.2 x 0.45 x 0.139 x 33 (2) = What happens when frontal area and velocity increase? FDRAG = 0.5 x 1.2 x 0.45 x 0.160 x 33 (2) =  = density of air CD = drag coefficient of athlete and sled Af = frontal area of athlete and sled V2 = velocity - 5 m/s = 11.18 mph (1) 5 m/s (2) 5.75 m/s 1.08 N (+15%) 1.24 N (+31 %) An increase in frontal area has a bigger impact on the forces acting against the athlete and sled than friction. 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 group’s answer to the rest of the class. 4. Check your answers against Kristan Bromley’s (2008 World Skeleton Champion and engineer). Kristan Bromley video clips: THE PARTS THAT MAKE A BOB SKELETON 22:34