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

•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

FactorsWeight

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

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?

The bob skeleton: kinetic energy gained during a run

0

20000

40000

60000

80000

100000

120000

140000

160000

180000

200000

5 10 15 20 25 30 35 40 45 50 55 60

Speed in metres per second (m/s)

Kin

etic

en

erg

y (J

ou

les)

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/s2

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 (0.2 - 0.6)

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).

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.

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.

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 CDLower CD

CD = 1.05

CD = 0.5CD = 0.47

CD = 0.42

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

= DensityCD = Drag coefficient

Af = Frontal area

V2 = Velocity

= 1000 kg/m3

= 19 300 kg/m3

CD = 1.05 CD = 0.47 Af = 0.139 m2

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 %)

1.43 N (+52 %)

TASKS1. 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).