Speed and Velocity Units

Acceleration Units

Angle Units

m/s or ms-1

m/s2 or ms-2

rad

Significant figures rule 2

Power = Force x Velocity

Peak power occurs at approx one third of maximum isometric force and an intermediate velocity of contraction

• Females - 6 to 8 times higher injury rate
• Landing with knees less flexed
• Poor Hamstrings Quadriceps balance
• Hamstrings protection of ACL reduced
• Hamstring co-activation defacit
• Slow activation of Hamstrings 

• Reduced knee and hip flexion angles
• Increased knee valgues
• Internal rotation of the femur on tibia

Landing/Pivoting with knee extended:

• Patella tendon shear load higher
• Hamstring co-activation less effecive in protecting ACL

Joint moment (Nm) = Muscle Force (N) x Moment arm (m)

M=Fxd

F=M/d

• Falls
• Balance
• Acciedents
• OA and Chronic Conditions
• Osteoporosis/Fractures
• Ergonomic Problems

Upper body (head and trunk) - 50%

Arms - 5% each - 10%

Legs - 20% each - 40%

Laying down on board with scales:

Measure and record the length of the board = Board length (D)

Record the mass measured on both scales in kg = M1 & M2

Calculate ratio of the two masses (R) = M1/M2

Divide the length of the board into two segments:

d1=D/R+1

d2=D-d1

Under static conditions, the vertical projection of the COM has to fall within the base of support

The COM can be outside the human body, which has important impliations for sporting performance (eg high jumping)

Specified by magnitude (number) only. May be positive or negative

Examples: Temperature (T) Volume (V) Mass (m) Time (t) Energy (E)

Specified by magnitude and direction

Examples: Displacement, Velocity, Acceleration, Force, Momentum

Can be graphically represented as an arrow

Example: 400m sprint

Distance 400m

Displacement 0m

Distance 200m

Displacement: 100m along straight 62 metres vertically across track. 

O = 62

A = 100m

Pythagoras. Square root of 62squared plus 100squared

=118

Calculate angle = O A = TOA = TAN = 62/100 = 0.62

Angle therefore = Inverse Tan (0.62) = 32 degrees

Change in Velocity

Change in Time

Therefore

V2-V1

Change in Time

A sprinter's velocity is 3m/s on leaving the blocks and 7m/s two seconds later

What is the acceleration?

7-3

2

=2m/s2

Projection height

Projection Angle

Projection Speed

For a given projection speed and a given projection height, the optimal angle for max horizonal distance is 45 degrees and max vertical distance is 90 degrees

However in sporting movements the optimum angle is not always 45 degrees - The reasons lie in the anatomical structure of the human body, humans are not machines

Eg optimum shot angle is around 32 degrees

optimum long jump angle is around 22 degrees

Is the difference in height from which the body is initially projected and the height at which it lands 

The greater the projection height, the greater the horizonal distance

The greater the projection speed the greater the horizontal and vertical distances

The horizontal distance increases in proportion to the square of the increase in projection speed

Long Jump example

Take of speed v is 9.81m/s and take of angle is 22.1 calculate the horizontal components of v - vx and vy

9.81 = h

to find vx which is a - we have h and need to find a = COS

?/H=COS(0)

?/9.81=COS(22.1) 

?=COS(22.1)x9.81 = 9.09 m/s

to find vy which is o - we have h and need to find o = SIN

?/H=SIN(0)

?/9.81=SIN(22.1)

?=SIN(22.1)x9.81 = 3.69 m/s

Position - m

Distance - m

Displacement - m

Velocity - m/s

Acceleration m/s2

Angular position - rad

Angluar distance - rad

Angular displacement - rad

Angular velocity - rad/s

Angular acceleration - rad/s2

Angular distance - The sum of all angular changes that have occured

Angular displacement - The distance between the initial and the final position of the pendulum:

Final angle - Initial angle

Linear:

change in displacement

change in time

Angular (w):

change in displacement (02-01)

change in time (t2-t1)

For any given anguar velocity, the linear velocity increases with an increase in the radius of rotation:

Linear velocity = radius of rotation x angular velocity

v=rw

The unit for angular velocity HAS to be rad/s

Example of relationship between linear and angular velocity

Known: r1 = 0.2m

r2 = 0.4m

w = 30 rad/s

0.4-0.2 = 0.2m 

0.2 x 30 = 6 m/s

Linear:

Change in velocity

Change in time

Angular:

Change in angular velocity (w2-w1)

Change in time (t2-t1)

Step: Right heel contact to Left heel contact

Stride: Right heel contact to Right heel contact 

Swing Phase: toe off to heel contact

Single support phase: one foot on the ground

Double support phase: both feet on the ground

Stance phase approx 60%

(20% contact 30% midstance 50% propulsion)

Swing phase approx 40%

In running there is no double support phase, rather a flight phase: both feet off the ground

Gait calculation exmple:

1.57s

Swing

Stance (contact, midstance, propulsive)

Double support

Swing = 0.628

Stance = 0.942

Contact = 0.1884

Midstance = 0.2826

Propulsion = 0.471

Double support is 20% in stride therefore 

= 0.314

Cadence: Quantifies the number of steps per minute steps.min-1

Number of steps x 60

Time (s)

Stride freequency: Quantifies the number of strides per second (steps.s-1) or Hz

number of strides

time (s)

Stride and cadence example:

Usain bolt 41 steps in 9.63s

Calculate: Average speed, Number of strides, Average stride length, Cadence, Stride frequency 

Average speed = 10.38m/s

Number of strides = 20.5

Average stride length = 2.44m

Cadence = 255.45 steps.min-1

Stride frequency = 2.13Hz

Normalise stride length by height (stature)

Stride length

height

Use anatomical landmarks:

Greater Trochanter of Femur

Lateral Femoral Epicondyle

Lateral Malleolus

Head of fifth metatarsal

Use 2-D angles to calculate joint angles properly

Calculating angles eg hip

Opposite - 0.26cm

Adjacent - 0.43cm

O and A = TOA = TAN

= TAN (0.26/0.43) 

= 31.16 degrees

Push or a pull

Tends to cause a body to accelerate or change shape

Vector (has magnitude and direction)

Kicking or throwing a projectile

pushing feet against the floor

Lifting a weight

Friction

Gravity

Force is a vector 

Unit: Newtons (N)

A meaaure of a bodys inertia

Depends on quantity of matter of which a body is composed 

Units: Kg

Scalar: Magnitude, but no direction

Weight

Gravitational force exerted on a body by the Earth

Weight = mass x gravitational acceleration (W=mg)

g is the acceleration due to gravity (9.81m/s2 on Earth)

On the moon it is only abut 1.62

Vector (magnitude and direction)

Mass is a measure of a bodys inertia whereas weight is a force due to gravity

Force = Mass x Acceleration

Units: N 

1N = 1kg.m/s2

A force applied to a body causes an acceleration of that body which is proportional to the force and inversely proportional to the bodys mass

Example for F=ma

Sprinter has horizontal velocity of 15m/s2 she has a mass of 58kg

How much force is the sprinter applying to the blocks?

For every action there is an equal and opposite reaction

When a body exerts a force on a second, the second body exerts a reaction force that is equal in magnitude and opposite in direction on the first body

Ground reaction force example

calculate the reaction force that the ground is exerting on a person of a mass of 80kg

walking 1.2 x BW

jogging 2.1 x BW

sprinting 4.8 x BW

Landing after countermovement jump 10 x BW

If a force is applied over a period of time, an impulse is applied

An impulse is the product of force and time:

Implse = Force x Time

Vector (magnitude and direction)

Unit: Ns

Calculating impulse

F = 100N

t = 10s

100 x 10 = 1000Ns

An impluse can be though of as the area under a force curve

Momentum, M, is a measure of the 'quantity of motion' of a body:

M = mv

m = mass of the body (m)

velocity of the body (v)

Vector

Units: kg.m/s or N.s

If the resulatant external force acting on a system is zero, the total momentum of a system remains constant

M1=M2

Total momentum of system at time t1 (M1)

Total momentum of system at time t2 (M2)

Example problem for momentum

Ice hockey players collide:

A 90kg ice hockey player travelling at 6.0m/s

A 80kg ice hockey player travelling at 7.0m/s

They entangle and continue to move, but at what velocity?

90 x 6 = 540 kg.m/s

80 x 7 = 560 kg.m/s

540 - 560 = 170kg (combined mass) (v)

-20 = 170 (v)

-20/170 = (v)

= 0.12 m/s is the velocity

The total momentum of an isolated system is conserved

If the resultant external force acting on the system is zero the total momentum of a system remains constant

The change in momentum is equal to the impulse applied to the body 

Impulse = change in momentum

Impulse = m2-m1

Ft=(Delta) M

We call this the impulse momentun relationship

A torque is a force that causes rotation

Torque = Force x Distance (T=Fd)

Vector

Counterclockwise is positive 

Units: Nm

Torque = Muscle force x Moment arm

Muscule strength depends on both muscular force and moment arm

The moment arm of a muscle with respect to a joint axis of rotation depends on the joint angle

Example: Biecps and tricps co-contraction

F biceps = 2000N

MA biceps = 0.05m

F triceps = 2500N

MA triceps = 0.04m

What movement will occur

2000 x 0.05 = 100Nm

-2500 x 0.04 = -100Nm

100-100 = 0Nm

Isomeric contraction

The moment of inertia is the property of a body to resist rotation

It is the angular equivalent for mass

Mass is the property of a body to resist linear acceleration

Depends on: 

• The mass of the body
• How the mass is distributed about the axis of rotation

I=mk2

I = moment of inertia

m = mass of the body

k = radius of gyration

Units: Kg.m2

Scalar (magnitude, no direction)

Force = Mass x Acceleration

Torque = Moment of inertia x Angular acceleration

Torque and force example

knee extensors 3cm from axis of rotation at knee

How much force must the knee extensors exert to produce an angular acceleration at the knee of 1rad/s2 at a given moment of inertia of 0.24kg.m2

Moment of inertia = 0.24

Acceleration = 1 rad/s2

Torque = 0.24 x 1 = 0.24 Nm

Required force therefore is:

Force = Torque/Distance

0.24/0.3 = 8.0 N

Angular momentum is a measure of angular motion possesed by a body

Angular momentum = Moment of inertia x Angular velocity

H = Iw

Units: Kg.m2/s

Vector

Angular momentum example

Moment of inertia of diver = 6 kg.m2

Angular velocity of diver = -9.2 rad/s

What is the angular momentum?

H = Iw

H = 6 x -9.2

H = -55 Kg.m2/s (clockwise)

The total (linear) momentum of an isoalted system is conserved

If the resultant external force acting on the system is zero the total momentum of a system remains constant 

The total angular momentum of an isolated system is conserved

If the resultant external torque acting on the system the total angular momentum of a system reamains constant

Conservation of angular momentum in a dive example

Moment of inertia at take of is 14 kg.m2

Angular velocity at take of is -2.6 rad/s

Moment of inertia in pike is 5.8 kg.m2

Angular velocity as take of is what?

H = Iw

14.0 x -2.6 = -36.4

Angular momentum must be conserved therefore angular momentum must remain as -36.4

rearrange equation to H/I = w

-36.4/5.8 = -6.3 rad/s

Linear --- Angular

Force --- Torque

Force = mass x acceleration --- Torque = Moment inertia x agular acceleration

Linear momentum (M=mv) --- Angular momentum (H=Iw)

Work = Fd --- Work = T0

Power = Fv --- Power = Tw

Linear Impulse Ft --- Angular Impulse Tt

A model can be defined as an artificial representation of reality 

Types of biomechanical models are as follows:

• Conceptual
• Statistial or regression 
• Mathmatical (computer)

Models can be used to increase knowledge and insight about reality and estimate or predict variables of interest

1. Knowledge of the system being modelled

2. Experimental data that constitute system inputs and/or outputs

In general simple is better, need to decide what should be neglected and included 

Direct or indirect

In direct use, the model proceeds from cause to effect, nd typically yields a unique solution

In inverse use, a model attempts to move from the effect to the cause and typically yields several possible solutions (not unique)

• Point mass (Athlete or implament)
• Rigid body
• Musculoskeletal

Simulation nvolves the performance of a series of controlled experiments using the model

First step: 54% hip 31% knee and 15% ankle

Stecond stance: knee only accounts for 9% total power and ankle up to 38%

Maximal velocity: 39% hip 17% knee and 44% ankle

The end of the chain is not freely moveable 

Characterised by:

• High force production
• Low or moderate movement velocity
• Push like movement pattern (all joint angles change simultaneously)

The end of the chain is freely moveable

Characterised by:

• Can operate push like
• High force or high accuracy 
• All joint angles move simultaneously 
• Can operate throw like 
• Joint angles occur in a sequential order
• This maximises movement velocity

By maximising projection speed

By rotating segments sequentially

Kinetic chain is all about angular momentum being transerred through segments

Eg trunk moves and angular momentum is transferred to arm, moment of inertia of the arm is smaller and so angular velocity must increase to conserve angular momentum

Baseball pitcher example

Trunk rotation of 0.43 rad/s

moment inertia of trunk 2.4 kg.m2

Trunk stops rotating and momentum transferred to arm with a moment of inertia of 0.023 kg.m2

Angular momentum generated = 0.43 x 2.4 = 1.032 kg.m2/s

Angular velocity of arm after momentum transerref = 1.032/0.023 = 45 rad/s

The function of mono-articular muscles is predominantly the generation of momentum 

The function of bi-articular muscles includes both the generation and transfer of momentum (also contribute to more than one motion)

If a force is applied over a distance mechanical work is performed 

Work is the product of force and displacement

Work = Force x Displacement

Scalar 

Units: J

Work example

How much work is performed? 

Mass of bar bells = 250kg

moved 0.75m

First must convert mass into weight 250 x 9.81 = 2452.5 N

2452.5 x 0.75 = 1839.375 J

Power is the rate at which work is done

Power = work/change in time

Scalar 

Units: Watts (W)

Power is the product of force and velocity

P = w/change in time = force x distance/time

= F x d/change in time = Force x Velocity

P = Force x Velocity

Power example 

A power lifter lifts 236kg over a distance of 0.62m in a time of 0.42 seconds

236 x 9.81 = 2315.16 N x 0.62 =

1435.3992 J

1435.3992/0.42 = 3417.6 W

Energy is a body's capacity to do work, forms of energy incule: kinetic, gravitational potential, elastic potential, chemical, thermal (heat)

Kinetic energy is the energy of motion

KE=1/2.m.v2

m = mass of body

v = speed of body

Scalar

Units: J

Kentic energy of pole vaulter example

Mass of pole vaulter = 80kg

Speed at end of run up = 9.5 m/s

0.5 x 80 x 9.5 sqaured = 3610 J

Gravitational energy is the enrgy due to a body's height above a reference surface

Graviational potential energy is the product of a body's weight and height

PEgrav = mgh

m = mass of body

g = acceleration due to gravity (9.81m/s2)

h = vertical height above reference level

Gravitational potential energy example

mass of diver 78kg

height of centre of mass above water 3.83m

Elastic potential energy is the energy stored in a spring

PEelastic = 0.5 . k . x2

k = stiffness of the spring

x = deformation of the spring

Scalar

Units: J

The total amount of enery is always consered, no energy is lost, simply transformed from one form to another

Pole vault example

Mass of vaulter 80kg

Height of vaulter = 5.50m

What is the vaulters velocity at touchdown?

Use conervation of energy

Grav potential energy = KE at touch down

80 x 9.81 x 5.50 = 0.5 x 80 x v2

4316 = 40 x v2

4316/40 = v2

107.9 = v2

sqaure root 107.9 = v

-10.39 m/s

-because he's falling

Hockey problem advanced

90kg at 6m/s

80kg at 7m/s

Calculate energy loss as a result of the collision

Before the impact = (0.5 x 90 x 6 squared) + (0.5 x 80 x 7 squared) = 3580 J

(90 x 6 = 540) - (80 x 7 = 560) = -20 = 170 (v)

0.12m/s

After the impact = 0.5 x 170 x 0.12 squared = 1.224 J

3580 - 1.224 = 3578.776 J

The resistance force that acts at the interface 

between two surfaces in contact

Arises due to applied force that produces relative motion (kinetic friction), or tends to produce relative motion (static friction)

Force therefore, units are: Newtons

Mechanical interaction between two bodies or surfaces (studs, cleats, spikes)

• Molecular interaction between two surfaces eg: sole of a sports shoe and the court such as basketball and netball
• Hand and a sports ball or implament eg: netball pass, discus throw, high bar or weight lifting
• Due to surface roughness, contact is only made at a few points
• This causes electrostatic force between atoms or moecules
• Leading to 'breaking off' of surface protrusions

Magnitude of the friction force, F, is given by

F = uR

u = coefficient of friction

R = normal reaction force

Vector: Direction of friction force is opposite to the direction of motion (or opposite to the direction of aplied force) 

Units: Newton (N)

Two surfaces are in relative motion (slding); 

Fk = ukR

Fk = Kinetic friction force 

Uk = coefficient of kinetic friction 

R = Normal reaction force 

Two surfaces are stationary;

Fs (less than or equal to) usR

Fm = usR

Fs = static friction force

Fm = maximum static friction force

us = coefficient of static friction

R = normal reaction force

• A dimensionless number (no units)
• An indicator of the ease of sliding of two surfaces (low value of u means easy sliding)
• Coefficent of static friction is greater than coefficient of kinetic friction (it is more difficult to get a body sliding than it is to keep sliding

Example problem (friction)

The coefficient of static friction between a sled and 

the snow is 0.18 with a coefficient of kinetic 

friction of 0.15. A 250 N boy sits on the 200 N 

sled.

a. How much force directed parallel to the 

horizontal surface is required to start the sled in 

motion?

b. How much force is required to keep the sled in 

motion?

To start the sled in motion, the applied force must exceed the force of maximum static friction:

Fm =usR

Fm = (0.18) (200+250)

=81N 

Therefore the applied force must be greater than 81 Newtons

To maintain motion the applied force must equal the force of kinetic friction:

Fk =ukR

Fk = (0.15) (200+250)

=67.5N 

Therefore the applied force must be at least 67.5 newtons

Example problem 2 (friction)

A sled including its passenger weighs 1200 N. It 

takes a horizontal force of 250 N to start the sled 

in motion. Once the sled is moving a force of 220 

N is required to keep the sled in motion. 

a. What is the coefficient of static friction?

b. What is the coefficient of kinetic friction?

a. Fm = usR

250 = (us)(1200)

250/1200 = us

= 0.21

b. Fk = ukR

220 = (uk)(1200)

220/1200 = uk

=0.18

Contact area eg size of shoe sole

Relative speed of two surfaces (the coefficient of kinetic friction is almost independent of the speed of the two surfaces

Practical ways to increase or decrease friction 

Increase/decrease the weight of the body 

 (Increase/decrease the normal reaction force)

Pull up/push down on the body 

(increase/decrease the normal reaction force)

Apply a lubricant to the surface(s)

- Water, oil, graphite powder, chalk

-this reduces the coefficent of friction between the two surfaces

Change one or both of the surfaces

-eg a rougher or smoother surface

(increases or decreases the coefficient of friction between the two surfaces)

– tests for playing surfaces in field sports 

(hockey, American football)

– ice surface in winter sports (bobsled)

– ball surface texture (basketball, rugby)

– gloves (Football, American football)

– chalk and glue (gymnastics, weightlifting, 

throwing, pole vaulting)

– body oil, sweat (wrestling)

Horizontal sliding of two surfaces

-measure the applied force with a spring balance or load cell)

Inclined plane

-applied force increases with increasing angle

-measure the angle at which the body begins to slide

Form drag (also called pressure drag or profile drag)

Skin friction

Mechanism of form drag

• fluid separates from the surface of the body
• crates a turbulent wake
• energy is lost from the body in creating the eddy currents
• This creates regions of higher and lower pressure

4 Factors affecting drag

1. Fluid density - Air = low density, low drag - Water = high density, high drag (form drag is important in water sports)
    
2. The frontal area - size of the body, ball or implament
    
3. Speed - Drag increases with speed (Flow separates earlier; greater turbulant wake = more drag
    
4. Shape - more streamlined the body the later the flow separates leading to less turbulant wake = less drag

Density of fluid (p)

air: 1.20 kg/m3; Water: 1000 kg/m3

Cross sectional (frontal) area (Ap)

Drag coefficient (Cd); depends on shape of the body

relative speed of the body and fluid (v)

The aerodynamic/hydrodynamic 'form drag' force FD, is given by

FD = ½ApCdv²

Vector: oppose the forward motion of the body in the fluid

Units: N

Example drag problem 

Usain Bolt (JAM) reached a

top speed of 12.2 m/s

Calculate the form drag force acting on Bolt at

this speed.

(Bolt has a frontal area of 0.50 m2 and a

drag coefficient of 0.60). 

What forward propulsive power was Bolt

generating to overcome this drag force?

Remember, the air density is 1.20 kg/m3

• Use ‘streamlined’ equipment (with a lower CD)

• use teardrop shapes

• Use ‘streamlined’ body position (with a lower AD)

– use teardrop shapes