Mechanical Work, Energy and Power (segment models …leyland/Kin201 Files/WorkEnergy.pdf · Mechanical Work, Energy and Power (segment models) ... Back to the Vertical Jump ... A

1

Mechanical Work,Energy and Power(segment models)

Hamill & Knutzen Chapters 10 & 11Winter 1979 Chapter 5

Work

Work is the product of force times thedistance through which that force moves aload.

Work = Force x distance W = F x d W = F x d x cosθ

Units: Joules (do not use N.m)

Work Work is a scalarWork is a scalar F x d => MLT-2 x L => ML2T-2

As the L’s are are the same, the square ofthem will always be positive.

Torque is a vectorTorque is a vector F x dperpendicular => MLT-2 x L As the L’s are perpendicular to each other

one could be positive the other negative,therefore torque has direction.

Power

Power = Δwork Δtime

= (force) x Δdistance Δtime

= force x velocity

Units => Watts (J/s)

Energy

Definition: “The ability to do work”

Kinetic Energy = ½mv2

Gravitational Potential Energy = mgh(h is measured from the objects position to ground

and therefore is negative, hence PE is positive)

Elastic Strain Energy = ½kx2

Units => Joules

Units

F x d => MLT-2 x L => ML2T-2

½mv2 => M(LT-1)2 => ML2T-2

mgh => MLT-2 x L => ML2T-2

What are the units of the spring constantin the equation for strain energy (½kx2)?

MT-2

2

Error in Hamill & Knutzen text? Force = kΔs Elastic Strain Energy = ½kΔx This is

wrong (see Andrew’s slides also). k is the same constant? The authors refer

to it as the stiffness constant in both thesection on elastic force and energy.

F => MLT-2 Therefore units of k => MT-2

Energy => MLT-2 ?????? Elastic Strain Energy = ½kx2

Conservation of Energy

The total energy of a closed system isconstant since energy does not enter orleave a closed system.

This only occurs in human movementwhen the object is a projectile and weneglect air resistance. Then the totalenergy of the system (TE) = PE + KE.

Note that gravity does not change thetotal energy of the system.

Work-Energy Relationship(staying with Linear Kinetics)

Work = Δtotal mechanical energy Assuming we are studying a rigid body

(one that cannot store elastic energy), theequation becomes.

Fd Fd = = Δ(Δ(½mv½mv2 2 + + mghmgh)) Fd Fd = = ΔΔ½mv½mv2 2 + + ΔΔmghmgh

Work-EnergyRelationship

This is not a newmechanical concept.It can be derivedfrom Newton’ssecond law.

( )vmWork

vdvmFds

dvvmdsF

dsdvvmF

dtds

dsdvmF

dtdvmF

amF

2

2

1=

=

!!=!

!!=

!!=

!=

!=

" "

Work-Energy RelationshipAngular Kinetics

If force is applied off centre (i.e. the line of action ofthe force vector does not pass through the centre ofrotation of the body) then rotation as well astranslation will occur.

(if we assume ahorizontal forcethrough the centreof gravity or if wejust ignore potentialenergy)

!

Work = "Energy

Fd = " 12m

2

v

#$ = " 12 I

2

%

3

Linear Force?

Fd = Δ(½mv2 + mgh)

Note: net forcewould have tofactor in gravity

Translation

Fd = Δ(½mv2 + mgh)

dd

Eccentric Force? Rotation and Translation

This is because:

The centre of mass of a rigid bodyinstantaneously accelerates in thedirection of the applied resultant externalforce, irrespective of where the force isapplied on the body.

The rotational effect is also instantaneous. However, you do not get something for

nothing.

Rotation and Translation

dd

You can breakout the linearand rotationalcomponents ofthe eccentricforce

Note that “d” is longer than in thecentric force scenario

!

Work = "Energy

Fd = " 12m

2

v

#$ = " 12 I

2

%

4

Remember this slide from theTennis Lecture?

Racket

Blue vector (action on C of g) Fd = Δ(½mv2 + mgh)

Green vector Green vector

ττθ θ = ΔΙω= ΔΙω22

Back to the Vertical Jump

This is the same data from the problem Igave you earlier.

Take-off velocity = 2.534 m/s Mass = 61.2 kg New Question: New Question: If the jumper’s centre of

gravity moved +0.5 m vertically from thebottom of the drop down to the take offposition, how much work did she do in thisphase of the jump and what was heraverage force production?

Vertical Jump Power (Kin 142 & 343)

If you used body mass (61.2 kg) instead ofbody weight (600 N) you should havecalculated and answer of 77.3 kgm.s-1

Where does the above equation come from?!

Power = 2.21"Wt " d

Power = 2.21" 600 " 0.327

Power = 758 #W (J /s)

Power = force x velocityFrom vf

2 = vi2 + 2ad we can calculate the velocity

of take-off and, as we started from zero velocity,the average velocity during take-off.

!

Vto = 2ad = 2a " d

Vto = 19.62 " d = 4.42 d

Average velocity # 2.21" d

Power = Force "Velocity

Power = 2.21"mass" g " d

Physiologists & Mechanical Units! You will come across a lot of physiology

texts that report the power output fromsuch tests in kg.m.s-1.

This is not a unit of power. Without being too pedantic, I wonder why

they cannot multiply the result by g (9.81m.s-2) to get the correct units of; kgm2.s-3,or Joules/sec (J/s) or Watts.

Fundamental units: ML2T-3

Sayers Equation Average power is not ideally the attribute we

wish to measure in a vertical jump. The Sayers equation is an estimate of peak

leg power. Peak Leg Power (Watts) = [60.7 x jump

height (cm)] + [45.3 x body mass (kg)] – 2055 Do it for the subject we just used (jump height

= 0.327 meters, body mass = 61.2 kg Compare to average power calc. (758 Watts)

5

Bowflex Treadclimber “Reduce your workout time - dual-

motion treadles let you step forward likea treadmill and up like a stair climber soyou get more exercise in less time”

“TreadClimber® machine burns up to2 TIMES more calories than atreadmill - at the same speed!”

“Studies were conducted at theprestigious Human PerformanceLaboratory at New York's AdelphiUniversity. The results were dramatic! In22 separate trials, the TreadClimber®machine burned up to 2 times morecalories in 30 minutes than a treadmill atthe same speed!”

Company Website Sep-2006

http://www.treadclimber.com/trc_microsite/fitnessbenefits.jsp

Work is Work (Power Output is…) Sure it is possible to burn twice the calories but

……………it would be twice as difficult TV commercial “burn twice the calories in one easy

motion” “What do you get when you combine the best aerobic

features of the stairclimber, treadmill, and ellipticaltrainer? Quite simply, you get a triple-charged cardioworkout “ Bowflex Website Sep-2006

Top CrossFit athletes ≅ 400 watts sustained for 2¾ min Approx equivalent to 80 RPM at 7.5 kp (kg-Force) on a

Monark Bike. (although using less muscle mass so itwould be very difficult to generate that much power forthat long on a bike.

Wingate test (30 seconds maximal output) topperformers ≅ 700 Watts.

Next Slide

The relationship of metabolic powerproduced in skeletal muscle to themechanical power of activity. (Adapted fromH.G. Knuttgen, Strength Training andAerobic Exercise: Comparison and Contrast,Journal of Strength and ConditioningResearch 21, no. 3 (2007): 973-978.)

40-milebike

2 hoursOxidative2301,000

2-km row6minutes

Oxidative4602,000

100-msprint

14seconds

Glycolytic9204,000

OlympicLifts

1 secondPhosphagen1,3806,000

Example ofactivity

Time toexhaustion

Predominantenergy system

Mechanicaloutput

power (W)

Metabolicpower(watts)

Olympic Lifting andPowerlifting Power Outputs

Jerk ≈ 2,140 W (56 kg)Jerk ≈ 4,786 W (110 kg)Second pull

Average power output from transition tomaximum vertical velocity ≈ 5,600 Watts (100kg male); 2,900 Watts (75 kg female). Peakpower over a split second would be higher.

Average Power (Powerlifting)• bench ≈ 300 W• squat ≈ 1,000 W• deadlift ≈ 1,100 W

Power to Weight Ratio

For events like the Tour de France it is a matter of wattsper kilogram of body weight, that is, the specific poweroutput at lactate threshold - the amount of power/weight thatthe body can sustainably generate. It turns out that 6.7 ismore or less a magic number - the power/weight ratiorequired to win the TDF.

In many sports it isnot just about howmuch power yououtput ….it is alsoabout how much youweigh.

6

Energy/Power Analysis The previous is OK for a

fitness test or an estimate ofworkrate (power) duringexercise.

However, to calculateenergy change (power)segment by segment weneed to do a dynamicanalysis.

We need to takeaccelerations into account ifthe movement is toodynamic for a static analysis

Inverse Dynamic Analysis

α ax

ΣFx = maxΣFy = mayΣM = Igαay

Muscle Moment Power

MusclePower

Ang.

Vel.

MuscleMoment

Flex.

Ex.

Flex.

Ex.

+

-

Mechanical Work of Muscles

Wm Pm

t

t

dt= !1

2

.

Wm M

t

t

dtj j= ! "1

2

.

Mechanical Energy TransferBetween Segments

Muscles can obviously do work on a segment(muscle moment power).

However, if there is translational movement ofthe joints there is mechanical energy transferbetween segments. (i.e. one segment doeswork on an adjacent segment by force-displacement through the joint centre).

Transfer of energy is very important inimproving the overall efficiency of humanmovement patterns.

Vj

Vj

θ1

Fj1

Fj2

Joint Force Power

Seg1

Seg2

Fj1Vjcosθ ispositive

θ2

Fj2Vjcosθ isnegative

7

Level

Level

Level

Uphill

Uphill

Uphill

Vastus Lateralis

Soleus

Gastrocnemius

Glycogen Usage

Human Energy Harvesting Biomechanical

Energy Harvesting:Generating ElectricityDuring Walking withMinimal User Effort

J. M. Donelan,1* Q. Li,1V. Naing,1 J. A.Hoffer,1 D. J. Weber,2A. D. Kuo3

Science 8 February2008: Vol. 319. no.5864, pp. 807 - 810

Rate of change of the energyof a segment (power) [Ps]

Muscle moment power for the proximal joint Muscle moment power for the distal joint Joint force power for the proximal joint Joint force power for the distal joint

s p d p p d dP M M F v F v = ! !+ + +

Total InstantaneousEnergy of a Body

ET = ½mv2 + mgh + ½Iω2

Ener

gy (J

)

15

10

5

0

Support Phase Swing Phase

Total K.E.P.E.R.K.E.

right heel contact right toe off right heel contact

Percent of Stride0 20 40 60 80 100

Energy of the Foot

Efficiency Metabolic efficiency is a measure of the

muscles ability to convert metabolic energyto tension.

A high metabolic efficiency does notnecessarily mean that an efficient movementis taking place (e.g. cerebral palsy).

The ability of the central nervous system tocontrol the tension patterns is whatinfluences the mechanical efficiency.

8

Overall Muscular EfficiencyMuscular Eff. = Net mechanical work

Net metabolic energy

Net mechanical work = Internal work + External work

Internal work: Work done by muscles inmoving body segments.

External work: Work done by muscles tomove external masses or work against externalresistance.

Aprrox. 20-25% efficiency.

Contraction timerelated to force -velocity curve

EfficiencyAll efficiency calculations involve somemeasure of mechanical output divided by ameasure of metabolic input. Metabolicwork is not too difficult to estimate if wedo gas analysis. External work also easyto calculate. But we need to calculateinternal mechanical work. Clearly we mustat least calculate absolute energy changes(negative work is still an energy cost to thebody). However, isometric contractionsagainst gravity still a problem.

Causes of Inefficient Movement Co-contraction Isometric Contractions Against Gravity

Example of hands out straight. Nomechanical work being done!

Jerky Movementshigh accelerations & decelerations waste

energy compared to gradual acceleration Generation of energy at one joint and

absorption at another (walking example) Joint friction (small)

Flow of Energy

MetabolicEnergy

Bodysegmentenergy

O2 uptake

CO2 expired Externalwork

mechanical energy(muscle tension)

maintenanceheat

heats ofcontraction

isometric workagainst gravity

loss due to co-contractionor absorption by negativework at another joint

joint friction

Burning Calories!

So does atreadmill,lifecycle or stairmaster give youan accuratevalue forcalories burnt orpower output?