Outline Kinetics – Forces in human motion – Impulse-momentum – Mechanical work, power, &...

Outline

• Kinetics– Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics

Linear momentum (G)

Product of the mass and linear velocity of an object

G = mv

Units: kg * m / sG: vector quantity

direction of velocity vector

mv

Impulse

Impulse = ∫F dt F = force applied to objectarea under a force-time curveproduct of the average force and time of applicationif constant force (F): Impulse = F * t Units: N * s

An impulse imparted on an object causes a change in momentum

∆G= Impulse ∆G= ∫F dt

if constant force (F): ∆G = F * t

Gfinal = Ginitial + F * t

if average force (F): ∆G = F * t

Gfinal = Ginitial + F * t

Gfinal = Ginitial + Impulse

A person (90 kg) on a bike (10 kg) increases v from 0 to 10 m/s. What

impulse was required?

Gfinal = Ginitial + Impulsemvf = mvi + Impulse Impulse = mvf - mvi

vi = 0; vf = 10 m/s; m = 90kg +10kg = 100 kg

Spiking a volleyball

What is the impulse applied to the ball?vinitial=3.6m/s (towards spiker)

vfinal=25.2m/s (away from spiker)

m=0.27 kgtcontact=18ms

A) 5.83 NsB) 324 NC) 7.776 NsD) 432 NE) I don’t understand

Vertical jump: impulse-momentum analysis

Stance: Vertical impulse increases momentum

Fy = Fg,y – mgmvtakeoff = mvi + ∫(Fg,y - mg)dt vi = 0 vtakeoff = ∫(Fg,y - mg)dt

To maximize jump height: maximize stance impulse

increase time of force application,increase Fg,y

mg

Fg,y

Stance time = 0.52 sFgy (ave) = 750 Nmg = 570 Nmvtakeoff = ∫(Fg,y - mg)dt

F g,y (

N)

A jumping person (m = 57 kg; vi = 0) has an average Fg,y of 750 N for 0.5 seconds.

What is the person’s vertical takeoff velocity?mvy,takeoff = mvi + ∫(Fg,y - mg)dt

57 * vy,takeoff = 0 + (Fg,y - mg) * t

57 * vy,takeoff = (750 – 559.17) * 0.5

vy,takeoff = 1.67 m/s

A jumping person (m = 57 kg; vi = 0) has an average Fg,y of 750 N for 0.5 seconds.

What is the person’s vertical takeoff velocity?mvy,takeoff = mvi + ∫(Fg,y - mg)dt

57 * vy,takeoff = 0 + (Fg,y - mg) * t

57 * vy,takeoff = (750 – 559.17) * 0.5

vy,takeoff = 1.67 m/s

How high did they jump?(1.67)2/2g=0.14m

Walking or running at a constant average speed

On average, forward velocity of body does not change during stance

∆ vx = 0

∫ Fg,x dt = 0

Walk: 1.25 m/s (constant avg. v)∫Fg,x dt = 0 ---> A1 = A2

A1

A2Fgx

(body weights)

A1

A2

Run: 3.83 m/s (constant avg, v)∫Fg,x dt = 0 ---> A1 = A2

Fgx

(body weights)

Run: 3.83 m/s (constant avg. v)

Accelerating: ∫Fg,x dt > 0

AcceleratingFg,x

A1

A2

A1 < A2

Time

0

Decelerating: ∫Fg,x dt < 0

Fg,x

A1

A2

A1 > A2

Decelerating

Time

0

A person (100 kg) on a bicycle (10 kg) can apply a decelerating force of 200N by maximally squeezing the brake levers. How long will it take for the bicyclist to stop if he is traveling at 13.4 m/sec (30 miles per hour) and the braking force is the only force acting to slow him down?

A soccer ball (4.17N) was travelling at 7.62 m/s until it contacted the head of a player and sent travelling in the opposite direction at 12.8 m/sec. If the ball was in contact with the player’s head for 22.7 milliseconds, what was the average force applied to the ball?

Outline

• Kinetics – Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics

Mechanical Work & EnergyPrinciple of Work and EnergyWork

to overcomefluid and friction forcesgravitational and elastic forces

Mechanical EnergyKinetic energyPotential energy Gravitational Elastic

Conservation of Energy

Units for Work and Mechanical energy Joule = Nm

Work (U) U = force * distanceU =|F| *|r| * cos (θ)

F: force appliedr: distance movedθ: angle between force vector and line of displacement

Scalar 1 N * m = 1 Joule

F

rU = Fr

Fr

θ = 0

Work (U) U = force * distanceU =|F| *|r| * cos (θ)

F: force appliedr: distance movedθ: angle between force vector and line of displacement

Scalar 1 N * m = 1 Joule

F

rU = Fr cos (θ)

F

r

θ = 30

θ = 30

Work (U) U = force * distanceU =|F| *|r| * cos (θ)

F: force appliedr: distance movedθ: angle between force vector and line of displacement

Can be positive or negativePositive work: Force and displacement in same directionNegative Work: Force and displacement in opposite directions

Scalar 1 N * m = 1 Joule

F

rU = Fr cos (θ)

θ = 30

Work against Resistive (Non-Conservative) Forces

Work to overcome resistances (friction, aero/hydro)

1 N * m = 1 JouleDissipative (lost as heat)

Which of the following is NOT and example of a non-conservative force?

A) FrictionB) Air ResistanceC) Water ResistanceD) GravityE) None of the above

Work against Conservative Forces

Work to overcome gravity or spring forces

Work leads to energy conservation

Potential energy

• When work on an object by a force can be expressed as the change in an object’s position.– Work done by gravitational forces• Gravitational potential energy

– Work done by elastic forces• Elastic (strain) potential energy

Potential energy arises from position of an object

Gravitational potential energy (Ep,g)

mg = weight of objectry = vertical position of

object

Ep,g = mgry

ry

mgU = F*r = mg*ry

Elastic potential energy: energy stored when a spring is stretched or compressed

Stretched(Energy stored)

Rest length(no energy stored)

Compressed(Energy stored)

Spring Ep,s = 0.5kx2

Kinetic energy (Ek,t)

m = massv = velocityk = kinetic, t = translational

m

Ek,t = 0.5 mv2

v

Kinetic energy is based on velocity of an object

Work-Energy TheoremMechanical work = ∆ Mechanical energy

When positive mechanical work is done on an object, its mechanical energy increases.

U=F*r=DEWhen negative mechanical work is done (e.g. braking)

on an object, its mechanical energy decreases. U=-F*r=DE

U= DE = DEk+DEp

∆ Mechanical Energy = Mechanical work

A 200 Newton objectis lifted up 0.5 meter.∆Ep,g = mg∆ry

∆Ep,g = 200 • 0.5 = 100 J

Ep,g = 0

Ep,g = 100 J

U = 100 J

∆ Mechanical Energy = Mechanical work

Ep,g = 0

Ep,g = 100 J

U = -100 J

A 200 Newton objectis lowered 0.5 meter. Negative work

Mechanical work in uphill walking

A person (mg = 1000 N) walks 1000m on a 45°uphill slope. How much mechanical work is required to lift the c.o.m. up the hill?

A) -1,000 kJB) 1,000 kJC) 707 kJD) – 707kJE) I am lost

45°

∆ry

1000 m

Law of Conservation of Energy

DE=UDEk+DEp=Uext

If only conservative forces are acting on the system:

DEk+DEp=0

Ek+Ep=Constant

Ek1+Ep1=Ek2+Ep2

A woman with a mass of 60kg dives from a 10m platform, what is her potential and kinetic energy 3m into the dive?

A) PE = 0 J, KE = 1765.8 JB) PE = 4120.2 J, KE = 0 JC) PE = 4120.2 J, KE = 1765.8 JD) PE= 1765.8 J, KE = 4120.2 J

Mechanical Power (Pmech): Rate of performing mechanical work

Pmech = U / ∆t

Pmech = (F * r * cos θ ) / ∆t

Pmech = F * v cos θ

UnitsJ / s = Watts (W)

A sprinter (80 kg) increases forward velocity from 2 to 10 m/s in 5 seconds. U & Pmech ?

A) U = 2560 J, P = 12.8 kWB) U = 3840 J, P = 19.2 kWC) U = 2560 J, P = 512 WD) U= 3840 J, P = 768 W

A sprinter (80 kg) increases forward velocity from 2 to 10 m/s in 5 seconds. U & Pmech ?

U = Ek,t(final) - Ek,t(initial) = 0.5m(vx,f2 - vx,i

2) vx,i = 2 m/s

vx,f = 10 m/s

U = (0.5)(80)(100 - 4) = 3840 JPmech = U / ∆t = 3840 J / 5 s = 768 W

Swimming: work & power to overcome drag forces

A person swims 100 meters at 1 m/s against a drag force of 150N.

Work:

Power:

Swimming: work & power to overcome drag forces

A person swims 100 meters at 1 m/s against a drag force of 150N.

Work:U = F * d = 150 N * 100 meters = 15, 000 JPower:Pmech = F * v = 150 N * 1 m/s = 150W

or you could calculate time (100 seconds) and use work/time

Mechanical power to overcome drag

Pmech = Fdrag * v

Pmech = -0.5CDAv2 * v = (-0.5CDA) * v3

Swimming, bicycling: most of the muscular power output is used to overcome drag

Summary

Work: result of force applied over distanceEnergy: capacity to do work

Kinetic Energy: energy based on velocity of an objectPotential Energy: energy arising from position of an object

Power: rate of Work production

Outline

• Kinetics (external)– Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics

Kinetic energy (Ek,t)

m = massv = velocityk = kinetic, t = translational

m

Ek,t = 0.5 mv2

v

Gravitational potential energy (Ep,g)

mg = weight of objectry = vertical position of

object

Ep,g = mgry

ry

mg

Elastic energy: energy stored when a spring is stretched or compressed

Stretched(Energy stored)

Rest length(no energy stored)

Compressed(Energy stored)

Spring

Mechanical energy in level walking

Some kinetic energySome gravitational potential energyLittle work done against aerodynamic dragUnless slipping, no work done against friction

Not much bouncing (elastic energy)

Mechanical energy fluctuations in level walking

Average Ek,t constant (average vx constant)

Average Ep,g constant (average ry constant)

HOWEVEREk,t and Ep,g fluctuate within each stance

Walkvx decreasesry increases

vx increasesry decreases

WALK

Walk: inverted pendulum• 1st half of stance: decrease vx & increase ry

– Ek,t converted to Ep,g

• 2nd half of stance: increase vx & decrease ry

– Ep,g converted to Ek,t

• KE & GPE “out of phase”• Energy exchange: as much as 95%recovered

during single stance phase

Ek,t (J)

Ep,g (J)

Etot (J)

Time (s)

WALK

Inverted pendulum model for walking

Leg

C.O.M.

Walk: inverted pendulum• 1st half of stance: decrease vx & increase ry

– Ek,t converted to Ep,g

• 2nd half of stance: increase vx & decrease ry

– Ep,g converted to Ek,t

• KE & GPE “out of phase”• Energy exchange: as much as 95%recovered

during single stance phase• But, energy is lost with each step as collision

vx & Ek,t decreasery & Ep,g decrease

vx & Ek,t increasery & Ep,g increase

RUN

RUN

Ek,t (J)

Ep,g (J)

Etot (J)

Time (s)

Run

But what about EE?

Run: spring mechanism

Ek,t & Ep,g are in phase. Elastic energy is stored in leg.

Leg (spring)

C.O.M.

Leg

C.O.M.

Leg

C.O.M.

WalkInverted pendulum

RunSpring mechanism

Attach some numbers to these ideas

For 70kg person, Walking, 1.5 m/sec:If the COM rises 4 cm, what is DGPE?

a) 2746.8 Jb) -2746.8 Jc) 27.468 Jd) -27.468 Je) I am lost

Attach some numbers to these ideas

For 70kg person, Walking, 1.5 m/sec:If the COM rises 4 cm, what is DGPE?How much must velocity decrease to have KE match that?

a) 1.74 m/sb) 1.2 m/sc) 0.3 m/sd) -0.3 m/se) 2.38 m/s

Attach some numbers to these ideas

Running, 3 m/sec: If COM sinks by 4 cm and velocity decreases by 10%How much energy could be stored elastically?

a) 87.3 Jb) -87.3 Jc) 32.382 Jd) -32.382 Je) I am lost

If there was no inverted pendulum

For 70kg person, Walking, 1.5 m/sec:If com rises 4 cm and they take 1 stride per secondHow much mechanical power would have to be produced?

a) 27.5 Wb) 54.9 Wc) 109.9 Wd) I am lost