Outline Kinetics – Forces in human motion – Impulse-momentum – Mechanical work, power, &...
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![Page 1: Outline Kinetics – Forces in human motion – Impulse-momentum – Mechanical work, power, & energy – Locomotion Energetics.](https://reader035.fdocuments.net/reader035/viewer/2022081519/56649f425503460f94c619b8/html5/thumbnails/1.jpg)
Outline
• Kinetics– Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics
![Page 2: Outline Kinetics – Forces in human motion – Impulse-momentum – Mechanical work, power, & energy – Locomotion Energetics.](https://reader035.fdocuments.net/reader035/viewer/2022081519/56649f425503460f94c619b8/html5/thumbnails/2.jpg)
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
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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
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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
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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
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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
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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
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Stance time = 0.52 sFgy (ave) = 750 Nmg = 570 Nmvtakeoff = ∫(Fg,y - mg)dt
F g,y (
N)
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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
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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
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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
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Walk: 1.25 m/s (constant avg. v)∫Fg,x dt = 0 ---> A1 = A2
A1
A2Fgx
(body weights)
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A1
A2
Run: 3.83 m/s (constant avg, v)∫Fg,x dt = 0 ---> A1 = A2
Fgx
(body weights)
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Run: 3.83 m/s (constant avg. v)
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Accelerating: ∫Fg,x dt > 0
AcceleratingFg,x
A1
A2
A1 < A2
Time
0
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Decelerating: ∫Fg,x dt < 0
Fg,x
A1
A2
A1 > A2
Decelerating
Time
0
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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?
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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?
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Outline
• Kinetics – Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics
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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
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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
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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
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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
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Work against Resistive (Non-Conservative) Forces
Work to overcome resistances (friction, aero/hydro)
1 N * m = 1 JouleDissipative (lost as heat)
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Which of the following is NOT and example of a non-conservative force?
A) FrictionB) Air ResistanceC) Water ResistanceD) GravityE) None of the above
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Work against Conservative Forces
Work to overcome gravity or spring forces
Work leads to energy conservation
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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
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Gravitational potential energy (Ep,g)
mg = weight of objectry = vertical position of
object
Ep,g = mgry
ry
mgU = F*r = mg*ry
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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
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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
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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
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∆ 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
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∆ Mechanical Energy = Mechanical work
Ep,g = 0
Ep,g = 100 J
U = -100 J
A 200 Newton objectis lowered 0.5 meter. Negative work
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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
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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
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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
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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)
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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
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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
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Swimming: work & power to overcome drag forces
A person swims 100 meters at 1 m/s against a drag force of 150N.
Work:
Power:
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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
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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
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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
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Outline
• Kinetics (external)– Forces in human motion– Impulse-momentum– Mechanical work, power, & energy– Locomotion Energetics
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Kinetic energy (Ek,t)
m = massv = velocityk = kinetic, t = translational
m
Ek,t = 0.5 mv2
v
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Gravitational potential energy (Ep,g)
mg = weight of objectry = vertical position of
object
Ep,g = mgry
ry
mg
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Elastic energy: energy stored when a spring is stretched or compressed
Stretched(Energy stored)
Rest length(no energy stored)
Compressed(Energy stored)
Spring
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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)
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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
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Walkvx decreasesry increases
vx increasesry decreases
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WALK
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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
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Ek,t (J)
Ep,g (J)
Etot (J)
Time (s)
WALK
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Inverted pendulum model for walking
Leg
C.O.M.
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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
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vx & Ek,t decreasery & Ep,g decrease
vx & Ek,t increasery & Ep,g increase
RUN
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RUN
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Ek,t (J)
Ep,g (J)
Etot (J)
Time (s)
Run
But what about EE?
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Run: spring mechanism
Ek,t & Ep,g are in phase. Elastic energy is stored in leg.
Leg (spring)
C.O.M.
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Leg
C.O.M.
Leg
C.O.M.
WalkInverted pendulum
RunSpring mechanism
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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
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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
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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
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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