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