Transcript Kinetics_Part3

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 / s
G: vector quantity
direction of velocity vector
mv
Impulse
Impulse = ∫F dt
F = force applied to object
area under a force-time curve
product of the average force and
time of application
if 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 + Impulse
mvf = 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 kg
tcontact=18ms
A) 5.83 Ns
B) 324 N
C) 7.776 Ns
D) 432 N
E) I don’t understand
Vertical jump: impulse-momentum analysis
Stance: Vertical impulse increases momentum
Fy = Fg,y – mg
mvtakeoff = mvi + ∫(Fg,y - mg)dt
vi = 0
vtakeoff = ∫(Fg,y - mg)dt
mg
To maximize jump height:
maximize stance impulse
increase time of force application,
increase Fg,y
Fg,y
mvtakeoff = ∫(Fg,y - mg)dt
Stance time = 0.52 s
Fgy (ave) = 750 N
mg = 570 N
1000
Fg,z (N)
Fg,y (N)
1500
mg
500
0
0
0.4
0.8
Time (s)
1.2
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
Fg,y (Bodyweights)
Walk: 1.25 m/s (constant avg. v)
∫Fg,x dt = 0 ---> A1 = A2
0.3
Fgx
0.0
(body
weights)
-0.3
0.0
A2
A1
0.2
0.4 0.6
Time (s)
0.8
Run: 3.83 m/s (constant avg, v)
∫Fg,x dt = 0 ---> A1 = A2
0.3
A2
Fgx
(body 0.0
weights)
-0.3
0.0
A1
0.1
0.2 0.3
Time (s)
0.4
Run: 3.83 m/s (constant avg. v)
Horizontal velocity
-1
of C.O.G. (m • s )
3.9
3.8
3.7
3.6
3.5
0.00
Right
stance
Left
stance
0.50
0.25
Time (s)
0.75
Accelerating: ∫Fg,x dt > 0
Fg,x
A2
0
Accelerating
A1 < A 2
A1
Time
Decelerating: ∫Fg,x dt < 0
Decelerating
A1 > A 2
Fg,x
A2
0
A1
Time
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 & Energy
Principle of Work and Energy
Work
to overcome
fluid and friction forces
gravitational and elastic forces
Mechanical Energy
Kinetic energy
Potential energy
Gravitational
Elastic
Conservation of Energy
Units for Work and Mechanical energy
Joule = Nm
Work (U)
U = force * distance
U =|F| *|r| * cos (θ)
F
r
F: force applied
U = Fr
r: distance moved
θ: angle between force vector and line of displacement
Scalar
1 N * m = 1 Joule
F
r
θ=0
Work (U)
F
U = force * distance
U =|F| *|r| * cos (θ)
θ = 30
r
U = Fr cos (θ)
F: force applied
r: distance moved
θ: angle between force vector and line of displacement
Scalar
1 N * m = 1 Joule
F
r
θ = 30
Work (U)
F
U = force * distance
U =|F| *|r| * cos (θ)
θ = 30
r
U = Fr cos (θ)
F: force applied
r: distance moved
θ: angle between force vector and line of displacement
Can be positive or negative
Positive work: Force and displacement in same direction
Negative Work: Force and displacement in opposite
directions
Scalar
1 N * m = 1 Joule
Work against Resistive
(Non-Conservative) Forces
Work to overcome resistances (friction, aero/hydro)
1 N * m = 1 Joule
Dissipative (lost as heat)
Which of the following is NOT and
example of a non-conservative force?
A) Friction
B) Air Resistance
C) Water Resistance
D) Gravity
E) 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)
U = F*r = mg*ry
mg
ry
Ep,g = mgry
mg = weight of object
ry = vertical position of
object
Elastic potential energy: energy stored when
a spring is stretched or compressed
Spring
Ep,s = 0.5kx2
Rest length
(no energy stored)
Stretched
(Energy stored)
Compressed
(Energy stored)
Kinetic energy (Ek,t)
m
v
Ek,t = 0.5 mv2
m = mass
v = velocity
k = kinetic, t = translational
Kinetic energy is based on velocity of an object
Work-Energy Theorem
Mechanical work = ∆ Mechanical energy
U= DE = DEk+DEp
When positive mechanical work is done on an object, its
mechanical energy increases.
U=F*r=DE
When negative mechanical work is done (e.g. braking)
on an object, its mechanical energy decreases.
U=-F*r=DE
∆ Mechanical Energy = Mechanical work
A 200 Newton object
is lifted up 0.5 meter.
∆Ep,g = mg∆ry
∆Ep,g = 200 • 0.5 = 100 J
Ep,g = 100 J
U = 100 J
Ep,g = 0
∆ Mechanical Energy = Mechanical work
A 200 Newton object
is lowered 0.5 meter.
Negative work
Ep,g = 100 J
U = -100 J
Ep,g = 0
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 kJ
B) 1,000 kJ
C) 707 kJ
D) – 707kJ
E) I am lost
1000 m
∆ry
45°
Law of Conservation of Energy
DE=U
DEk+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 J
B) PE = 4120.2 J, KE = 0 J
C) PE = 4120.2 J, KE = 1765.8 J
D) 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
Units
J / 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 kW
B) U = 3840 J, P = 19.2 kW
C) U = 2560 J, P = 512 W
D) 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,i2)
vx,i = 2 m/s
vx,f = 10 m/s
U = (0.5)(80)(100 - 4) = 3840 J
Pmech = 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 J
Power:
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 = -0.5CDAv2 * v = (-0.5CDA) * v3
Swimming, bicycling: most of the muscular
power output is used to overcome drag
Summary
Work: result of force applied over distance
Energy: capacity to do work
Kinetic Energy: energy based on velocity of an
object
Potential 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
Ek,t = 0.5 mv2
m = mass
v = velocity
k = kinetic, t = translational
v
Gravitational potential energy (Ep,g)
mg
ry
Ep,g = mgry
mg = weight of object
ry = vertical position of
object
Elastic energy: energy stored when a
spring is stretched or compressed
Spring
Rest length
(no energy stored)
Stretched
(Energy stored)
Compressed
(Energy stored)
Mechanical energy in level walking
Some kinetic energy
Some gravitational potential energy
Little work done against aerodynamic drag
Unless 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)
HOWEVER
Ek,t and Ep,g fluctuate within each stance
Walk
vx decreases
ry increases
vx increases
ry decreases
Gravitational
potential energy
of C.O.G. (J)
Kinetic energy
of C.O.G. (J)
100
WALK
50
0
-50
100
-50
Right
stance
0
Left
stance
50
0
0.4
0.8
Time (s)
1.2
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)
WALK
Ep,g (J)
Etot (J)
Time (s)
Inverted pendulum model for walking
C.O.M.
Leg
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 decrease
ry & Ep,g decrease
vx & Ek,t increase
ry & Ep,g increase
RUN
Kinetic energy
of C.O.G. (J)
800
Right
stance
Left
stance
Right
stance
Left
stance
RUN
700
600
500
Gravitational
potential energy
of C.O.G. (J)
150
75
0
-75
-150
0.00
0.25
0.50
Time (s)
0.75
Run
Ek,t (J)
Ep,g (J)
Etot (J)
Time (s)
But what about EE?
Run: spring mechanism
Ek,t & Ep,g are in phase. Elastic energy is stored in
leg.
C.O.M.
Leg (spring)
Walk
Inverted pendulum
C.O.M.
Leg
Run
Spring mechanism
C.O.M.
Leg
Attach some numbers to these ideas
For 70kg person,
Walking, 1.5 m/sec:
If the COM rises 4 cm, what is DGPE?
a)
b)
c)
d)
e)
2746.8 J
-2746.8 J
27.468 J
-27.468 J
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)
b)
c)
d)
e)
1.74 m/s
1.2 m/s
0.3 m/s
-0.3 m/s
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)
b)
c)
d)
e)
87.3 J
-87.3 J
32.382 J
-32.382 J
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 second
How much mechanical power would have to be produced?
a)
b)
c)
d)
27.5 W
54.9 W
109.9 W
I am lost