Download Mechanical Work, Energy and Power (segment models)

Work
Mechanical Work,
Energy and Power
(segment models)
Hamill & Knutzen Chapters 10 & 11
Winter 1979 Chapter 5
Work is the product of force times the
distance through which that force moves a
load.
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 scalar
 F x d => MLT-2 x L => ML2T-2
 As the L’s are are the same, the square of
them will always be positive.
 Torque 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 constant
in the equation for strain energy (½kx2)?
 MT-2
1
Error in Hamill & Knutzen text?
Conservation of Energy
 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 the
section on elastic force and energy.
 F => MLT-2 Therefore units of k => MT-2
 Energy => MLT-2 ??????
 Elastic Strain Energy = ½kx2
 The total energy of a closed system is
constant since energy does not enter or
leave a closed system.
 This only occurs in human movement
when the object is a projectile and we
neglect air resistance. Then the total
energy of the system (TE) = PE + KE.
 Note that gravity does not change the
total 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), the
equation becomes.
 Fd = Δ(½mv
Δ(½mv2 + mgh)
mgh)
2
 Fd = Δ½mv + Δmgh
Work-Energy
Relationship
This is not a new
mechanical concept.
It can be derived
from Newton’s
second law.
F = m!a
F = m ! dv
Work-Energy Relationship
Angular Kinetics
dt
F = m ! dv
ds
F = m ! v ! dv
! ds
 If force is applied off centre (i.e. the line of action of
the force vector does not pass through the centre of
rotation of the body) then rotation as well as
translation will occur.
dt
ds
F ! ds = m ! v ! dv
Work = "Energy
Fd = " 1 2 m v 2
" Fds = m" vdv
Work = m( v )
1
2
#$ = "
2
1
2
I%2
(if we assume a
horizontal force
through the centre
of gravity or if we
just ignore potential
energy)
!
2
Linear Force?
Translation
Fd = Δ(½mv2 + mgh)
d
Note: net force
would have to
factor in gravity
Eccentric Force?
This is because:
 The centre of mass of a rigid body
instantaneously accelerates in the
direction of the applied resultant external
force, irrespective of where the force is
applied on the body.
 The rotational effect is also instantaneous.
 However, you do not get something for
nothing.
Fd = Δ(½mv2 + mgh)
Rotation and Translation
Rotation and Translation
Note that “d” is longer than in the
centric force scenario
d
You can break
out the linear
and rotational
components of
the eccentric
force
Work = "Energy
Fd = " 1 2 m v 2
#$ = "
1
2
I%2
!
3
Remember this slide from the
Tennis Lecture?
Back to the Vertical Jump
 This is the same data from the problem I
gave you earlier.
 Take-off velocity = 2.534 m/s
 Mass = 61.2 kg
 New Question: If the jumper’s centre of
gravity moved +0.5 m vertically from the
bottom of the drop down to the take off
position, how much work did she do in this
phase of the jump and what was her
average force production?
Green vector
τθ = ΔΙω2
Blue vector (action on C of g)
Fd = Δ(½mv2 + mgh)
Racket
Vertical Jump Power
Power = force x velocity
(Kin 142 & 343)
From vf2 = vi2 + 2ad we can calculate the velocity
of take-off and, as we started from zero velocity,
the average velocity during take-off.
Power = 2.21" Wt " d
!
Power = 2.21" 600 " 0.327
Vto = 2ad = 2a " d
Power = 758 # W (J /s)
Vto = 19.62 " d = 4.42 d
Average velocity # 2.21" d
 If you used body mass (61.2 kg) instead of
body weight (600 N) you should have
calculated and answer of 77.3 kgm.s-1
 Where does the above equation come from?
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 from
such 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 (9.81
m.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)
4
Bowflex Treadclimber
 “Reduce your workout time - dualmotion treadles let you step forward like
a treadmill and up like a stair climber so
you get more exercise in less time”
 “TreadClimber® machine burns up to
2 TIMES more calories than a
treadmill - at the same speed!”
 “Studies were conducted at the
prestigious Human Performance
Laboratory at New York's Adelphi
University. The results were dramatic! In
22 separate trials, the TreadClimber®
machine burned up to 2 times more
calories in 30 minutes than a treadmill at
the same speed!”
 Company Website Sep-2006
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 elliptical
trainer? Quite simply, you get a triple-charged cardio
workout “ 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 it
would be very difficult to generate that much power for
that long on a bike.
 Wingate test (30 seconds maximal output) top
performers ≅ 700 Watts.
http://www.treadclimber.com/trc_microsite/fitnessbenefits.jsp
Next Slide
 The relationship of metabolic power
produced in skeletal muscle to the
mechanical power of activity. (Adapted from
H.G. Knuttgen, Strength Training and
Aerobic Exercise: Comparison and Contrast,
Journal of Strength and Conditioning
Research 21, no. 3 (2007): 973-978.)
Olympic Lifting and
Powerlifting Power Outputs
Jerk ≈ 2,140 W (56 kg)
Jerk ≈ 4,786 W (110 kg)
Second pull
Average power output from transition to
maximum vertical velocity ≈ 5,600 Watts (100
kg male); 2,900 Watts (75 kg female). Peak
power over a split second would be higher.
Average Power (Powerlifting)
• bench ≈ 300 W
• squat ≈ 1,000 W
• deadlift ≈ 1,100 W
Metabolic Mechanical Predominant Time to
Example of
power
output
energy system exhaustion
activity
(watts)
power (W)
6,000
1,380
Phosphagen 1 second Olympic
Lifts
4,000
920
Glycolytic
14
seconds
2,000
460
Oxidative
6
2-km row
minutes
1,000
230
Oxidative
2 hours
100-m
sprint
40-mile
bike
Power to Weight Ratio
 In many sports it is
not just about how
much power you
output ….it is also
about how much you
weigh.
 For events like the Tour de France it is a matter of watts
per kilogram of body weight, that is, the specific power
output at lactate threshold - the amount of power/weight that
the body can sustainably generate. It turns out that 6.7 is
more or less a magic number - the power/weight ratio
required to win the TDF.
5
Energy/Power Analysis
Inverse Dynamic Analysis
 The previous is OK for a
fitness test or an estimate of
workrate (power) during
exercise.
 However, to calculate
energy change (power)
segment by segment we
need to do a dynamic
analysis.
 We need to take
accelerations into account if
the movement is too
dynamic for a static analysis
Muscle Moment Power
Flex.
ΣFx = max
ΣFy = may
ΣM = Igα
ay
Mechanical Work of Muscles
Muscle
Moment
t2
Wm =
Ex.
Flex.
Ang.
Vel.
ax
α
! Pm. dt
t1
t2
Ex.
Wm =
+
Muscle
Power
! M " . dt
j
j
t1
-
Mechanical Energy Transfer
Between Segments
 Muscles can obviously do work on a segment
(muscle moment power).
 However, if there is translational movement of
the joints there is mechanical energy transfer
between segments. (i.e. one segment does
work on an adjacent segment by forcedisplacement through the joint centre).
 Transfer of energy is very important in
improving the overall efficiency of human
movement patterns.
Joint Force Power
Seg1
Fj1
Fj1Vjcosθ is
positive
θ1
Vj
Vj
θ2
Fj2
Seg2
Fj2Vjcosθ is
negative
6
Human Energy Harvesting
Vastus Lateralis
Level
Uphill
 Biomechanical
Energy Harvesting:
Generating Electricity
During Walking with
Minimal User Effort
 J. M. Donelan,1* Q. Li,1
V. Naing,1 J. A.
Hoffer,1 D. J. Weber,2
A. D. Kuo3
 Science 8 February
2008: Vol. 319. no.
5864, pp. 807 - 810
Gastrocnemius
Level
Uphill
Level
Uphill
Soleus
Glycogen Usage
Rate of change of the energy
of a segment (power) [Ps]
Total Instantaneous
Energy of a Body
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




P
s
=
M ! +M ! +F v +F v
p
p
p
d
right heel contact
d
right toe off
d
right heel contact
Swing Phase
Support Phase
15
Energy (J)
Total
K.E.
P.E.
R.K.E.
5
0
0
20
40
Efficiency
 Metabolic efficiency is a measure of the
muscles ability to convert metabolic energy
to tension.
 A high metabolic efficiency does not
necessarily mean that an efficient movement
is taking place (e.g. cerebral palsy).
 The ability of the central nervous system to
control the tension patterns is what
influences the mechanical efficiency.
Energy of the Foot
10
ET = ½mv2 + mgh + ½Iω2
60
Percent of Stride
80
100
7
Overall Muscular Efficiency
Contraction time
related to force velocity curve
Muscular Eff. = Net mechanical work
Net metabolic energy
Net mechanical work
= Internal work + External work
 Internal work: Work done by muscles in
moving body segments.
 External work: Work done by muscles to
move external masses or work against external
resistance.
 Aprrox. 20-25% efficiency.
Causes of Inefficient Movement
Efficiency
All efficiency calculations involve some
measure of mechanical output divided by a
measure of metabolic input. Metabolic
work is not too difficult to estimate if we
do gas analysis. External work also easy
to calculate. But we need to calculate
internal mechanical work. Clearly we must
at least calculate absolute energy changes
(negative work is still an energy cost to the
body). However, isometric contractions
against gravity still a problem.
Flow of Energy
maintenance
heat
Metabolic
Energy
CO2 expired
mechanical energy
(muscle tension)
 Example of hands out straight. No
mechanical work being done!
 Jerky Movements
 high accelerations & decelerations waste
energy compared to gradual acceleration
 Generation of energy at one joint and
absorption at another (walking example)
 Joint friction (small)
Burning Calories!
isometric work
against gravity
O2 uptake
 Co-contraction
 Isometric Contractions Against Gravity
joint friction
Body
segment
energy
heats of
contraction
loss due to co-contraction
or absorption by negative
work at another joint
External
work
So does a
treadmill,
lifecycle or stair
master give you
an accurate
value for
calories burnt or
power output?
8