Chapter 4. MAE002-Kinetics of particles-Energy methods.pdf

Solve particle kinetics problems using the principle of work
and energy.
• Calculate the power and efficiency of a mechanical system.
• Solve particle kinetics problems using conservation of energy.
• Solve particle kinetic problems involving conservative central
forces

1. MAE002-Dynamics
Kinetics of Particles: Energy methods
Agolfballwilldeformuponimpactasshownbythishigh-speedphoto.The
maximumdeformationwilloccurwhentheclubheadvelocityandtheballvelocity
arethesame.
Chapter 4

2. Objectives
•Calculatetheworkdonebyaforce.
•Calculatethekineticenergyofaparticle.
•Calculatethegravitationalandelasticpotentialenergyofa
system.
•Solveparticlekineticsproblemsusingtheprincipleofwork
andenergy.
•Calculatethepowerandefficiencyofamechanicalsystem.
•Solveparticlekineticsproblemsusingconservationofenergy.
•Solveparticlekineticproblemsinvolvingconservativecentral
forces.

3. Contents
•Introduction
•Workandenergy
•Conservationofenergy

4. Introduction
Thepogostickallowstheboytochangebetweenkinetic
energy,potentialenergyfromgravity,andpotentialenergyin
thespring.

5. Introduction
•Previously,problemsdealingwiththemotionofparticleswere
solvedthroughthefundamentalequationofmotion,
•Thecurrentchapterintroducesoneadditionalmethodofanalysis..F ma
•Methodofworkandenergy:directlyrelatesforce,mass,velocity
anddisplacement.

6. Introduction
2 -6
Forces and
Accelerations
Velocities and
Displacements
Newton’s Second Law
(last chapter)
Work-Energy2211 TUT 
 G
F ma

7. Work of a Force
•Differential vectoris the particle displacement.rd

•Work of the forceis dzFdyFdxF
dsF
rdFdU
zyx 


cos

•Work is a scalarquantity, i.e., it has magnitude
and sign but not direction.force. length 
•Dimensions of work are Units are  J 1.356lb1ftm 1N 1 J 1 joule

8. Work of a Force
•Work of a force during a finite displacement, 
 






2
1
2
1
2
1
2
1
cos
21
A
A
zyx
s
s
t
s
s
A
A
dzFdyFdxF
dsFdsF
rdFU


•Workisrepresentedbytheareaunderthe
curveofF
tplottedagainsts.
•F
tistheforceinthedirectionofthe
displacementds

9. Work of a Force
What is the work of a constant force in rectilinear motion? 
12
cosU F x 

  
12
sinU F x 

 12
U F x

 12
0U


a)
b)
c)
d)

10. Work of a Force
•Work of the force of gravity,  yWyyW
dyWU
dyW
dzFdyFdxFdU
y
y
zyx





12
21
2
1
•WorkoftheweightisequaltoproductofweightWandvertical
displacementy.
•In the figure above, when is the work done by the weight positive?
a) Moving from y
1to y
2b) Moving from y
2to y
1
c) Never

11. Work of a Force
•Magnitudeoftheforceexertedbyaspringis
proportionaltodeflection, lb/in.or N/mconstant spring 

k
kxF
•Work of the force exerted by spring,2
2
2
12
1
2
1
21
2
1
kxkxdxkxU
dxkxdxFdU
x
x



•Workoftheforceexertedbyspringispositive
whenx
2<x
1,i.e.,whenthespringisreturningto
itsundeformedposition.
•Workoftheforceexertedbythespringisequalto
negativeofareaundercurveofFplottedagainstx, xFFU 
 21
2
1
21

12. Work of a Force
As the block moves from A
0to A
1, is
the work positive or negative?
Positive Negative
As the block moves from A
2to A
o, is
the work positive or negative?
Positive Negative
Displacementis
intheopposite
directionofthe
force

13. Forces which do notdo work (ds= 0 or cos  0:
•Weight of a body when its center of gravity moves horizontally.
•Reaction at a roller moving along its track, and
•Reaction at frictionless surface when body in contact moves along
surface.
•Reaction at frictionless pin supporting rotating body,
Work of a Force

14. Particle Kinetic Energy: Principle of Work & EnergydvmvdsF
ds
dv
mv
dt
ds
ds
dv
m
dt
dv
mmaF
t
tt



•Consider a particle of mass m acted upon by force F

•Integrating from A
1to A
2,energykineticmvTTTU
mvmvdvvmdsF
v
v
s
s
t




2
2
1
1221
2
1
2
12
2
2
1
2
1
2
1
•The work of the force is equal to the change in kinetic energy of
the particle.F

•Units of work and kinetic energy are the same:JmNm
s
m
kg
s
m
kg
2
2
2
2
1












mvT

15. Applications of the Principle of Work and Energy
•Thebobisreleasedfrom
restatpositionA
1.
Determinethevelocityof
thependulumbobatA
2
usingwork&kinetic
energy.
•Force acts normal to path and does
no work.P
 glv
v
g
W
Wl
TUT
2
2
1
0
2
2
2
2211




•Velocityisfoundwithoutdetermining
expressionforaccelerationand
integrating.
•Allquantitiesarescalarsandcanbe
addeddirectly.
•Forceswhichdonoworkare
eliminatedfromtheproblem.

16. Applications of the Principle of Work and Energy
•Principleofworkandenergycannotbe
appliedtodirectlydeterminethe
accelerationofthependulumbob.
•Calculatingthetensioninthecord
requiressupplementingthemethodof
workandenergywithanapplicationof
Newton’ssecondlaw.
•As the bob passes through A
2,W
l
gl
g
W
WP
l
v
g
W
WP
amF
nn
3
2
2
2


 glv 2
2
If you designed the rope to hold twice the
weight of the bob, what would happen?

17. Power and Efficiency
13 -18
• rate at which work is done.vF
dt
rdF
dt
dU
Power






•Dimensionsofpowerarework/timeorforce*velocity.Unitsfor
powerareW746
s
lbft
550 hp 1or
s
m
N 1
s
J
1 (watt) W 1 


•inputpower
outputpower
input work
koutput wor
efficiency




18. Sample Problem 13.1
13 -19
Anautomobileweighing4000
lbisdrivendowna5
o
inclineat
aspeedof60mi/hwhenthe
brakesareappliedcausinga
constanttotalbreakingforceof
1500lb.
Determinethedistancetraveled
bytheautomobileasitcomesto
astop.
SOLUTION:
•Evaluatethechangeinkinetic
energy.
•Determinethedistancerequired
fortheworktoequalthekinetic
energychange.

19. SOLUTION:
•Evaluate the change in kinetic energy.  lbft481000882.324000
sft88
s 3600
h
mi
ft 5280
h
mi
60
2
2
12
1
2
1
1
1




















mvT
v  0lb1151lbft481000
2211



x
TUT ft 418x
•Determine the distance required for the work
to equal the kinetic energy change.   
 x
xxU
lb1151
5sinlb4000lb1500
21


 00
22  Tv
Sample Problem 13.1

20. Sample Problem 13.2
13 -21
Twoblocksarejoinedbyan
inextensiblecableasshown.Ifthe
systemisreleasedfromrest,
determinethevelocityofblockA
afterithasmoved2m.Assume
thatthecoefficientoffriction
betweenblockAandtheplaneis
m
k=0.25andthatthepulleyis
weightlessandfrictionless.
SOLUTION:
•Applytheprincipleofwork
andenergyseparatelytoblocks
AandB.
•Whenthetworelationsare
combined,theworkofthecable
forcescancel.Solveforthe
velocity.

21. Sample Problem 13.2
SOLUTION:
•Applytheprincipleofworkandenergy
separatelytoblocksAandB.  
 

  
2
2
1
2
2
1
2211
2
kg200m2N490m2
m2m20
:
N490N196225.0
N1962sm81.9kg200
vF
vmFF
TUT
WNF
W
C
AAC
AkAkA
A






mm   
 
  
2
2
1
2
2
1
2211
2
kg300m2N2940m2
m2m20
:
N2940sm81.9kg300
vF
vmWF
TUT
W
c
BBc
B






22. Sample Problem 13.2
13 -23
•When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.  
2
2
1
kg200m2N490m2 vF
C    
2
2
1
kg300m2N2940m2 vF
c     
 
2
2
1
2
2
1
kg500J 4900
kg300kg200m2N490m2N2940
v
v

 sm 43.4v

23. Sample Problem 13.3
Aspringisusedtostopa60kgpackage
whichisslidingonahorizontalsurface.
Thespringhasaconstantk=20kN/m
andisheldbycablessothatitis
initiallycompressed120mm.The
packagehasavelocityof2.5m/sinthe
positionshownandthemaximum
deflectionofthespringis40mm.
Determine(a)thecoefficientofkinetic
frictionbetweenthepackageand
surfaceand(b)thevelocityofthe
packageasitpassesagainthroughthe
positionshown.
SOLUTION:
•Applytheprincipleofworkandenergy
betweentheinitialpositionandthe
pointatwhichthespringisfully
compressedandthevelocityiszero.
Theonlyunknownintherelationisthe
frictioncoefficient.
•Applytheprincipleofworkand
energyforthereboundofthepackage.
Theonlyunknownintherelationis
thevelocityatthefinalposition.

24. Sample Problem 13.3
13 -26
SOLUTION:
•Apply principle of work and energy between initial
position and the point at which spring is fully compressed.  0J5.187sm5.2kg60
2
2
2
12
1
2
1
1  TmvT  
  
kk
kf
xWU
mm
m
J377m640.0sm81.9kg60
2
21


   
   
   
   J0.112m040.0N3200N2400
N3200m160.0mkN20
N2400m120.0mkN20
2
1
maxmin
2
1
21
0max
0min




 xPPU
xxkP
kxP
e    J112J377
212121 
 kef
UUU m  0J112J 377-J5.187
:
2211



k
TUT
m 20.0
km

25. Sample Problem 13.3
13 -27
•Apply the principle of work and energy for the rebound
of the package. 
2
3
2
12
3
2
1
32 kg600 vmvTT    
J36.5
J112J377
323232


 kef
UUU m 
2
3
2
1
3322
kg60J5.360
:
v
TUT


 sm103.1
3v

26. Sample Problem 13.4
13 -28
A2000lbcarstartsfromrestatpoint1
andmoveswithoutfrictiondownthe
trackshown.
Determine:
a)theforceexertedbythetrackon
thecaratpoint2,and
b)theminimumsafevalueofthe
radiusofcurvatureatpoint3.
SOLUTION:
•Applyprincipleofworkandenergyto
determinevelocityatpoint2.
•ApplyNewton’ssecondlawtofind
normalforcebythetrackatpoint2.
•Applyprincipleofworkandenergyto
determinevelocityatpoint3.
•ApplyNewton’ssecondlawtofind
minimumradiusofcurvatureatpoint3
suchthatapositivenormalforceis
exertedbythetrack.

27. Sample Problem 13.4
13 -29
SOLUTION:
•Applyprincipleofworkandenergytodetermine
velocityatpoint2.

  sft8.50ft2.32ft402ft402
2
1
ft400:
ft40
2
1
0
2
22
2
2
22211
21
2
2
2
2
2
1
21






vsgv
v
g
W
WTUT
WU
v
g
W
mvTT
•Apply Newton’s second law to find normal force by
the track at point 2.:
nnamF 
WN
g
g
Wv
g
W
amNW
n
5
ft20
ft402
2
2
2


 lb10000N

28. Sample Problem 13.4
•Apply principle of work and energy to determine velocity
at point 3.
  sft1.40sft2.32ft252ft252
2
1
ft250
3
2
3
2
33311



vgv
v
g
W
WTUT
•ApplyNewton’ssecondlawtofindminimumradiusof
curvatureatpoint3suchthatapositivenormalforceis
exertedbythetrack.:
nnamF 
33
2
3 ft252

g
g
Wv
g
W
amW
n

 ft50
3

29. Sample Problem 13.5
13 -31
The dumbwaiter Dand its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
Determine the power delivered by the
electric motor Mwhen the dumbwaiter
(a)is moving up at a constant speed of
8 ft/s and (b)has an instantaneous
velocity of 8 ft/s and an acceleration of
2.5 ft/s
2
, both directed upwards.
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
equal to Fv
D, v
D= 8 ft/s.
•In the first case, bodies are in uniform
motion. Determine force exerted by
motor cable from conditions for static
equilibrium.
•In the second case, both bodies are
accelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.

30. Sample Problem 13.5
13 -32
•In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium. 
slbft1600
sft8lb 200


DFvPower   hp 91.2
slbft550
hp 1
slbft1600 

Power
Free-body C::0yF lb 4000lb8002  TT
Free-body D::0yF lb 200lb 400lb 600lb 600
0lb 600


TF
TF

31. Sample Problem 13.5
13 -33
•In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
2
2
12
sft25.1sft5.2
DCD aaa
Free-body C::
CCy amF  lb5.38425.1
2.32
800
2800  TT
Free-body D::
DDy amF 
lb 1.2626.466005.384
5.2
2.32
600
600


FF
TF   slbft2097sft8lb 1.262 
DFvPower   hp 81.3
slbft550
hp 1
slbft2097 

Power

32. Thepotentialenergystoredat
thetopoftherollercoasteris
transferredtokineticenergyas
thecarsdescend.
Theelasticpotentialenergy
storedinthetrampolineis
transferredtokineticenergyand
gravitationalpotentialenergyas
thegirlfliesupwards.
Potential Energy

33. Potential Energy2121 yWyWU 

•Work of the force of gravity ,W

•Work is independent of path followed; depends
only on the initial and final values of Wy. 
WyV
g
potential energyof the body with respect
to force of gravity.
21
21 ggVVU 

•Units of work and potential energy are the same:JmNWyV
g
•Choice of datum from which the elevation yis
measured is arbitrary.

34. Potential Energy
13 -41
•Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,2
2
2
12
1
2
1
21 kxkxU 

•The potential energy of the body with respect
to the elastic force,
2121
2
2
1
ee
e
VVU
kxV



•Note that the preceding expression for V
eis
valid only if the deflection of the spring is
measured from its undeformed position.

35. Conservation of Energy
•Work of a conservative force,2121 VVU 

•Concept of work and energy,1221 TTU 

•Follows thatconstant
2211


VTE
VTVT
•When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.

WVT
WVT


11
110 


WVT
VWg
g
W
mvT


22
2
2
2
2
1
2 02
2
1
•Frictionforcesarenotconservative.Total
mechanicalenergyofasysteminvolving
frictiondecreases.
•Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.

36. Sample Problem 13.6
A 20 lb collar slides without friction
along a vertical rod as shown. The
spring attached to the collar has an
undeflected length of 4 in. and a
constant of 3 lb/in.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 in. to position 2.
SOLUTION:
•Apply the principle of conservation of
energy between positions 1 and 2.
•The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial kinetic
energy is zero.
•Solve for the kinetic energy and velocity
at 2.

37. Sample Problem 13.6
SOLUTION:
•Apply the principle of conservation of energy between
positions 1 and 2.
Position 1:  
0
lbft20lbin.24
lbin.24in. 4in. 8in.lb3
1
1
2
2
12
1
2
1



T
VVV
kxV
ge
e
Position 2:  
 
2
2
2
2
2
2
2
1
2
2
2
2
12
2
2
1
311.0
2.32
20
2
1
lbft 5.5lbin. 6612054
lbin. 120in. 6lb 20
lbin.54in. 4in. 01in.lb3
vvmvT
VVV
WyV
kxV
ge
g
e




Conservation of Energy:lbft 5.50.311lbft 20
2
2
2211


v
VTVT  sft91.4
2v

38. Sample Problem 13.7
The0.5lbpelletispushedagainstthe
springandreleasedfromrestatA.
Neglectingfriction,determinethe
smallestdeflectionofthespringfor
whichthepelletwilltravelaroundthe
loopandremainincontactwiththe
loopatalltimes.
SOLUTION:
•Sincethepelletmustremainincontact
withtheloop,theforceexertedonthe
pelletmustbegreaterthanorequalto
zero.Settingtheforceexertedbythe
looptozero,solvefortheminimum
velocityatD.
•Applytheprincipleofconservationof
energybetweenpointsAandD.Solve
forthespringdeflectionrequiredto
producetherequiredvelocityand
kineticenergyatD.

39. Sample Problem 13.7
SOLUTION:
•Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.:
nnmaF  
222
2
sft4.64sft32.2ft 2 

rgv
rvmmgmaW
D
Dn
•Apply the principle of conservation of energy between
points Aand D. 
0
18ftlb360
1
22
2
12
2
1
1


T
xxkxVVV
ge 
  lbft5.0sft4.64
sft2.32
lb5.0
2
1
lbft2ft4lb5.00
22
2
2
2
1
2
2


D
ge
mvT
WyVVV 25.0180
2
2211


x
VTVT in. 47.4ft 3727.0 x