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Notes on Dynamics
by
Stephen F. Felszeghy CSULA Prof. Emeritus of ME
These notes are a supplement to FE Reference Handbook, 9.4 Version, for Computer-Based Testing, NCEES, June 2016, pp. 72-79. These notes were prepared for the FE/EIT Exam Review Course class meeting held on Oct. 22, 2016, 1:00 p.m. to 4 p.m.
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Dynamics
Kinematics–dealswithmotionaloneapartfromconsiderationsofforceandmass.Kinetics–relatesunbalancedforceswithchangesinmotion.
KinematicsofParticlesRectilinearMotionofaParticle
Suppose
€
v = v x( ) ;apply“ChainRule”:
€
dvdt
= a =dvdx
dxdt
→ a =dvdxv
DeterminationofMotionofaParticleIntegratedifferentialrelations:
Motion
Kinematics
KineticsUnbalancedForces
Positioncoordinate(Rectilineardisplacement):
€
x = f (t)→ x = x(t)
Velocity:
€
v =dxdt
= ˙ x
Acceleration:
€
a =dvdt
=d2xdt 2 = ˙ ̇ x
€
dx = v dtdv = adtv dv = adx
PO+x
x
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3
AngularMotionofaLine
Differentialrelations:
€
dθ =ω dtdω =α dtω dω =α dθ
Noteanalogywithrectilinearmotion.
Twocommoncases:
1.Accelerationa=constant,orα=constant2.Acceleration
€
a = f (t) ,or
€
α = f (t) Seemotionequationsinthe9.4Handbookonpp.73‐74.
CurvilinearMotionofaParticleVectorswillbedenotedbyuprightboldfaceletters,e.g.,r.Vectorswillbedenotedbyunderlinedlettersinhandwriting,e.g.,r.Scalarcomponentofvectorrwillbedenotedbyitalicr.
ReferenceAxis
Rigidbodymovinginplane
θ
+θ
Angularpositioncoordinate(Angulardisplacement):
€
θ = f (t)
Angularvelocity:
€
ω =dθdt
= ˙ θ
Angularacceleration:
€
α =dωdt
=d2θdt 2 = ˙ ̇ θ
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RectangularComponentsApplication:Seeprojectilemotioninthe9.4Handbookonp.74.MotionRelativetoTranslatingReferenceAxes
v(tangenttopath)patpath)
a
r
P
y
xO
Path
Positionvector:
€
r = r(t)(Vectorfunction)
Velocity:
€
v =drdt
= ˙ r
Acceleration:
€
a =dvdt
=d2rdt 2 = ˙ ̇ r
y
z
x
j
ki
O
Positionvector:
€
r = x i + y j+ zkVelocity:
€
v = ˙ x i + ˙ y j+ ˙ z k Acceleration:
€
a = ˙ ̇ x i + ˙ ̇ y j+ ˙ ̇ z k
Wewrite:
€
vx = ˙ x , etc.ax = ˙ ̇ x , etc.
y
O
B
A
€
rA
€
rA /B
x
y’
x’
€
rB
“Translating”meansx’–y’axesmovebutremainparalleltox–yaxes.
€
rA = rB + rA /B
˙ r A = ˙ r B + ˙ r A /B
vA = vB + vA /B
aA = aB + aA /B
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TangentialandNormalComponents
€
v =drdt
=dsdte t = v e t
a =dvdt
=d2rdt 2
=d2sdt 2e t +
v 2
ρen =
dvdte t +
v 2
ρen
€
= ate t + anen = a t + an RadialandTransverseComponents
€
a = ˙ v = ˙ ̇ r = ˙ ̇ r − r ˙ θ 2( )er + r ˙ ̇ θ + 2˙ r ˙ θ ( )eθ
€
vr = ˙ r
€
vθ = r ˙ θ
€
ar = ˙ ̇ r − r ˙ θ 2
€
aθ = r ˙ ̇ θ + 2˙ r ˙ θ
€
en
r
P
y
xO
Path
s
€
e t
C
Centerofcurvature
€
CP = ρ = radius of curvaturee t = unit vector tangent to pathen = unit vector normal to path pointing to Cs = directed distance along path
PolarcoordinatesofP:
€
r, θ( )
€
er = unit vector in r direction
€
eθ = unit vector perpendicular to r in direction of increasing θ
€
r = rer
€
v = ˙ r = ˙ r er + r ˙ θ eθ
y
Ox
€
eθ
€
er
€
θ Path
€
r = rerP
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KineticsofParticles:Newton’sSecondLaw(Equationofmotion)where
€
F = resultant force∑m = mass of particlea = absolute acceleration, measured in a newtonian frame of reference (inertial system)
GraphicalRepresentationofNewton’s2ndLawUnitsQuantitySystem
Length Time Mass Force
SI m s kg
€
N = kg ⋅ ms2
USCS ft s
€
slug = lb ⋅ s2
ft lb
€
v = r ˙ θ eθa = −r ˙ θ 2er + r ˙ ̇ θ eθ
€
F = ma∑
€
F2
€
F1€
F3
€
ma
P P
Free‐bodydiagram(FBD)
Kineticdiagram(KD)(Mass‐accelerationdiagram)
Ifpathisacircle,thenr=constant,
€
˙ r = ˙ ̇ r = 0,
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Ineithersystem,W=mg,whereW=weightg=accelerationduetogravityAtsurfaceofearth:(SI)g=9.807m/s2(USCS)g=32.174ft/s2AVOID:lbf,lbmEquationsofMotion:RectangularComponents
€
Fx∑ = max
Fy∑ = may
EquationsofMotion:TangentialandNormalComponents
y y
O Ox x
€
Fy∑
€
Fx∑ P P€
ma y
€
ma x
FBD KD
€
Ft∑ = mat
Fn∑ = man
FBD KD
y y
x xO O
€
Fn∑
€
Ft∑
P P€
man
€
ma t
Path
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EquationsofMotion:RadialandTransverseComponentsKineticsofParticles:EnergyMethodsAboveresultistheprincipleofworkandenergy.Units:(SI)N⋅m=J;(USCS)ft⋅lb
y
O Ox x
r r
€
θ
€
θ
Path
y
€
Fθ∑
€
Fr∑
P P€
maθ
€
ma r
€
Fr∑ = mar
Fθ∑ = maθ
FBD
KD
y
x
Position2
Position1
€
r2
€
r
€
r1
O
P
€
F
sPath
TheworkdonebyFontheparticleduringafinitemovementoftheparticlealongacurvedpathfromposition1toposition2is
€
U1→2 :
€
U1→2 = F ⋅ drr1
r2∫ (Line integral)
Itcanbeshown:
€
U1→2 = Fts1
s2∫ ds
=12
mv22 −
12
mv12
Let T =12
mv 2 = kinetic energy of particle
Then, U1→2 = T2 −T1
= ΔT or T2 = T1 + U1→2
Position2
Position1
€
v2
€
v1€
Ft
€
Fn
y
xO
P
s
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WorkDoneonParticlebyGravitationalForce
Note
€
U1→2isindependentofpathfrom1to2.Forthisreason
€
W iscalledaconservativeforce.
WorkDoneonParticlebyaLinearly‐ElasticSpringForce
Note
€
U1→2isindependentofpathfrom1to2.Forthisreason
€
Fsiscalledaconservativeforce.
€
U1→2 = − W dyh1
h2∫ = − Wh2 −Wh1( )
€
Let Vg = Wy = mgy = gravitational potential energy of particle
€
Then, U1→2 = − Vg( )2− Vg( )1[ ]
= −ΔVg
Letk=springconstantx=springelongation
€
Fs = k x = spring force Then,
€
U1→2 = − Fsx1
x2∫ dx = − kxdxx1
x2∫
= −12kx2
2 −12kx1
2
€
Let Ve =12
kx 2 = elastic potential energy
of particle
€
Then, U1→2 = − Ve( )2 − Ve( )1[ ] = −ΔVe
2
P
1
€
x2
€
x1
€
Fs = k x
Undeformedlengthofspring
Path
1
2
h1
h2
P
y
xO
€
F =W
g
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SummaryThework‐energyequationcannowbewrittenas:
€
U1→2 = ΔT + ΔVg + ΔVe where
€
U1→2istheworkdoneontheparticlebyforcesotherthangravitationalandspringforces.If
€
U1→2 aboveiszero,then:
€
T2 + Vg( )2 + Ve( )2 = T1 + Vg( )1 + Ve( )1Thisisthelawofconservationoftotalmechanicalenergy.
PowerandEfficiency
Poweristhetimerateofdoingworkbyaforceonaparticle.
€
Power = F ⋅ v Units:
€
SI( ) N ⋅m s = J s = W; USCS( ) hp = 550 ft ⋅ lbs
€
η =power outputpower input
= mechanical efficiency
KineticsofParticles:MomentumMethods
Defineangularmomentum
€
HO ofparticleaboutO:
€
HO = r ×mv
y
x
€
v
€
t1
€
t2
O
€
F
P
Path
€
r
RecallNewton’s2ndlaw:
€
F = ma =ddt
mv( )
where
€
F = resultant force
€
mv = linear momentum of particle
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Then,
€
˙ H O = r ×ma = r ×F = MO
or
€
MO = ˙ H O
€
where MO = sum of the moments about O of all forces acting on particle
EquationsofImpulseandMomentum
Whatisthecumulativeeffectofintegrating
€
F and
€
MO withrespecttotimeoveranintervalfrom
€
t1 to
€
t2?
€
Fdtt1
t2∫ = d mv( ) = mv2mv1
mv 2∫ −mv1
or
€
mv1 + Fdtt1
t2∫ = mv2
Graphicalinterpretation:
or
€
mvx( )1 + Fxt1
t2∫ dt = mvx( )2
€
mvy( )1 + Fyt1
t2∫ dt = mvy( )2
Units:
€
SI( ) kg ⋅ ms
= N ⋅ s; USCS( ) lb ⋅ s
Recall
€
MO =dHO
dt
€
MOt1
t2∫ dt = dHO = HO( )2HO( )1
HO( )2∫ − HO( )1
y
xO €
mv1
€
Fdtt1
t2∫
€
mv2
Initiallinear
momentum
Linearimpulse
Finallinear
momentum
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or
€
HO( )1 + MOt1
t2∫ dt = HO( )2
Units:
€
SI( ) kg ⋅ m2
s= N ⋅m ⋅ s; USCS( ) lb ⋅ ft ⋅ s
ExtensiontoSystemofnParticles
Let
€
mv∑ = mii=1
n
∑ vi
€
mv1∑ + Fdtt1
t2∫∑ = mv2∑
Note:Linearimpulsesfrominternalforcesofactionandreactioncancel.Ifnoexternalforcesactfromtime
€
t1to
€
t2,then
€
mv1 = mv2∑∑
andthetotallinearmomentumoftheparticlesisconserved.
Let
€
HO∑ = rii=1
n
∑ ×mivi
€
HO( )1∑ + MO dtt1
t2∫∑ = HO( )2∑
Sumofallinitiallinear
momenta
Sumofallfinallinear
momenta
Sumofalllinear
impulsesfromexternal
forces
Sumofallinitialangularmomenta
Sumofallfinal
angularmomenta
Sumofallangularimpulses
fromexternalforces
Initialangular
momentum
Angularimpulse
Finalangular
momentum
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Note:Angularimpulsesfrominternalforcesofactionandreactioncancel.Ifnoexternalforcesactfromtime
€
t1to
€
t2,then
€
HO( )1 = HO( )2∑∑
andthetotalangularmomentumoftheparticlesisconserved.Graphicalinterpretation:
DirectCentralImpactBeforeImpactAfter
Totallinearmomentumisconservedduringimpact:
€
m1v1 + m2v2 = m1 ′ v 1 + m2 ′ v 2
O O Ox x x
y y y
€
m1v1( )1
€
m2v2( )1
€
m3v3( )1
€
m1v1( )2
€
m2v2( )2
€
m3v3( )2
€
Fdtt1
t2∫∑
€
MO dtt1
t2∫∑
€
m1
€
m1
€
m1
€
m2
€
m2
€
m2
€
v1
€
v2
€
′ v 1
€
′ v 2
€
′ v 1 < ′ v 2
€
′ v 1 < ′ v 2
€
v1 > v2
u
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€
Coefficient of restitution : e =velocity of separationvelocity of approach
=′ v 2 − ′ v 1
v1 − v2
Iftotalkineticenergyisconserved,impactissaidtobeperfectlyelasticand
€
e =1.Ifparticlessticktogetherafterimpact,
€
′ v 1 = ′ v 2,impactissaidtobeperfectlyplastic,and
€
e = 0.Forallotherimpactcases,
€
0 ≤ e ≤1.Aspecialcaseoccurswhen
€
m1 = m2,collisioniselastic,
€
v1 > 0 ,and
€
v2 = 0.Then,
€
′ v 1 = 0 and
€
′ v 2 = v1.KinematicsofRigidBodies
Typesofplanemotion:Rectilineartranslation
CurvilineartranslationFixed‐axisrotation
€
A1
€
B1
€
A2
€
B2
€
A1
€
B1
€
A2
€
B2
€
A1
€
A2
€
θ
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GeneralplanemotionCombinationoftranslationandrotationTranslation
Recallanalysisof“MotionRelativetoTranslatingReferenceAxes”:
€
rA = rB + rA B
NowAandBareanytwoparticlesinthetranslatingrigidbody.Therefore,
€
rA B = constant vector ,and
€
vA = vBaA = aB
RotationAboutaFixedAxis
Recallanalysisof“AngularMotionofaLine”:
Then,thevelocityofparticlePisv = ω × r and the acceleration is a = α × r + ω × (ω × r ) = α × r - ω2r
€
A1
€
B1€
A2
€
B2
P
€
r
y
x
Fixedaxis€
θ €
ω =dθdt
= ˙ θ
α =dωdt
=d2θdt 2 = ˙ ̇ θ
Defineangularvelocityvectorω and angular acceleration vector α as follows: ω = ωk α = αk
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Note:Inrθcoordinates,
€
vr = 0
€
vθ =ωr
€
ar = −ω 2r
€
aθ =α r Intnaxes,
€
v =ωr
€
an =ω 2r
€
at =α r
General Plane Motion – Absolute and Relative Velocity and Acceleration
Graphicalinterpretation:
PlaneMotionTranslationwith
€
vB RotationaboutBwithω PlaneMotionTranslationwith
€
aB RotationaboutBwithω andα
Axesx–ytranslatewiththeiroriginattachedtoparticleB.rA = rB + rrel vA = vB + ω × rrel
aA = aB + α × rrel + ω × (ω × rrel )= aB + α × rrel − ω2rrel
Y
X€
rA
O
y
xB
A
€
rB €
rrel
α ω
ω × rrel
ω × rrel
€
vA
€
vA
€
vB
€
vB €
vB
€
vB
A A A
B B B
€
aA
€
aB
€
aB €
aB
€
aA
€
aB
α × rrel
α × rrel−ω2rrel
B B B
A A A
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InstantaneousCenterofRotationinPlaneMotion
Suppose
€
vB = 0 inthepreviousanalysis.Then,
vA=ω ×rrel.
ThisresultimpliesthebodyisrotatingforaninstantaboutpointB.Suchapointiscalledaninstantaneouscenterofrotation(I.C.R.).Suchapointcanbedetermined,asfollows,ifthevelocitiesoftwodifferentparticlesinabodyareknown.
Note:ThelocationoftheI.C.R.changeswithtimeingeneral.Hence,
€
a ICR ≠ 0 ingeneral!
PlaneMotionofaParticleRelativetoaRotatingFrame
Y
XO
x
y
B
A
€
rA
€
rB€
rrel
ω α Axesx–yarebody‐fixedaxes,whichhaveangularvelocityω andangularaccelerationα .ParticleAmovesrelativetothebody‐fixedaxesx–y.TherelativepositionvectorofAreferencedtothex–yaxesis
€
rrel = xi + yj
A€
vA
€
vB BI.C.R €
vA
€
vB
A
B I.C.R€
vA A
€
vB B
I.C.R
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TherelativevelocityofAwithrespecttothex–yaxesis:
€
vrel = ˙ x i + ˙ y jTherelativeaccelerationofAwithrespecttothex–yaxesis:
€
a rel = ˙ ̇ x i + ˙ ̇ y j
TheabsolutepositionvectorofAintheX–Yinertialaxesisgivenby:
€
rA = rB + rrel
TheabsolutevelocityofAintheX–Yinertialaxesisgivenby:
vA = vB + ω × rrel + vrelTheabsoluteaccelerationofAintheX–Yinertialaxesisgivenby:
aA = aB + α × rrel + ω × (ω × rrel ) + 2ω × vrel + arelTheterm2ω × vrel is known as Coriolis acceleration.
KineticsofRigidBodies:ForcesandAccelerationsEquationsofMotionforBodyinPlaneMotionFBDKD
€
F = ma c∑
€
Mc∑ = Icα or
€
M p∑ = Icα + ρ pc ×ma c
y y
x xO O
c c
p p€
F1
€
F2
€
F3
€
ma c
ρpcIcα
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where:m=totalmassc=centerofmassIc=massmomentofinertiaaboutaxisthroughcparalleltozaxisp=anymomentcenterinx–yplane
Incomponentform:
€
Fx∑ = macx
€
Fy∑ = macy
€
Mc∑ = Icα NoncentroidalRotationFBDKD
€
Mq∑ = Icα + ρqc ×ma ct
€
= Iqα where
€
Iq =massmomentofinertiaaboutaxisthroughqparalleltozaxis.LawsofFriction
N F
P
Wg
BlockisinitiallyatrestwhenforcePisappliedanditsmagnitudeisprogressivelyincreasedfromzero.Aslongas
€
P = F < µsN ,theblockwillnotslide.
€
µs = coefficientofstaticfriction.
c c
€
F1
€
Q
€
F2
Icα
ρqc
(bearingforce)
Centerof
rotation
y
xO
y
xO
q q€
ma ct
€
ma cn
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When
€
P = F = µsN ,theblockstartstoslide,andFbecomes:
€
F = µk N where
€
µk = coefficientofkineticfriction,
€
µk < µs .KineticsofRigidBodies:EnergyMethods
Forabodyinplanemotion,theworkdoneonthebodybyallexternalforces
€
Fiis
€
U1→2 = Firi( )1
ri( )2∫∑ ⋅ dri
whenthebodyisdisplacedfromposition1toposition2.Forabodyinplanemotion,thekineticenergyis
€
T =12mvc
2 +12Icω
2
Forabodyinplanemotion,theworkdoneonthebodybyacoupleMis
€
U1→2 = Mdθθ1
θ 2∫
whenthebodyisdisplacedfromposition1toposition2.
Ingeneral,
€
U1→2 =12m vc( )2
2+12Icω2
2 −12m vc( )1
2+12Icω1
2
€
= T2 −T1= ΔT
or
€
T2 = T1 +U1→2 IfagravitationalforceWactsonthebody,and/oralinearly‐elasticspringforce,thenthework‐energyequationcanbewrittenas:
€
U1→2 = ΔT + ΔVg + ΔVe
where
€
U1→2nowexcludesgravitationalandspringforces.If
€
U1→2 aboveiszero,totalmechanicalenergyisconserved.
Page 21
21
NoncentroidalRotation
€
T =12Iqω
2 whereqisthecenterofrotation.
PowerdevelopedbyacoupleM
€
Power = Mω
KineticsofRigidBodies:MomentumMethods
Forabodyinplanemotion,theequationsofimpulseandmomentumare;
€
m vc( )1 + Fit1
t2∫∑ dt = m vc( )2
€
Icω1 + Mct1
t2∫∑ dt = Icω 2
Graphicalinterpretation
If
€
Fit1
t2∫∑ dt = 0 ,then
€
m vc( )1 = m vc( )2 andwesaylinearmomentumisconserved.If
€
M pt1
t2∫∑ dt = 0 aboutsomepointp,then
€
Icω1 + ρ pc ×m vc( )1 = Icω 2 + ρ pc ×m vc( )2
andwesaytotalangularmomentumaboutpointpisconserved.
c c c€
m vc( )1
€
m vc( )2
€
F1dtt1
t2∫
€
F2dtt1
t2∫
€
F3dtt1
t2∫
€
Icω 2
€
Icω1
Initiallinearandangularmomenta
Sumofalllinearandangularimpulses(aboutc)fromexternalforces
Finallinearandangularmomenta
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Vibrations
Whenamassmovesbackandforthaboutanequilibriumposition,themotionisdescribedasvibration.Asimpleexampleofvibrationisthemotionofamassmconnectedtoamasslessspringwithspringconstantk.
MassmdisplacedfromitsequilibriumpositionFBDKD
€
Fx∑ = m ˙ ̇ x
€
m ˙ ̇ x + k x = 0 (1)(Equationofmotion)
Let
€
ωn2 =
km,thenEq.(1)becomes:
€
˙ ̇ x +ωn2x = 0(2)
g k
m
€
δst
x
Unstretchedpositionofspring
Staticequilibriumpositionofmass
Note:
€
kδst = mg
€
m ˙ ̇ x
€
mg
€
k x + δst( )
€
−k x + δst( ) + mg = m ˙ ̇ x
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23
ThesolutionofEq.(2)is
€
x = x 0( )cos ωnt( ) +˙ x 0( )ωn
sin ωnt( )
where
€
x 0( )and
€
˙ x 0( )areinitialconditions.
Themotioniscalledsimpleharmonicmotion,and
€
ωn =km
=gδst
isknown
asthenatural(circular)frequency(rad/s).Sincenoforcingfunctionappearsintheequationofmotion(1),thevibrationaboveiscalledfreevibration.Foranexampleoftorsionalvibration,seep.78inthe9.4Handbook.
x
O t
€
τ n =2πωn
,calledperiod
€
scycle