1

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.

2

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

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

4

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

5

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

6

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,

7

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

8

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

9

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

10

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

11

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

12

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

13

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

14

€

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

15

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

16

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

17

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

18

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

19

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

20

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.

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

22

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

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