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ME 274 – Spring 2021 – Krousgrill sections Exam 3 – Wednesday, April 14

8:00-9:30 PM (alternate time: 8:00-9:30 AM) Topicscovered:

• Work/energyequationforparticles(Section4.Bofthelecturebook)• Linearimpulse/momentumequationforparticles(Section4.Cofthelecture

book)• Impactofparticles(Section4.Cofthelecturebook)• Angularimpulse/momentumequationforparticles(Section4.Dofthe

lecturebook)• Newton/Eulerequationsforrigidbodies(Section5.Aofthelecturebook)• Work/energyequationforrigidbodies(Section5.Bofthelecturebook)

Preparation:Itisveryimportantthatyoureviewthefundamentalsonthesetopicspriortoworkinglotsofexamples.Attachedarethelecturesummarysheetsforthetopicscoveredonthisexam.Goovertheseindetail,andmakenoteofthingsthatarenotclear.Contactyourinstructorforhelponthosetopics.Onceyouhavereviewedthefundamentals,thenusethelearningresourcesavailabletoyou:

• Solutionvideosforallexampleproblemsfromthelecturebookarefoundontheblog.

• Solutionvideosforallhomeworkproblemsarefoundontheblog.

• Solutionsforallquizzesgiventhissemesterarefoundontheblog.

• Hintsforallconceptualquestionsarefoundontheblog.

• Pre-lectureandlecturevideosforallsections(includingProfessorDavies’andGibert’ssections)arefoundontheblog.

• TheWeeklyJoyswebsitecanprovideyouwithalargenumberofexams(w/solutions)frompastsemesters.Usetheselast,onlyafteryouhavereviewedallothermaterial.

• Recommendation:Holdbackoneortwosampleexams.Taketheseexamsunderthepressureoftimelimits.Quitattheendof90minutes,andreviewyourwork.Takenoteofproblemsthattookyoualongtimetothinkthrough.Identifyanyweaknessesyouhaveatthispoint.Donotwaituntiltheexamtofindoutwhereyoumayneedsomehelp.

• Contactyourinstructorifyouneedhelp.

Kinetics: Four-Step Problem Solving Method

The suggested plan of action for solving kinetics problems:

1. Free body diagram(s). Draw appropriate free body diagrams (FBDs) for the problem. Yourchoice of FBDs is problem dependent. For some problems, you will draw an FBD for eachbody; for others, you will draw an FBD for the entire system. An integral part of your FBDsis your choice of coordinate systems. For each FBD, draw the unit vectors corresponding toyour coordinate choice.

2. Kinetics equations. At this point, you will need to choose what solution method(s) thatyou will need to use for the particular problem at hand. In this section of the course wewill study four basic methods: Newton/Euler, work/energy, linear impulse/momentum andangular impulse/momentum. Based on your choice of method(s), write down the appropriateequations from your FBD(s) from Step 1.

3. Kinematics. Perform the needed kinematic analysis. A study of the equations in Step 2above will guide you in deciding what kinematics are needed to find a solution to the problem.

4. Solve . Count the number of unknowns and the number of equations from above. If you donot have enough equations to solve for your unknowns, then you either: (i) need to drawmore FBDs, OR (ii) need to do more kinematic analysis. When you have su�cient equationsfor the number of unknowns, solve for the desired unknowns from the above equations.

Method Body model Fundamental equations

Newton-Euler (relating forces to accelerations)

particle

rigid body (G = c.m. and A = any point on body)

Work-energy (relating change in speed to change in position)

particle

rigid body (G = c.m. and A = any point on body)

Linear impulse-momentum (relating change in velocity to change in time)

particle

rigid body (G = c.m.)

Angular impulse-momentum (relating change in angular velocity to change in time)

particle (O = fixed point)

rigid body (A = fixed point or c.m.)

!F! = m!a

!F! = m!aG

!

M A! = I A!" + m!rG / A # !aA

T1 +V1 +U1!2nc( ) = T2 +V2

where T =

12

mv2

T1 +V1 +U1!2nc( ) = T2 +V2

where T =

12

mvA2 +

12

I A!2 + m!vAi

!! " !rG / A( )

!F! dt

t1

t2

" = m!v2 # m!v1

!M A! dt

t1

t2

" =!H A2 #

!H A1

where!H A = I A

!!

!MO! dt

t1

t2

" =!

HO2 #!

HO1

where!HO = m!rP /O ! !vP

!F! dt

t1

t2

" = m!vG2 # m!vG1

Kinetics Table

FUNDAMENTALequation:thework-energyequation T1 +V1 +U1→2

nc( ) = T2 +V2

Summary:Work-EnergyEquation1

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where:

T = 1

2mv2 ≥ 0( ) = kinetic energy

U1→2nc( ) =

!F∑( ) i et ds

1

2

∫ = work done by non− conservative forces

=!F∑( )x

dx +!F∑( )y

dy⎡⎣⎢

⎤⎦⎥

1

2

∫

!F1

!F2

!F3

P et

x

pathofP

y

en

s

V =Vgr +Vsp = potential energy

Vgr = mgh sign depends on h( )

Vsp = 1

2kΔ2 ≥ 0 ALWAYS( )

SIGNofworkterm:Thedirectionofaforcerelativetothedictatesthesignofthework,U.

mg

P pathofP

datumline

h

!F : U1→2

F( ) > 0P

et pathofP

en

!F : U1→2

F( ) = 0

!F : U1→2

F( ) < 0 et

FUNDAMENTALequation:thework-energyequation

T1 +V1 +U1→2nc( ) = T2 +V2

Summary:Work-EnergyEquation2

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!F1

!F2

!F3

P et

x

pathofP

y

en

s

SYSTEMCHOICE:Makeyourchoiceofsystemas“large”asreasonable–youwanttomakeworklessforcesINTERNALtothesystem.

SOLUTIONPROCESS:1.  Drawfreebodydiagram(FBD)forsystemofyourchoice(seecommentbelowonsystem

choice).

2.  Writedownthework-energyequation.

3.  Writedowntheappropriatekinematics(velocity)equationsfortheproblem.

4.  Ifyouhaveenoughequations,solveforthedesiredunknowns.Ifyoudonothaveenoughequations,thenyouhaveprobablymissedsomeinformationfromkinematics.

CONSERVATION:Ifnoworkisdoneonthesystem,,thenenergyisconserved.

U1→2nc( ) = 0

FUNDAMENTALequation:thelinearimpulse-momentumequation:

m!v2 = m!v1 +

!F∑( ) dt

1

2

∫

Summary:LinearImpulse-MomentumEquation1

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!F1

!F2

!F3

P

x

y

SYSTEMCHOICE:Makeyourchoiceofsystemas“large”asreasonable–youwanttomakeasmanyforcesaspossibleINTERNALtothesystem.

CONSERVATION:Ifthereisnonetforceactingofthesysteminagiven

direction(sayx),,thenlinearmomentuminthatdirection

isconserved.

!F∑( )x

dt1

2

∫ = 0

FUNDAMENTALequation:thelinearimpulse-momentumequation:

m!v2 = m!v1 +

!F∑( ) dt

1

2

∫

Summary:LinearImpulse-MomentumEquation2

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!F1

!F2

!F3

P

x

y

SOLUTIONPROCESS:1.  Drawfreebodydiagram(FBD)forsystemofyourchoice.Asmentionedbefore,make

yoursystemas“large”aspossible.2.  Writedowntheimpulse-momentumequation.3.  Writedowntheappropriatekinematics(velocity)equationsfortheproblem.4.  Ifyouhaveenoughequations,solveforthedesiredunknowns.Ifyoudonothave

enoughequations,thenyouhaveprobablymissedsomeinformationfromkinematics.

FUNDAMENTALequations:thelinearimpulse-momentumequationsandcoefficientofrestitution(COR)equation

Summary:CentralImpactProblems

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n

A

t

B

mA mB

!vA2

!vA1

n

A

t

B

FAB

n

A

t

B

FAB

A : Ft∑ = 0 ⇒ mAvA1t = mAvA2t ⇒ vA2t = vA1t

B : Ft∑ = 0 ⇒ mBvB1t = mBvB2t ⇒ vB2t = vB1t

A+ B : Fn∑ = 0 ⇒ mAvA1n + mBvB1n = mAvA2n + mBvB2n

COR : e = −vB2n − vA2nvB1n − vA1n

⎛

⎝⎜⎞

⎠⎟

COMMENTS:•  TheCORequationisvalidforONLYthen-componentsofvelocity.

•  EnergyisNOTconservedduringimpact,exceptfore=1.

FUNDAMENTALequation:

COMPUTINGangularmomentum:

NOTE:OmustbeaFIXEDpoint!Why?

!HO( )2 =

!HO( )1 +

!MO dt∑

1

2

∫

!HO = m!rP/O × !vP ; general

= m xi + y j( )× "xi + "y j( ) = m x"y − y "x( ) k ; Cartesian

= m rer( )× "rer + r "θ eθ( ) = mr2 "θ k ; polar

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pathofP

eR

R

O

eθ

θ

P

x

y

Summary:AngularImpulse/MomentumEquation1

Summary:Angularimpulse/momentumequation2

WHENshouldIusethisequation?Thinkcentral-forceproblems… When,angularmomentumaboutOisconserved.

FUNDAMENTALequation:

!HO( )2 =

!HO( )1 +

!MO dt∑

1

2

∫

!

MO∑ =!0

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pathofP

eR

R

O

eθ

θ

P

x

y

IMPORTANT:ThisequationcanNOTgiveinformationontheradialcomponentofvelocityfortheparticle.Why?Whyisthisimportant?Lookattheaboveequationforcomputingangularmomentum.Typically,usework/energyfortheadditionalequation.

eR

O

eθ

θ

P !F

centralforceproblem:forceFactsdirectlytowardpointO

whereOisaFIXEDpoint.

Summary:Newton/EulerEquations1

FUNDAMENTALequations:

CRITICALISSUES:•  ForNEWTON(1):Gmustbethecenterofmassofthebody

•  ForEULER(2):AisANYpointonthebody.Thesamepoint“A”mustbeusedacrosstheboardintheequation–youcannotmixandmatchpointsA.

SIMPLIFICATION:IfAis:EITHERthecenterofmassGORafixedpoint(zeroacceleration)ORisparallelto,thentheEulerequation(2)reducesto:

TERMINOLOGY:IAisknownasthe“massmomentofinertia”ofthebodyaboutpointA.ThesizeofIAisdependentonthelocationofA.

1( ) !F∑ = m!aG

2( ) !M A∑ = I A

!α + m!rG / A × !aA

!

M A∑ = I A!α

SAMEpoint“A”!

!aA

!rG / A

Wewillusethisformoftheequationmostofthetime

G

!F1

!F2

!F3

A



MASSMOMENTOFINERTIA:,withrmeasuredfrompointA

•  circulardiskofradiusR,massmandc.m.atG:

•  thinbaroflengthL,massmandc.m.atG:

PARALLELAXESTHEOREM:

•  YouMUSThaveIGontheright-handsideoftheP.A.T.equation.

•  Themassmomentofinertiaaboutthec.m.isthesmallestforall

pointsonthebody.

•  Whendoyouneedtousethis?Usefulwhenanon-centroidalpointisusedinEuler’sequation.

RADIUSOFGYRATION:.Dependentonlyonthe“shape”,NOTonthemass.

1( ) !F∑ = m!aG

2( ) !M A∑ = I A

!α ;A=c.m.ORfixedpointOR

!rG / A "!aA

I A = r2 dm∫

IG = mR2 / 2

IG = mL2 / 12

I A = IG + md2

kA ! I A / m



SOLUTIONMETHOD:thefour-stepplan

1( ) !F∑ = m!aG

2( ) !M A∑ = I A

!α ;A=c.m.ORfixedpointOR

!rG / A "!aA

SAMEpoint“A”!







DrawINDIVIDUALfree-bodydiagrams

forNewton/Euler.

Besuretousethecorrectmassmomentof

inertiaforyourchoiceofpoint“A”.UsePATifnecessary.

Typicallythemostdifficultstep.RecalltherigidbodykinematicsfromChapter2.

Ifyouareshortequations,gobacktoStep3–Kinematics.

G

!F1

!F2

!F3

A

Summary:Work/EnergyEquation1

FUNDAMENTALequation:with:

SPECIALCHOICESFORPOINT“A”:IfAisEITHERthec.m.ORafixedpoint,thenthekineticenergyequationreducesto:

PARALLELAXISTHEOREM:AswiththeNewton/Eulerequation,youwillneedtousethePATifyouchooseAtobeanythingotherthanthec.m.

SYSTEMCHOICE:MakeyoursystemBIG!IncludeasmanycomponentswithinsystemtomakeworklessforcesINTERNAL(noworkonsystem).DifferentchoicethanforNewton/Euler.

T2 +V2 = T1 +V1 +U1→2nc( )

T = 1

2mvA

2 + 12

I Aω2 + m!vA i

!ω × !rG / A( )

T = 1

2mvA

2 + 12

I Aω2

rotationtranslation

G

!F1

!F2

!F3

A

ROLLINGWITHOUTSLIPPING:Thefrictionforceatano-slippointdoesnotwork.Why?Recallthattheno-slippointisstationary–noworkisdoneonastationarypoint.

Summary:Work/EnergyEquation2

FUNDAMENTALequations:with:withA=EITHERc.m.ORfixedpointSOLUTIONMETHOD:thefour-stepplan







Drawfree-bodydiagramofentire

systemforwork/energy.

Besuretousethecorrectmassmomentof

inertiaforyourchoiceofpoint“A”.Use

PATifnecessary.

Typicallythemostdifficultstep.Recallthe

rigidbodykinematicsfromChapter2.

Ifyouareshortequations,goback

toStep3–Kinematics.

T2 +V2 = T1 +V1 +U1→2nc( )

T = 1

2mvA

2 + 12

I Aω2

G

!F1

!F2

!F3

A