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1|Introduction
37
byanelectriceldbehaves sodierentlyfromtheeectby agravitationaleld? ? Therearelogically
twodierenttypesofmass.Oneisinertialmass,describingtheresistancetoacceleration;theotheris
gravitationalmass,describingthecouplingtothegravitationaleld.
~a
=
~
F
total
=m
i
;
and
~
F
gravitational
=
m
g
~g
Whosaysthatthesetwomassesarethesame?
Itis an experimental question. . Are e they? ? The e question was rstasked by Newton, but the
real creditgoes toEotvos. . In n thelate1800’s and continuingintothe rstdecade ofthe1900’s,he
carriedoutsomemiraculouslypreciseexperimentstomeasuretheratioofthetwomasses,
m
i
=m
g
. Is
8
. Later
experiments by Dicke improved this toone part in n 10
11
1.3ConservationLaws
Classical mechanics is not just about forces s and accelerations. . Even n if f the e problems you examine
to approach those e problems, , and d the conservation laws s are at t least as s fundamental l as
~
F
=
m~a
.
Conservationofenergy,ofmomentum,ofangularmomentumcanallbederivedfromthatequation,
butdid you knowthatyou can derive
~
F
=
m~a
from energy? ? If f so,then which of the two is more
fundamental.
Work,Energy
Inonedimensionwithpointmasses,thework-energytheoremisverysimpleandisnotatalldicult
toderivefromNewton’sequations. Thereareacoupleofwaystodoso,oneinvolvesthefullpowerof
calculusmanipulationsandafewlinesofalgebra;theotherismoreintuitive,butmoretedious.
Startwiththerstway(andifyouhaven’treviewedsection0.5thengobackanddoitnow).
F
x
=
ma
x
=
m
dv
x
dt
=
m
dv
x
dx
dx
dt
=
mv
x
dv
x
dx
(1
:
1)
Thisusedthechainrulefordierentiationandthen itused the denition ofvelocity. . Nowintegrate
thisequationwithrespectto
x
betweensomespeciedinitialandnallimits.
W
=
Z
x
f
x
i
F
x
dx
=
Z
x
f
x
i
mv
x
dv
x
dx
dx
=
Z
x
=
x
f
x
=
x
i
mv
x
dv
x
=
m
2
v
2
x
x
=
x
f
x
=
x
i
=
m
2
v
2
f
m
2
v
2
i
=
K
(1
:
2)
Striptheinterveningmaterialfromthisandyouhavethework-energytheorem:
W
=
Z
x
f
x
i
F
x
dx
=
m
2
v
2
f
m
2
v
2
i
=
K
(1
:
3)
This took two o (long) lines, but you need to have a a complete understanding of f the e chain rule e for
dierentiation and d of f the methods s to o change variables s in n an integral. . There e are two parts to this
derivation:First,knowinghowtodothemanipulations.Second,understandingwhatthemanipulations
meanandwhytheyarevalid.
Asecondwaytoreachthisresultislongerbutsimpler. Itdoesn’tinvolveany y specialcalculus
tricks,onlythedenitionofanintegral. Itstartswiththespecialcaseofaconstantforce,thensince
a
x
=
F
x
=m
theaccelerationisaconstant. That’safamiliarcase:
a
x
=constant  !
v
x
=
a
x
t
+
C;
then
x
=
1
2
a
x
t
2
+
Ct
+
D
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1|Introduction
38
Usetheinitialconditionsthatat
t
=0positionis
x
=
x
0
andvelocityis
v
x
=
v
0
.
v
0
=
a
x
.
0+
C;
x
0
=
1
2
a
x
.
0
2
+
C
.
0+
D
so
v
x
=
a
x
t
+
v
0
;
and
x
=
1
2
a
x
t
2
+
v
0
t
+
x
0
Eliminatethevariable
t
betweenthelasttwoequations
t
=
v
x
v
0
=a
x
;
so
x
=
1
2
a
x
v
x
v
0
2
=a
2
x
+
v
0
v
x
v
0
=a
x
+
x
0
Rearrangethelastresult,simplifyingthealgebratoget
x
=
1
2
v
2
x
v
2
0
=a
x
+
x
0
or
ma
x
(
x
x
0
)=
1
2
mv
2
x
1
2
mv
2
0
Allofthisrearrangementwasjustelementaryalgebra,andthereasonforthenalmanipulation,mul-
tiplyingby
m
wastogetthecombination
ma
x
intherstterm. Thatbecomes
F
x
.
(
x
x
0
)=
1
2
mv
2
x
1
2
mv
2
0
(1
:
4)
Nowfortherealcase. Constantforcesaretextbookidealizationsoftherealworld. Atbestyou
doincalculusandthensneakupontheresult. Approximateaposition-dependentforceasasequence
ofsteps.
F
x
(
x
)=
8
>
>
>
>
>
<
>
>
>
>
>
:
F
1
(
x
0
<x<x
1
)
F
2
(
x
1
<x<x
2
)
F
3
(
x
2
<x<x
3
)
F
4
(
x
3
<x<x
4
)
:::
F
N
(
x
N
1
<x<x
N
)
x
0
x
1
x
2

Fig.1.2
Ineachoftheseintervals,theequation(1.4)applies.Ateachpoint
x
0
,
x
1
,
:::
thespeedis
v
0
,
v
1
,etc.
F
1
(
x
1
x
0
)=
1
2
mv
2
1
1
2
mv
2
0
F
2
(
x
2
x
1
)=
1
2
mv
2
2
1
2
mv
2
1
F
3
(
x
3
x
2
)=
1
2
mv
2
3
1
2
mv
2
2
F
4
(
x
4
x
3
)=
1
2
mv
2
4
1
2
mv
2
3
F
5
(
x
5
x
4
)=
1
2
mv
2
5
1
2
mv
2
4

F
N
(
x
N
x
N
1
)=
1
2
mv
2
N
1
2
mv
2
N
1
Thesumofalltheseequationsis
F
1
(
x
1
x
0
)+
F
2
(
x
2
x
1
)+
F
3
(
x
3
x
2
)++
F
N
(
x
N
x
N
1
)=
1
2
mv
2
N
1
2
mv
2
0
becausethetermsontherighttelescope: Allthetermsexcepttherstandthelastcancelinpairs. In
conventionalnotationthisis
N
X
k
=1
F
k
x
k
=
N
X
k
=1
F
x
(
x
k
)
x
k
=
1
2
mv
2
f
1
2
mv
2
i
where
x
k
=
x
k
x
k
1
x
N
=
x
f
x
0
=
x
i
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1|Introduction
39
Thelimitofthisequationas
x
k
!0isjustthedenitionoftheword\integral",sothisreproduces
Eq.(1.3).
These twoderivations of the sameequation,
W
= 
K
, lookcompletely y dierent, , but they
aren’t. When n you lookclosely y at t them and compare them step by step they are much h more e alike
thanyourstthink. Thesecond d derivationtellsyouthatthecombinationofforcetimesdistanceis
important|especiallyforce timesalittlebit ofdistance:
F
x
dx
. Writedownthis s combinationand
manipulateit.
F
x
dx
=
ma
x
dx
=
m
dv
x
dt
dx
=
m
dv
x
dt
dx
=
mdv
x
dx
dt
=
mv
x
dv
x
Howeveryoudoit,yougetthework-energyrelationstatingthattheworkonapointmass(
R
F
x
dx
)
equalsthechangeinitskineticenergy.
Whathappensifyoudon’thaveapointmass? Whatifyouaren’toperatinginonedimension?
inchapters8,4,and2respectively.
MechanicalEnergy
Forthissamesimplecaseofapointmassinonedimension,andif theforceisafunctionofposition
only,youcanrearrangethework-energytheoremtogetaconservationlaw.
Theworkisanintegralof
F
x
(
x
)
dx
. Whenyouevaluateintegrals,themostcommonwayyou
useistond an anti-derivative and then toevaluateitatthe endpointsofthe integrationinterval,
thefundamentaltheoremofcalculus: If
f
hasanantiderivativeandif
f
isintegrablethenthetwoare
related.
If
f
(
x
)=
dF
(
x
)
=dx
then
Z
b
a
f
(
x
)
dx
=
F
(
b
F
(
a
)
(1
:
5)
In the present case the function you’re integrating is
F
x
(
x
) and Iwill denote its anti-derivative by
U
(
x
).Thatis,
F
x
(
x
)=
dU
(
x
)
dx
;
then
Z
x
f
x
i
F
x
(
x
)
dx
U
(
x
f
)+
U
(
x
i
)
(1
:
6)
Nowapplythistothework-energytheorem,Eq.(1.3).
W
=
K
is
Z
x
f
x
i
F
x
(
x
)
dx
U
(
x
f
)+
U
(
x
i
)=
m
2
v
2
f
m
2
v
2
i
whichis
m
2
v
2
i
+
U
(
x
i
)=
m
2
v
2
f
+
U
(
x
f
)
(1
:
7)
Thisis the reasonfortheminussignin the denition of
U
inEq.(1.6). Ifyou u don’tputtheminus
signthereyouwouldnotgetaplussignhere,inthisconservationofenergyequation.
U
iscalledthepotentialenergycorrespondingtotheforce
F
x
(
x
),andthisequationsays that
thesumofthekineticandthepotentialenergieshasthesamevaluesthroughoutthemotion;thesum
isconserved.
Can’tyoualwaysdothismanipulationtogetconservationofmechanicalenergyoutofthework-
energy theorem? ? Can’tyou u always nd an anti-derivative (maybe in abig table of integrals)? ? For
F
fr
k
F
N
.
Thefrictional forceasanobjectslidesoverasurfaceisthecoecientof(kinetic)friction timesthe
normalcomponentoftheforceon the surface. . It’sjustaconstant,soyou u cancertainly integratea
constantandgetapotentialenergy
U
andthengetconservationofmechanicalenergy.
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1|Introduction
40
No.
Inthisexampleofdryfriction,thefrictionalforceisvelocitydependenteventhoughitdoesn’t
looklike it|there’snovelocityinthe expression
k
F
N
. True,but t that’s becausethisexpression
is wrong. . Thefrictional l force is velocity dependentbecause itisalways opposite the velocity. . This
commonlyusedexpressionfortheforceissimplynotright. Amorecorrectwaytowriteitis
~
F
fr
k
F
N
^
v
(1
:
8)
where^
v
=
~v=v
istheunitvectorinthedirectionofthevelocityoftheslidingmass. Ifitslidesinthe
reversedirection,theforcefromfrictionisreversedandyoucanhavetwooppositevaluesoftheforce
atthesamevalueoftheposition. Itisnotafunctionof
x
aloneandno
U
exists. Whydon’tpeople
Notice that I I keep using the clumsy y phrase \conservation n of mechanical energy" instead d of
\conservation of energy". . Why? ? That’s s tied in to the factthat conservation of energy was one of
themostdicultlaws to sortout. . Newtonmay y have written hisbasicequations in thelate1600s,
butconservationofenergydidnotbecomeanacceptedlawofphysicsuntilthemiddle1800s. These
two expressions,kineticenergy and potential energy,seem easy to derive now, buthistorically even
thesewerediscovered through tortuousand tortious routes.* * Theproblemisthatenergy is s notjust
mechanicalenergy.
Energy is s an n abstraction, a a bookkeeping device much h like money. . It t is s aprescription n saying
confusionaroseintryingtogureoutwhattermstoincludeinthis sum. . Itwas s resolvedwhenheat
wasseentobeanotherformofenergy(aterminthesum). Thenlight,sound,chemicalenergy,and
DidIjustsaythatmoneyisanabstraction? Isn’titjustthebillsandcoinsthatyoucarrywith
coinsarenowherenearconserved. Ifyouincludewhat’sinyourcheckingaccount,that’snotbillsand
coinsstoredinavault. Itisasetofbitsstoredonacomputerdisk. Asavingsaccountismoreofthe
same. Thenthere e are whatevermysteriousmanipulationsthe FederalReserveBoardcanimplement.
The total \money"isacombination ofmany dierentthings,ofwhich justatinyamountisinany
M
1
,
:::
,
M
5
).Thedierencewiththeabstractioncalledenergyisthatwethinkweknowwhatgoesintoit.Or
dowe?Cosmologistsweresurprisedbythediscoverythattheexpansionoftheuniverseisaccelerating,
causedbysomethingcalled\darkenergy"forlackofabettername. Nooneknowswhatitis.
Derive
F
=
ma
AfewparagraphsbackIsaidthatyoucanderive
F
=
ma
fromenergy. How? Takeconservationof
mechanicalenergyanddierentiateit:
\
E
=
1
2
mv
2
x
+
U
(
x
) isconserved"
means
dE
dt
=0
d
dt
1
2
mv
2
x
+
U
(
x
)
=
1
2
m
2
v
x
dv
x
dt
+
dU
dx
dx
dt
=
v
x
m
dv
x
dt
+
dU
dx
=0
or,
m
dv
x
dt
dU
dx
* butnottorturous
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1|Introduction
41
andthatis
F
x
=
ma
x
. This s isnotjustaprettytheorem,itis sometimesausefulwaytoderivethe
andI’llregularlyusethismethodlaterinthetext.
Momentum
ThereareotherwaystomanipulateNewton’sequation. Therstandleastinterestingwayistostart
fromNewton’sequationforasinglepointmassandintegrateitwithrespecttotime.
~
F
=
m~a
!
Z
t
2
t
1
~
Fdt
=
Z
t
2
t
1
=
m~v
(
t
2
m~v
(
t
1
)
Whenyouhavetwoormoreinteractingmasses,yougetsomethingmoreuseful
~
F
on1by2
=
m
1
~a
1
;
~
F
on2by1
=
m
2
~a
2
(1
:
9)
0=
m
1
d~v
1
dt
+
m
2
d~v
2
dt
=
d
dt
m
1
~v
1
+
m
2
~v
2
(1
:
10)
Thisisconservationofmomentum.Itdoesn’tmatterhowcomplicatedtotwoforcesareaslongasthe
thirdlawissatised.
Ifyouhavethreeorthreemillionparticlestheresultisthesame,onlythenotationchanges.Put
indices
i
and
j
forcesfromalltheother particles
m
i
~a
i
=
X
j
6=
i
~
F
on
i
by
j
i
.
X
i
m
i
~a
i
=
X
i
X
j
6=
i
~
F
on
i
by
j
=0
Writethisoutforthreemasses! Really
(1
:
11)
Becauseaccelerationisthetime-derivativeofvelocity,thisequationsaysthat
d
dt
X
i
m
i
~v
i
=0
;
or
X
i
m
i
~v
i
=aconstant
(1
:
12)
CanyoureallygofromEq.(1.9)to(1.10)?Whatifmassisn’tconstant?Then
~
F
6=
m~a
anyway
andyoushouldhavebeenusing
~
F
=
d~p=dt
. Thatalsomakesthewholeprocessmuchmorenatural.
(Gobackanddoitthatway. It’seasy.)
AngularMomentum
Anothermanipulationof Newton’sequationusesthecrossproduct. . Pickan n originandlet
~r
bethe
coordinatevectorofasinglepointmassfromthatorigin.Noticethedierencehere:Thisresultdepends
onhavingchosenanorigin.Itdoesn’tmatterwhichoriginyoupick,butyouhavetopickone.
~r
~
F
=
~r
d~p
dt
=
~r
d~p
dt
+
d~r
dt
~p
=
d
dt
~r
~p
(1
:
13)
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1|Introduction
42
itissimply
~v
~p
=
~v
m~v
=0. Thenaltermistheordinaryproductruleforderivatives,andthe
termsinthisequationgetnames,torqueandangularmomentum.
~
=
d
~
L
dt
is
torque=time-derivativeofangularmomentum
(1
:
14)
~
L
=
~r
1
~p
1
+
~r
2
~p
2
+
(1
:
15)
Again, when n you u have e many particles, , you u put t indices s on n the e masses and sum m over r all l the
particles,butunlikethecaseoflinearmomentumtheresultsaresucientlycomplexthattheydeserve
acompletechapter(eight)tothemselves.
Intheabsence ofexternalforces,you havethe conservationlaw:
~
L
total
=constant. Howto
derivethis? Theforceonasingleparticle,the
i
th
one,is
~
F
i
=
F
i;
external
+
X
j
6=
i
~
F
on
i
by
j
Thisrepresentsthetotalforceonasingleparticleas causedby everythingelse intheuniverse. . The
index
j
isforalltheotherparticlesinthebody.Inthepresentcase,thereisnoexternalforce,and
~
=
X
i
~r
i
~
F
i
=
X
i
~r
i
X
j
6=
i
~
F
on
i
by
j
Lookatoneparticularpairoftermsin thissum,theonesforwhich
i
=1
;j
=2and
i
=2
;j
=1.
Thoseare
~r
1
~
F
on1by2
+
~r
2
~
F
on2by1
Thesetwoforcesareopposite,sothisis
~r
1
~
F
on1by2
~r
2
~
F
on1by2
=
~r
1
~r
2
~
F
on1by2
Iftheforcesactingbetweenthesemassesactalongthelinebetweenthem thenthiscrossproductis
zero.Thesamewillapplytoallpairsofforces.Iftheforceisnotalignedthisway,thenthissumisnot
zero,andtheexpressionforangularmomentumpresentedhereisnotconserved. Seeproblem4.47for
anexampleofthis,thoughtheresolutionoftheproblemisleftforelsewhere.
Mass
Massisconserved. Iftwoobjectscollideandtheyhavemasses
m
1
and
m
2
,thenafterthecollisionthe
totalmasswillbethesame.
m
1
+
m
2
=
m
3
+
m
4
Thisdoesn’tmeanthat
m
3
=
m
1
and
m
4
=
m
2
. Masscanbemovedfromoneobjecttotheother,
perhapsbychippingoorbecausealoosepartofonemassbecomesattachedtotheother.
Youcanseewhythisistrueifyouassumethatmomentumconservationisvalid.Ifyoumeasure
thevelocityofsomemasstobe
~v
andafriendofyoursismovingbyatvelocity
~u
thenyourfriendwill
concludethatthevelocityofthemassis
~v
~u
. Ifmomentumconservationholdsforyouthenitshould
holdforyourfriend. Thetwoequationsare
m
1
~v
1
+
m
2
~v
2
=
m
3
~v
3
+
m
4
~v
4
and
m
1
(
~v
1
~u
)+
m
2
(
~v
2
~u
)=
m
3
(
~v
3
~u
)+
m
4
(
~v
4
~u
)
1|Introduction
43
Subtracttheseandyouhave(forall
~u
)
m
1
~u
+
m
2
~u
=
m
3
~u
+
m
4
~u
=)
m
1
+
m
2
=
m
3
+
m
4
(1
:
16)
Theideahereappearsagaininmuchmoredetailinchapternine,especiallysection9.11. There,inthe
contextofspecialrelativityyouwillseethatmassconservationisnotquiterightafterall.
Thisderivationshowedthatifyoustartfrommomentumconservationandlookatitfromanother
pointofview,yougetaveryinterestingresult. Whatifkineticenergyisconserved,acompletelyelastic
1
2
m
1
v
2
1
+
1
2
m
2
v
2
2
=
1
2
m
3
v
2
3
+
1
2
m
4
v
2
4
=)
1
2
m
1
(
~v
1
~u
)
2
+
1
2
m
2
(
~v
2
~u
)
2
=
1
2
m
3
(
~v
3
~u
)
2
+
1
2
m
4
(
~v
4
~u
)
2
Ifthisistrueforallvaluesof
~u
,thenthecoecientsof
u
2
andof
~u
itselfmustagree.Theseare
m
1
+
m
2
=
m
3
+
m
4
and
m
1
~v
1
+
m
2
~v
2
=
m
3
~v
3
+
m
4
~v
4
Thissaysthatonceyou’veassumedthatyouhaveacompletelyelasticcollision,youautomaticallyget
conservationofmomentumandofmass. Thatis\automatically"ifyouassumethatdierentlymoving
observerswillallhave the samebasicequationsformechanics. . Galileowenttogreatpainstoargue
whythatistrue. Lookonpage294foraquotefromhistextonthesubject.
1.4TheTools
Ismechanicsacollectionoftrickstosolvevariouslycontrivedproblemsorisitasystematicapproach
to analyzingcomplex systems? ? I I think it is s thelatter,butit often appears more like the formerin
introductorytexts. Thereare ahandfulofrulesthat,systematicallyfollowed,willallowyoutosetup
evenvery complicated problems. . The e resulting equations can stillbe hard to solve, butthat’s only
becauseofthemathematics,notbecauseofthephysics.
Thesearemechanicalsystemsthatwe’redealingwithhere,andthatlimitsthesortofthingsyou
havetodealwith. Iwilllayoutasetofrulesandthenshowby y examplejustwhattheymean. . The
wholepointisthateveryproblemisattackedthesameway. Youdon’tlearnonemethodnowanda
dierentmethod later. . Youwillnotbelieveme e whenIsaythis,butfollowingthesesystematicrules
willsaveyouahugeamountoftime.
anideaofwhatishappening.
2. There e are typically several masses in eachproblem, so for r eachone e of them listin
ordinarylanguagethethingsthatareactingonit. Aforceisnotathing;anacceleration
isnotathing;evenan
m~a
isnotathing. Atableisathing;ahandisathing;aropeis
athing;theEarthisathing. Remember: Hereathingcanactonamassonlybyeither
beingincontactwiththemassorbyitsgravitationalpull.
3. Foreachmass,godownthelistofthings s acting onitand drawthatmasstogether
anaccelerationvectorforeachmass. Labeleachmassandeachvector.Usesymbolsthat
conveysomemeaning.
4.Pickabasisintermsofwhichyoucanwritethevectors.Notethatyouarenotrequired
tousethesamebasisasyouturnyourattentionfromonemasstoanother. Youcanthink
1|Introduction
44
ofthemasseparateproblems,andthemathematicswillunitethemforyou. Labelyour
coordinates.
5. Foreachmasswrite
~
F
=
m~a
in thebasischosen. . Or,
~
F
=
d~p=dt
if you need the
moregeneralform.
6. Breakthe e vectorequations intoequations fortheirvarious components,sothatyou
don’thavevectorstocarrythroughtherestofthealgebra.
7. DidyouneedtouseNewton’sthirdlaw? ? Isthemagnitudeofthisforceequaltothe
magnitudeofthatforce?
8.Countthenumberofequationsandthenumberofunknowns. Iftheymatch,you(may)
haveawinner.
9.Solve.
10. Analyzetheresults.
Example
Twomassesareincontactandaresittingonahorizontaltable. Pushtheleftonetowardtheright
andndtheircommonaccelerationandtheforcesbetweenthem.Assumezerofriction.
1.Sketch:
2.Actingontheleftmass:
1.you, 2.table e 3. . Earth(gravity) ) 4. . othermass
Actingontherightmass: 1.table, , 2. . Earth h 3. . othermass
Notice: Youarenotactingontherightmass.
3.
~
F
you
~
F
tabl
;
1
m
1
~g
~
F
contact
~
F
contact
~
F
tabl
;
2
m
2
~g
~a
(both)
4.
^
y
^
x
Thesamebasisforbothmasses
5.
~
F
1
=
F
you
^m
1
g^y
+
F
table
;
1
^F
contact
^x=m
1
a
x
^x
~
F
2
m
2
g
^
y
+
F
table
;
2
^
y
+
F
contact
^
x
=
m
2
a
x
^
x
6.
F
you
F
contact
=
m
1
a
x
F
contact
=
m
2
a
x
The
y
-componentsareoflittleuse.
7.Newton’sthirdlawwasusedinapplying
F
contact
tothetwomasses.
8.Theunknownsare
F
contact
and
a
x
,andtherearetwoequations.
9.
a
x
=
F
you
=
(
m
1
+
m
2
)
and
F
contact
=
F
you
m
2
=
(
m
1
+
m
2
)
10. Thedimensionsarecorrect,butyoudohavetolook.
If
m
1
m
2
thenthecontactforceisapproximately
F
you
.
If
m
1
m
2
thenthecontactforceismuchsmallerthen
F
you
.
F
you
m
1
m
2
Fig.1.3
Shouldyoubelievethis?Dotheexperiment: Useabookandapenor
andpushononesideortheother. Howdoyoumeasurethecontactforce?
Putangerofyourotherhandbetweenthem. Thenreverseyourhandsand
1|Introduction
45
pushontheothermasstofeelthecontactforceagain.
Doyouhavetodothisevery timethatyousetupaproblem? ? Yes,untilthetimecomesthat
you nolongermakemistakes. . Then n youcan startskippingsteps. . (I’m still l waiting.) ) It’s s alsotrue
thatfollowingall these steps isagreattime-saver. . Youdon’tbelievethis. . Noonebelievesitatthe
beginning.
Example
introductoryphysicsbook(probablybyan actofCongress). . You u canseepicturesofrealversionsof
thisapparatusatthiswebsiteofearlyscienticapparatus:
physics.kenyon.edu/EarlyApparatus/Mechanics/Atwoods
Machine/Atwoods
Machine.html
m
1
m
2
^
y
^
y
m
1
~g
T
1
m
2
~g
T
2
Fig.1.4
Mustyouexecutethissystematicprocedurebythenumbers?Maybenot.Hereitisanarration,
butallthestepsarepresent:
Hangtwomassesonstringsandrunthestringoveraxedpulley. Theheaviermasswillstarttodrop
andthelighteronetorise,andtheproblemistogureouttheaccelerationofeither. CanIneglect
themassofthepulley? Probablynot,butIwilldoitanywayfornow. CanIneglectthemassofthe
string? Again,maybenot,butdoitanywaythistime. Lateryoucanreturntothesubjectwhenyou
y
upforonemassandwithitdown
fortheother. Isthisnecessary? No,butit’saconvenientoptionbecauseyoucantheneasilyuseone
coordinate
y
forbothmasses,andaslongasthestringbehaves,
a
y
isthesameforboth.
The things s that t act on
m
1
are gravity and the string. . Acting g on
m
2
are again gravity y and
thestring. The e single coordinate
y
applies to themotion of eachmassifyou make the reasonable
assumptionthatthestringdoesn’tstretch(conservationofstring). ApplyNewton’slawtoeachmass,
treatingthetwomassesastwoseparateproblemshavingnothingtodowitheachother.
~
F
1
=
m
1
~g
+
~
T
1
=
m
1
~a
1
;
~
F
2
=
m
2
~g
+
~
T
2
=
m
2
~a
2
(1)
F
1
y
=
m
1
g
T
1
=
m
1
a
y
;
(2)
F
2
y
m
2
g
+
T
2
=
m
2
a
y
(1
:
17)
Thesearetwoequationsinthethreeunknowns(
a
y
,
T
1
,
T
2
),soyouneedanotherequation. Thatis
T
1
=
T
2
. Why y is this so? ? Itcomesfrom m atorqueequation. . Ifthemassofthepulley y is zero,any
non-zerotorque on it wouldgiveitinnite angularacceleration(Eq.(8.5)), and thatcan’thappen.
Withthesethreeequationsyoucansolveforeverything.
a
y
=
m
1
m
2
m
1
+
m
2
g;
T
1
=
T
2
=
2
m
1
m
2
m
1
+
m
2
g
(1
:
18)
Inanalyzingthissolutiontherearethreecasesthatpushittoitslimit:
m
1
=
m
2
,
m
1
m
2
,
and
m
1
m
2
.
First:
m
1
=
m
2
:
a
y
=0
; T
=
2
mm
m
+
m
g
=
mg
1|Introduction
46
Itbalances,givingzeroaccelerationandwithjustenoughtensioninthestringtomakethetotalforce
oneachmasszero.
Second:
m
1
m
2
:
a
y
m
1
m
1
g
=
g; T
2
m
1
m
2
m
1
g
=2
m
2
g
With
m
1
dominant,thatmassacceleratesat+
g
,causingtheothermasstoaccelerateupat+
g
. As
m
2
m
2
g
,thisrequiresthestringtopullupwith twice
thisforce,anditdoes.
Third:
m
1
m
2
:
a
y

m
2
m
2
g
g; T
2
m
1
m
2
m
2
g
=2
m
1
g
Now
m
2
acceleratestoward
y
andpulls
m
1
upagainstgravity,requiringatensionof2
m
1
g
todoso.
Why did Atwood invent his machine and why is it important enough even to have aname?
Perhaps heinventeditas away toharassphysics students. . Perhaps s abetterreason is as adevice
g
invented,andthismachineslowedthemotionenoughtomakemeasurementsofaccelerationeasier. If
machinedoesnotlooklikethesimplepicturehere;ithascomplicationsdesignedtoreducetheeects
offrictionbecausegood,low-frictionbearingsdidn’texistthen.
Whatdoesthishavetodowithenergy?Inchaptertwoyouwillndmuchmoreonthesubject
of energy,but fornowI’ll assume that you have seen somethingofthesubjectbefore and thatthe
development of mechanical energy y in the e last t section is familiar. . You u remember r the e gravitational
potentialenergytobe
mgh
,andifyoudon’t,thentheequation(1.6)tellsyouthat
F
y
mg
dU
dy
;
so
U
(
y
)=
mgy
IntheAtwood machine asdrawn,Iwill assumethateverythingstartsatrestwhenthemasseshave
coordinates
y
=0. Takethezero-pointofpotentialenergy y tobezeroatthatpointtoo. . Laterthe
masseswillhavepickedupspeedandtheirpositionswillhavechanged.Writedownthetotalmechanical
energy.
E
=
1
2
m
1
v
2
y
+
1
2
m
2
v
2
y
+
m
2
gy
m
1
gy
Thetwokineticenergiesarepositiveand(forpositive
y
)the
m
2
getspositivepotentialenergyandthe
potentialenergyfor
m
1
isnegative.
Energyisconserved. Thatmeansthatthederivativeofthetotalenergywithrespecttotimeis
zero.
E
=
1
2
m
1
v
2
y
+
1
2
m
2
v
2
y
+
m
2
gy
m
1
gy
dE
dt
=0=
d
dt
1
2
m
1
v
2
y
+
1
2
m
2
v
2
y
+
m
2
gy
m
1
gy
=
m
1
v
y
dv
y
dt
+
m
2
v
y
dv
y
dt
+
m
2
g
dy
dt
m
1
g
dy
dt
0=(
m
1
+
m
2
)
dv
y
dt
+(
m
2
m
1
)
g
(1
:
19)
Thethirdlineusedthechainruleforderivativestodierentiate
v
2
y
withrespectto
t
. Thefourthline
used thedenitionof velocity,
v
y
=
dy=dt
,tocancelsomefactors. Solve e the nalequationforthe
acceleration
a
y
=
dv
y
=dt
andyouhavethe resultEq.(1.18). . Oratleastpartofit. Usingenergy
methodsisoftenaneasierwaytogettotheresultyouwant,butitdoesnotalwaysgiveyouall the
resultsyouwant. Inthiscaseitdoesn’tprovideanequationforthetensioninthestring. Ifyouneed
that,youcanuse
~
F
=
m~a
ononeofthetwomassesandgetthetensionfromthatsingleequation.