3|SimpleHarmonicMotion
117
Thetwocapacitorsarethesame,
C
,andthetwoinductancesontheendsare
L
.Theoneinthemiddle
is
L
0
. Theresultingequationsforthetwopartsofthecircuitare
L
L
0
L
I
1
I
2
1
C
I
1
+
L
d
2
I
1
dt
2
+
L
0
d
2
dt
2
I
1
+
I
2
=0
1
C
I
2
+
L
d
2
I
2
dt
2
+
L
0
d
2
dt
2
I
1
+
I
2
=0
Findthemodesofoscillationofthecurrents
I
1
and
I
2
,thenwritethegeneralformof
I
1
(
t
)and
I
2
(
t
).
Drawpicturesofhowthecurrentsmoveineachmode.
3.61 Thecarbonmonoxidemolecule,CO,canbemodeledastwomassesontheendsofaspring.Solve
fortheoscillationsofthismolecule,assumingthatthemotionisalongthesinglelongaxisbetweenthe
atoms. Compareyourresulttowhatappearsinsection6.7,especiallyEq.(6.46),derivedbydierent
methods.
3.62 Two masses,
m
and
m
, are e sitting on a a table e with h no friction and
initiallyatrest.Theyareconnectedbyaspring.Attime
t
=0,starttoapply
aconstantforce
F
0
tooneofthemasses,alongthelineconnectingthemasses.
Themotionsofthetwomassesare(maybe)
x
1
;
2
(
t
)=
F
0
t
2
=
4
m
(
F
0
=
4
k
)
1 cos
!
0
t
(
!
2
0
=2
k=m
)
Analyzetheseequationstodetermineiftheyareplausible.
3.63_Derivethemotion,
x
1
(
t
)and
x
2
(
t
),intheprecedingproblem.
3.64_Fromtheresultoftheprecedingproblem,(a)computetheworkdonebytheforceuptotime
t
.
(b)Computethetotalmechanicalenergyinthesystematthattimeandcomparethetworesults.
3.65_Twoequalmasses
m
areattachedtothreespringsasshownonpage100.Thespringconstants
ontheendsare
k
andtheoneinthemiddleis
k
0
.Aconstanthorizontalforce
F
0
isnowappliedtothe
massontheleft.
(a)Writeallthedierentialequationsofmotionforthecoordinates
x
1
and
x
2
.
(b)Findaninhomogeneoussolutiontotheequations.
(c)Findthegeneralsolutionsforthehomogeneouspartoftheequations.
(d)Writethetotalsolution.
(e)Attime
t
=0both massesareattheirequilibriumpositionwithzerovelocity. . Find d theirfuture
positions.
3.66 Forthecasethattheharmonicoscillatorisdamped(underdamped),ndGreen’sfunctionforthe
solution,derivingEq.(3.74).
3.67The motion n in the potential energy y function n of problem 3.45 can be e solved, , resulting in an
anharmonicoscillator. Thesolutionis
x
(
t
)=(
cos
2
t
+
sin
2
t
)
1
=
2
forappropriatevaluesofthe
constants. Verifythatthisissobysubstitutingitintotheconservationofenergyequation
K
+
U
=
E
and showingthat forappropriate
,
,and
itiscorrect. Youcan n solvefortheseintermsof
E
,
a
,and
b
,butitissimplerif you workbackwardsandlet
a
,
b
,and
bethecontrollingparameters,
expressing
E
,
and
intermsofthem. Andofcourseanalyzetheresults. Isthefrequencycorrect?
Anddrawgraphs. Whatdoesthegraphof
x
(
t
)looklikeforsmallenergyandforlargeenergy? And
canyouanticipatewhatthegraphshould looklikeevenbeforetryingtousetheequationtosketchit?
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3|SimpleHarmonicMotion
118
Thereasonthisproblemcanbesolvedwillbeclearafterdoingproblem6.50.
Ans:
=
p
2
a=m
,
E
=
a
+
b=
,
=(
b=a
),
x
min
=
p
,
x
max
=
p
.
3.68 Apotentialenergyisspeciedtobe(
>
0)
U
(
x
)=
0
(
x
0)
x
2
(
x>
0)
Amass
m
startsatapoint
x
d
withaspeed
v
0
totheright.Howmuchtimedoesittaketoreturn
toitsinitialpoint?Also,sketchagraphoftheforcefunctionthatcomesfromthispotential.
3.69 InEq.(3.56)thereisaspecialcasethatyoucandoapproximatelyinordertochecktheanalysis.
Assumethat
0
=
2
0
andthat
0
issmall. Expand d thecosinesnear
=
2anddotheresulting
(nowprettyeasy)integral.Thisisaresultthatyoucancomparetoasimple
at
2
=
2typeofcalculation,
whereyoujustletthemassattheendofthependulumdropashortdistance.
3.70Forthosewhowanttotryanumericalintegral,lookatthecomparisonoftheexactintegraland
thelinearapproximationinEqs.(3.54)and(3.56)tocheckthenumericalresultsstatedthere. Todo
thisintegralnumericallyyoumustrstrecognizethattheintegrandissingularat
=
0
.Ifyoumake
thesubstitution
0
=
x
2
,thentheintegrandinthe
x
variableisnolongersingular,butyoumust
evaluateitas
x
!0inordertonditsvalueatzero. Thenintegrate
dx
from
x
=0towhere
=
=
2.
3.71_TheAtwood’smachinedescribedinsection1.4hasthemasses
m
1
and
m
2
asindicated,butnow
thecordhasaconstantlinearmassdensity
. Neglectthemassofthepulley. Thesystemstartsat
time zero with velocity equals zero and the initial coordinateis
y
(0)=
y
0
. Theposition
y
ofmass
m
1
isclaimed tobeone oranotheroftheseresults. . Checkthedimensionsoftheproposedanswers,
thenconsidervariousspecialvaluesoftheparameters
m
1
,
m
2
,
,
y
0
anddemonstratethatallofthese
proposedanswersareimpossible. Donot solve e theproblemandtry tocomparetheanswertothese
proposed. (Thatcomesnext.)
y
(
t
)=
y
0
cosh
!t
with
!
2
=
m
1
m
2
m
1
+
m
2
+
L
y
(
t
)=
y
0
cosh
!t
+
m
1
m
2
2
1 cosh
!t
with
!
2
=
2
g
m
1
+
m
2
+
L
3.72 Solvetheequation (2.40) forthemotion of Atwood’s machine. . AssumeItstarts s from restat
initial
y
=
y
0
.Manycasestoanalyzehere: (a)
=0,(b)
m
1
=
m
2
=0,(c)there’sanexponentialin
thissolution.Cantheaccelerationeverbelargerthan
g
? Thespecialcase(b)willsuceforthisone.
3.73 (a) Writethe equation forconservation ofenergy forthepotentialfunction
U
(
x
)= 
kx
2
=
2,
andsketchthegraphof
U
.
(b)Separate variables s to get
dt
and write the integral that will allow w you u to nd
t
in terms s of
x
.
(c)Evaluatetheintegralforthespecialcasethat
E
=0andapplytheinitialconditionthat
x
(0)=
x
0
tosolvefor
x
(
t
).
3.74_Twomasses,
m
1
and
m
2
,aresittingonatablewithnofriction.Theyareconnectedbyaspring,
andaforce
F
0
cos
!t
isappliedto
m
1
alongthelineconnectingthemasses.(a)Findthesteady-state
solution(theinhomogeneoussolution)forthemotion.(b)Graphtheamplitudeforthemotionofeach
massasafunctionoftheapplied
!
.
3.75 For the e simple harmonic oscillator, , the period is s independent of f the e energy. . For r the e quartic
potential,
U
/
x
4
,howdoestheperioddependontheenergy?Alsocomparethesetothesolutionsof
problems2.19and3.32.Isthereapattern?
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3|SimpleHarmonicMotion
119
3.76 Twopendulumsarecoupledwithalightspring. Thetwolengthsofthependulums
arethesame,butthemassesaredierent.Thespringisattachedatacommondistance
fromthesupportandthelengthsofthependulumsare
L
. Themassesare
m
1
and
m
2
.
Whenthemassesarehangingverticallythespringisunstretched. Setuptheequationsof
motionandndthefrequenciesandmodesofoscillation. Ifyouthinkthatyou’regetting
involvedinsolvingaquarticequation,lookmoreclosely;it’snotthathard. Verifythat
thesesolutionsareorthogonalinthesenseofEq.(3.69). [Thisisaneasydemonstration
tosetupandtoseethatthemassmatrixisimportantinmakingthesemodesorthogonal.]
3.77 ThemanipulationsthatledfromEq.(3.75)toEq.(3.77):dothesameforthesimpleharmonic
oscillator.
3.78 Dothenumericalintegral,Eq.(3.77)from0to2. Usethemidpointrulewith
N
=1andwith
N
=2(orSimpson|lookitup).Themoreaccurateanswerforthis
u
when
z
=2is1
:
311029.
3.79 Inthesame waythatyou plotpositionversustimeingure3.16,plotthemotionforproblem
3.32.
3.80 InEq.(3.70)youseetheorthogonalityrelationfornormalmodes. Considerthesametwo-mass,
three-springexampleyouusedinsection 3.9,buttakeallthesprings tobeequalandtakeonemass
tobemuchlargerthantheother,say
m
2
=100
m
1
orso.Youcangothroughthewholesolutionofa
quadraticequation,thedeterminantofEq.(3.62),butwithoutthesymmetry.Butdon’t. Instead,draw
picturesof whatyouexpectthemodesoughttolooklikeandthenapplytheorthogonality relation.
Yourrstguessmaynotberight,butusethefactthattheyhavetobeorthogonaltoadjustthefactors.
Whatthenarethetwofrequencies,atleastapproximately? Justifyyourreasoningfortheshapeofthe
modes.Inotherwords,solvethisproblemapproximatelywithoutsolvingtheequations.
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ThreeDimensionalMotion
.
Readsections 0.3 0.4 4 0.6 0.11
Theworldisn’tonedimensional. Theaddedcomplexities s yougetwhenyouleavethestraight
lineleadtosomeprettyresults,somesurprisingresults,someperplexingresults,andsomejustplain
hardresults.
Staywiththecaseofconstantmassfornow,sothat
~
F
=
m~a
forasingleparticle. Startwith
thesamesortofspecialcasesthatyouhaveinonedimension,forwhichtheforceisafunctionoftime
orvelocityorposition,andthenputthemtogether.
When
~
F
is a a function n of time alone,there’s seems to be notmuch dierence from the one-
dimensionalcaseexcepttoturneverythingintovectors:
d
2
~r=dt
2
=
~
F
(
t
)
=m
. Canthatbeallthereis
tothiscase?Prettynearlyso.Youhavethreetimesasmuchcalculationtodo,andthepicturesofthe
resultswillbehardertodraw,butthere’snonewconcepthere|atleastinrectangularcoordinates!
Inpolarcoordinates,thedierentialequationfor
r
willinvolvetheangle
(
t
)andtheequationfor
willinvolve
r
. Notnecessarilysimple.
4.1ProjectileMotion
A standard problemthat every introductory physics text handles is thatofguringout motion ina
uniformgravitationaleld,
~
F
=
m~g
. You u do acouple ofeasyintegrals and the restisinterpreting
thealgebra. Ofcourseyouneglectairresistancebecauseyouhavetostartwiththeeasiestproblems
rst. What t if you don’tneglect it? ? Howdo o you describe airresistance mathematically? ? Todoso
fullyisquitedicultandcomplicated,becauseitdependsonmanyfactors.Whydoesagolfballhave
dimples? Because e if you use asmooth ball you can’thit it nearly as far. . Thereasons s forthis are
complex,involvingthechangeinairresistancewhenturbulenceisinducedandtheliftcausedbythe
interactionoftheball’sspinwiththeair|theMagnusforcethatappearsinEq.(4.51).
Airresistance with any object is strongly dependenton velocity. . At t low enough speeds it is
typically alinear r functionof
v
,then atsomewhathigherspeeds it’s morenearly quadratic. . Atstill
higherspeedsitcanevendecreasewithincreasingspeedforsomerangesof
v
.
Assumethattheresistanceislinearin
v
,notbecause itisagood approximationforordinary
speeds, but because it is s the only assumption thatallows youtouse straight-forward mathematics.
Everyothermodelismorediculttohandle.Qualitativelythough,it’sstillprettygood.
Inthismodeltheequationtoexamineis
~
F
=
m~g
b~v
=
m~a;
or
mg^y
b~v
=
m
d
2
~r
dt
2
(4
:
1)
whereIpickthe
y
-axistobeup. Taketheinitialconditionstobe
~r
(0)=0
;
and
~v
(0)=
v
x
0
^
x
+
v
y
0
^
y
=
v
0
cos
^
x
+
v
0
sin
^
y
Withtheseconditionsthe
z
-coordinatestaysatzero.
m
d
2
x
dt
2
+
b
dx
dt
=0
;
m
d
2
y
dt
2
+
b
dy
dt
mg
(4
:
2)
Thesearelinearconstantcoecientdierentialequations,oneinhomogeneous. Youcansolvethemin
severaldierentways,andIchoosetousethemethodofsection0.9.
120
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4|ThreeDimensionalMotion
121
BothofEqs.(4.2)havethesamehomogeneouspart,sodothatrst. Assumeanexponential.
x
=
Ae
t
=)
mA
2
e
t
+
bAe
t
=0
;
so
=0
or
b=m
The homogeneous s solution n is then
x
(
t
) =
A
+
Be
bt=m
. The e same e for
y
, though with dierent
constants.
Theinhomogeneouspartofthedierentialequationisaconstant,sotryaconstanttimestime
forasolution:
y
=
Ct
=)
m
.
0+
bC
mg;
or
C
mg=b
Putthesetogetherandthetotalsolutionis
x
(
t
)=
A
+
Be
bt=m
;
y
(
t
)=
A
0
+
B
0
e
bt=m
mgt=b
Applytheinitialconditionstoevaluatetheconstants.
x
(0)=0=
A
+
B;
y
(0)=0=
A
0
+
B
0
;
and
v
x
(0)=
v
0
cos
b
m
B;
v
y
(0)=
v
0
sin
b
m
B
0
mg
b
!
x
(
t
)=
m
b
v
0
cos
e
bt=m
y
(
t
)=
m
b
h
v
0
sin
+
mg
b
i
e
bt=m
mgt
b
(4
:
3)
Firstcheckthatthedimensionsarecorrect. Thishastobeare ex,soI’llleaveittoyou. . [Quickly:
Lookattheexponentin
e
bt=m
anddeterminetheunitsfor
m=b
. Thenlookatalltheotherplaces
that
m=b
appears.] Ofcourseyoushouldalsogobacktotheveryrstequationwhere
b
showsupand
checkthattheunitsyoufoundfor
m=b
matchthere.
Backinchapter2youlookedatwhatlinearviscositydidtothemotionofaparticle,butthat
waswithoutgravity,andthesolutionappearsintheequations(2.13)through(2.16)insection2.2(b).
Theresultshere,Eqs.(4.3),shouldagreewiththosepreviousequationsifyousimplyturnogravity.
Checkitout.
Thenextstepistocheckthatthesenewresultsareplausible. Ifthereisnoairresistanceafter
all,thenyouknowtheanswerisasimple
1
2
gt
2
sortofresult. Doesthisgivetherightanswerif
b
!0?
Startwiththe
e
bt=m
term. You’vegot0
=
0and1 1,sousepowerseriesexpansions. . Startwith
the
e
bt=m
term;theexponentialis,fromEq.(0.1)
e
bt=m
=1 
bt
m
+
b
2
t
2
2
m
2
b
3
t
3
6
m
3
+
Substituteinto
x
(
t
)andyouhave
x
(
t
)=
m
b
v
0
cos
bt
m
+
b
2
t
2
2
m
2
b
3
t
3
6
m
3
+

=
m
b
v
0
cos
bt
m

=
v
0
cos
t
+
(4
:
4)
Alltherestofthetermsgotozeroas
b
!0,andthisisthecorrectresultforzeroresistance.
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4|ThreeDimensionalMotion
122
Nextexamine
y
:
y
(
t
)=
m
b
h
v
0
sin
+
mg
b
i
bt
m
+
b
2
t
2
2
m
2
b
3
t
3
6
m
3
+

mgt
b
=
m
b
h
v
0
sin
+
mg
b
i
bt
m
b
2
t
2
2
m
2
+
b
3
t
3
6
m
3
+
mgt
b
=
v
0
sin
t
1
2
gt
2
+
(4
:
5)
Again,therestofthetermsvanishas
b
!0. Thesetwolimitingequationsfor
x
and
y
arewhatyou
mustndoryouhavetogobackanddiscoveryourmistake.
Go totheoppositeextremenow. . Whathappens s tothistrajectory afteralong time? ? Thisis
prettysimple:Intheequations(4.3),theexponentialfactor
e
bt=m
willgotozeroafteralongenough
time,andallthatisleftafterthatis
x
(
t
)
m
b
v
0
cos
and
y
(
t
)
m
b
h
v
0
sin
+
mg
b
i
mgt
b
(4
:
6)
Ithasgonesomedistancehorizontallyandisthendroppingverticallyatconstantvelocity.Theterminal
speedis
mg=b
. Itmayofcoursehitthegroundbeforethis\longenough"timehasbeenreached.You
cangetintothisterminalspeedzoneifyouresomethingfromtheedgeofacli,whichallowsenough
timebeforehittingtheground. Youcanalsogetintothisdomainifyouretheprojectileintoalarge
barrelofhoney,increasing
b
suciently.
We’renotdoneyet.Justaswiththezeroviscositycaseyoucaneliminate
t
betweentheequations
(4.3)for
x
and
y
togetanon-parametricequationforthetrajectory.
y
=
x
tan
+
mg
bv
0
cos
+
m
2
g
b
2
ln
bx
mv
0
cos
(4
:
7)
Thathelpsalot,doesn’tit?
x
y
Fig.4.1
Therereallyaresomethingsthatyoucan digoutof this equation.
As
x
starts at zero,thisis zero. . When
x
increases, theargument of the
logarithmdropsfromonetozero.Theloggoesto 1there,andthatpoint
is
x
=
mv
0
cos
=b
,thesamevalueascomputedfromEq.(4.6). Whatis
thebehaviorforsmall
x
?Isitwhatyouexpect?Againyoudoapowerseries
expansion,rememberingtheseriesforalogarithm,ln(1+
x
)=
x
+.Then
thetermsin1
=
cos
canceleach otherandyou are leftwith
y
=
x
tan
,
preciselyasneeded.
The vertical dashed line in the gure shows avertical asymptote
x
=
mv
0
cos
=b
, and the
trajectory approachesthatverticalmotionas timegoestoinnity (oruntilithitstheground). . The
curveddashedlineisthetrajectorytheofthesamemasswiththesameinitialconditionsbutwithout
friction. Theshapeofthemotionwithfrictionisnolongerthesymmetricparabolathatyou’veseen
soofteninelementarywork,andyoucannotmakesomeofthesimplifyingassumptionsthatyoumay
havethentakenforgranted. Doesittakethesametimetocomedownastogoup? No. Isthepeak
ofthecurvehalfwaybetweentheleftandrightinterceptswiththe
x
-axis?No.
Example
TogetaqualitativefeelforhowequationssuchasEq.(4.7)behave,examineitforsmall notzero
valuesoftheairresistance. Useaseries s expansion toseetheeectoftheair. . Therst
b
is inthe
denominator,soitlooksliketherightsideisapproachinginnityas
b
approacheszero,butcarryon.
Theseriesforthelogarithmfrompage1isallthat’sneeded.
4|ThreeDimensionalMotion
123
y
=
x
tan
+
mg
bv
0
cos
+
m
2
g
b
2
ln
bx
mv
0
cos
=
x
tan
+
mg
bv
0
cos
+
m
2
g
b
2
"
bx
mv
0
cos
1
2
bx
mv
0
cos
2
1
3
bx
mv
0
cos
3

#
=
x
tan
gx
2
2
v
2
0
cos2
gbx
3
3
mv
3
0
cos3
gb
2
x
4
4
m
2
v
4
0
cos4

(4
:
8)
andthe1
=b
termshavedisappeared.Wheredoesitland?Assumethesimplestcase,thatitisredon
alevelsurfacesothattheequationforthegroundis
y
=0.Yousolvethatasasimultaneousequation
with(4.8). Onerootfactorsinstantly,
x
=0,andtherestis
tan
gx
2
v
2
0
cos2
gbx
2
3
mv
3
0
cos3
gb
2
x
3
4
m
2
v
4
0
cos4
=0
(4
:
9)
Youcouldlookupthecubicformulaandndthesolution,butlet’snot. Youcouldtryexaminingthe
eectoftherstnewterm,theonein
bx
2
,usingthemorefamiliarquadraticformulaforthat. Butno.
Iftheairresistanceissmall,thenthisequationisalmostlinear,consistingofthersttwoterms. This
callsfortheiterativemethods describedinsection 0.11. Ifyou’veskippedit,thenread d itnow. . You
willseeitagainseveraltimes.
ApplytheiterationmethodtoEq.(4.9)toseetheeectofthefrictionalfactor
b
.
tan
gx
2
v
2
0
cos2
gbx
2
3
mv
3
0
cos3
=0  !
x
=
2
v
2
0
g
cos
2
tan
2
b
3
mv
0
cos
x
2
(4
:
10)
Inthelowestorderapproximationneglectthenalterm,proportionalto
b
. Thisgivesthezerothorder
approximation
x
0
=
2
v
2
0
g
cos
sin
=
v
2
0
g
sin2
showingtherangein avacuum. . Thisis s thecaseinwhich you ndthatthemaximumrangeoccurs
foraringangleof45
,wheresin2
=1. Thecorrectiontothisascausedbyairresistanceiswhat
comesnext. Iteratethequadraticequationtogetanimprovedroot,
x
1
,byputting
x
0
intotheright
handsideofEq.(4.10).
x
1
=
2
v
2
0
g
cos
sin
2
b
3
mv
0
cos
2
v
2
0
g
cos
sin
2
=
v
2
0
g
sin2
8
bv
3
0
3
mg
2
cos
sin
2
=
v
2
0
g
sin2
4
bv
3
0
3
mg
2
sin2
sin
(4
:
11)
Youseethattherangehasdecreasedfromthesimplevacuumresult. (Surprise!) ) Lookatthecoecient
ofthesecondterm.Itis
bv
0
=mg
timestherstterm(times
4
3
sin
).Thisisjusttheratiooftheforce
bytheairtotheforcebygravity,andthatisjustwhatyouexpect.Orisit?Didyouanticipatethatit
wouldcomeoutthisway?Maybenexttime. Doesthemaximumrangestilloccurforaringangleof
45
?No.
4|ThreeDimensionalMotion
124
Whydoestheaddedfactorinthesecondtermhaveasin
init? Isitplausible? ? Youseethat
when
isverysmall,thiscorrectionisverysmallandwhen
isupnear90
thecorrectionisbigger.
Thatmakes sensebecauseatsmallangles,itisnotintheairforalongtime;itstaysin theairfar
longerwhenredatlargerangles. Thecorrectionshouldthenbebiggerforlarger
thanforsmaller,
anditis. Comparethesizeofthecorrectionfor
=1
and
=89
.
Canyougetastill betterresultby repeatingthisiteration,getting
x
2
by putting
x
1
intothe
righthandsideofEq.(4.10)? No! Thatwouldbewrong. Canyougureoutwhy? Andwhatwould
youhavetodotondthehigherordercorrectioncorrectly?Don’tworry;wewon’t.
Withoutderivingtheresult,Iwillstatethatthemaximumrange,foundbysetting
dx
1
=d
=0
fromEq.(4.11),occursforanangleslightlylessthan45
:
maxrange
4
2
bv
0
mg
p
2
So you see whatyou getintowhen you start to add the slightest bitof reality to these previously
elementaryproblems.
4.2GeneralResults
If you read Newton’s Principia you will not t nd d the equation
~
F
=
m~a
. You u will not t even n nd
theequation
~
F
=
d~p=dt
. His s presentation of physics was nothing likethe modern way to develop
thesubject,and itis close to unreadabletoday.* * The e treatment we are accustomedtowasmostly
developedbyLeonardEuler,andifyoulookupEuler’scollectedworksyouwillndthelanguageand
thenotationremarkablemodern. ThatisbecausewemostlyuseEuler’snotation.
Starting from the basic equation describing the motion of a particle, there are some general
resultstoderive.Startwithonedimension.
F
x
=
m
dv
x
dt
!
F
x
v
x
=
mv
x
dv
x
dt
!
F
x
v
x
=
m
2
dv
2
x
dt
Readingfromrighttoleft,thissaysthatthepower,therateofchangeofenergyis
dW
dt
=
d
dt
1
2
mv
2
x
=
F
x
v
x
Stillanotherwaytomanipulateitistointegratewithrespectto
t
.
Z
t
f
t
i
F
x
v
x
dt
=
Z
t
f
t
i
m
2
dv
2
x
dt
dt
=
Z
t
=
t
f
t
=
t
i
m
2
d
(
v
2
x
)=
1
2
mv
2
f
1
2
mv
2
i
(4
:
12)
Thisisthesimplestformofthework-energytheorem,
W
total
=
Z
t
f
t
i
F
x
v
x
dt
=
Z
x
f
x
i
F
x
dx
=
K
=
K
f
K
i
(4
:
13)
Doesthislookfamiliar?Ifso,thatmaybebecauseyou’vestudiedsections1.3and2.3,butIthought
itworthrepeating.
Inonedimension,
F
x
v
x
dt
=
F
x
dx
,andiftheforce
F
x
isafunctionofthecoordinate
x
alone,
thentheintegraldependsontheinitialandnalvaluesof
x
andnotonhowfastorhowslowlyyouwent
* though h there e is an n amazing book by y S. Chandrasekhar that t attempts s to o make it t accessible:
\Newton’sPrincipiaforthecommonreader".
4|ThreeDimensionalMotion
125
fromthestarttothenish.
F
x
(
x
) has ananti-derivative, , and the fundamentaltheoremof calculus
saysthatyouevaluatethisanti-derivativeattheendpointsandsubtract. Callthisanti-derivative 
U
.
F
x
(
x
)= 
dU
dx
;
then
Z
x
f
x
i
F
x
(
x
)
dx
U
(
x
)
x
f
x
i
U
(
x
f
)+
U
(
x
i
)
(4
:
14)
Putthisintotheprecedingequationandrearrangeit.
U
(
x
f
)+
U
(
x
i
)=
1
2
mv
2
f
1
2
mv
2
i
!
1
2
mv
2
i
+
U
(
x
i
)=
1
2
mv
2
f
+
U
(
x
f
)
(4
:
15)
This is in theform of aconservation law. . Somethingevaluated d atone time equals thesame thing
evaluatedatalatertime:Conservationofmechanicalenergy.Thisisthereasonforchoosingtheminus
signon
U
.Itproducesanicerresult.
Staying withone dimensionforamoment,whatcan go wrong? ? A A simple form of forcethat
you’veseeninanintroductorycourseisfriction. Commondryfrictionisvelocitydependent,violating
theassumptionsleadingtothisconservationlaw. Whenanobjectslidesalongasurface,dryfrictionis
representedby
F
fr
=
k
F
N
,where
F
N
istheperpendicular(normal)componentoftheobject’sforce
onthesurface and
k
isthecoecientofkinetic(sliding)friction. Assumingthatthis s coecientis
independentofthespeedofsliding,thecomponentofforce,
F
x
,dependsonthedirectionofthemotion.
Itmaynotdepend onthemagnitude ofthevelocity,butitdependson ^
v
. Atonevalueof
x
,ifthe
objectslidestotherightthentheforceistotheleft;ifitslidesleft,theforceistotheright. Thereis
no
F
x
(
x
)andsono
U
(
x
).AmorecompleteformfortheequationofdryfrictionappearsinEq.(1.8).
Example
x
Fig.4.2
Forasimpleexampleofthis,lookatamassslidingupandbackdown
aslope. Itstartsuphillwithspeed
v
0
. Howfastisitmovingwhenit
comesbacktoits startingposition (assumingthatitdoes)? ? Forthis
exampleyouhavetotaketwocases,correspondingtothedirectionof
motionandsotothedirectionoffriction.
(up)
F
x
mg
sin
k
mg
cos
;
then
W
=
Z
x
top
0
dxF
x
=
mg
sin
k
mg
cos
x
top
=
K
=0 
1
2
mv
2
0
Thatwasonthewayup. Comingdownitis
(down)
F
x
mg
sin
+
k
mg
cos
;
and
W
=
Z
0
x
top
dxF
x
mg
sin
+
k
mg
cos
x
top
=
K
=
1
2
mv
2
f
0
Dividetheseequationstoeliminate
x
top
,leavingarelationbetween
v
f
and
v
0
. Thensimplifytheresult.
v
2
f
=
v
2
0
sin
k
cos
sin
+
k
cos
(4
:
16)
Ifthefrictionistoolargeortheangle toosmall,of courseit doesn’t return. . Thentoothereis s the
peskyfactthatstaticfrictionisnotthesameaskinetic. Thiswillaectwhetheritsticksatthetopor
not.Inanycase,youdonothaveconservationofmechanicalenergy.Thereisno
U
.
4|ThreeDimensionalMotion
126
Eveninthisexamplewithfrictionthereisapartoftheforce(gravity)forwhichpotentialenergy
exists.Inthework-energytheoremitisusefultodividetheforceintotwotypes: Conservative(thereis
apotentialenergy)andNonconservative(thereisn’t).
F
x
=
F
x;
cons
+
F
x;
other
;
then
F
x;
cons
(
x
)= 
dU
dx
and
W
total
=
Z
x
f
x
i
F
x
dx
U
(
x
)
x
f
x
i
+
Z
x
f
x
i
F
x;
other
dx
Putthisinto(4.12)andrearrange.
Z
x
f
x
i
F
x;
other
(
x
)
dx
=
1
2
mv
2
f
+
U
(
x
f
)
1
2
mv
2
i
+
U
(
x
i
)
W
other
=
E
(4
:
17)
andthis
E
isthetotalmechanicalenergy,
K
+
U
. Thisformofthework-energytheoremisequivalent
totheothers,butitincludestheotherformsasspecial cases. . Ifthereis s no\other"forcethenyou
haveconservationofmechanicalenergy. Ifyoudecidenottousethepotentialenergythenyousimply
make\other"thewholething.
PotentialEnergy,3-d
Eqs.(4.12)and(4.17)appliestoonedimension,andintwoorthreedimensionstherearemorecom-
plexities. Even n if the forceisafunction ofpositionalone,the resultingworkintegralmay not be a
functionofpositionalone(i.e.notafunctionofitsendpointsalone). Eq.(4.12)relatesthechangein
kineticenergytoanintegraloftheforce,andthatintegralwillusuallydependonthepaththatyou
tooktogofromtheinitialpositiontothenalposition. Itisafunction,notoftheendpointsofthe
path,butofthewholepath.Didyougoonastraightline?Didyoumovealongthearcofacircle? In
Eq.(4.14),ifthereisapotentialenergyfunctionofpositiontheinitialvalueoftheenergy hassome
determinedvalue,sothenalvaluedoestoo. Theworkintegralcannotthendependonthewayyou
wentfromtheinitialtothenalposition.Inthreedimensionsthesameconditionmusthold.
Thereareconditionsontheforceneededinorderthatthepotentialenergyexist.Whentheforce
isafunctionoftherectangularcoordinates,
~
F
(
x;y;z
),thenecessaryconditionsare
@F
x
@y
@F
y
@x
=0
(4
:
18)
Youalsoneedthesameequations,butwith(
y;z
)andwith(
z;x
)insteadof(
x;y
),givingatotalof
threeequations.Thepartialderivativenotation(
@
insteadof
d
)meansthattheothertwocoordinates
aretreatedasconstantsduringthedierentiation.Toseewhythisequationisneeded,lookatthecase
wherethepotentialenergyintegralgoesaroundaloop,sothatthenalpointisthesameastheinitial
point. Seeproblem4.30foraquickderivation.
In three dimensions the dierential relation between force and energy simply y extends s thatof
Eq.(4.14)tomorecomponents.
F
x
(
x;y;z
)= 
@U
@x
;
F
y
(
x;y;z
)= 
@U
@y
;
F
z
(
x;y;z
)= 
@U
@z
(4
:
19)
AndyouseefromthesethatEq.(4.18)isthestatementthatyoucandopartialderivativesineither
order:
@
2
U
@x@y
=
@
2
U
@y@x
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