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8|RigidBodyMotion
277
Picturethesemotions. Forthe
0
=1case,therotationisaboutthelinefromtheorigintothe
farcornerofthesquare.Onlytwomassesaremovingandtheyaresymmetricallyplacedwithrespectto
therotationaxisatadistance
r
?
=
a=
p
2.Theangularmomentumisobviouslyinthesamedirection,
andthemomentofinertiaaboutthisaxisis2
m
a=
p
2
2
=
ma
2
.That’spreciselytheeigenvalue
for
thiscase.
x
y
x
y
x
y
~!
~!
~!
out
0
=1
0
=3
0
=4
Fig.8.11
Forthe
0
=3case,therotationaxisisstillinthe
x
-
y
plane,butalong ^
y
^
x
,perpendicular
to the preceding one. . Now w three e masses s are moving, and d the rotational l inertia a about this axis s is
2
m
a=
p
2
2
+
m
a
p
2
2
=3
ma
2
aspromised.Thisdirectionstillhasenoughsymmetrythatitiseasy
tobelievethat
~
L
isinthesamedirectionas
~!
.
For
0
= 4, theeigenvectoris along ^
z
, andagain three masses s are in motion, , with moment
2
ma
2
+
m
a
p
2
2
= 4
ma
2
. This s is a suciently y non-symmetric c case e that it is not immediately
obviousthattheangularmomentumisinthesamedirectionas
~!
,butdoacoupleofcrossproducts
andyoucaneasilypersuadeyourselfthatit’strue. Again,theeigenvalue
=4
ma
2
isthismomentof
inertiaaboutthe
z
-axis.
Thebasis ^
x
^
y
,^
z
,inwhichIcomputedthecomponentsof
I
inthisexamplewerearbitrary. I
pickedthemfortheirobviousconvenienceincomputingtheanswer. WhatifIpickanotherbasis,one
consistingofthe three orthogonaleigenvectors thatIjust found? ? I’llcall l them
~e
1
,
~e
2
,and
~e
3
as a
commonnotationusedforbasisvectors.
~e
1
=
^x+^y
=
p
2
;
~e
2
=
^^x
=
p
2
;
~e
3
=^
z
(8
:
38)
Doesitmatterthattheseareunitvectors? No,aslongastheyareindependentthatisgoodenough.
You could drop the
p
2 and d compute e the matrix for
I
, butforother manipulations you should be
consistent. InkeepingwiththechangeIwillusesubscripts1,2,3insteadof
x
,
y
,
z
onthecomponents
I
(
~e
1
)=
X
j
I
j
1
~e
j
=
I
11
~e
1
+
I
21
~e
2
+
I
31
~e
3
=
1
~e
1
Thisdeterminestherstcolumnofthematrix for
I
inthis basis. . The e othercolumnsarefoundthe
sameway.
I
(
~e
2
)=
X
j
I
j
2
~e
j
=
I
12
~e
1
+
I
22
~e
2
+
I
32
~e
3
=
2
~e
2
I
(
~e
3
)=
X
j
I
j
3
~e
j
=
I
13
~e
1
+
I
23
~e
2
+
I
33
~e
3
=
3
~e
3
Thematrixfor
I
isnow
(
I
)=
0
@
1
0
0
0
2
0
0
0
3
1
A
=
0
@
ma
2
0
0
0
3
ma
2
0
0
0
4
ma
2
1
A
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8|RigidBodyMotion
278
and you u see e that in this basis the e matrix x is s diagonal. . That t can n provide e signicant t advantages s in
manipulationforfurthercalculations. Noticethatyouwouldgetthesamematrix x even ifyoudonot
putinthe
p
2factorsinEq.(8.38).
Isitalwayseasy tondtheeigenvectors? ? No,thisexampleisvery y easyandyoudidn’treally
havetoconfrontacubicequationtosolvefortheeigenvalues. Alas,itisnottypical.
8.5PropertiesofEigenvectors
Therearesomegeneralresultstobeprovedabouttheseeigenvectorsandeigenvalues,andthemethods
ofproofusedherewillshowupinseveralotherimportantcontexts,sotheyareworthlearning. First
anidentityabouttheinertiatensor.
I
(
~!
)=
Z
dm~r
(
~!
~r
)=
Z
dm
r
2
~!
~r
(
~!
.
~r
)
~!
*
1
.
I
(
~!
2
)=
~!
*
1
.
Z
dm
r
2
~!
2
~r
(
~!
2
.
~r
)
=
Z
dm
r
2
~!
*
1
.
~!
2
~!
*
1
.
~r
(
~!
2
.
~r
)
=
I
(
~!
*
1
)
.
~!
2
(8
:
39)
Whatisacomplexconjugationdoingherewhenallthesevectorsaresupposedtobereal? Andwhat
doesit meananyway? ? Here e Iam allowingforthepossibility thatthese maybe complexinorderto
provethattheyaren’t. Howcanavectorbecomplex? Intheusualbasis,justallowthecomponents
tobecomplexnumbers. Youwon’trepresentitbyasinglearrowanymore(youcouldusetwo),but
youwon’tneedtodothatanyway.
Thissymmetrypropertyoftheinertiatensor,
~!
*
1
.
I
(
~!
2
)=
I
(
~!
*
1
)
.
~!
2
(8
:
40)
playsakeyroleinmuchofthisanalysis. Verysimilaridentitiesappearinverydierent-lookingcontexts
such as dierential equations,andthere they willlead to results much like theoneshere. . Atensor
satisfyingthis equationiscalled \Hermitian"or\symmetric"dependingon whetheraphysicistora
mathematicianrespectivelyistalking.Itparallelstheresultsintheequations(7.48)to(7.50).
Inthecalculationofeigenvaluesandeigenvectorsintheexampleoftheprecedingsection,the
eigenvalues(rootsofacubic)wererealandtheeigenvectorsareorthogonal.Isthatageneralproperty?
Fortensors thatsatisfythe identityderived in Eq.(8.40) it is. . Toprove e this the manipulationsare
simplebutnotobvious.Assumethatyouhavetwoeigenvectors:
I
(
~!
1
)=
1
~!
1
and
I
(
~!
2
)=
2
~!
2
Takethescalarproductoftherstwith
~!
*
2
andthesecondwith
~!
*
1
.
~!
*
2
.
I
(
~!
1
)=
1
~!
*
2
.
~!
1
and
~!
*
1
.
I
(
~!
2
)=
2
~!
*
1
.
~!
2
Takethecomplexconjugateofthesecondequation:
~!
1
.
I
(
~!
*
2
)=
*
2
~!
1
.
~!
*
2
Subtractthisfromtherstoftheprecedingequations.
~!
*
2
.
I
(
~!
1
~!
1
.
I
(
~!
*
2
)=(
1
*
2
)
~!
1
.
~!
*
2
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8|RigidBodyMotion
279
TheidentityEq.(8.40)saysthattheleftsideofthisequationiszero,andtherightsideis
0=
1
*
2
~!
*
2
.
~!
1
(8
:
41)
Nowtakethespecialcasethat
~!
1
=
~!
2
.Thesetwovectorsarethesamevector,sonecessarilythetwo
eigenvaluesarethesamealso
0=
1
*
1
~!
*
1
.
~!
1
Thesecondfactorisnotzero,andthisimpliesthattherstfactormustbe. Thatinturnsaysthat
isreal.Withreal
andrealcomponentsoftheinertiatensor,allthecomponentsof
~!
arerealtoo.
NowEq.(8.41)is
0=
1
2
~!
2
.
~!
1
(8
:
42)
and this implies that fortwodierent eigenvalues, i.e.
1
6=
2
, the correspondingeigenvectors s are
orthogonal,asexempliedintheexampleofFigure8.11:(^
z
)
?
(^
y
^
x
)
?
(^
x
+^
y
).
Theseeigenvaluesarethemomentsofinertiaabouttheaxisdenedbytheeigenvector,andthey
areofcoursenon-negative. Ofcourse? Yes,foraneigenvector,
!
2
=
~!
.
I
(
~!
)=
Z
dm
r
2
!
2
(
~r
.
~!
)
2
=
!
2
Z
dmr
2
?
where
r
?
is theperpendiculardistancefrom
dm
totheaxisdenedby
~!
. Thisintegralwillbezero
onlyinthespecialcasethatallthemassisontheaxis|whenthebodyisaline. Inallothercasesit
ispositive. Thisnalintegralisthemomentofinertiaaboutthe
~!
direction.
8.6Dynamics
Thebasicequationforrotatingobjectsis
~
=
d
~
L=dt
,Eq.(8.6).Now
~
L
isnothingmorethanatensor
expressioninvolving
~!
.
~
=
d
dt
I
(
~!
)=
I
d~!
dt
+
dI
dt
(
~!
)
Writtenthiswayallthecomponentsoftheinertiatensoraretime-dependent,andthat’snotafruitful
approach. The e trickis totransform tothe rotating coordinatesysteminwhichthe body is atrest.
Thenthethetensor
I
isconstant. Fromtheequation(5.6),
~
=
d
~
L
dt
=
_
~
L
+
~!
~
L
=
I
_
~!
+
~!
I
(
~!
)
andthisagainusesthechapterveconventionthatthedotnowreferstothetimederivativeinthe
rotatingsystem. Inthisrotatingsystem,youcanpickthecoordinatessothatthecomponentsof
I
areadiagonalmatrix|thebasisusingeigenvectorsof
I
.
~
=
I
_
~!
+
^
x!
x
+^
y!
y
+^
z!
z
I
xx
^
x!
x
+
I
yy
^
y!
y
+
I
zz
^
z!
z
x
=
I
xx
_
!
x
+
!
y
!
z
(
I
zz
I
yy
)
y
=
I
yy
_
!
y
+
!
z
!
x
(
I
xx
I
zz
)
(8
:
43)
z
=
I
zz
_
!
z
+
!
x
!
y
(
I
yy
I
xx
)
ThesearetheEulerequations. (Asifhedidn’thaveenoughthingsnamedforhim.Heevenusedtobe
ontheSwisstenfrancnote.)
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8|RigidBodyMotion
280
Example
Ifyourautomobiletireismisaligned,itsaxisofrotationdoesn’tlineupwithitsaxisofsymmetry.
Callthe
z
-axisthesymmetryaxisofthetire,then
I
xx
=
I
yy
. Asyoudriveatconstantvelocitywhat
torquedoesthiscauseonyourtire?Intheinertialsystem
~!
isaconstantbecauseitisdenedbythe
axleofthecar. Whathappensto
~!
intherotatingsystem?Again,Eqs.(5.6)and(5.13):
~!
z
-axis
x
-axis
~
Fig.8.12
d~!
dt
=
_
~!
+
~!
~!
=
_
~!
=0
(8
:
44)
andthismeansthatitisconstantintherotatingsystemtoo.Theequations(8.43)arenow
x
=
!
y
!
z
(
I
zz
I
yy
)
;
y
=
!
z
!
x
(
I
xx
I
zz
)
;
z
=0
Ifthemisalignmentbetween
~!
andtheaxis ofthetireis theangle
then pickthe
x
-
y
coordinates
(rotating)sothat
~!
=
!
^
z
cos
!
^
x
sin
.
x
=0
;
y
=
!
2
cos
sin
(
I
zz
I
xx
)
;
z
=0
Thisisaconstanttorqueaboutthe
y
-axis,sowhydoesitcausethewheeltoshake?Thisistherotating
systemremember. Intheinertialsystem(thedriver)thistorquevector,
y
^
y
,isspinningabouttheaxle
atarate
!
,andthesizeofthistorquevariesasthesquareof
!
.
Howcanyoupicturethis?Imagineyourselfstandingonthisdisk(tire?) andhanging
ontothe
z
-axisashardasyoucan. Yourfeetarestraddlingthe
x
-axis,whichisstationary
betweenyourlegs. Ifyoulookstraightaheadyouseetheangularvelocityvectorstanding
still infrontofyou (
_
~!
= 0). But if f youlookinthe distance you may get dizzy from
watchingtheworldspinningaround. Ifyoulookdownandtoyourrightyouwillseethe
torquevectoratyourfeet,pointingalongthe
y
-axis. (Noticethat
I
zz
>I
xx
.) Andwhere
is
~
L
?
~
L
=
I
(
~!
)=
I
(
!
^
z
cos
!
^
x
sin
)=
!
I
zz
^
z
cos
I
xx
^
x
sin
Thispointsinadirectionclosertothe
z
-axisthanthe
~!
vectordoes. Asyoulookstraight
aheadyoucan seeitxedin frontofyouonthenearsideof
~!
. Remember,
_
~
L
=0. I
didn’tdrawthisvectorbecausethepictureisclutteredenoughalready.
FreeRotation
Whatifthereisnotorque,andthe object is freeto rotateinspacelike afrisbee oraplanet. . The
equations(8.43)(with
~
=0)aredierentialequationsfor
~!
. Theycanbesolved,butexceptforthe
caseofasymmetricrigidbodythesolutionisnon-linearandtough. Forthatreasonpick
I
xx
=
I
yy
.
0=
I
xx
_
!
x
+
!
y
!
z
(
I
zz
I
xx
)
0=
I
xx
_
!
y
+
!
z
!
x
(
I
xx
I
zz
)
(8
:
45)
0=
I
zz
_
!
z
+0
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8|RigidBodyMotion
281
Thelastlinesaysthat
!
z
isaconstant.Thatinturnmakesthersttwoequationslinearequationsfor
!
x
and
!
y
.Youknowhowtosolvethose:
e
t
.
!
x
(
t
)=
Ae
t
;
!
y
(
t
)=
Be
t
0=
I
xx
A
+
!
z
(
I
zz
I
xx
)
B
0=
I
xx
B
!
z
(
I
zz
I
xx
)
A
Anon-zerosolutionfor
A
and
B
requiresthatthedeterminantvanish.
I
2
xx
2
+
!
2
z
(
I
zz
I
xx
)
2
=0  !
=
i!
z
I
zz
I
xx
I
xx
=
i!
0
(8
:
46)
Putthesebackintooneoftheequationsfor
A
and
B
toget
0=
I
xx
A
i!
z
I
zz
I
xx
I
xx
+
!
z
(
I
zz
I
xx
)
B
! 
iA
+
B
=0
Asolutionisnow
!
x
=
Ae
i!0t
+
A
0
e
i!0t
;
!
y
iAe
i!0t
+
iA
0
e
i!0t
Chosesomeinitialconditions:
!
x
(0)=
!
0
;
!
y
(0)=0
then
!
x
(
t
)=
!
0
cos
!
0
t;
!
y
(
t
)=
!
0
sin
!
0
t
(8
:
47)
Therotationaxisprecessesaboutthe
z
-axisatanangularvelocity
!
0
.
Example
Earth.Itisfreelyspinninginspaceanditisslightlyellipsoidalsothatthemomentsofinertiaaren’t
thesameaboutallaxes. Theequatorialbulgemakes
I
zz
>I
xx
=
I
yy
. Iftheangularvelocityofthe
planetexactlylinedupwithitsaxisofsymmetry,theconstant
!
0
inEq.(8.47)wouldbezero. Itwould
betoomuchofacoincidenceforthealignmenttobeperfectanditisn’t.Itmissesbyanamountsuch
thatifyougototheNorthpoleandlookfortheangularvelocityvectoryouwillnditseveralmeters
away. Thenitwandersaroundthepoleatarate
!
0
2
=
400days. Itsmotionisnotasregularasthis
rigidbodyanalysiswouldleadyoutobelieve,buttheEarthisn’tperfectlyrigideither.
ThattheEarthisn’trigidshouldleadtodampingofthisoscillationwithinyearsorcenturies,but
it’sstillhere.IthasbeenapuzzlewhatkeepsthisChandlerwobblegoing,butrecentanalysispointsto
uctuatingpressureonthebottomoftheoceanasthelargestsourceoftheexcitation.
Thisprecessioncangivegeologistsinformationabouttheinterioroftheplanet.Equation(8.46)
tellsyouabout
I
zz
I
xx
andthatgivessomeconstraintsonthedistributionofmasswithintheEarth.
And notjust Earth; Mars too. . One e of the measuring devices senttoMars looked at thatplanet’s
wobble,and thatsayssomethingabouttheinteriorstructureof Mars. . Why y hasn’tthewobble been
completelydampedinthecaseofMars? Afterall,ithasnooceanstoexcitetheoscillation.Unknown.
MaybetheideathattheoceanscauseitonEarthiswrong.Youwillhavetodoasearchofthecurrent
literatureonthesubjecttogetsomeideasofthecomplexityoftheproblem.
StabilityofRotations
Whathappenswhenyoutossahammeroratennisracketupandsetitspinning?Theanswerdepends
verymuchonhowyoudoit,andthemotioncanbeverysmoothorverywild.Ifyoudon’thaveeither
ahammerorarackethandy,perhapsyouhaveaheavyrubberband. Thenyoucanwrapitarounda
booksothatthepagesdon’topenupwhenyoutossitupandspinning. Dependingontheaxisabout
whichitisspinningyouwillgetverydierentresults.
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8|RigidBodyMotion
282
To analyze this, startagain from Euler’s equations (8.43). . This s time assume that the three
momentsofinertiaaredierentfromeachother,butassumethattheangularvelocityvectorisalmost
along oneofthe principle axes. . Thiscalculation n will involve aseries expansionand alinearization.
Assumethattherotationisalmostalongthe
z
-axis. Thetorqueiszerobecauseyouhavetossedthe
objectupanditisthenfreetorotateasitwill.
~!
(
t
)=
!
0
^
z
+
~
(
t
)
;
with
!
0
Startfromtheequations (8.43)with
~
=0,andasusualwiththese expansions,keep only the rst
ordertermsin
.
0=
I
xx
_
x
+
y
!
0
(
I
zz
I
yy
)
0=
I
yy
_
y
+
x
!
0
(
I
xx
I
zz
)
0=
I
zz
_
z
Every place that an
2
appeared I dropped d it, and the resulting g equations s are e linear. . In n the third
equation,both
!
x
and
!
y
arerstorderin
sothatkilleditsnalterm.
The third equation says that
z
is aconstant, , and Imay y as well take ittobe zero because
anon-zerovaluewould justbearedenition of the originalrotationrate
!
0
aboutthe
z
-axis. The
othertwoequationsarelineardierentialequationsfor
x
and
y
. Youknowhowtohandlethose: an
exponential.
x
=
Ae
t
;
y
=
Be
t
then
0
0
=
I
xx
!
0
(
I
zz
I
yy
)
!
0
(
I
xx
I
zz
)
I
yy

A
B
Togetanon-zerosolutionthedeterminantmustvanish.
I
xx
I
yy
2
!
2
0
(
I
xx
I
zz
)(
I
zz
I
yy
)=0
(8
:
48)
The nature of thesesolutions depends on the sign of
2
. Ifitis s negative then
is imaginary and
you haveoscillations aboutthe ^
z
direction;if
2
ispositivethen
isrealandyouhaveexponential
movementawayfromtherethe^
z
-axis:thatrotationisunstable.
2
/(
I
xx
I
zz
)(
I
zz
I
yy
)
If
z
hasthelargestmoment,
I
zz
>I
xx
and
I
zz
>I
yy
;
then
2
<
0
and this says that
is imaginary and thatthe motionis stable. . If
I
zz
is thesmallest ofthe three
momentsyouhavethesame result:
is imaginary and the motionisstable. . Itisonlyinthe e third
case,when
I
zz
isintermediatebetween the othertwomomentsofinertia,that
2
ispositiveand
comesouttobereal. Thatmotionisunstable. Takeabookandwrapaheavyrubberbandaboutit.
Nowtossitintheair,spinningaboutoneofthethreeaxesofsymmetry. Doitforeachaxis,andthe
dierenceintheresultswillbeveryobvious. SeeforexampletheYouTubevideoSolidBodyRotation,
doneinorbit.
There is more to learn about the stability y of f these rotations. . What t is the kinetic energy y of
rotationabouteachaxis?UseEq.(8.33)abouteachaxis:
K
=
1
2
~!
.
~
L
.
~!
=
!
0
^
x
!
K
=
1
2
I
xx
!
2
0
;
~!
=
!
0
^
y
!
K
=
1
2
I
yy
!
2
0
;
~!
=
!
0
^
z
!
K
=
1
2
I
zz
!
2
0
8|RigidBodyMotion
283
Theangularmomentumineachcaseiseasybecausetheseareeigenvectorsoftheinertiatensorand
arerespectively
L
x
=
I
xx
!
0
;
L
y
=
I
yy
!
0
;
L
z
=
I
zz
!
0
Writetheenergyintermsoftheseangularmomenta.
~!
=
!
0
^
x
!
K
=
L
2
2
I
xx
; ~!
=
!
0
^
y
!
K
=
L
2
2
I
yy
; ~!
=
!
0
^
z
!
K
=
L
2
2
I
zz
Thisassumesthesamemagnitudefortheangularmomentumineverycase,andtheseequationssay
thatforagivenangularmomentumtherotationabouttheaxiswiththelargestmomentofinertiahas
thesmallestkineticenergy.
Whetherthisobjectisarigidbodyornot,iftherearenotorquesonittheangularmomentum
isconserved. Ifyouhaveatrulyrigidbody,thenbothangularmomentumandmechanicalenergyare
conserved,butcompletelyrigidbodiesareamathematicalction. Anyrealobjectwill exslightly,and
thishasanimportantconsequence. Ifasatelliteissettorotatingaboutoneofitsprincipleaxesand
if thatis oneofthe twostable axes,then youmightexpecttherotationtoremainaboutthataxis
forever. But
:::
Atiny exibilitywillcauseatinyfriction|awaytodissipateenergy. Theonlytruly
stablerotationinthiscasewillbeabouttheaxiswiththelargestmomentofinertia,theonethathas
thesmallestkineticenergyforthegivenangularmomentum.
~!
Fig.8.13
EarlyinthespaceprogramthesatelliteExplorer-1wasplacedinorbitaroundEarthandsetspinning
aboutitslongaxis,theonewiththesmallestmoment,andonetheythoughtwouldprovideastable
rotation.Thesatellitehowever,hadwhipantennaswithmorethanenough exibilitytodestabilizethis
rotationandtosetittumblingaboutitsreallystableaxis|endoverend|muchtotheembarrassment
oftheengineersandphysicistsinvolved.
Ifyouhaveseenthe motionpictures 2001andits sequel 2010, aspace ship wasleftinorbit
aroundJupiter’smoonIoattheendoftherstpicture,andwhentheyreturnedinthesequelitwas
tumbling end-over-end. . Theymadenocommentinthemovie,but t they got the physics right. . The
sameappliestothemorerecentmovie,\Gravity". WhenSandraBullockwasrstcutloosefromthe
spacecraft,shewastumblingwildly.Afterabrieftimeshewasrotatingsmoothlyheadoverfeet.Was
thatreallyabouttheaxiswiththelargestmomentofinertia?Shoulditbesidewaysorasshown,front
toback? Thatdoesdependonthemassofherbackpack,anditappearsthattheydiditcorrectlyin
themovie.
8.7PerturbationTheory_
Iftheequationfortheeigenvaluesofaninertiatensoris easytosolvethenyou’relucky. . Whatifit
isn’t?Atworstyoumayhavetoresorttosolvingitnumerically,butthereisanintermediatesituation
thatcomesupsurprisinglyoftenandthatfallsbetweentheseextremes. Iftheproblemisclose toone
thatis easilysolved,thenthe method ofperturbationsisavailable. . Thisinvolves s aseries expansion
thatcanbeveryusefulinboththequantitativeandqualitativeanalysisoftheseproblems.
Forexample,ifyouhaveamassofuniformdensityandintheshapeofarectangularboxthen
it’seasytondtheeigenvectorsandeigenvaluesoftheinertiatensorusingthecenteroftheboxasthe
origin. Theprincipleaxesareparalleltotheedgesofthebox,andtheeigenvaluesareeasilycomputed.
8|RigidBodyMotion
284
Whatifyounowaddasmallmasstothecornerofthebox? Suddenlythesymmetryislostandyou
don’thaveanyintuitivefeelingforthedirectionsoftheeigenvectors,butinsomesensetheeigenvectors
oughttobeclosetotheoriginalonesthatwereparalleltotheedges.
Example
Computethecomponentsoftheinertiatensorintheexamplejustmentioned. Thenwhatarethe
eigenvectorsandeigenvaluesofthetensor? Callthesidesofthebox
a
,
b
,
c
. Firstcompute
R
dmx
2
.
Afterthatit’salldownhill.For
dm
useasliceofthewholemass:
dm
=
M
abc
bcdx
!
Z
dmx
2
=
Z
a=
2
a=
2
M
abc
bcdxx
2
=
M
a
x
3
3
a=
2
a=
2
=
Ma
2
12
The integrals of
y
2
dm
and
z
2
dm
just change
a
to
b
to
c
,and the o-diagonalelementsarezero.
Thisproduces(
I
0
). Forlittle
m
,itisjustapointmass,soEq.(8.24)producestheresultwithoutany
integration. Thisis(
I
1
).
(
I
0
)=
M
12
0
@
b
2
+
c
2
0
0
0
a
2
+
c
2
0
0
0
a
2
+
b
2
1
A
(
I
1
)=
m
4
0
@
b
2
+
c
2
ab
ac
ab
a
2
+
c
2
bc
ac
bc
b
2
+
c
2
1
A
Fig.8.14
Thetotaltensorcomesfrombothofthemassesandthecomponentsarethesum
ofthesetwomatrices. Suddenly,ndingtheeigenvectorsandeigenvaluesishard.Ifyou
wantageneralanswerforthiscase,lookuptheformulaforthesolutionofageneralcubic
equation. YouwillthenseewhyI’mnotgoingtouseit.
If
m
M
, the eigenvectors and eigenvalues s will be close to those of
I
0
, and
because that matrix isdiagonal, you already knowwhatthese approximate values are.
Thenextproblemistondthecorrectionstothem.
Theideaistoassumethatthereisapowerseriesexpansionfortheseresultsandthentocompute
thetermsofthatexpansiononeatatime.Thisis
I
=
I
0
+
I
1
and
I
(
~!
)=
~!;
with
~!
=
~!
0
+
~!
1
+
2
~!
2
+
and
=
0
+

1
+
2
2
+
(8
:
49)
Thisintroductionof\
"isaconvenientwaytokeeptrackofthemanyterms. Noneofthese
~!
0
,
1
,
2
,
:::
orthesimilarlysubscripted
’sareknown. Thatdoesn’tmatterbecauseyouwilllettheequations
tellyouwhattheyare. Plugin:
I
0
+
I
1
 
~!
0
+
~!
1
+
=
0
+

1
+
 
~!
0
+
~!
1
+
Multiplyeverythingandlookforthevariouscoecientsofpowersof
.
0
:
I
0
(
~!
0
)=
0
~!
0
1
:
I
0
(
~!
1
)+
I
1
(
~!
0
)=
0
~!
1
+
1
~!
0
2
:
I
0
(
~!
2
)+
I
1
(
~!
1
)=
0
~!
2
+
1
~!
1
+
2
~!
0

Whenyouhave twopowerseriesthatequaleachother,this requires thatthe individualtermsmust
matchtoo,andthatisallthatIdidtoextracttheseparateequations.
8|RigidBodyMotion
285
Nowrearrangetheseequationstoputthemintoastandardform.
I
0
(
~!
0
0
~!
0
=0
(0)
I
0
(
~!
1
0
~!
1
=
1
~!
0
I
1
(
~!
0
)
(1)
I
0
(
~!
2
0
~!
2
=
1
~!
1
+
2
~!
0
I
1
(
~!
1
)
(2)

(8
:
50)
The most importantproperty thattheseequationspossessthatthatunderalmost all circumstances
theycannotbesolved.Theyhavenosolutions. Ifyouplungeaheadandtrytosolvethemanywaythen
atsomepointyouwillndyourselfdividingbyzero. Thisdoesn’tmeanthattheproblemishopeless,
itjustmeansthatyouhavetondtheveryspecialcasesinwhichtheycanbesolved. Therearetwo
waystodothis,bruteforceorcleverness. Bothareinstructive,butthesecondoneiswhatyoushould
masterbecauseitisatechniquethatyouwillseemanymoretimes. Youmayhaveseenitalreadyif
you haveworkedthroughthesections 7.9and 7.10. Workproblem8.52forthebruteforcemethod.
It’snothardinthiscase.
The0
th
equationaboveistheonethatyousupposedlyknowhowtosolve. It’stheonewithout
theextracomplicatingtermcomingfromthepointmassonthecorner.Fortherest
:::
The clever method is to use Eq. (8.40), but now w all the e numbers are e real:
~!
1
.
I
(
~!
2
) =
I
(
~!
1
)
.
~!
2
. Applyittotheseperturbationequations. . TakethescalarproductofEq.(8.50)(1)with
~!
0
.
~!
0
.
I
0
(
~!
1
0
~!
1
=
~!
0
.
1
~!
0
I
1
(
~!
0
)
(8
:
51)
Applythesymmetrypropertyoftheinertiatensortochangetheleftsideofthisto
I
0
(
~!
0
)
.
~!
1
0
~!
0
.
~!
1
=
I
0
(
~!
0
0
~!
0
.
~!
1
=0
ThisimpliesthatthescalarproductontherightsideofEq.(8.51)mustbezeroalso.Thatis
~!
0
.
1
~!
0
~!
0
.
I
1
(
~!
0
)=0  !
1
=
!
0
.
I
1
(
~!
0
)
~!
0
.
~!
0
and you u have found d the e correction n to o the e original eigenvalue without t doing g any additional matrix
manipulation|onlytakingascalarproduct.Themagnitudeof
~!
0
cancelsfromthisequation,soyou
canmakeitssizeanythingyouwant. Translatethisintothenotationofmatricesandyouhave
~!
0
=
0
@
!
0
0
0
1
A
!
1
=(1 0 0)
m
4
0
@
b
2
+
c
2
ab
ac
ab
a
2
+
c
2
bc
ac
bc
b
2
+
c
2
1
A
0
@
1
0
0
1
A
=
m
4
(
b
2
+
c
2
) (8
:
52)
astherstordercorrectiontothersteigenvalue. Youwilleasilyseethattherstordercorrectionsto
theothereigenvaluesaretheotherdiagonalelementsof(
I
1
).
Whatistherstordercorrectiontotheeigenvector?
~!
1
? ThatrequiressolvingEq.(8.50)(1).
Writeitoutinallitsgorydetail,andthetheequationsthatyouhavetosolvearesurprisinglysimple.
I
0
(
~!
1
0
~!
1
=
1
~!
0
I
1
(
~!
0
)
M
12
0
@
b
2
+
c
2
0
0
0
a
2
+
c
2
0
0
0
a
2
+
b
2
1
A
0
@
!
1
x
!
1
y
!
1
z
1
A
M
12
(
b
2
+
c
2
)
0
@
!
1
x
!
1
y
!
1
z
1
A
=
m
4
(
b
2
+
c
2
)
0
@
!
0
0
0
1
A
m
4
0
@
b
2
+
c
2
ab
ac
ab
a
2
+
c
2
bc
ac
bc
b
2
+
c
2
1
A
0
@
!
0
0
0
1
A
(8
:
53)
8|RigidBodyMotion
286
!
M
12
0
@
0
(
a
2
b
2
)
!
1
y
(
a
2
c
2
)
!
1
z
1
A
=
m
4
0
@
0
ab
ac
1
A
!
0
(8
:
54)
Andhereyouhavetheequationsfor
!
1
y
and
!
1
z
.Theirsizesareproportionalto
m=M
,asbetstheir
beingsmallcorrections.
Areyoudonehere? Imaybebutyouaren’t. Whatdoesthisresultlooklike? If
a
,
b
,and
c
are
somethinglike1,2,3insomeunits,thenhowdoesthedirectionoftheeigenvectorchangewhenthis
correctionisadded?Isittowardorawayfrom
m
? Andifthesidesareinstead2,3,1,thendoesyour
answertothischange? Playaroundwithspecialcases.
Whatabout
!
1
x
,theothercomponentofthecorrection.Takethattobezero,becauseincluding
itwouldjustredenetheoriginal
!
0
. (Andthere’satechnicalreason,becauseifyougointothehigher
ordercorrectionsitwillreallymessthingsupifyoudon’tdothis.)
Exercises
Threeequal pointmasses
m
areatthecornersoftherighttriangle(
x;y
)=(0
;
0),(
a;
0),(0
;b
).
Whereisthecenterofmass?
Removethemassesfromtheprecedingtriangleandreplacethesideswiththinrodsofconstantlinear
massdensity
.Whereisthecenterofmass?
Removetherodsfromtheprecedingtriangleandreplacetheareawithauniformsheetofconstant
areamassdensity
. Whereisthecenterofmass?
Writeouttheequations(8.4)explicitly for2+2=4pointmassesandverifythatitdoeswhatit
says.
Atensor
f
isdenedbytheequation
f
(
~v
)=
~v
forall
~v
.Computethecomponentsof
f
.
Avector-valuedfunction
F
actsonvectorsintheplane,reversingthedirectionofeach. Compute
thecomponentsofthisfunction,a22matrix.
Withrespecttotheorigin,whatarethecomponentsofthetensorofinertiaforapointmass
m
at
thepoint(
x;y;z
)=(
a;
0
;
0)?
Withrespecttotheorigin,whatarethecomponentsofthetensorofinertiaforapointmass
m
at
thepoint(
x;y;z
)=(
a;a;
0)? andwhatifitscoordinatesare(
a;a;a
)?
Use the summation convention andnotationof Eq.(0.56) to writethe equations (8.17),(8.21),
(8.22).
10 Forthespringsthatappearingure8.6,callthem
k
1
leftandrightand
k
2
topandbottom.What
arethecomponentsofthe
f
in
~
d
=
f
(
~
F
)?
11 Whatarethedimensionsof
I
(^
x
)?
12 StartingatEq.(8.38),dropthedenominators
p
2fromthebasisvectors,sothat
~e
1
=^
x
+^
y
etc.
Nowcomputethecomponentsofthesameinertiatensorasdonethere.
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