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asp.net mvc 4 and the web api pdf free download : Scan multiple pages into one pdf control Library platform web page .net winforms web browser mechanics27-part1298

8|RigidBodyMotion
267
Example
waist? Thehoopis s acircle,but you are notspinningit about its centerbut about its
edge. Ifyouareworkingreallyhardatit,thentoagoodapproximationapointonthe
edgeisnearlystandingstill.Whatisthehoop’sangularmomentum?Usetheparallelaxis
theoremandstartbyndingtheangularmomentumassumingthatthehoopisspinning
~!
isalongthisaxis.
~!
~r
~
L
=
R
dm~r
~!
~r
,
~!
~r
aretangenttothecirclewith
magnitude
!r
.Inturn,
~r
thisvectorisalongtheaxiswithmagnitude
!r
2
. Thatmakes
~
L
=
Mr
2
~!
alongtheaxis(outofthepageandparalleltotheangularvelocity).
~r
cm
pointsfromtherightedgetothecenter
Mr
2
~!
to
thetotal.
~
L
=2
Mr
2
~!
.
8.3TensorComponents
Computingwithtensorsislikecomputingwithvectors|thegeometrygetsoutofhandquickly. For
vectorsyoucanusethetoolofcomponentstodocalculations,andthesamethingistruehere.Tensors
Godirectlyforthecomponentsoftheinertiatensor.Othertensorsareconceptuallynodierent.
Startwiththevectors
~!
and
~
L
,whichhavecomponentswithrespecttoabasisyou’vechosen
~!
=
!
x
^
x
+
!
y
^
y
+
!
z
^
z
and
~
L
=
L
x
^
x
+
L
y
^
y
+
L
z
^
z
Nowrelatethemwiththefunction
I
.
~
L
=
I
(
~!
)
is
L
x
^
x
+
L
y
^
y
+
L
z
^
z
=
I
(
!
x
^
x
+
!
y
^
y
+
!
z
^
z
)
(8
:
14)
UsethelinearitypropertyofEq.(8.9),whichisthedeningpropertyofatensor.
I
(
!
x
^
x
+
!
y
^
y
+
!
z
^
z
)=
!
x
I
(^
x
)+
!
y
I
(^
y
)+
!
z
I
(^
z
)
(8
:
15)
Theexpression
I
(^
x
)isavector.Assuchithasthreecomponents. Denotetheseas
I
(^
x
)=
I
xx
^
x
+
I
yx
^
y
+
I
zx
^
z
(8
:
16)
Inthesameway,theothertermsare
I
(^
y
)=
I
xy
^
x
+
I
yy
^
y
+
I
zy
^
z
and
I
(^
z
)=
I
xz
^
x
+
I
yz
^
y
+
I
zz
^
z
(8
:
17)
Theindicesdenotewhichoutputbasisvectorandwhichinputbasisvectoryouarereferringto,
andtheorderoftheindicesiscarefullychosenforlaterconvenience.Theseninenumbers(e.g.
I
xy
)are
thecomponentsofthetensor
I
. Insertequations(8.15),(8.16),and(8.17)intoequation(8.14)toget
L
x
^
x
+
L
y
^
y
+
L
z
^
z
=
!
x
I
xx
^
x
+
I
yx
^
y
+
I
zx
^
z
+
!
y
I
xy
^x+I
yy
^y+I
zy
^z
+
!
z
I
xz
^
x
+
I
yz
^
y
+
I
zz
^
z
* Therearegeneralizationsofthisstatementformorecomplicatedkindsoftensors,butfortoday
thisisgoodenough.
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8|RigidBodyMotion
268
Forthesevectorstobeequal,theirrespectivecomponentsmustmatch.Equatethecoecientof^
x
on
thelefttothesimilarcoecientontheright.Nowrepeattheprocessfor^
y
and^
z
. Thisexpressesthe
componentsof
~
L
intermsofthecomponentsof
~!
.
L
x
=
I
xx
!
x
+
I
xy
!
y
+
I
xz
!
z
L
y
=
I
yx
!
x
+
I
yy
!
y
+
I
yz
!
z
(8
:
18)
L
z
=
I
zx
!
x
+
I
zy
!
y
+
I
zz
!
z
Inmatrixnotationthisis
0
@
L
x
L
y
L
z
1
A
=
0
@
I
xx
I
xy
I
xz
I
yx
I
yy
I
yz
I
zx
I
zy
I
zz
1
A
0
@
!
x
!
y
!
z
1
A
(8
:
19)
andthisisthereasonforthepeculiar-lookingwaythatyouaretoldtomultiplymatrices,acrosstherow
oftherstfactoranddownthecolumnsofthesecond.Itallcomesfromtheequations(8.14)-(8.17),
dening thecomponentsofa tensor. . The e meaningof theexpression (8.19is thesetof equations
(8.18). The e matrix indices are arrangedas
I
(whichrow)
;
(whichcolumn)
. Compare e the placementofthe
indicesinEq.(8.18)andinEq.(8.17).Thenotationisdesignedtocomeoutinthewayconventionally
usedwithmatrices.
Anabbreviatedformfortheseequations iseasiertowriteifyouuseauniednotationforthe
basisvectors.
~e
i
,withtheindex
i
runningovertheset
x
,
y
,
z
(ormoreoften1,2,3).
~e
i
replacesthe
threevectors ^
x
,^
y
,^
z
,andtheequations(8.16)and(8.17)becomeoneequation.
I
(
~e
i
)=
X
j
I
ji
~e
j
(8
:
20)
RecallthediscussionatEq.(0.27).
Themanipulationsthattookyoufromequation(8.14)through(8.19)becomefarmorecompact
ifyouusethenotationofEq.(8.20).
~
L
=
X
i
L
i
~e
i
=
I
(
~!
)=
I
X
j
!
j
~e
j
=
X
j
!
j
I
(
~e
j
)=
X
j
!
j
X
i
I
ij
~e
i
(8
:
21)
The basis vectors
~e
i
are independent, so the coecients of
~e
i
mustagree on the two sides of the
equation. Thatis,
L
i
=
X
j
I
ij
!
j
(8
:
22)
Thesetwolinesaresomuchmorecompactthatyoushouldcomparetheircontentline-by-linewiththe
precedingequationstoverifythattheyarewhattheyclaimtobe.
Tocomputethesecomponents,gobacktothedenitionoftheinertiatensor,Eq.(8.7). You
needavectoridentityforthetriplecrossproducttomanipulatethis,Eq.(0.24).
~
A
(
~
B
~
C
)=
~
B
(
~
A
.
~
C
~
C
(
~
A
.
~
B
) =)
I
(
~!
)=
Z
dm~r
(
~!
~r
)=
Z
dm
r
2
~!
~r
(
~!
.
~r
)
(8
:
23)
Nowcompute
I
(^
x
).
I
(^
x
)=
Z
dm
r
2
^~r(^x
.
~r
)
=
Z
dm
(
x
2
+
y
2
+
z
2
)^
x
(
x^x
+
y^y
+
z^z
)
x
=
Z
dm
(
y
2
+
z
2
)^
x
yx
^
y
zx
^
z
=
I
xx
^
x
+
I
yx
^
y
+
I
zx
^
z
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8|RigidBodyMotion
269
of^
x
,^
y
,and ^
z
togettherstcolumnofthematrixofcomponents,andthecalculationsfortheother
sixcomponentsareidentical,givingthesecondandthirdcolumns ofthematrix.
I
xx
=
Z
dm
(
y
2
+
z
2
)
I
xy
Z
dmxy
I
xz
Z
dmxz
I
yx
Z
dmyx
I
yy
=
Z
dm
(
x
2
+
z
2
)
I
yz
Z
dmyz
I
zx
Z
dmzx
I
zy
Z
dmzy
I
zz
=
Z
dm
(
x
2
+
y
2
)
Thethreediagonalcomponents,
I
xx
,
I
yy
,and
I
zz
axes,and theothercomponents are called the products of inertia. . There’s s some symmetry y among
theninecomponents,as
I
xy
=
I
yx
etc,makingthematrixinEq.(8.19)symmetric. Thisisaspecial
propertyoftheinertiatensor,anditisnottrueforalltensors.Youcanwritethismatrixas
(
I
)=
Z
dm
0
@
y
2
+
z
2
xy
xz
xy
x
2
+
z
2
yz
xz
yz
x
2
+
y
2
1
A
(8
:
24)
morecompact? Yes. Use
x
1
,
x
2
,and
x
3
x
,
y
,and
z
,thenapplythe
summationconventionasinEq.(0.28),thenEq.(8.23)becomes
L
i
~e
i
=
I
(
~!
)=
Z
dm
x
j
x
j
!
i
~e
i
x
i
~e
i
!
j
x
j
L
i
=
Z
dm
x
j
x
j
!
i
x
i
!
j
x
j
=
Z
dm
x
j
x
j
ik
x
i
x
k
!
k
I
ik
=
Z
dm
x
j
x
j
ik
x
i
x
k
L
i
=
I
ik
!
k
(8
:
25)
Recallthatnowarepeatedindexwithinasingletermisautomaticallysummed. Also,suchasummed
alsosimpliestheappearanceofEq.(8.21)byremovingalltheexplicitsummationsymbols. Theyare
unnecessarynow.
Example
Athinuniformrectangularplateisplacedwithonecornerattheoriginandwiththesidesalongthe
x
and
y
axes: (0
<x<a
),(0
<y<b
). Whatarethecomponentsofthetensorofinertia?
LookatEq.(8.24)andyouseethatalotoftermsarezero|everywherethere’sa
z
.Thismeans
thattheonlyintegralstodoare
Z
dm
of
x
2
; y
2
; xy
Theareamassdensityis
=
m=ab
,and
Z
a
0
dx
Z
b
0
dyx
2
=
ba
3
3
;
Z
a
0
dx
Z
b
0
dyy
2
=
ab
3
3
;
Z
a
0
dx
Z
b
0
dyxy
=
a
2
b
2
4
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8|RigidBodyMotion
270
Multiply eachofthetheseexpressions by the areamassdensityandputthemintotheirappropriate
slotsinthematrixforthetensorcomponents.
(
I
)=
m
ab
0
@
1
3
ab
3
1
4
a
2
b
2
0
1
4
a
2
b
2
1
3
ba
3
0
0
0
1
3
(
ab
3
+
ba
3
)
1
A
=
m
0
@
1
3
b
2
1
4
ab
0
1
4
ab
1
3
a
2
0
0
0
1
3
(
a
2
+
b
2
)
1
A
(8
:
26)
Whathappenstothisresultif
a
or
b
equalszero? Doesthatresultmakesense? Perhapsrelateitto
someequationappearingearlierinthischapter.
Inanearlierstudy ofrotations youencounteredthemomentofinertiaasasubjectinitsown
~
L
=
I~!
. Youcanseenowthatthiswasonlyarstapproximationtothe
subject.Butthenwhatis thatold
I
Z
r
2
?
dm
or
~!
.
I
(
~!
)
=!
2
(8
:
27)
where
r
?
istheperpendiculardistancetoyouraxis. Therstexpressioniswhatyouhavelikelyseen
before, though perhaps not in this s notation. . If f you use acoordinate system where this axis is the
z
-axis,then
r
?
=
p
x
2
+
y
2
andthisis
I
zz
,thebottomrightelementoftheabovematrix.Theother
x
-and
y
-axes.Thesecond,
more complicatedlookingexpressioninEq.(8.27) is amore generalway torelate thetensortothe
moment. Youcanderiveitinproblem8.11. Noticehoweasyitistoabusethenotation,using
I
for
themomentofinertiaand
I
Example
~
L
~!
^y
^z
^xout
1
2
Fig.8.8
Repeatapreviousexample,Figure8.4,onlynowexpressitinanewlanguage.
Twopointmassesareattheendsofalightrodlyinginthe
y
-
z
planeasshown
on theright. . The e integral is just asum over r two terms s this time. . For r both
masses the value of
x
is zero, , and when you evaluate this s sum, , the
y
2
+
z
2
factorisjust
r
2
1
or
r
2
2
dependingonwhichmassyou’redealingwith.
I
xx
=
m
1
r
2
1
+
m
2
r
2
2
I
yy
=(
m
1
r
2
1
+
m
2
r
2
2
)sin
2
;
I
zz
=(
m
1
r
2
1
+
m
2
r
2
2
)cos
2
I
xy
=
I
xz
=0
;
I
yz
= (
m
1
r
2
1
+
m
2
r
2
2
)sin
cos
Asamatrixthisis
(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
1
0
0
0
sin
2
cos
sin
cos
sin
cos
2
1
A
(8
:
28)
Iftheangularvelocityisalongthe
z
-axis,theangularmomentumcomponentsare
0
@
L
x
L
y
L
z
1
A
=(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
1
0
0
0
sin
2
cos
sin
cos
sin
cos
2
1
A
0
@
0
0
!
1
A
=(
m
1
r
2
1
+
m
2
r
2
2
)
!
0
@
0
cos
sin
cos
2
1
A
=)
~
L
=(
m
1
r
2
1
+
m
2
r
2
2
)
!
cos
^
y
sin
+^
z
cos
)
(8
:
29)
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8|RigidBodyMotion
271
Theangularmomentumispointingtowardtheupperleft,perpendiculartotheaxisbetweenthemasses,
andnotinthedirectionof
~!
. Thisbehavioristypical,becauseitisonlyinthecaseofasymmetrical
rotatingbodyorincasesofaspecial
~!
thatthesetwovectorswilllineup.Theimpressionthatyoumay
havegotteninyourrstintroductiontothissubjectwasnodoubtrestrictedtosuchselectexamples.
Iftheangularvelocityisalongtheaxisoftherod,
~!
=^
y
sin
+^
z
cos
,thencompute
~
L
as
(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
1
0
0
0
sin
2
cos
sin
cos
sin
cos
2
1
A
0
@
0
!
cos
!
sin
1
A
=
0
@
0
0
0
1
A
~!
^
z
^
y
Around this line of rotation, all the mass s is on the axis, , so there’s s no angular momentum in this
approximationthatthesearepointmasses.
If the axis s of f rotation is in the direction n ( ^
y
sin
+^
z
cos
), the direction n of f the result t in
Eq. (8.29), and d the direction perpendicular to o the axis between n the e masses, , compute e the e angular
momentum:
(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
1
0
0
0
sin
2
cos
sin
cos
sin
cos
2
1
A
0
@
0
!
sin
!
cos
1
A
=(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
0
!
sin
!
cos
1
A
This
~
L
isinthesamedirectionas
~!
,andifyoustartwith
~!
=
!
^
x
(outofthepage),yougetasimilar
result,withthesameproportionalityfactor,(
m
1
r
2
1
+
m
2
r
2
2
).Alongthesedirectionsthevectors
~!
and
~
L
arealigned.
PerpendicularAxisTheorem
Itapplieswhenamassisdistributedinaplane,sothatitisessentiallytwo-dimensional. Make
thatthe
x
-
y
plane,then
I
zz
=
I
xx
+
I
yy
.Theproofinvolvesnothingmorethanwritingthevaluesof
thecomponentsfromEq.(8.24):
I
zz
=
Z
dm
(
x
2
+
y
2
)
;
but
I
xx
=
Z
dm
(
y
2
+
z
2
)=
Z
dmy
2
and
I
yy
=
Z
dm
(
x
2
+
z
2
)=
Z
dmx
2
(8
:
30)
because
z
=0forall
dm
,andthat’sallthereistoit.
Example
x
y
z
R
Fig.8.9
A thindiskofmass
M
R
hasitscenterattheoriginandhas
z
=0.Computetheinertiacomponents:
I
zz
=
Z
dm
(
x
2
+
y
2
)=
Z
R
0
2
rdr
M
R
2
r
2
=
MR
2
2
Use theperpendicularaxis theorem,and becausethesymmetry ofthe disk
impliesthat
I
xx
=
I
yy
,youseethatbothoftheseare
MR
2
=
4.Theproducts
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8|RigidBodyMotion
272
ofinertiaareallzerobecausetheyallinvolveintegratinganoddfunctionover
asymmetricdomain:
R
a
a
xdx
=0.
(
I
)=
MR
2
4
0
@
1 0 0
0 1 0
0 0 2
1
A
(8
:
31)
Suppose that at some instant the e angular velocity is between the ^
z
and ^
y
directions:
~!
=
!
(^
z
cos
+^
y
sin
).Then
~
L
=
MR
2
!
(2^
z
cos
+^
y
sin
)
=
z
than
does
~!
~!
direction,theangularmomentumwillrotatearoundit.
~
L
and
~!
,andthat’sallIneed.
~
L
willtraceoutaconearound
~!
,
andthatmeansthatyouneedtoapplyatime-varyingtorquetodothis:
~
=
d
~
L=dt
.
Theanglebetweenthetwovectorsis(problem8.26).
sin
=cos
sin
=
p
3cos2
+1
(8
:
32)
~
L
~!
z
y
x
~
L
~!
Fig.8.10
~!
?Forgetthecoordinate
system;thatwilljustgetintheway now. . Thevector
~
L
~!
.
Thelengthof
~
L
isconstant,soitstimederivativeisperpendiculartoitself. Itisalsoperpendicularto
~!
,anditsvalueisthen
d
~
L=dt
=
~!
~
L
=
~
Themagnitudeofthistorqueis
!L
sin
,andyoucangetthemagnitude
L
fromthevector
~
L
inthe
precedingparagraph.
!L
sin
=
1
4
!MR
2
!
4cos
2
+sin
2
1
=
2
cos
sin
p
3cos2
+1=
1
4
M
2
!
2
cos
sin
Therewassomuchcancellationofthecomplicatedfactorsinthiscalculationthatyoushouldsuspect
thatthere’saneasierway.Thereis. Themagnitudesandanglesareconstant,sowhynotevaluatethe
magnitudeoftheproductattheinitialtime?
~!
~
L
=
!
(^
z
cos
+^
y
sin
)
MR
2
!
(2^
z
cos
+^
y
sin
)
=
4=
1
4
MR
2
!
2
^
x
cos
sin
Canyoudothiscalculationofthetorquewithoutreferringtoaspeciccoordinatesystem,just
manipulatingtheoriginalformfor
I
(
~!
)?Yes,butit’snotmuchhelp:
~!
~
L
=
~!
I
(
~!
)=
~!
Z
dm~r
(
~!
~r
)
=
~!
Z
dm
r
2
~!
~r
(
~!
.
~r
)
Z
dm
(
~!
~r
)(
~!
.
~r
)
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8|RigidBodyMotion
273
Ifyoucanndauseforthisequation,you’rewelcometoit.
Thiscalculationhasaverypracticalapplication. Whenyouhavedrivenyourcarforalongtime
it mayneedadjustment. . If f you hitacurbwith the wheel of yourcaryou can knockthewheelout
wheels so that theirangularvelocity and angularmomentum vectors areparallel, , hoping thereby y to
eliminatethetorque
~!
~
L
Arelatedproblemwilloccuriftheaxisofrotationdoesnotpassthroughthecenterofmassof
thewheel.Itisthenoutofbalanceandwillalsocausevibrations,callingforawheelbalancing.
Example
itscenter.Theresultis2
MR
2
=
5. Onceyouknowthis,youimmediatelyknowallthe
componentsofitsinertiatensorbecauseoftheball’ssymmetry: this s samemoment
ofinertiaappearsallalongthematrixdiagonalandalltheo-diagonalelementsare
zero.Itisamultipleoftheunitmatrix. Whatarethecomponentswhentheoriginis
apointonthesurfaceoftheball? Choosethecoordinatesystemsothatthe
z
-axis
passesthroughthecenterandyouhave
I
(
~!
)=
I
cm
(
~!
)+
M~r
cm
~!
~r
cm
!
(
I
)=
2
MR
2
5
0
@
1 0 0
0 1 0
0 0 1
1
A
+
M
0
@
y
2
+
z
2
xy
xz
xy
x
2
+
z
2
yz
xz
yz
x
2
+
y
2
1
A
atcenter
=
2
MR
2
5
0
@
1 0 0
0 1 0
0 0 1
1
A
+
M
0
@
R
2
0
0
0
R
2
0
0
0
0
1
A
=
MR
2
5
0
@
7 0 0
0 7 0
0 0 2
1
A
KineticEnergy
calculation.
K
=
1
2
Z
v
2
dm
=
1
2
Z
dm
(
~!
~r
)
2
=
1
2
Z
dm
(
~!
~r
)
.
(
~!
~r
)
=
1
2
Z
dm~!
.
~r
(
~!
~r
)
=
1
2
~!
.
I
(
~!
)
(8
:
33)
Thesolevectoridentitythatyouneedforthisisthatyoucaninterchangethedotandthecrossproduct
inthetriplescalarproduct:
~
A
.
~
B
~
C
=
~
A
~
B
.
~
C
.
If
~
L
and
~!
arealigned,sothat
~
L
=
I~!
,thenthiskineticenergyis
1
2
I!
2
.
Whatifabody is both rotating and moving? ? Forexample e rollingmotion. . Thecalculationis
~!
movingorigin,andlet
~v
0
bethevelocityofthisoriginwithrespecttoyou. Thevelocityofamass
dm
isthen
~v
0
+
!
~r
,wherethis
~r
isfromthemovingoriginto
dm
.
~v
0
~!
dm
K
=
1
2
Z
dmv
2
=
1
2
Z
dm
~v
0
+
!
~r
2
=
1
2
Z
dmv
2
0
+
1
2
Z
dm
!
~r
2
+
~v
0
.
~!
Z
dm~r
=
1
2
mv
2
0
+
1
2
~!
.
I
~!
+
m~v
0
.
~!
~r
cm
(8
:
34)
8|RigidBodyMotion
274
Iftheoriginforcomputingtheinertiatensoristhecenterofmass,thenthenaltermiszero,andthe
kineticenergyofapointmassatthecenterofmass.
Inthesamespirit,whatistheangularmomentumforanobjectthatisbothmovingandrotating?
~
L
=
Z
dm~r
~v
=
Z
dm~r
~v
0
+
!
~r
=
Z
dm~r
~v
0
+
Z
dm~r
!
~r
=
~r
cm
m~v
0
+
I
~!
(8
:
35)
Thisis thesumoftheangularmomentum ofapointmassat thecenterofmassplustherotational
Having looked d at t kinetic c energy and angular momentum from dierent origins, what t about
torque? Thetorqueonapieceofmass
dm
is
~r
d
~
F
positionoftheoldoriginis
~
R
withrespecttotheneworigin,thenthecoordinateof
dm
is
~r
new
=
~r
+
~
R
so
~
new
=
Z
~r
new
d
~
F
=
Z
~r
+
~
R
d
~
F
=
~
old
+
~
R
Z
d
~
F
Ifthetotalforceiszero,thetorqueisindependentoftheorigin.
8.4PrincipalAxes
Asidefromspinningwithzerotorque,thereareotherreasonstobeinterestedinlininguptheangular
velocityandtheangularmomentum;itmakescalculationseasier.Whencalculatingthecomponentsof
theinertiatensorfortheexampleoftwopointmasses,Eq.(8.28),thematrixwouldhavebeendiagonal
y
isalongthelineconnectingthemassesthen
~!
diagonal.
~
L
~!
^
y
^
z
Theintegralfortheinertiacomponentsisthesameasbefore,beinga
sumoftwoterms,butthistimethesumiseasier.
I
xx
=
m
1
r
2
1
+
m
2
r
2
2
;
I
zz
=(
m
1
r
2
1
+
m
2
r
2
2
)
Allothercomponentszero
Theequation
~
L
=
I
(
~!
)translatesintocomponentsas
0
@
L
x
L
y
L
z
1
A
=(
m
1
r
2
1
+
m
2
r
2
2
)
0
@
1 0 0
0 0 0
0 0 1
1
A
0
@
0
!
sin
!
cos
1
A
=)
~
L
=(
m
1
r
2
1
+
m
2
r
2
2
)^
z!
cos
ThisagreesexactlywithEq.(8.29).Lookandsee.
Whenyoucomputethecomponentsof
I
,thedenitionis
I
(
~e
i
)=
I
ji
~e
j
(recalltheimpliedsum).
~!
and
~
L
areparallel,choose
~e
1
alongthatdirection. Then
~
L
=
I
(
~!
)
and
I
(
~e
1
)=(amultipleof)
~e
1
=
I
11
~e
1
. Therstcolumnofthematrixhasexactlyonenon-zeroentry
in the upperleft cornerofthematrix. . The e matrix issymmetric,so the upperrowhas only asingle
entryaswell.
Ifyoucanndotherdirectionsalongwhich
~!
isparallelto
~
L
,youcanusethosedirectionsfor
basisvectorstoo. Iftherearethreesuchdirections,thenyouhaveacompletebasisandthematrixis
diagonal. Canyoudothis? Yes. It’sguaranteed.
8|RigidBodyMotion
275
EigenvectorsandEigenvalues
Thestatementthat
~!
isparallelto
~
L
meansthatonevectorisamultipleoftheother:
~
L
=
~!
.This
istheequation
I
(
~!
)=
~!
(8
:
36)
Ifthisissatised,then
~!
isan\eigenvector"of
I
,and
isan\eigenvalue".Severalquestionsarise.
inertiatensor
alltensors
1.Dotheyalwaysexist?
yes
yes
2.Canyoualwaysmakeabasisoutofthem?
yes
no
3.Ifyoucanmakeabasis,isitorthogonal?
yes
no
4.Howdoyoundthem?
Thereinliesatale.
Translatetheproblemintocomponents. InthenotationofEq.(8.19)thelastequation(8.36)
becomes
0
@
I
xx
I
xy
I
xz
I
yx
I
yy
I
yz
I
zx
I
zy
I
zz
1
A
0
@
!
x
!
y
!
z
1
A
=
0
@
!
x
!
y
!
z
1
A
I
,therearefourunknowns
in these three equations:
and thethreecomponents of
~!
. Itdoesn’t t sound promising. . Thereis
alwaysonesolution,
~!
=0,soallthreecomponentsofthisvectorarezeroand
isarbitrary. Thatis
notaveryinterestingsolutionthough,andyoucan’tstarttoconstructabasisoutofit. Arethereany
non-zerosolutions?No,atleastnotforarbitrary
. But,therearealwayscertainvaluesof
forwhich
thereisanon-zerosolutionfor
~!
.
Nowwouldbeagoodideatoreviewsection0.10,especiallythematerialon
Rewritethissetofequationsbymovingeverythingtooneside.
2
4
0
@
I
xx
I
xy
I
xz
I
yx
I
yy
I
yz
I
zx
I
zy
I
zz
1
A
0
@
1 0 0 0
0 1 1 0
0 0 0 1
1
A
3
5
0
@
!
x
!
y
!
z
1
A
=
0
@
0
0
0
1
A
Furtherrewriteitas
0
@
I
xx
I
xy
I
xz
I
yx
I
yy
I
yz
I
zx
I
zy
I
zz
1
A
0
@
!
x
!
y
!
z
1
A
=
0
@
0
0
0
1
A
(8
:
37)
This matrix acting on the components of
~!
gives the zero vector. . Under r almost all circumstances
thatrequires
~!
to be zero, but there can be an exception. . Thatis s that the matrix be singular|
itsdeterminant is zero. . If f you wantanon-zero solution(andyou do),thenthe determinantof the
coecientsofthethreelinearequationsmustbezero*. Thatisanalgebraicequationfor
,acubic
equation,anditwillalwayshaveasolution. Inthecaseathand,thetensorofinertia,itwillalwayshave
threerealsolutions. Foreachsolution
(aneigenvalue),thereisacorresponding
~!
(aneigenvector).