Physics140
Torqueandangularmomentum
Chapter8
Rota<onalkine<cenergy
Rememberthatmovingobjectshavekine<cenergy.
1 2
K = mv
2
1
2
Rota<ngbodiesaremoving
ètheymusthavekine<cenergy
1
1
K = m1v12 + m2 v22
2
2
Bothpointscoverangleθoverthesame
<me
èconstantangularvelocity(ω)
Linearvelocityisthen
Prof.M.Nikolic,Phys140
v1 = r1ω
v2 = r2ω
Rota<onal
iner<a
1
1
1
K = m1r12ω 2 + m2 r22ω 2 = (m1r12 + m2 r22 )ω 2
2
2
2
1 2 Rota<onalkine<cenergy.
K = Iω
Units:[J]
2
2
Rota<onaliner<a
Rota)onaliner)aalsocalledmomentofiner)a:
N
2
2
2
2
I = ∑ mi ri = m1r1 + m2 r2 + m3r3 + ... + mN rN
i =1
2
Units:[kgm2]
Alwayscalculatethedistancefromthecenterofrota<on.
Example:Whatisthemomentofiner<aforthisobject,rota<ngaboutanaxisoutof
theplanethroughtheblackpoint?
I = m1r12 + m2 r22 + m3r32
I = 1 kg ⋅ (3 m) 2 + 3 kg ⋅ (1 m) 2 + 2 kg ⋅ (2 m) 2
I = 20 kg m 2
Prof.M.Nikolic,Phys140
3
Examples:Rota<onaliner<a
BiggerRota<onalIner<aI:
hardertomakesomethingstartrota<ng
hardertomakesomethingstoprota<ng
TightRopeWalkerwithalongpole.
Thelongpolehasalargerota<onaliner<a
èitishardertomakehimstartrota<ngandfalloff.
HoldanobjectatitsendèmoreI
HoldanobjectatitscenterèlessI
è“chokeup”onabaseballbattolowerthe
rota<onaliner<aandspeeduptheswing
hWps://www.youtube.com/watch?v=n5y5PEKy0J8
Prof.M.Nikolic,Phys140
hWp://forum.orioleshangout.com/forums/showthread.php/111696-Why-isNick-choking-up/page2
4
Rota<onaliner<aofsolidbodies
Therota<onaliner<aofasolidbodycanbecalculatedby“addingup”allthepar<clesit
ismadefrom(technicallyan‘integral’incalculus)
We’lljustusetheresultsofthesecalcula<ons.
Prof.M.Nikolic,Phys140
5
Conceptualques<on–Rota<onaliner<a
Q1
Thetwocylindersshownbelowhavethesamemassandradius.Whichonehasthe
greatermomentofiner<a?
Accordingtothetable:
→Ihollow=MR2
→Ifull=½MR2
→Ihollow>Ifull
✓A.  Thehollowobjectontheright.
B.
C.
Thesolidobjectontheled.
Theyshouldhavethesame
momentsofiner<a.
D.  Needmoreinfo.
Prof.M.Nikolic,Phys140
6
Example:Rota<ng<re
1 2
K = Iω
2
Whatistherota<onalkine<cenergyofa<re(mass10kgandradius20cm)onacar
movingataspeedof20m/s?Assumethatthe<reisauniformsoliddisk(cylider).
Whatisgiven:
m=10kg
r=20cm=0.2m
v=20m/s
K=
1 2
Iω
2
Andweneedtofindrota<onal
1 2
I
=
mr
iner<aofthe<re(solidcylinder)
2
1
I = 10 kg ⋅ (0.2 m)2 = 0.2 kg m 2
2
Angularvelocitycanbefoundfromlinearvelocity
v
ω=
r
ω=
20m/s
= 100 rad/s
0.2 m
1
1
K = Iω 2 = 0.2 kg m 2 ⋅ (100 rad/s)2 = 1000 J
2
2
Prof.M.Nikolic,Phys140
7
Rota<onalForce(Torque)
ArchimedesofSyracuse(c.287BC–c.212BC)
"GivemealeverandaplacetostandandIwillmovetheEarth."
Prof.M.Nikolic,Phys140
8
Torque
Arota<ng(spinning)bodywillcon<nuetorotateunlessitisacteduponbyatorque.
Torque=Leverarm<mesForce
Torqueisavectorandhasmagnitudeanddirec<on.
Prof.M.Nikolic,Phys140
9
Torque
Onlythecomponentoftheforcethatisperpendiculartotheradiusproducesthe
torque.
Hinge
end
FI
r
τ = ±rF⊥
θ
FII
Unitsoftorque:[Nm]
NotJoules!
risthedistancefromtherota<onaxis(hinge)
tothepointwheretheforceFisapplied.
o Posi)vetorque–appliedforcecausestheobjecttorotatecounterclockwise
(CCW)
o Nega)vetorque–appliedforcecausestheobjecttorotateclockwise(CW)
Radial(parallel)componentoftheforceproducesnotorque.
Prof.M.Nikolic,Phys140
10
Conceptualques<on–Rota<onaliner<a
Q2
Youareusingawrenchandtryingtoloosenanut.Whichofthearrangements
shownismosteffec<veinlooseningthenut?Listinorderofdescendingefficiency
thefollowingarrangements:
A.
✓B.
C.
D.
1,2,3,4
2,1=4,3
2,1,4,3
1=2=4,3
Addingthelengthtothewrenchparalleltotheforceappliedwillnotincreasethetorque.
Prof.M.Nikolic,Phys140
11
Example:Torque
Thepullcordofalawnmowerengineiswoundaroundadrumofradius6.00cm,
whilethecordispulledwithaforceof75.0Ntostarttheengine.Whattorquedoes
thecordapplytothedrum?
F
r
Whatisgiven:
r=6cm=0.06m
F=75N
τ = ±rF⊥
Direc<onofthetorque:theforcewillmakethe
lawnmowerrotateCCW→posi<vetorque
τ = rF
τ = 0.06 m ⋅ 75 N = 4.5 Nm
Prof.M.Nikolic,Phys140
12
Example:Acoupleofforces
Calculatethetorqueduetothethreeforcesshownabouttheledendofthebar(the
redX).Thelengthofthebaris4mandF2actsinthemiddleofthebar.
F2=30N 300
450
Weneedtofindthetorquedueto
everyappliedforceandsumthemup
F2I=F2cos300
X
100
F3=20N
τ net = ∑τ i =τ 1 + τ 2 + τ 1...
3
F1=25N
TorquefromF1:r1=0→τ1=0
TorquefromF2:
2 = ±r2 F2⊥
• r2isinthemiddleofthebar
→r2=4m/2=2m
• Direc<on:F2pullsup→CCWrota<on
→posi<vetorque
τ
Prof.M.Nikolic,Phys140
τ 2 = r2 F2 cos 300
τ 2 = 2 m ⋅ 30 N ⋅ cos300 = 51.9 Nm
13
Example:Acoupleofforces
Calculatethetorqueduetothethreeforcesshownabouttheledendofthebar(the
redX).Thelengthofthebaris4mandF2actsinthemiddleofthebar.
F2=30N 300
450
F2I=F2cos300
X
100
F3=20N
F1=25N
F3I=F3sin100
TorquefromF3: τ = ±r F
3
3 3⊥
• r3isattheandofthebar
→r3=4m
• Direc<on:F3pullsdown→CWrota<on
→nega<vetorque
τ 3 = −r3 F3 sin 10 0
τ 3 = −4 m ⋅ 20 N ⋅ sin10 0 = −13.9 Nm
Thenettorque(sum): τ net
= τ1 +τ 2 +τ1
τ net = 0 + 51.9 Nm + (−13.9 Nm) = +38 Nm
Thebarwillgoup-CCW
Prof.M.Nikolic,Phys140
14
Rota<onalformofNewton’ssecondlaw
!
!
Newton’ssecondlaw: ∑ F = ma
Rota<onalformofNewton’ssecondlaw: ∑τ = Iα
α–angular
accelera<on
Example:Findangularaccelera<onofthewheelwithamassof1kgthatstartsfromrestand
hastheforceof1Nappliedtoit.Assumethehubsandspokesaremassless,sothatthe
rota<onaliner<aisI=mR2.
τ net = ±rF⊥ = Iα
I = mr 2
α=
Prof.M.Nikolic,Phys140
I = 1 kg ⋅ (0.5 m)2 = 0.25 kg m 2
−rF
0.5 m ⋅1 N
2
=−
=
−2
rad/s
I
0.25 kg m 2
15
Calcula<ngworkdonefromthetorque
Theworkdonebyatorqueis:
W = τΔθ
Δθistheangle(inradians)theobject
turnsthrough
Andallequa<onsfromChapter5s<llapply!
Δω = ω f − ω i = αΔt
ω 2f − ωi2 = 2αΔθ
Prof.M.Nikolic,Phys140
1
Δθ = θ f − θ i = ω i Δt + αΔt 2
2
Δθ =
1
(ωi + ω f )Δt
2
16
τ net = ∑±rF⊥ = Iα
Example:Spinningdisk
Arota<ng,12-kgdiskisspinningatanangularvelocityof200rad/sandisactedon
bytwoforces,oneof30Nthatis0.2mtothewestofthecenterofthediskand
poin<ngstraightdownandtheotherof50Nontherimofthedisk,0.3mawayeast
ofthecenterandac<ngstraightup.
a)Whatisthenettorqueonthediskbyeachoftheforces?
TorquefromF1:
1 = ±r1F1⊥
F2 • r1=0.2mandF1=30N
• Direc<on:F1pullsdown→CCW
rota<on→posi<vetorque
τ
r1
r2
TorquefromF2:
= ±r2 F2⊥
2
• r2=0.3mandF2=50N
• Direc<on:F2pullsup→CCW
rota<on→posi<vetorque
τ
F1
Thenettorque(sum): τ net = 6 Nm +15 Nm = +21 Nm
Prof.M.Nikolic,Phys140
τ 1 = r1F1
τ 1 = 0.2 m ⋅ 30 N = 6 Nm
τ 2 = r2 F2
τ 2 = 0.3 m ⋅ 50 N = 15 Nm
17
Example:Spinningdisk
Arota<ng,12-kgdiskisspinningatanangularvelocityof20rad/sandisactedonby
twoforces,oneof30Nthatis0.2mtothewestofthecenterofthediskand
poin<ngstraightdownandtheotherof50Nontherimofthedisk,0.3mawayeast
ofthecenterandac<ngstraightup.
b)Whatisthetotalworkdoneonthediskbythenetforceoverthefirst10seconds?
W =τ netΔθ
and
1
Δθ = ω i Δt + αΔt 2
2
Weneedtofindangular
accelera<on!
τ net = Iα
Forthedisk:
1
I = mr 2
2
1
I = 12 kg ⋅ (0.3 m)2 = 0.54 kg m 2
2
Δθ = 20 rad/s ⋅10 s +
α=
21 Nm
= 38.9 rad/s2
2
0.54 kg m
1
2
38.9 rad/s2 ⋅ (10 s) = 2145 rad
2
W = τ net Δθ = 21 Nm ⋅1590 rad = 45 kJ
Prof.M.Nikolic,Phys140
18
Example:Spinningdisk
Arota<ng,12-kgdiskisspinningatanangularvelocityof20rad/sandisactedonby
twoforces,oneof30Nthatis0.2mtothewestofthecenterofthediskand
poin<ngstraightdownandtheotherof50Nontherimofthedisk,0.3mawayeast
ofthecenterandac<ngstraightup.
c)Whatisthetotalpowerappliedtothediskoverthefirst10seconds?
Power(Chapter6):
P=
ΔE W
=
Δt Δt
P=
45,00 J
= 4,500 W
10 s
Whensolvingproblems,alwaystrytousetheequa<onwiththemostinforma<onprovided.
Forexample,ifyouweregivenini<alandfinalangularvelocity,youcouldfindangular
displacementbyusing:Δθ=½(ωf+ωi)Δtwithoutevensolvingforangularaccelera<on.You
couldalsousetheworkenergytheorem:Wnet=ΔK=½Iωf2-½Iωi2.
Prof.M.Nikolic,Phys140
19
Equilibrium
Thecondi<onsforequilibriumare:
Transla<onalequilibrium
!
∑F = 0
∑F
=0
∑F
=0
x
x
Rota<onalequilibrium
!
∑τ = 0
Notethatallthecondi<onshavetobesa<sfiedforequilibrium!
Prof.M.Nikolic,Phys140
20
Rulesforsolvingequilibriumproblems
1.  Drawafreebodydiagram.
2.  Set-upthecoordinatesystem(specifyxandyaxes).
3.  Iden<fyalltheforcesthatareexertedontheobjectandlabelthem
appropriately.
4.  Drawvectorarrowsrepresen<ngalltheforcesac<ngontheobject.Illustrate
thedirec<onsoftheforces.
5.  Toapplyforcecondi<on→findxandycomponentsforeveryforcedrawn
andsumthemuptozero.
6.  Toapplytorquecondi<on
→Chooseaconvenientrota)onaxis–itshouldgothroughatleastone
unknownforce
→Determinethedirec<onofeachtorqueandsumthemuptozero
Some<mesyouwillnotneedall3condi<onstosolvetheproblem→choosethe
oneswiththemostinforma<onprovided.
Prof.M.Nikolic,Phys140
21
Conceptualques<on–Seesaw
Q3
Whatisthemassoftheboyiftheseesawisbalanced?
Girl: τ g
= ±rg Fg⊥
CCWmo<on
Posi<vetorque
τ g = rg mg g
τ b = ±rb Fb⊥
Boy:
m=25kg
τ b = −rb mb g
m=??
✓A.  50kg
B.
C.
D.
25kg
50N
10kg
Prof.M.Nikolic,Phys140
CWmo<on
Nega<vetorque
Condi<onforbalance:
∑τ = τ
g
rg mg g − rb mb g = 0
+τb = 0
rg mg
mb =
rb
mb =
3 m ⋅ 25 kg
= 50 kg
1.5 m
22
Example:Suppor<ngasign
Asignissupportedbyauniformhorizontalboomoflength3.00mandweight80.0N.
Acable,inclinedata350anglewiththeboom,isaWachedatadistanceof2.38m
fromthehingeatthewall.Theweightofthesignis120.0N.Whatisthetensionin
thecableandwhatarethehorizontalandver<calforcesexertedontheboombythe
hinge?
Iden<fyalltheforcesac<ngonthebar:
StartwiththeFBDforthebar.
y
Fy
X
T
350
F x
x
mbarg
1.  Weightofthebar–gravita<onalforceactsat
thecenterofmass–foruniformobjectsat
thecenteroftheobject
2.  Weightofthesign–pushingthebardown
3.  Tensionforce–ac<ngalongacable
4.  Horizontalandver<calforceexertedbythe
hinge
msg
Chooseaconvenientrota<onaxis
→hinge–youwilleliminateFxandFyfromthetorqueequa<on
Prof.M.Nikolic,Phys140
23
Example:Suppor<ngasign
Asignissupportedbyauniformhorizontalboomoflength3.00mandweight80.0N.
Acable,inclinedata350anglewiththeboom,isaWachedatadistanceof2.38m
fromthehingeatthewall.Theweightofthesignis120.0N.Whatisthetensionin
thecableandwhatarethehorizontalandver<calforcesexertedontheboombythe
hinge?
Applythecondi<onsforequilibriumtothebar.
y
Fy
X
∑τ = 0
T
bar
+τ s +τ T = 0
Bar: τ bar = −mBar grbar
350
F x
x
mbarg
∑τ = τ
msg
Sign: τ s = −ms grs
Tension: τ T = +T⊥ rT = +TrT sin 350
− 80 N ⋅1.5 m − 120 N ⋅ 3 m + T ⋅ 2.38 m ⋅ sin350 = 0
120 Nm + 360 Nm
T=
= 352.9 N
1.36 m
Prof.M.Nikolic,Phys140
24
Example:Suppor<ngasign
Asignissupportedbyauniformhorizontalboomoflength3.00mandweight80.0N.
Acable,inclinedata350anglewiththeboom,isaWachedatadistanceof2.38m
fromthehingeatthewall.Theweightofthesignis120.0N.Whatisthetensionin
thecableandwhatarethehorizontalandver<calforcesexertedontheboombythe
hinge?
Applythecondi<onsforequilibriumtothebar.
y
Fy
X
∑F
x
T
350
F x
Fx − Tx = 0
Fx = T cos 35 0 = 352 .9 Ncos 35 0 = 289 .1 N
Ty
T x
mbarg
=0
x
msg
∑F
y
=0
Fy + Ty − mbar g − ms g = 0
Fy = −352.9 Nsin350 + 80 N +120 N = −2.4 N
Nega<vesign–ver<calforcepointsdown.
Prof.M.Nikolic,Phys140
25
Example:Equilibriuminthehumanbody
Apersonisstandingwithhisarmoutstretchedinahorizontalposi<on.Theweightof
thearmis30Nanditscenterofgravityisattheelbow,27.5cmfromtheshoulder
joint.Thedeltoidpullstheupperarmatanangleof150abovethehorizontalandat
distance12cmfromthejoint.Whatisthemagnitudeoftheforceexertedbythe
deltoidmuscle.
Chooseaconvenientrota<onaxis
→shoulderjoint–youwilleliminateFs
(scapula)thattheproblemisnotaskingfor
y
x
Applythecondi<onsforequilibriumto
thearm.
∑τ = 0
Deltoidmuscle: τ m = + Fm⊥ rm = + Fm rm sin 15
0
Armweight: τ arm = −marm grarm = −30 N ⋅ 0.275 m
∑τ = τ
m
+ τ arm = 0
Fm ⋅ 0.12 m ⋅ sin150 − 30 N ⋅ 0.275 m = 0
Fm =
8.25 Nm
= 266 N
0.031 m
Notethatwhenyouaddweights,deltoidmusclehastoexertmoreforcetofightthe
addi<onalweightofweights.
Prof.M.Nikolic,Phys140
26
Themo<onofrollingobjects
Work–EnergytheoremaswerememberitfromChapter6:
Wtot = ΔK tot= K f − K i
Forarollingobjectthereare2typesofkine<cenergy
1.  Transla<onalkine<cenergy–linearmo<onofthecenterofmass
2.  Rota<onalkine<cenergy
K tot = K T + K rot
1 2
KT = mvCM
2
K rot
1 2
= Iω
2
Iftheobjectrollswithoutslippingthenvcm=Rω.
Prof.M.Nikolic,Phys140
28
Example:Rollingsphere
Asphereofradius12.2cmrolls2.50mdowna25oinclinewithoutslipping.How
fastinm/sisitmovingattheboWomifitisini<allyatrest?Ignorefric<on.
Drawadiagram.
y
yi
25o
x
Work–kine<cenergytheorem
Wtotal = Wcons + Wnc = ΔKtot
Wnc = ΔKT + ΔK rot + ΔU g + ΔUel
Thereisnofric<onandspring.
Prof.M.Nikolic,Phys140
1
1
1
1
0 = mv 2f − mvi2 + Iω 2f − Iω i2 + mg(y f − yi )
2
2
2
2
Ini<alcondi<ons(atthetopofincline)
yi=dsin250=2.5mx0.42=1.05m
vi=0
ωi=0
Finalcondi<ons(attheboWomofincline)
yf=0m
vf=?
ωf=vf/Rèvf=Rωf
Rota<onaliner<aofthesphere I =
2
mR 2
5
1
12
0 = mv 2f − 0 +
mR 2ω 2f − 0 + mg(0 − yi )
2
25
R 2ω 2f = (Rω f )2 = v 2f
29
Example:Rollingsphere
Asphereofradius12.2cmrolls2.50mdowna25oinclinewithoutslipping.How
fastisitmovingattheboWomifitisini<allyatrest?
y
yi
25o
x
1
12 2
0 = mv 2f +
mv f − mgyi
2
25
1
1
7
gyi = v 2f + v 2f = v 2f
2
5
10
10 gyi
vf =
7
10 ⋅ 9.8 m/s2 ⋅1.05 m
vf =
= 3.83 m/s
7
Prof.M.Nikolic,Phys140
30
Conceptualques<on–Rota<onaliner<a
Q4
Threeuniformobjectshavingthesamemassanddiameterarereleased
simultaneouslyfromrestatthesamedistanceabovetheboWomofahillandroll
downwithoutslipping.Theobjectsareasolidsphere,asolidcylinder,andathinwalledhollowcylinder.WhichofthefollowingobjectswillreachtheboWomofthe
hillfirst?
✓A.  Sphere
B.
C.
D.
Hollowcylinder
Solidcylinder
Allatthesame<me.
Morerota<onaliner<a→theslowerobjectis
2
4
1
5
10
mR 2 = mR 2 < mR 2 = mR 2 < mR 2 = mR 2
5
10
2
10
10
Prof.M.Nikolic,Phys140
31
Angularmomentum
Linearmomentum
!
!
p = mv
Angularmomentum
L = Iω
Units:[kgm2/s]
SecondNewton’slaw
! Δp!
∑ F = Δt
ΔL
∑τ = Δt
Direc<onofangularmomentum:
Counterclockwise(CCW)mo<on
èAngularmomentumposi<ve
Clockwise(CW)mo<on
èAngularmomentumnega<ve
Prof.M.Nikolic,Phys140
32
Conserva<onofangularmomentum
Noexternaltorque
ΔL
∑τ = 0 ⇒ Δt = 0
Prof.M.Nikolic,Phys140
Li = L f
33
Example:Aturntable
Aturntableofmass5.00kghasaradiusof0.100mandspinsat30rpm.Whatisthe
angularmomentum?Assumethatthemassofaturntableisuniformlydistributed.
Whatisgiven:
m=5kg
R=0.1m
f=30rpm
L = Iω
Wecanassumethatturntableisauniformdisk
→fullcylinder
I=
1
MR 2
2
Angularvelocitycanbefoundfromfrequency: ω
ω = 2π ⋅ 30
1
2
I = 5 kg ⋅ (0.1 m) = 0.025 kg m 2
2
= 2πf
rev 1 min
⋅
= 2π ⋅ 0.5 Hz = 3.14 rad/s
min 60 s
L = 0.025 kg m2 ⋅ 3.14 rad/s = 0.079kg m2 / s
Prof.M.Nikolic,Phys140
34
Example:Afigureskater
Afigureskaterisspinningatarateof1rev/swithherarmsoutstretched.Shethen
drawsherarmsintoherchests,reducingherrota<onaliner<ato67%ofitsoriginal
value.Whatishernewrateofrota<on?
Whatisgiven:
fi=1rev/s
If=67%Ii
Thereisnoexternalforce→angularmomentumisconserved
Li = L f
ωf =
Ii
ωi
If
I iω i = I f ω f
2πf f =
ff =
Prof.M.Nikolic,Phys140
Ii
2πf i
0.67 I i
1
1 rev/s
fi =
= 1.5 rev/s
0.67
0.67
35
Review
Rota<onalkine<cenergy
K=
1 2
Iω
2
Rota<onaliner<a
N
I = ∑ mi ri
2
i =1
Torque
τ = ±rF⊥
Newton’ssecond
lawforrota<on
∑ τ = Iα
Prof.M.Nikolic,Phys140
Workdonebytorque
W = τΔθ
Angularmomentum
L = Iω
ΔL
∑τ = Δt
Conserva<onof
angularmomentum
ΔL
∑τ = 0 ⇒ Δt = 0
Li = L f
36
Review
Equilibrium
Work–kine<cenergytheorem
!
∑F = 0
∑F
x
=0
!
Wtotal = Wcons + Wnc = ΔKtot
∑F
x
=0
− ΔU g = ΔKT + ΔK rot
∑τ = 0
Prof.M.Nikolic,Phys140
37