Multi-level parametric reduced models of rotating bladed disk ...
Chapter(8 · Example:A*hanging*mass*rotating*a*solid*disk.*As#seen#in#the#figure,#a...
Transcript of Chapter(8 · Example:A*hanging*mass*rotating*a*solid*disk.*As#seen#in#the#figure,#a...
Chapter(8
Rotational(Motion
Newton�s(Second(Law(for(Rotational(Motion(About(a(Fixed(Axis
Example:((Hoisting(a(CrateA"motor"is"used"to"lift"a"crate"with"the"dual"pulley"system"shown"below.The"combined"moment"of"inertia"of"the"dual"pulley"is"46.0"kg·m2.""The"crate"has"a"mass"of"451"kg.""A"tension"of"2150"N"is"maintained"in"the"cableattached"to"the"motor.""Find"the"angular"acceleration"of"the"dual"pulleyand"the"tension"in"the"cable"connected"to"the"crate.
Newton�s(Second(Law(for(Rotational(Motion(About(a(Fixed(Axis
ατ ITT =−=∑ 2211 yy mamgTF =−"=∑ 2
equal
ymamgT +=2ay = aT = ℓ2α
T11 − mg+may( )2 = Iα
Newton�s(Second(Law(for(Rotational(Motion(About(a(Fixed(Axis
α2=ya
T11 − mg+may( )2 = Iα
T11 − mg+m2α( )2 = Iα
α =T1ℓ1 −mgℓ2
I +mℓ22
=2150 N( ) 0.600 m( )− 451 kg( ) 9.80m s2( ) 0.200 m( )
46.0 kg ⋅m2 + 451 kg( ) 0.200 m( )2 = 6.34rad s2
Newton�s(Second(Law(for(Rotational(Motion(About(a(Fixed(Axis
Tension'in'the'cable'connected'to'the'crate:
T2 = mg + may = mg + ml2!
= (451 kg)(9.80 m/s2) + (451 kg)(0.200 m)(6.34 rad/s2)
= 4990 N
Newton�s(Second(Law(for(Rotational(Motion(About(a(Fixed(Axis
Maximum mass that could be lifted with this system:
α =T1ℓ1 −mmaxgℓ2
I +mmaxℓ22 = 0 , i.e. m <mmax to lift the crate
T1ℓ1 −mmaxgℓ2 = 0 ⇒ mmax =T1ℓ1
gℓ2
=2150 N( ) 0.600 m( )
9.80m s2( ) 0.200 m( )= 658 kg
T2 =mmaxg+mmaxℓ2α =mmaxg+ 0 = 658 kg( ) 9.80m s2( ) = 6450 N
Rotational(Work(and(Energy
θFrFsW ==
θrs =
Fr=τ
τθ=W
Consider)the)work)done)in)rotatinga)wheel)with)a)tangential)force,)F,by)an)angle)!.
Rotational(Work(and(Energy
DEFINITION(OF(ROTATIONAL(WORK
The(rotational(work(done(by(a(constant(torque(in(turning(an(object(through(an(angle(is(
τθ=RW
Requirement:(The(angle(mustbe(expressed(in(radians.
SI(Unit(of(Rotational(Work:(joule((J)
Rotational(Work(and(Energy
KE = 12mr
2ω 2( )∑ = 12 mr2∑( )ω 2 = 1
2 Iω2
22212
21 ωmrmvKE T ==
ωrvT =
According)to)the)Work/Energy)theorem:)))))W = KEf - KE0
So)WR should)be)able)to)produce)rotational)kinetic)energy.
Calculate)the)kinetic)energy)of)a)mass)m undergoing)rotational)motionat)radius)r and)moving)with)tangential)speed)vT
For)a)system)of)rotating)masses,)the)total)kinetic)energy)is)the)sum)overthe)kinetic)energies)of)the)individual)masses,
Rotational(Work(and(Energy
221 ωIKER =
DEFINITION(OF(ROTATIONAL(KINETIC(ENERGY
The(rotational(kinetic(energy(of(a(rigid(rotating(object(is
Requirement:(The(angular(speed(mustbe(expressed(in(rad/s.
SI(Unit(of(Rotational(Kinetic(Energy:(joule((J)
Thus,(the(rotational(version(of(the(Work0Energy(theorem is:
WR = KERf - KER0 where(((({ 221 ωIKER =τθ=RW
Example: A*hanging*mass*rotating*a*solid*disk.*As#seen#in#the#figure,#a2.0#kg#mass#attached#to#a#string#is#rotating#a#solid#disk#of#mass#10.0#kg#and#radius#0.20#m#pivoting#around#its#center.#If#the#system#is#initially#at#rest,#what#is#the#angular#velocity#of#the#disk#after#the#mass#falls#0.70#m?#
M
m
� T��T
m = 2.0 kg M =10.0 kg r = 0.20 m d = 0.70 m→ Find ω f
⇒ Work done on the disk:WR = ΔKER ⇒ τθ = Trθ = 1
2 Iω f2 − 1
2 Iω02
Since: rθ = d, Idisk = 12 Mr
2,ω0 = 0 ⇒ Td = 14 Mr
2ω f2
⇒ Work done on the hanging mass:
WNC = ΔE⇒−Td = 12 mvf
2 +mghf( )− 12 mv0
2 +mgh0( )
r
{dSince: v0 = 0 vf = rω f hf − h0 = −d⇒ −Td = 1
2 mr2ω f
2 −mgd
Add disk eq. + hanging mass eq. ⇒ 0 = 14 Mr
2ω f2 + 1
2 mr2ω f
2 −mgd
∴ω f = −2r
mgdM + 2m
= −2
0.202.0( ) 9.8( ) 0.70( )10.0+ 2 2.0( )
= −9.9 rad/s
Total&energy&of&a&rotating&and&translating&rigid&body&ina&gravitational&field
!X""c.m.
vCM
Mg
MI
total energy = E = Erotationabout CM
+Etranslationof CM
= 12 Iω
2 + 12 MvCM
2 +MghCM
Since a gravitational field is a conservative force ⇒ Ef = E0
Rotational(Work(and(Energy
Example:((Rolling'Cylinders
A"thin'walled"hollow"cylinder"(mass"="m,"radius"="r)"anda"solid"cylinder"(also,"mass"="m,"radius"="r)"start"from"rest"atthe"top"of"an"incline.
Determine"which"cylinder"has"the"greatest"translationalspeed"upon"reaching"the"bottom.
Rotational(Work(and(Energy
mghImvE ++= 2212
21 ω
iiifff mghImvmghImv ++=++ 2212
212
212
21 ωω
iff mghImv =+ 2212
21 ω
ENERGY&CONSERVATION
rv ff =ω
o oo
o
Rotational(Work(and(Energy
iff mghrvImv =+ 22212
21
2
2rIm
mghv of +=
The$cylinder$with$the$smaller momentof$inertia$will$have$a$greater final$translationalspeed.
o
Since$$$$$Isolid = ½mr2 and$$$$$Ihollow = mr2
Then,$$$$$Isolid < Ihollow ! vf solid > vf hollow
Angular(Momentum
DEFINITION(OF(ANGULAR(MOMENTUM
The(angular(momentum(L of(a(body(rotating(about(a(fixed(axis(is(the(product(of(the(body�s(moment(of(inertia(and(its(angular(velocity(with(respect(to(thataxis:(
ωIL =Requirement:(The(angular(speed(mustbe(expressed(in(rad/s.
SI(Unit(of(Angular(Momentum:(kg·m2/s
τ EXT = Iα = I ΔωΔt
=Δ Iω( )Δt
=ΔLΔt∑
∴ τ EXT∑( )Δt = ΔL
⇒ "angular impulse-angular momentum theorm"
If τ EXT∑( ) = 0
⇒ ΔL = 0 ⇒ Lf = L0
⇒ Conservation of angular momentum
Consider)the)rotational)version)of)Newton’s)2nd Law:
Angular(Momentum
PRINCIPLE(OF(CONSERVATION(OF(ANGULAR(MOMENTUM
The(angular(momentum(of(a(system(remains(constant((is(conserved)(if(the(net(external(torque(acting(on(the(system(is(zero.
Lf = L0
Angular(Momentum
Conceptual(Example:((A"Spinning"Skater
An#ice#skater#is#spinning#with#botharms#and#a#leg#outstretched.##Shepulls#her#arms#and#leg#inward#andher#spinning#motion#changesdramatically.
Use#the#principle#of#conservationof#angular#momentum#to#explainhow#and#why#her#spinning#motionchanges.
Angular(Momentum
Example:((A"Satellite"in"an"Elliptical"Orbit
An#artificial#satellite#is#placed#in#an#elliptical#orbit#about#the#earth.##Its#pointof#closest#approach#is#8.37#x#106#mfrom#the#center#of#the#earth,#andits#point#of#greatest#distance#is#25.1#x#106#m#from#the#center#ofthe#earth.
The#speed#of#the#satellite#at#the#perigee#is#8450#m/s.##Find#the#speedat#the#apogee.
Angular(Momentum
ωIL =Since&no&external&torques&are&presentin&this&case,&we&have&angular&momentum&conservation
PPAA II ωω =
rvmrI == ω2
P
PP
A
AA r
vmrrvmr 22 =
Angular(Momentum
PPAA vrvr =
vA =rPvPrA
=8.37×106 m( ) 8450m s( )
25.1×106 m= 2820m s
P
PP
A
AA r
vmrrvmr 22 =
Example: A"potter’s"wheel"is"rotating"around"a"vertical"axis"through"its"centerat"a"frequency"of"2.00"rev/s."The"wheel"can"be"considered"a"uniform"disk"of"mass"4.80"kg"and"diameter"0.360"m."The"potter"then"throws"a"3.10"kg"chunk"of"clay,"approximately"shaped"as"a"flat"disk"of"radius"11.0"cm,"onto"the"center"of"the"rotating"wheel."(a)"What"is"the"frequency"of"the"wheel"after"the"clay"sticks"to"it?"(b)"What"fraction"of"the"original"mechanical"energy"of"the"wheel"is"lost"to"friction"after"the"collision"with"the"clay?
r ="0.110"m
R ="0.180"m
M ="4.80"kg
m ="3.10"kg
!0 ="2.00"rev/s"="12.6"rad/s"a) Lf = L0 ⇒ I fω f = I0ω0
ω f =I0
I fω0 =
12 MR
2
12 MR
2 + 12 mr
2 ω0
=4.80( ) 0.180( )2
4.80( ) 0.180( )2+ 3.10( ) 0.110( )2 12.6( )
=10.2 rad/s =1.62 rev/s
b) KE0 −KEf
KE0
=1−KEf
KE0
=1−12 I fω f
2
12 I0ω0
2 =1−ω f
ω0
=1−10.212.6
= 0.190
The$Vector$Nature$of$Angular$Variables
Right6Hand$Rule:$$Grasp&the&axis&of&rotation&with&your&right&hand,&so&that&your&fingers&circle&the&axisin&the&same&sense&as&the&rotation.
Your&extended&thumb&points&along&the&axis&in&thedirection&of&the&angular&velocity.
∴we can express L as a vector in the direction of ω:L = I ω
and write conservation of angular momentum in vector form:Lf =
L0
Example: A"person"sitting"on"a"chair"that"can"rotate"is"initially"at"rest"and"holding"a"bicycle"wheel"which"is"spinning"with"its"angular"momentum"vector"in"the"vertically"up"direction"and"with"magnitude"20"rad/s."The"mass"andradius"of"the"bicycle"wheel"are"5.0"kg"and"0.30"m,"respectively,"approximatedas"a"solid"disk."The"mass"and"average"radius"of"the"person"through"a"vertical"axis"are"90"kg"and"0.35"m,"respectively,"approximated"as"a"solid"cylinder."If"the"person"now"flips"the"spinning"wheel"so"that"the"angular"momentum"vector"is"vertically"down,"what"is"the"angular"velocity"of"the"person?
Lf =
L0 ⇒ −
L1 +L2 =
L1
∴L2 = 2
L1 , in upward direction
I2ω2 = 2I1ω1 ⇒ ω2 =2I1I2
ω1
ω2 = 212 5.0( ) 0.30( )2
12 90( ) 0.35( )2 20( ) =1.6 rad/s
E
E