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