ChapterChapter 11A11A –– AngularAngular MotionMotion

AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

© 2007 WIND TURBINES such as these can generate significant energy in a way that is environ- mentally friendly and renewable. The concepts of rotational acceleration, angular velocity, angular displacement, rotational inertia, and other topics discussed in this chapter are useful in describing the operation of wind turbines. Objectives:Objectives: AfterAfter completingcompleting thisthis module,module, youyou shouldshould bebe ableable to:to:

•• DefineDefine andand applyapply conceptsconcepts ofof angularangular displacement,displacement, velocity,velocity, andand acceleration.acceleration. •• DrawDraw analogiesanalogies relatingrelating rotationalrotational--motionmotion parametersparameters ((,, ,, )) toto linearlinear ((x,x, v,v, aa)) andand solvesolve rotationalrotational problems.problems. •• WriteWrite andand applyapply relationshipsrelationships betweenbetween linearlinear andand angularangular parameters.parameters. Objectives:Objectives: (Continued)(Continued)

•• DefineDefine momentmoment ofof inertiainertia andand applyapply itit forfor severalseveral regularregular objectsobjects inin rotation.rotation. •• ApplyApply thethe followingfollowing conceptsconcepts toto rotation:rotation: 1.1. RotationalRotational work,work, energy,energy, andand powerpower 2.2. RotationalRotational kinetickinetic energyenergy andand momentummomentum 3.3. ConservationConservation ofof angularangular momentummomentum RotationalRotational Displacement,Displacement, 

ConsiderConsider aa diskdisk thatthat rotatesrotates fromfrom AA toto B:B: B AngularAngular displacementdisplacement ::  A MeasuredMeasured inin revolutions,revolutions, degrees,degrees, oror radians.radians.

11 revrev == 360360 0 == 22 radrad

TheTheThe bestbestbest measuremeasuremeasure forforfor rotationrotationrotation ofofof rigidrigidrigid bodiesbodiesbodies isisis thethethe radianradianradian... DefinitionDefinition ofof thethe RadianRadian

OneOne radianradian isis thethe angleangle  subtendedsubtended atat thethe centercenter ofof aa circlecircle byby anan arcarc lengthlength ss equalequal toto thethe radiusradius RR ofof thethe circle.circle.

s s   R

R 1 rad = = 57.30 R ExampleExample 1:1: AA roperope isis wrappedwrapped manymany timestimes aroundaround aa drumdrum ofof radiusradius 5050 cmcm.. HowHow manymany revolutionsrevolutions ofof thethe drumdrum areare requiredrequired toto raiseraise aa bucketbucket toto aa heightheight ofof 2020 mm?? s 20 m    == 4040 radrad R 0.50 m R

Now,Now, 11 revrev == 22 radrad

1 rev   40 rad  h = 20 m 2 rad

 == 6.376.37 revrev ExampleExample 2:2: AA bicyclebicycle tiretire hashas aa radiusradius ofof 2525 cmcm.. IfIf thethe wheelwheel makesmakes 400400 revrev,, howhow farfar willwill thethe bikebike havehave traveled?traveled?

2 rad   400 rev  1 rev  == 25132513 radrad

ss == RR == 25132513 radrad (0.25(0.25 m)m)

ss == 628628 mm AngularAngular VelocityVelocity AngularAngular velocityvelocity,, isis thethe raterate ofof changechange inin angularangular displacement.displacement. (radians(radians perper second.)second.)   AngularAngular velocityvelocity inin rad/s.rad/s. tt AngularAngular velocityvelocity cancan alsoalso bebe givengiven asas thethe frequencyfrequency ofof revolution,revolution, ff (rev/s(rev/s oror rpm):rpm):

 fff AngularAngularAngular frequencyfrequencyfrequency fff(rev/s).(rev/s).(rev/s). ExampleExample 3:3: AA roperope isis wrappedwrapped manymany timestimes aroundaround aa drumdrum ofof radiusradius 2020 cmcm.. WhatWhat isis thethe angularangular velocityvelocity ofof thethe drumdrum ifif itit liftslifts thethe bucketbucket toto 1010 mm inin 55 ss?? s 10 m    == 5050 radrad R 0.20 m R

 50 rad   t 5 s h = 10 m == 10.010.0 rad/srad/s ExampleExample 4:4: InIn thethe previousprevious example,example, whatwhat isis thethe frequencyfrequency ofof revolutionrevolution forfor thethe drum?drum? RecallRecall thatthat  == 10.010.0 rad/srad/s..  2ff or 2 R 10.0 rad/s f 1.59 rev/s 2 rad/rev Or,Or, sincesince 6060 ss == 11 min:min: rev 60 s rev h = 10 m f 1.59 95.5 s 1 min min

ff == 95.595.5 rpmrpm AngularAngular AccelerationAcceleration AngularAngular accelerationacceleration isis thethe raterate ofof changechange inin angularangular velocity.velocity. (Radians(Radians perper secsec perper sec.)sec.)

   Angular acceleration (rad/s2 ) t TheThe angularangular accelerationacceleration cancan alsoalso bebe foundfound fromfrom thethe changechange inin frequency,frequency, asas follows:follows:

2( f )   Since  2 f t ExampleExample 5:5: TheThe blockblock isis liftedlifted fromfrom restrest untiluntil thethe angularangular velocityvelocity ofof thethe drumdrum isis 1616 rad/srad/s afterafter aa timetime ofof 44 ss.. WhatWhat isis thethe averageaverage angularangular acceleration?acceleration? 0  fof  or R tt

16 rad/s rad  4.00 4 s s2 h = 20 m == 4.004.00 rad/srad/s22 AngularAngular andand LinearLinear SpeedSpeed

From the definition of angular displacement:

s = R Linear vs. angular displacement

sR vR  v = R tt  t

LinearLinear speedspeed == angularangular speedspeed xx radiusradius AngularAngular andand LinearLinear Acceleration:Acceleration:

From the velocity relationship we have:

v = R Linear vs. angular velocity

vvRv vR  a = R tt  t

LinearLinear accel.accel. == angularangular accel.accel. xx radiusradius Examples:Examples: R1 A ConsiderConsider flatflat rotatingrotating disk:disk: B

 = 0; f = 20 rad/s R2 t = 4 s R = 20 cm What is final linear speed 1 R = 40 cm at points A and B? 2

vAf = Af R1 = (20 rad/s)(0.2 m); vvAf == 44 m/sm/s

vAf = Bf R1 = (20 rad/s)(0.4 m); vvBf == 88 m/sm/s AccelerationAcceleration ExampleExample R ConsiderConsider flatflat rotatingrotating disk:disk: 1 A

 = 0; f = 20 rad/s B t = 4 s R2 WhatWhat isis thethe averageaverage angularangular R1 = 20 cm andand linearlinear accelerationacceleration atat B?B? R2 = 40 cm

 f 0 20 rad/s   = 5.00 rad/s22 t 4 s = 5.00 rad/s

a = R= (5 rad/s2)(0.4 m) a=a= 2.002.00 m/sm/s22 AngularAngular vs.vs. LinearLinear ParametersParameters

RecallRecall thethe definitiondefinition ofof linearlinear vvf  0 a  accelerationacceleration aa fromfrom kinematicskinematics.. t

But,But, aa == RR andand vv == RR,, soso thatthat wewe maymay write:write: vv RR   a  f 0 becomes R  f 0 t t AngularAngular accelerationacceleration isis thethe timetime     f 0 raterate ofof changechange inin angularangular velocity.velocity. t AA Comparison:Comparison: LinearLinear vs.vs. AngularAngular

vv0  f 0  f s vt t tt 2 2

vvatfo  fo  t

1 2 1 2 svtat0 2  0tt2

1 2 1 2 svtatf 2  f tt2

22 22 2as vf v0 2f 0 LinearLinear Example:Example: AA carcar travelingtraveling initiallyinitially atat 2020 m/sm/s comescomes toto aa stopstop inin aa distancedistance ofof 100100 mm.. WhatWhat waswas thethe acceleration?acceleration?

Select Equation: 100100 mm

22 2as vf v0 vvo == 2020 m/sm/s vvf == 00 m/sm/s

2 2 0 - vo -(20 m/s) a = = aa == -2.00-2.00 m/sm/s22 2s 2(100 m) AngularAngular analogy:analogy: A disk (R = 50 cm), rotating at 600 rev/min comes to a stop after making 50 rev. What is the acceleration?

o = 600 rpm Select Equation: R 22 f = 0 rpm 2f 0  = 50 rev rev 2 rad1 min 600 62.8 rad/s 50 rev = 314 rad min 1 rev 60 s

2 -(62.8 rad/s)2 0 - o = = == -6.29-6.29 m/sm/s22 2 2(314 rad) ProblemProblem SolvingSolving Strategy:Strategy:

.. DrawDraw andand labellabel sketchsketch ofof problem.problem.

.. IndicateIndicate ++ directiondirection ofof rotation.rotation.

.. ListList givensgivens andand statestate whatwhat isis toto bebe found.found.

Given: ____, _____, _____ (,,f ,,t) Find: ____, _____ . Select equation containing one and not the other of the unknown quantities, and solve for the unknown. ExampleExample 6:6: AA drumdrum isis rotatingrotating clockwiseclockwise initiallyinitially atat 100100 rpmrpm andand undergoesundergoes aa constantconstant counterclockwisecounterclockwise accelerationacceleration ofof 33 rad/srad/s2 forfor 22 ss.. WhatWhat isis thethe angularangular displacement?displacement?

Given:Given: o = -100 rpm; t = 2 s  2  = +2 rad/s R rev 1 min 2 rad 100 10.5 rad/s min 60 s 1 rev

1122 ott 22 ( 10.5)(2)  (3)(2)  = -20.9 rad + 6 rad  == -14.9-14.9 radrad Net displacement is clockwise (-) SummarySummary ofof FormulasFormulas forfor RotationRotation

vv0  f 0  f s vt t tt 2 2

vvatfo  fo  t

1 2 1 2 svtat0 2  0tt2

1 2 1 2 svtatf 2  f tt2

22 22 2as vf v0 2f 0 CONCLUSION:CONCLUSION: ChapterChapter 11A11A AngularAngular MotionMotion