Physics 111 Lecture 10. SJ 8th Ed.: Chap Torque, Energy, Rolling. Copyright R. Janow Spring basics, energy methods, 2nd law problems)

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1 hysics Lectue 0 Toque, Enegy, Rolling SJ 8th Ed.: Chap Recap and Oveview Toque Newton s Second Law fo Rotation Enegy Consideations in Rotational Motion Rolling Enegy Methods Second Law Applications 0.6 Toque 0.7 The Rigid Body Unde a Net Toque (Rotational Second Law 0.8 Enegy Consideations in Rotational Motion (Wok, owe, and Enegy 0.9 Rolling Motion of a Rigid Object (coves basics, enegy methods, nd law poblems Add example 0.8, 0., 0. Copyight R. Janow Sping 0

2 Key Angula Dynamics Concepts Moment of netia measues BOTH mass and its distibution depends on the choice of otation axis p is pependicula distance fom mass point to axis masses on axis have p 0, no contibution to m i,i dm cm + Mh x fo dm point paallel masses fo continuous axis theoem bodies same mass small lage Toque τ TWST - measues applied foce AND whee it acts MOMENT ARM (leve am p pependicula distance fom axis to line of action of foce τ foce x leve am sin( Rotational nd Law: Moment of inetia plays the ole of mass Toque plays the ole of foce Rotational wok: owe used when doing otational wok: W x W o v t net ma W ot ot τ net τ W t ot τω τω p p Copyight R. Janow Sping 0

Toque and Moment am (leve am Toque τ measues the effect of a foce applied at a point as it causes otation about some axis Toque is a vecto, magnitude is: τ sin φ d is the foce φ is the angle the foce

3 Toque and Moment am (leve am Toque τ measues the effect of a foce applied at a point as it causes otation about some axis Toque is a vecto, magnitude is: τ sin φ d is the foce φ is the angle the foce makes with the displacement fom the otation axis d is the moment am (o leve am of the foce The leve am d sin(φ, is the pependicula distance fom the axis of otation to a line dawn along the diection of the foce Diection: counteclockwise toque is positive clockwise toque is negative Copyight R. Janow Sping 0

Toque in 3D oce applied at a point a vecto τ foce x moment am sin( φ Toque is a vecto pependicula to the plane containing and Axis of otation along z, though Moment am sin(φ Toque sin(φ (CCW about z

4 Toque in 3D oce applied at a point a vecto τ foce x moment am sin( φ Toque is a vecto pependicula to the plane containing and Axis of otation along z, though Moment am sin(φ Toque sin(φ (CCW about z Tansvese component of the foce sin(φ Moment am is simply Toque is still sin(φ The toque is the same if is applied anywhee along its line of action Vecto pictue: toque vecto is pependicula to plane of and, (along z-axis in sketch, CCW sense z y 90 o φ φ φ 90 o x ad line of action of Toques about the same axis add (supeposition Copyight R. Janow Sping 0

5 Effect of a Constant Net Toque 0. A constant non-zeo net toque is exeted on a wheel. Which of the following quantities must be changing?. angula position. angula velocity 3. angula acceleation 4. moment of inetia 5. kinetic enegy 6. the mass cente location A.,, 3 B. 4, 5, 6 C.,, 5 D.,, 3, 4 E., 3, 5 Copyight R. Janow Sping 0

6 Examples: Evaluating toques Massless, hoizontal ba: φ 90 o Moment am l sin(90 o l Toque τ - mgl clockwise l mg φ CCW + CW - Ba as above but NOT hoizontal Moment am l sin(φ Toque τ + m g l sin(φ CCW moment am φ l φ mg Toques can add (they must efe to the same otation axis (Supeposition Hoizontal see saw, unequal ams m N Calculate toques aound pivot point L m L W m g W m g ee Body Diagam τ net τ +τ + τ N Moment am 0 fo foce N about pivot τ net + m gl m gl Copyight R. Janow Sping 0

Rotational Second Law nfe nd law fo a foce acting on a single point mass: T axis p m ma T T T (definitio n toque τ net m a T (otational τ inetia net τ ma T ALES ALSO TO RGD BODES AXS XED m net τ net

7 Rotational Second Law nfe nd law fo a foce acting on a single point mass: T axis p m ma T T T (definitio n toque τ net m a T (otational τ inetia net τ ma T ALES ALSO TO RGD BODES AXS XED m net τ net tot nd Law Example: ind & a T. oce acts at the im of a igid disk. T axis p Moment of inetia τ net Apply Second Law T Tangential of im : (linea m a T T m m acceleation (otational inetia Resulting Angula Acceleat ion T m comes fom Copyight R. Janow Sping 0

8 Example: Angula acceleation a fo an unbalanced ba Ba is massless and oiginally hoizontal Rotation axis though fulcum point N has zeo toque ind angula acceleation of ba and the linea acceleation of m just afte you let go +y L N L m g fulcum m g a - L a + L Use: whee: τ net tot τ net tot tot + m L m L + τ net τ o,i + m gl m gl Whee did sin( factos in moment am go? net toque total about pivot m gl m gl m L + m L What if ba is not hoizontal? Constaints: Evaluation: Let m m m L 0 cm, L 80 cm gl L + L ad/s 8.65 gl Clockwise a L +.7 m/s Acceleates U Copyight R. Janow Sping 0 g( Does answe change if axis changes?

9 See Saw 0.. Suppose eveything is as it was in the peceding example, but the ba is NOT hoizontal. Assume both masses ae equal. Which of the following is the coect equation fo the angula acceleation? A B (L L L + L (L L L + L g g cos( mg L N fulcum L mg C (L L gsin( L + L D (L L gcos( L + L E (L L L + L g sin( τ net tot Copyight R. Janow Sping 0

10 See Saw - Solution 0.. Suppose eveything is as it was in the peceding example, but the ba is NOT hoizontal. Assume both masses ae equal. Which of the following is the coect equation fo the angula acceleation? τ net / tot τ net τ o,i + m gl cos( m gl cos( m g L N fulcum L m g tot + m L m L + [m L m L + m L ] g cos( m L o equal masses m m m (L L L f ba is balanced when it s gcos( hoizontal, does that change + L when it is not hoizontal? Copyight R. Janow Sping 0

11 Method fo solving Second Law poblems Systems with many components have many unknowns. need an equal numbe of independent equations Method: Daw o sketch system. Adopt coodinates, name the vaiables, indicate otation axes, list the known and unknown quantities, Daw fee body diagams of key pats. Show foces at thei points of application. Weights of bodies act as if they ae at the mass cente Calculate toques about a (common axis. May need to apply both foms of Second Law to each pat Note: can have Tanslation: net.eq. 0 net i ma but τ net.ne. 0 Rotation: net τ i τ Geneate equations using Second Law. Thee may be constaint equations (exta conditions connecting unknowns. Make sue thee ae enough (N equations. Simplify and solve the set of (simultaneous equations. ntepet the final fomulas. Do they make intuitive sense? Refe back to the sketches and oiginal poblem Calculate numeical esults, and sanity check anwes (e.g., ight ode of magnitude? Copyight R. Janow Sping 0

12 Example: Angula Acceleation of a Wheel Standad appoach: Beak into two sub-systems wheel is acceleated angulaly by tension T block is acceleated linealy by weight mg, with tension opposing Daw fee body diagams The wheel is otating and so we apply Στ Ι The tension (tangential supplies the toque The mass m moves in a staight line, so apply Newton s Second Law Σ y ma y mg -T How to connect the two poblems above? A constaint links linea acceleation to a a mg Copyight R. Janow Sping 0

13 Method using the Second Law Cod does not slip o stetch (constaint Disk s otational inetia slows acceleations Let m. kg, M.5 kg, 0. m ind acceleation of mass m, find fo disk BD fo mass m: T y T mg y ma mg BD fo disk, with axis at o : N Mg T m (g a Unknowns: T, a τ 0 + T T M T m(g a Eliminated T M Unknowns: a, So fa: Equations, 3 unknowns Need a constaint: Substitute and solve: mg M - m M + m M ( mg M a (m (m a + mg ( 4 ad/s + M/ mg + M/ a ( mg N suppot foce N at axis O has zeo toque fom no slipping assumption popotional to g 4.8 m/s Copyight R. Janow Sping 0

14 Same poblem using system point of view (caution Define the system (dashed line ignoe intenal toques & foces (e.g., Tensions detemine total system otational inetia (about O Apply linea Second Law (no new info: x 0 y 0 no foces N Mg + m g Apply otational Second Law to system: τ net,ext sys τ net,ext τ mass + sys pul + mass M m + mg a pul N Mg M mg M + m a T M m g + m mg Copyight R. Janow Sping 0

15 Wok & Enegy fo Linea and Rotational Motion Linea Vesion of Wok-Kinetic Enegy Theoem K W...wok done by extenal foces (including Wok : dw Kinetic o ds...intega te... W o d s path K mv point mass only, no otation enegy : potentials Linea Vesion of Mechanical Enegy Consevation E mech K + U Mechanical E K + U Enegy mech W non-consevative only U otential enegy diffeence; e.g., gavity U mg owe o o v dw dt ds dt h Modifications fo Rotating Bodies Add Rotational Kinetic Enegy K K cm + K ot...so... K K cm + K ot K ot ω f Toques do otational wok even when zeo wok is done on the mass cente: dw ot τ d (toque x angula distance W ot τ d limits Wok-KE Theoem is the same as befoe but KE and Wok include otation ω Copyight R. Janow Sping 0 0

16 Wok Done by ue Rotation Apply foce to mass at point, causing otation-only about axis. Displacement is only along (tansvese diection d W o ds cos(90 - sin( as φ d fo φ tanslation ds ds d How much wok does foce do as it otates though an infinitesimal distance Only the tansvese component of the foce (along the displacement does wok the same component that contibutes to the toque: τ sin( φ dw τ d The adial component of the foce does no wok because it is pependicula to the displacement Copyight R. Janow Sping 0

17 nstantaneous owe: Divide both sides of dw τd by dt dw dt τ d dt τω Example: The powe output of a cetain ca is 00 hp at 6000 pm. What is the coesponding toque? Convet the powe to Watts: hp 746 Watts 00 hp 00 hp 746 Watts hp Convet the angula velocity ω to ad/s: ω π ad ev.49 x 0 5 min 60 s Watts 6000 ev/min 6000 ev/min τω Apply and solve fo the toque: 68 ad/s τ ω.49 x 0 N.m/s 68 ad/s 5 37 N.m Copyight R. Janow Sping 0

18 Example: An electic moto attached to a gindstone exets a constant toque of τ 0 N.m. The system stats fom est. The moment of inetia of the gindstone is.0 kg.m. a ind the kinetic enegy afte 8 seconds. b ind the wok done by the moto duing this time. c ind the aveage powe deliveed by the moto. a τ ω τ / 0. 0 / ad/ s f ω 0 + t 5. 0 x ad/ s K K ω x. 0 x ( J. f f b ist, angle tuned though: W ω 0 t + t x 5.0 x (8.0 Definition of wok, constant toque f i τ d τ 0.0 x ad 600 J. 600 J. 8 s. c Aveage powe: av 00 Watts. but instantaneous powe is not constant: τω W t { 0 W. at t 0 00 W. at t 400 W. at t Should equal wok W done by moto 4 8 s. s. Copyight R. Janow Sping 0

19 Enegy Consevation: fo tanslation + otation about mass cente (cm W includes both consevative and K tot K ot + K cm W OR E mech K tot + U W nc non-consevative foces, teated as extenal to system W nc includes only non-consevative foces, U contains the consevative foces Summay: ue Tanslation (ixed Diection ue Rotation (ixed Axis osition coodinate x Angula position Velocity v dx/dy Angula velocity ω d/dt Acceleation a dv/dt Angula acceleation dω/dt Mass m Rotational inetia Newton's second law net ma Newton's second law τ net Wok dw dx Wok dw τ d Kinetic enegy K cm (m/v Kinetic enegy K ot (/ω owe (constant foce.v owe (constant toque τω Wok KE theoem W K Same, include K ot W K Copyight R. Janow Sping 0

The ed cuve shows the path (called a cycloid swept out by a point on the im of the wheel.

20 Rolling: A wheel olls without slipping on a table The geen line above is the path of the mass cente of a wheel. The ed cuve shows the path (called a cycloid swept out by a point on the im of the wheel. When thee is no slipping, the elationships between the tanslational (mass cente and otational motion ae simple: ω v cm s R v ω R a cm cm R Copyight R. Janow Sping 0

21 Rolling pue otation aound CM + pue tanslation of CM The bottom point of the wheel is stationay (no slipping The top point T is moving at speed v com, (fastest Someone moving with the CM of the wheel sees tangential speed v tang ω v com fo no slipping (im a ue otation + As seen fom the lab with otation tuned off, all points on the wheel would move at speed v com b ue tanslation Total velocity fo a point on the im v lab v tang + v c Rolling motion com At cente : v lab 0 + v com At bottom: v lab v com + v com At top : v lab v com + v com v 0 com Copyight R. Janow Sping 0

Stationay obseve sees ROLLNG as pue otation about contact point, which is constantly changing Snapshot in time: Complementay view to pevious is the instantaneous cente of

22 Stationay obseve sees ROLLNG as pue otation about contact point, which is constantly changing Snapshot in time: Complementay view to pevious is the instantaneous cente of otation v tan g ω R R ω v A ω ϕ ω cm Rcos( A v cm ω R φ φ ω v tang 0 Angula velocity and acceleation ae the same about contact point o about CM. v cm ω R p cm Copyight R. Janow Sping 0

23 Example: ind the speed of the bowling ball as it olls w/o slipping to the bottom of the amp. Use enegy consevation Rotation acceleates due to fiction between the sphee and the amp iction foce supplies the net toque that causes angula acceleation. Contact point is always at est elative to the suface, so no wok is done against fiction The mechanical enegy does not change Thee is now a otation tem in the KE K E + mech Let U constant f 0 at the bottom K i + U of i the K plane f U K i 0 at the top (stat fom est U i Mgh at the top, U f 0 Rolling condition: f v f ω R cm f cm ω f + Mv f + M v Equate kinetic & potential enegy changes K cm + M v f R f Mgh R f f Suppose ball is a solid sphee cm Mv f MR Copyight R. Janow Sping 0 Mgh v f gh Note: Need Second Law appoach to find intenal foces, like fiction /

24 Example: A wheel acceleating while olling without slipping ulled by constant hoizontal foce applied at the CM to acceleate wheel iction foce f s makes it oll: applies toque to wheel, causes angula acceleation CCW + f s substitute: cm Linea nd Law motion of cm: i 0 x,i f s ma cm y, N τ Rotational nd Law axis though cm: cm,i f s cm cm No slipping constaint: a cm cm ( cm a cm cm cm ma cm - m cm f s + f s cm a also cm mg : cm + m solve fo find a cm : a cm cm cm cm + m m + cm cm cm cm f s cm + m Minimum µ s fo f s f mg no slipping: s µ s µ s,min mg find f s : /m Same esult by placing otation axis at : nstantaneous Rotation Axis /m is the no-fiction acceleation a cm /m also if cm 0 a cm /m fo a hoop ( cm m a cm 5/7m fo solid sphee, f s 0 if cm 0 f 0, 0, ω is constant, f s 0 f s is to the left Copyight R. Janow Sping 0

Example: Rolling on a amp w/o slipping again, via Second Law lace otation axis at mass cente - amp applies toque to the edge of olling object iction foce f s is up the amp, o/w wheel slides.

toque a cm is in x diection fo CCW (+ cm No slipping constaint: a R x,cm cm R (also Solve fo fiction foce in tems of a cm : g Mg f x,cm R cm cm f Solve fo mass cente acceleation: cm s M(a + gsin a

25 Example: Rolling on a amp w/o slipping again, via Second Law lace otation axis at mass cente - amp applies toque to the edge of olling object iction foce f s is up the amp, o/w wheel slides. toque is CCW aound axis, cm Apply linea nd Law to cm motion: x y τ Ma Ma x,cm y,cm f s N - Mgsin Mgcos 0 Apply otational nd f R Law fo axis at cm: cm cm cm s all othe foces have zeo moment am zeo toque a cm is in x diection fo CCW (+ cm No slipping constaint: a R x,cm cm R (also Solve fo fiction foce in tems of a cm : g Mg f x,cm R cm cm f Solve fo mass cente acceleation: cm s M(a + gsin a x,cm (M + Mgsin R s f cm a s R x,cm a x,cm ( gsin + MR cm + MgRsin,net cm cm MR cm + Same esult using instantaneous otation axis τ a cm g sin if we let cm zeo a cm is negative, so f s is positive (up Copyight R. Janow Sping 0

Dynamics of Rotational Motion

Dynamics of Rotational Motion Toque: the otational analogue of foce Toque = foce x moment am τ = l moment am = pependicula distance though which the foce acts a.k.a. leve am l l l l τ = l = sin φ = tan