Prof. Dr. I. Nasser Chapter0_I November 6, 07 Important Terms Chapter 0 Rotatonal moton Angular Dsplacement s, r n radans where s s the length of arc and r s the radus. Angular Velocty The rate at whch changes. Angular Acceleraton; Constant Angular Acceleraton The rate at whch the angular velocty changes. Instantaneous speed, or pont s lnear speed (or, tangental speed) ds d v t r r. dt dt Lnear acceleraton (or, tangental acceleraton) dvt d at r r. dt dt Centrpetal acceleraton dvt d at r r. dt dt Equatons and Symbols s r r = radus Basc Requrements: d lm t 0 t dt d lm t 0 t dt ds d v t r r dt dt dvt d at r r dt dt v r t a c r r r s = length of arc v = velocty = angle n radans = Angular velocty = angular acceleraton v = Lnear (tangental) velocty t a t = tangental acceleraton a c = Radal (Centrpetal) acceleraton. To be famlar wth the angular termnologes, such as Angular: dsplacement, velocty and acceleraton.. To relate the lnear and angular varables. 3. Master the knematc equatons n case of rotaton. 4. Dfferentate between tangental and centrpetal acceleraton
Prof. Dr. I. Nasser Chapter0_I November 6, 07 0- ROTATIONAL VARIABLES So far n our study of physcs we have (wth few exceptons) dealt wth partcles, objects whose spatal dmensons were unmportant for the questons we were askng. We now deal wth the (elementary!) aspects of the moton of extended objects, objects whose dmensons are mportant. The objects that we deal wth are those whch mantan a rgd shape (the mass ponts mantan ther relatve postons) but whch can change ther orentaton n space. They can have translatonal moton, n whch ther center of mass moves but also rotatonal moton, n whch we can observe the changes n drecton of a set of axes that s glued to the object. Such an object s known as a rgd body. We need only a small set of angles to descrbe the rotaton of a rgd body. Stll, the general moton of such an object can be qute complcated. Translaton vs. Rotaton Snce ths s such a complcated subject, we specalze further to the case where a lne of ponts of the object s fxed and the object spns about a rotaton axs fxed n space. When ths happens, every ndvdual pont of the object wll have a crcular path, although the radus of that crcle wll depend on whch mass pont we are talkng about. And the orentaton of the object s completely specfed by one varable, an angle whch we can take to be the angle between some reference lne panted on the object and the x axs (measured counter-clockwse, as usual). Because of the nce mathematcal propertes of expressng the measure of an angle n radans, we wll usually express angles n radans all through our study of rotatons; on occason, though, we may have to convert to or from degrees or revolutons. Revolutons, degrees and radans are related by: revoluton = 360 o = radans revoluton radans [Later, because of ts mportance, we wll deal wth the moton of a (round) object whch rolls along a surface wthout slppng. Ths moton nvolves rotaton and translaton, but t s not much more complcated than rotaton about a fxed axs.]
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Angular Dsplacement As a rotatng object moves through an angle from the startng poston, a mass pont on the object at radus r wll move a dstance s ; s length of arc of a crcle of radus r, subtended by the angle. When s n radans, these are related by s, n radans () r If we thnk about the consstency of the unts n ths equaton, we see that snce s and r both have unts of length, s really dmensonless; but snce we are assumng radan measure, we wll often wrte rad next to our angles to keep ths n mnd. Notes: We wll assume that s + f t s counterclockwse from the + x axs. Although has both magntude and drecton t s not generally consdered a vector quantty because addton of angular dsplacements s not communcatve. Only n the lmtng case of can an angular dsplacement be consdered a vector. Normally we are nterested n as a functon of tme or. revoluton = 360 0 = π radans radan = 57.3 0 = 0.59 revolutons A complete revoluton s some multple ntegers of π radans, e.g. ( ): π, 4π, 6π, () t n etc. If a body rotates about a fxed axs then all the partcles wll have same angular dsplacement (although lnear dsplacement wll dffer from partcle to partcle n accordance wth the dstance of partcles from the axs of rotaton). -------------------------------------------------------------------------- Smple Example: a- What angle n radans s subtended by an arc that has length.80 m and s part of a crcle of radus.0 m? b- Express the same angle n degrees. c- The angle between two rad of a crcle s 0.60 rad. What arc length s subtended f the radus s.40 m? a- The equaton s relates arc-length, radus and subtended angle. We fnd: r s.80 m.50 rad r.0 m b- To express ths angle n degrees use the relaton: 360 deg = rad (or, 80 rad ). Then we have: 80.50 rad (.50 rad) 85.9 rad c- We can fnd the arc length subtended by an angle by the relaton: s r. Then for an angle of 0.60 rad and radus.40m, the arc-length s s r (.40 m)(0.60).49 m. 3
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Angular Velocty : s the rate at whch changes. If n a tme perod the object has rotated through an angular dsplacement defne the average angular velocty for that perod as: t avg t then we () A more nterestng quantty s found as we let the tme perod be vanshngly small. Ths gves us the nstantaneous angular velocty, : d lm (3) t 0 t dt Angular velocty has unts of rad/s, or equvalently, /s or s. In more advanced studes of rotatonal moton, of a rotatng object s defned n such a way that t s a vector quantty. For an object rotatng counterclockwse about a fxed axs, ths vector has magntude and ponts outward along the axs of rotaton. For our purposes, though, we wll treat as a number whch can be postve or negatve, dependng on the drecton of rotaton. t Notes: has the same value for all partcles n a rotatng system. Tangental velocty, whch depends upon dstance from the rotatonal axs, vares dependng upon radus. avg rad/s t Angular velocty s a psuedovector. The drecton s determned from the rght hand rule (RHR). If one curls ther rght hand around the axs of rotaton wth ther fngers pontng n the drecton of rotaton, ther thumb then gves the drecton of the angular momentum vector. Note that the drecton of the angular velocty vector s along the axs of rotaton rather than n the drecton of moton. -------------------------------------------------------- In D, velocty v has a sgn (+ or ) dependng on drecton. Lkewse, for fxed-axs rotaton, has a sgn conventon, dependng on the sense of rotaton. ---------------------------------------------------------- 4
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Smple Example: What s the angular speed n radans per second of b- the Earth n ts orbt about the Sun and c- the Moon n ts orbt about the Earth? b- The Earth goes around n a (nearly!) crcular path wth a perod of one year. In seconds, ths s: In one year ts angular dsplacement s π radans (all the way around) so ts angular speed s c- How long does t take the moon to go around the earth? any good reference source wll tell you that t s 7.3 days. Convertng to seconds, we have: In that length of tme the angular dsplacement of the moon s π so ts angular speed s -------------------------------------------------------------- Angular Acceleraton; Constant Angular Acceleraton The rate at whch the angular velocty changes s the angular acceleraton of the object. If the object s (nstantaneous) angular velocty changes by wthn a tme perod, then the average angular acceleraton for ths perod s avg t (4) But as you mght expect, much more nterestng s the nstantaneous angular acceleraton, defned as: d lm t 0 t dt (5) Notes: Angular acceleraton has the same value for all partcles n a rotatng system. avg rad/s t Angular acceleraton s another psuedovector and ts drecton s also determned from the RHR. t 5
Prof. Dr. I. Nasser Chapter0_I November 6, 07 0- ROTATION WITH CONSTANT ANGULAR ACCELERATION We can derve smple equatons for rotatonal moton f we know that s constant. (Later we wll see that ths happens f the torque on the object s constant.) Then, f s the ntal angular dsplacement, angular acceleraton, then we fnd: s the ntal angular velocty and s the constant o t o o o o t t o (6) (7) (8) o o o t (9) where and are the angular dsplacements and velocty at tme t 0. t. o and o are the values of the angle and angular velocty at These equatons have exactly the same form as the knematc equatons for one dmensonal lnear moton gven n Chapter. The correspondences of the varables are: x, v, a 0 It s almost always smplest to set n these equatons, so you wll often see Eqs. 6 9 wrtten wth ths substtuton already made. o ------------------------------------------------------------ ----------------------------------------------------------- Example: Calculate the requred tme for a wheel, ntally at rest, to turn through 0 full revolutons f t can accelerate at a rate of rad/s. gven that rad/s, 0 rad, and 0 rad/s, then t t 0 rad 0 t rad/s t t 6.3 s. 6
Prof. Dr. I. Nasser Chapter0_I November 6, 07 ---------------------------------------------------- Example: A wheel spns at a rate of 30 revs/sec 30 revs/s 60 rad/s comes to a complete stop n 0 seconds. Fnd: a) the angular acceleraton of the wheel b) the number of revolutons the wheel undergoes before t comes to a stop gven that, 60 rad/s, 0, and take a) b) use t 0 s f 0 f 0 60 rad f t 6 t 0 s 0 60 o o o o 300 rad ( 6 ) Or, we can use, then t t 60 0 6 0 300 rad. Snce there are radans per revoluton, ths yelds 50 revolutons of the wheel. --------------------------------------------------------------------------- Example: A car engne s dlng at ω0= 500 rev/mn at a traffc lght. When the lght turns green, the crankshaft rotaton speeds up at a constant rate to ω = 500 rev/mn over an nterval of 3.0 s. The number of revolutons the crankshaft makes durng these 3.0 s s: f 500 500 rev t 3 75 rev 60 ----------------------------------------------------------- Example: The angular poston of a pont on the rm of a rotatng wheel s gven by 3 ( t) 4.0t 3.0t t, where s n radans f t s gven n seconds. a- What are the angular veloctes at t =.0 s and t = 4.0 s? b- What s the average angular acceleraton for the tme nterval that begns at t =.0 s and ends at t = 4.0 s? c- What are the nstantaneous angular acceleratons at the begnnng and end of ths tme nterval? a- In the problem we are gven the angular poston as a functon of tme. To fnd the (nstantaneous) angular velocty at any tme, use Eq. 3 and fnd: where, f t s gven n seconds, s gven n rad/s. The angular veloctes at the gven tmes are then b- Snce we have the values of and t =.0 s and t = 4.0 s, Eq. 4 gves the average angular acceleraton for the nterval: 7
Prof. Dr. I. Nasser Chapter0_I November 6, 07 The average angular acceleraton s.0 rad/s. c- We fnd the nstantaneous angular acceleraton from Eq. 5: where, f t s gven n seconds, s gven n rad/s. Then at the begnnng and end of our tme nterval the angular acceleratons are: ------------------------------------------------------- 8
Prof. Dr. I. Nasser Chapter0_I November 6, 07 H.W. Sample problem 0.0 The angular poston of a pont on the rm of a rotatng wheel s gven by ( t).00 0.600t 0.50t, where s n radans f t s gven n seconds. H.W. At what tme, t mn, does () t reach the mnmum value? What s () t at t mn? Calculate to fnd t mn.0 s, and ( tmn ).36 rad 77.9 ------------------------------------------------------------------- Example: An electrc motor rotatng a grndng wheel at 00 rev/mn s swtched off. Assumng constant negatve angular acceleraton of magntude.00 rad/s, (a) How long does t take the wheel to stop? (b) Through how many radans does t turn durng the tme found n (a)? (a) Convert the ntal rotaton rate to radans per second: ( t) 0 When the wheel has stopped then of course ts angular velocty s zero. Snce we know and we can use Eq. 6 to get the elapsed tme: o o t t and we get: o, The wheel takes 5.4 s to stop. (b) We want to fnd the angular dsplacement durng the tme of stoppng. Snce we know that the angular acceleraton s constant we can use Eq 9, and t mght be smplest to do so. Then we have: The wheel turns through 7.5 radans n comng to stop. ------------------------------------------------------------------------ 9
Prof. Dr. I. Nasser Chapter0_I November 6, 07 0-3 RELATING THE LINEAR AND ANGULAR VARIABLES As we wrote n Eq., when a rotatng object has an angular dsplacement pont on the object at a radus r travels a dstance s r, then a. Ths s a relaton between the angular moton of the pont and the lnear moton of the pont (though here lnear s a bt of a msnomer because the pont has a crcular path). The dstance of the pont from the axs does not change, so takng the tme dervatve of ths relaton gve the nstantaneous speed of the partcle as: ds d v t r r (0) dt dt whch we smlarly call the pont s lnear speed (or, tangental speed ) to dstngush t from the angular speed. Note, all ponts on the rotatng object have the same angular speed but ther lnear speeds depend on ther dstances from the axs. Smlarly, the tme dervatve of the Eq. 0 gves the lnear acceleraton of the pont: dvt d at r r () dt dt Here t s essental to dstngush the tangental acceleraton from the centrpetal acceleraton that we recall from our study of unform crcular moton. It s stll true that a pont on the wheel at radus r wll have a centrpetal acceleraton gven by: r vt a c r r r These two components specfy the acceleraton vector of a pont on a rotatng object. (Of course, f s zero, then at 0 and there s only a centrpetal component.) --------------------------------------------------------------- Example: What s the angular speed of a car travelng at 50 km/h and roundng a crcular turn of radus 0 m? To work consstently n SI unts, convert the speed of the car: () Example: A dsk, of radus 6.0 cm, s free to rotate at a constant rate of 00 rpm about ts axs. Fnd: a- the radal acceleraton b- the tangental acceleraton. a- 00 rpm 00 5.7 rad/s 60 v ar R R (0.06) (5.7) 948 m/s ; snce s constant. b- at 0 ----------------------------------------------------------------- 0
Prof. Dr. I. Nasser Chapter0_I November 6, 07
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Example: An astronaut s beng tested n a centrfuge. The centrfuge has a radus of 0 m and, n startng, rotates accordng to ( t) 0.30 t, where t n seconds gves n radans. When t = 5.0 s, what are the astronaut s (a) Angular velocty, (b) Lnear speed, (c) Tangental acceleraton (magntude only) and (d) Radal acceleraton (magntude only)?
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Extra Problems Q: A dsk a horzontal rotatng platform of radus r s ntally at rest, and then begns to accelerate constantly untl t has reached an angular velocty after complete revolutons. What s the angular acceleraton of the dsk durng ths tme? Ans: Gven the quanttes: 0, 0,, then, the angular acceleraton of o o f the dsk can be determned by usng rotatonal knematcs: /8 ------------------------------------------------------------------ Q: A rotatng wheel moves unformly from rest to an angular speed of 0.6 rev/s n 33 s. a) Fnd ts angular acceleraton n rad/s. b) Would doublng the angular acceleraton durng the gven perod have doubled fnal angular speed? Gven the quanttes: 0, 0, 0.6 rad/s, and t = 33 s, then, the o o f angular acceleraton of the dsk can be determned by a) Usng the knematc equaton: a) For double the angular acceleraton we should have: The angular speed wll be doubled as well --------------------------------------------------------------------------- 3
Prof. Dr. I. Nasser Chapter0_I November 6, 07 Q3: A racng car travels on a crcular track of radus 75 m. Suppose the car moves wth a constant lnear speed of 5.5 m/s. a) Fnd ts angular speed. b) Fnd the magntude and drecton of ts acceleraton. Gven that r 75 m, v v 5.5 m/s t a) Angular and lnear (tangental) speed are always related through : t v v r b) Wth a constant lnear speed the acceleraton s radal dv at 0) dr v a a v r as r ( r t -------------------------------------------------------------- Q4: A wheel.65 m n dameter les n a vertcal plane and rotates about ts central axs wth a constant angular acceleraton of 3.70 rad/s. The wheel starts at rest at t = 0, and the radus vector of a certan pont P on the rm makes an angle of 57.3 o wth the horzontal at ths tme. At t.00 s, fnd the followng: a) the angular speed of the wheel. b) the tangental speed of the pont P. c) the total acceleraton of the pont P. d) the angular poston of the pont P. Gven the quanttes: o 0, then: a) The wheel started at rest., therefore: o 80 57.3 rad, 3.70 rad/s, at t = s, b) The tangental speed of pont P located on the rm: c) To calculate the total acceleraton of the pont P, we need to calculate both the radal and tangental components And fnally: Its drecton wth respect to the radus to P can be evaluated from 4
Prof. Dr. I. Nasser Chapter0_I November 6, 07.e. 3.86 d) ------------------------------------------------------------------ Q: The angular poston of a pont on the rm of a rotatng wheel of radus R s gven by: θ (t) = 6.0 t + 3.0 t.0 t 3, where θ s n radans and t s n seconds. What s the average angular acceleraton for a pont at R/ for the tme nterval between t = 0 and t = 5 s? 4 rad/s Answer θ (t) = 6.0 t + 3.0 t.0 t 3 ω (t) = 6.0 + 6.0 t 6.0 t ω (0) = 6.0, ω (5) = -4 4 6 4 t 50 ------------------------------------------------------------------ Q: A unform dsk starts from rest and rotates, about fxed central axs, wth a constant angular acceleraton. It reaches an angular velocty of 3.7 rad/s when t has completed 5.00 revolutons. What s the angular velocty when t has completed 9.00 revolutons? 8.4 rad/s Frst calculate the acceleraton f (3.7) 0.987 rad/s 5 Second (9 revolutons) 0.987 9 8.38 rad/s ------------------------------------------------------------------------------ Q: A phonograph turntable rotatng at 33.3 rev/mn slows down and stops n 30 s after the motor s turned off. (a) Fnd ts (unform) angular acceleraton n unts of rev/mn. (b) How many revolutons dd t make n ths tme? (a) Here we are gven the ntal angular velocty of the turntable and ts fnal angular velocty (namely zero, when t stops) and the tme nterval between them. We can use Eq. 6 to fnd, whch we are told s constant. We have: 5
Prof. Dr. I. Nasser Chapter0_I November 6, 07 o o t t We don t need to convert the unts of the data to radans and seconds; f we watch our unts, we can use revolutons and mnutes. Notng that the tme for the turntable to stop s t = 30 s = 0.50mn, and wth = 33.3 rev/mn and = 0 we fnd: o The angular acceleraton of the turntable durng the tme of stoppng was 66.7 rev/mn. (The mnus sgn ndcates a deceleraton, that s, an angular acceleraton opposte to the sense of the angular velocty.) (b) Here we want to fnd the value of at t = 0.50 mn. To get ths, we can use ether Eq. 7 or Eq. 9. Wth, Eq. 9 gves us: 0 o The turntable makes 8.33 revolutons as t slows to stop. ----------------------------------------------------------------- Q: A dsk, ntally rotatng at 0 rad/s, s slowed down wth a constant angular acceleraton of magntude 4.0 rad/s. (a) How much tme elapses before the dsk stops? (b) Through what angle does the dsk rotate n comng to rest? (a) We are gven the ntal angular velocty of the dsk, = 0 rad/s. (We let the postve sense of rotaton be the same as that of the ntal moton.) We are gven the magntude of the dsk s angular acceleraton as t slows, but then we must wrte o The fnal angular velocty (when the dsk has stopped!) s = 0. Then from Eq. 6 we can solve for the tme t: o o t t and we get:. We can now use any of the constant (b) We ll let the ntal angle be o 0 contanng to solve for t; let s choose Eq. 8, whch gves us: o o ( ) and we get: equatons The dsk turns through an angle of 800 radans before comng to rest. ---------------------------------------------------------------- Q: A wheel, startng from rest, rotates wth a constant angular acceleraton of.00 rad/s. Durng a certan 3.00 s nterval, t turns through 90.0 rad. 6
Prof. Dr. I. Nasser Chapter0_I November 6, 07 (a) How long had the wheel been turnng before the start of the 3.00 s nterval? (b) What was the angular velocty of the wheel at the start of the 3.00 s nterval? (a)we are told that sometme after the wheel starts from rest we measure the angular dsplacement for some 3.00 s nterval and t s 9.00 rad. Suppose that we start measurng tme at the begnnng of ths nterval; snce ths tme measurement sn t from the begnnng of the wheel s moton, we ll call t t o. Now, wth the usual choce o 0 we know that at 3.00 s t we have = 90.0 rad. Also =.00 rad/s. Usng Eq. 7 to get: o whch we can use to solve for o : so that (Lookng ahead, we can see that we ve already answered part (b)!) Now suppose we measure tme from the begnnng of the wheel s moton wth the varable t. We want to fnd the length of tme requred for to get up to the value 7.0 rad/s. For ths perod the ntal angular velocty s = 0 and the fnal angular velocty s 7.0 rad/s. Snce we have we can use Eq. 6 to get t : o o t t whch gves o Ths tells us that the wheel had been turnng for 3.5 s before the start of the 3.00 s nterval. (b) In part (a) we found that at the begnnng of the 3.00 s nterval the angular velocty was 7.0 rad/s. ------------------------------------------------------------------ 7
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Summary of last lecture: Chapter 0 Rotatonal moton moton Classfcaton dsplacemet velocty acceleraton Lnear Rotaton d s d v s v a dt dt ------------------------------------------------------------------------ Important Terms Moment of nerta (rotatonal nerta) I n kg.m : a quantty expressng a body's tendency to resst angular acceleraton. For a pont mass the moment of nerta s just the mass tmes the square of perpendcular dstance to the rotaton axs, I = m r. That pont mass relatonshp becomes the bass for all other moments of nerta. Total moment of nertal s the sum of the products of the mass of each partcle n the body wth the square of ts dstance from the axs of rotaton Rotatonal knetc energy Krot I Parallel axs theorem: can be used to determne the mass moment of nerta of a rgd body about any axs, gven the body's moment of nerta about a parallel axs through the object's center of gravty and the perpendcular dstance between the axes. Equatons and Symbols s r d lm, t 0 t dt d lm t 0 t dt ds d v t r r dt dt dvt d at r r dt dt v r t ac r r r I mr Krot I CM I I MD s = length of arc r = radus v = velocty = angle n radans = Angular velocty = angular acceleraton v t a t = Lnear (tangental) velocty = tangental acceleraton a = Radal (Centrpetal) acceleraton c I moment of nerta (rotatonal nerta) m = mass of partcles. M = total mass D = dstance Basc Requrements:. Master the calculaton of the moment of nerta of a system of partcles.. Use the Parallel axs theorem to calculate the moment of nerta of a system of partcles.
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 0-4 KINETIC ENERGY OF ROTATION Because a rotatng object s made of many mass ponts n moton, t has knetc energy; but snce each mass pont has a dfferent lnear speed our formula from translatonal partcle moton, m K m v rotatng object as, havng ndvdual (dfferent!) lnear speeds energy of the rotatng object s If r The sum K v r no longer apples. If we label the mass ponts of the m v rot s the dstance of the th mass pont form the axs, then mr v v r K rot m v m r m r, then the total knetc and we then have: s called the moment of nerta for the rotatng object (whch we dscuss further n the next secton), and usually denoted I. (It s also called the rotatonal nerta n some books.) It has unts of kg m n the SI system. I of a body s a measure of the rotatonal nerta of the body. Wth ths smplfcaton, our last equaton becomes Krot (3) I ------------------------------------------------------------------ Example: Calculate the rotatonal nerta of a wheel that has a knetc energy of 4, 400 J when rotatng at. 60 revs/mn Frst, fnd the angular speed of the wheel n rad/s: 60 revs/mn 63.0 rad/s Fnally, we have Krot I I. K rot ----------------------------------------------------
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 0-5 CALCULATING THE ROTATIONAL INERTIA For a rotatng object composed of many mass ponts, the moment of nerta I I s gven by I m r (4) has unts of kg m n the SI system, and as we use t n elementary physcs, t s a scalar (.e. a sngle number whch n fact s always postve). More frequently we deal wth a rotatng object whch s a contnuous dstrbuton of mass, and for ths case we have the more general expresson (not requred n our course) I r dm (5) Here, the ntegral s performed over the volume of the object and at each pont we evaluate, where r s the dstance measured perpendcularly from the rotaton axs. The evaluaton of ths ntegral for several cases of nterest s a common exercse n multvarable calculus. In most of our problems we wll only be usng a few basc geometrcal shapes, and the moments of nerta for these are gven n the Appendx. Appendx: r Moments of Inerta for some shapes M s the total mass, a, b, and L are lengths, and R s the radus. ---------------------------------------- ---------------------------------------------------------------------- 3
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Example: As shown n the fgure, three masses, of.5 kg each, are fastened at fxed poston to a very lght rod pvoted at one end. Fnd the moment of nerta for the rotaton axes shown Apply the equaton 3 I m r m r m r m r m( r r r ).5( 3 ) kg.m 3 3 3 ---------------------------------------------- Example: The fgure shows a rgd body conssts of two partcles attached to a rod of neglgble mass. The masses are m =.00 kg and m =.00 kg and they are separated by a dstance r = r + r. (a) Fnd the moment of nerta of the body. Assume r = 0.33 m and r = 0.67 m are the dstance between m and the rotaton axs and m and the rotaton axs (the dashed, vertcal lne) respectvely. (b) What s the moment of nerta f the axs s moved so that s passes through m? (c) What s your comment on the two calculated values? Whch one wll be easy to rotate? (a) Apply the formula (4) of the moment of nerta, we can have I mr mr mr.0(0.33 ).0(0.67 ) 0.67 kg.m (b) I m r mr mr.0(0.00 ).0(.00 ).00 kg.m (c)??? --------------------------------------------------------------- Example: A rgd body conssts of two partcles attached to a rod of neglgble mass. The rotatonal nerta of the system about the axs shown n Fgure s 0 kg m. What s x? I m r m r m r 0 x x.4 m ---------------------------------------------------------------------- Example: A hoop rolls wthout sldng on a horzontal floor. The rato of ts translatonal knetc energy to ts rotatonal knetc energy (about ts central axs) s The rato s K mv mv edge K center I mr v / R --------------------------------------------------------------------------- 4
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Example: A sold sphere of mass m s fastened to another sphere of mass m by a thn rod wth a length of 3x. The spheres have neglgble sze, and the rod has neglgble mass. What s the moment of nerta of the system of spheres as the rod s rotated about the pont located at poston x, as shown? Moment of nerta for a system of dscrete masses s calculated as follows: I mr m( x) m( x) 9mx ------------------------------------------------------------------------- Q3: Rgd rods of neglgble mass lyng along the y axs connect three partcles. The system rotates about the x axs wth an angular speed of.0 rad/s. a) Fnd the moment of nerta about the x axs. b) Fnd the total rotatonal knetc energy evaluated from c) Fnd the tangental speed of each partcle. d) Fnd the total knetc energy evaluated from e) Your comment for b and d. a) K Krot m v rot I b) c) Dfferent lnear speeds for dfferent radus. However, all partcles are rotatng at same v angular speed: r 5
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Mass : Mass : Mass 3: d) The total knetc energy s: e) Both expressons lead to the same value. ----------------------------------------------------------------------------- Q4: A par of long, thn, rods, each of length L and mass M, are connected to a hoop of mass M and radus L/ to form a 4-spoked wheel as shown n the fgure. Express all answers n terms of the gven varables and fundamental constants. Calculate the moment of nerta for the entre spoked-wheel assembly for an axs of rotaton through the center of the assembly and perpendcular to the plane of the wheel. The moment of nerta for the spoked wheel s smply the sum of the ndvdual moments of nerta of ts three components: the two long thn rods and the hoop around the outsde: --------------------------------------------------------------------------------------- Example: Three pont masses,.e. they have no moment of nerta, each of mass m are placed at the corners of an equlateral trangle of sde a. Calculate the moment of nerta of ths system about an axs passng along one sde of the trangle. Choose the sde AB. The moment of nerta of system about AB sde of trangle a 3 ------------------------------------------------ Isystem IA IB IC 0 0 mx mx m ma 3 4 6
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Parallel Axs Theorem Moment of nerta of a body about a gven axs I s equal to the sum of moment of nerta of the body about an axs parallel to gven axs and passng through center of mass of the body s CM and, where s the mass of the body and D s the perpendcular dstance between the two axes. Ths stuaton s shown n the fgure. In symbol, the new moment of nerta of the object about the new axs wll have a new value I, gven by I ICM MD (6) Eq. 6 s known as the Parallel Axs Theorem and s sometmes handy for computng moments of nerta f we already have a lstng for a moment of nerta through the object s center of mass. Relates (axs through center-of-mass) to I w.r.t. some other axs: I I MD M I CM (See proof n text.) --------------------------------------------------------------- I MD CM ---------------------------------------------------------------------- Example: Calculate the moment of nerta for a rod about ts end pont. Rod of length L, mass M I CM Pvot, d = L/ CM M R 4 3 I I M d M L M L M L Example: Moment of nerta of a dsc about an axs through ts center of mass and perpendcular to ts plane s IG MR. Calculate the moment of nerta about an axs through ts tangent perpendcular to the plane The moment of nerta about an axs through ts tangent perpendcular to the plane s gven by: 3 I IG MR MR MR MR ----------------------------------------------------------- 7
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Example: Four thn rods of same mass M and same length l, form a square as shown n fgure. Calculate the moment of nerta of ths system about an axs through center O and perpendcular to ts plane. CM I ( rod) M M.I. of rod AB about pont P M M.I. of rod AB about pont [by the theorem of parallel axs] 3 O M M M and the system conssts of 4 rods of smlar type so by the symmetry Isystem ------------------------------------------------------ Example: Three rngs each of mass m and radus R are arranged as shown n the fgure. Calculate the moment of nerta of the system about YY '. CM I ( rng) mr M.I of system about YY ' Isystem I I I3, where I = moment of nerta of rng about CM, I I3 MI.. of nerta of rng about a tangent n a plane 7 Isystem ------------------------------------------------------------ mr mr mr mr mr mr Example: Three dentcal thn rods, each of length L and mass M, are welded perpendcular to one another as shown n the fgure. They are placed along X, Y and Z-axes n such a way that one end of each of the rod s at the orgn. The moment of nerta of ths system about Z axs s ICM( rod) ML Moment of nerta of the system about z-axs can be fnd out by calculatng the moment of nerta of ndvdual rod about z-axs I I ML because z-axs s the edge of rod and and 3 0 3 Isystem I I I3 ML ML 0 ML 3 3 3 --------------------------------------------------------------------- Example: Fnd the moment of nerta of a unform rng of radus R and mass M about an axs R from the center of the rng as shown n the Fgure. I o ICM Md MR M ( R ) 5MR ---------------------------------------------------- 4 M 3 I because rod n lyng on z-axs 8
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 Extra Problems Q: A unform slab of dmensons: a = 60 cm, b = 80 cm, and c =.0 cm (see Fgure) has a mass of 6.0 kg. Its rotatonal nerta about an axs perpendcular to the larger face and passng through one corner of the slab s: Use the equaton: I I +MD CM M a b M I= ( a b ) M ( ) ( ) ( a b ).0 kg.m 3 ------------------------------------------------------------------------ Q: Calculate the rotatonal nerta of a meter stck wth mass 0.56 kg, about an axs perpendcular to the stck and located at the 0 cm mark. A pcture of ths rotatng system s gven n the fgure. The stck s one meter long (beng a meter stck and all that) and we take t to be unform so that ts center of mass s at the 50 cm mark. But the axs of rotaton goes through the 0 cm mark. Now f the axs dd pass through the center of mass (perpendcular to the stck), we would know how to fnd the rotatonal nerta; from Fgure we see that t would be ICM M L ( 0. 56)(. 00) 4. 7 0 kg.m The rotatonal nerta about our axs wll not be the same. ----------------------------------------------------- Q: In the Fgure, two partcles, each wth mass m 0.85 kg, are fastened to each other, and to a rotaton axs at O, by two thn rods, each wth length d = 5.6 cm and mass M =. kg. The combnaton rotates around the rotaton axs wth the angular speed = 0.30 rad/s. Measured about O, what are the combnaton s (a) rotatonal nerta and (b) knetc energy? 9
Prof. Dr. I. Nasser Chapter 0-II November 6, 07 The partcles are treated pont-lke n the sense that Eq. 0-33 yelds ther rotatonal nerta, and the rotatonal nerta for the rods s fgured usng Table 0-(e) and the parallel-axs theorem (Eq. 0-36). (a) Wth subscrpt standng for the rod nearest the axs and 4 for the partcle farthest from t, we have I I I I I O, O, O 3, O 4, O I, O Md M d I, O Md M d I 3, O 4, O 3 md I m d ( ).,, 8 8 IO Md 5 md (. kg)(0.056 m) +5(0.85 kg)(0.056 m) 3 3 =0.03 kg m. (b) Usng Eq. 0-34, we have 8 4 5 K rot IO M 5 md (. kg) (0.85 kg) (0.056 m) (0.30 rad/s) 3 3 3.0 J. ------------------------------------------------- 0
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Rotatonal moton Basc Requrements:. Master the calculaton of the rotatonal torque.. Calculate the power n case of rotatonal. Basc Prncples: Why s the handle on a door located far away from the hnge? Why s t easer to loosen a nut usng a long wrench? Why are long wheel-base cars more stable than short wheel-base cars? handle on a door s located far away from the hnge loosen a nut usng a long wrench The ablty of a force to rotate an object about an axs depends on two varables:. The magntude of the force F.. The dstance r between the axs of rotaton and the pont where the force s appled. rotaton axs r F Try openng a door by applyng the same force F at dfferent ponts: r = outer edge, mddle, near hnge. You wll quckly realze that the resultng moton of the door the acceleraton depends on F and r. It turns out that the turnng ablty of a force s smply the product of F and r. The techncal name for ths turnng ablty s torque. Torque comes from Latn and means to twst. ------------------------------------------------
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 0-6 Torque Is a force that causes a rotatonal acceleraton of a rgd body about an axs or moton of a sngle partcle relatve to some fxed pont. - Torque s a vector. - Torque s postve when the body rotate counterclockwse (conventon) 3- Torque s negatve when the body rotate clockwse (conventon) Suppose the force F (whose drecton les n the plane of rotaton) s appled at a pont (relatve to the rotaton axs whch s the pvot). Suppose that the (smallest) angle between r and F s. Then the magntude of the torque exerted on the object by ths force s rf ( sn ) (7) By some very smple regroupng, ths equaton can be wrtten as r( F sn ) r F r F r F moment arm of F SI unt of torque s N.m (same as the work); but Never use Joules as a unt of torque, because Joules s a unt of work. ---------------------------------------------------- In Summary: Force causes lnear acceleraton. Torque causes angular acceleraton. ------------------------------------------------------------- If you want to easly rotate an object about an axs, you want a large lever arm r and a large perpendcular force F: r axs bad better best no good! (F = 0) no good! (r = 0) ------------------------------------------------------------------------- ------------------------------------------------------
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Example: Calculate the net torque (magntude, n N.m, and drecton) on a unform beam shown n the Fgure about a pont O passng through ts center. -------------------------------------------------------------------- Example: The pull cord of an engne s wound around a drum of radus 6.00 cm. The cord s pulled wth a force of 75.0 N by the engne. What magntude torque does the cord apply to the drum? ------------------------------------------------------------- Q: A seres of wrenches of dfferent lengths s used on a hexagonal bolt, as shown below. Whch combnaton of wrench length and Force apples the greatest torque to the bolt? The correct answer s c. Torque, the turnng effect produced by a force appled to a moment-arm, s calculated accordng to r F r( F sn ), where s the angle between the vectors r and F. Here, each combnaton of wrench length and force produces a net torque of LF except for answer c: r F r( F cos ) L(F cos30 ) 3LF 3
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 0-7 NEWTON S SECOND LAW FOR ROTATION =I s the rotatonal analogue of F = ma. Newton s Second Law Lnear Acceleraton of an object s drectly proportonal to the net force actng on t and nversely proportonal to ts mass (nerta). a F, a / m a F / m Newton s Second Law for Rotaton Angular Acceleraton of an object s drectly proportonal to the net force torque actng on t and nversely proportonal to ts mass rotatonal nerta., / I / I ------------------------------------------------- In Summary: we are gong to use the followng expressons for the torque and the equaton I () = r F () a R (3) 4
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Examples Example: The flywheel of a statonary engne has a moment of nerta 0f 30 kg.m. What constant torque s requred to accelerate the flywheel to an angular velocty of 900 rpm n 0 seconds, startng from rest? f 900 rpm 900 94. rad/s; 0 60 d 94. 0 I I 30 8.7 N.m; dt 0 -------------------------------------------------------------- Example: A unform thn rod of mass M = 3.0 kg and length L =.0 m s pvoted at one end O and acted upon by a force F = 8.0 N at the other end as shown n Fgure. Calculate the angular acceleraton of the rod at the moment the rod s n the horzontal poston as shown n ths fgure. Frst calculate the M.I. about the pvot O usng the PAT: L I0 ICM Md ML M ( ) ML ; () 3 Then, use the torque equatons as follows: Io ML r F = Fr 3 Equatng () = (3) mples FL 8 rad 4 Counterclockwse I 0 (3) s 3 -------------------------------------------------------------------------------- Example: A unform dsk of radus 50 cm and mass 4 kg s mounted on a frctonless axle, as shown n Fgure. A lght cord s wrapped around the rm of the dsk and a steady downward pull of 0 N s exerted on the cord. Fnd the tangental acceleraton of a pont on the rm of the dsk. Apply Newton s second law gves: () FL (3) ma F T 0 F T () Then, use the torque equatons as follows: ICM MR r F = Fr Equatng () = (3) and usng () TR (3) a R mples 0 5 m/s Clockwse Ma T a T M 4 ------------------------------------------------------------------------ 5
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Example: A rgd bar wth a mass M and length L s free to rotate about a frctonless hnge at a wall, see fgure a. The bar has a moment of nerta I = /3 ML about the hnge, and s released from rest when t s n a horzontal poston as shown. What s the nstantaneous angular acceleraton when the bar has swung down so that t makes an angle of 30 to the vertcal? a b The bar s acceleratng angularly n response to the torque due to the force of gravty actng on the center of mass. Its angular acceleraton due to ths torque at the fnal poston, see fgure (b), can be calculate as follows: L 0 I ML () r F = Fr Mg sn30 () 3 () = () mples Note that: 3g 4L Clockwse - At the horzontal poston, we have the maxmum torque: L Mg 0 - At the vertcal poston, we have the mnmum torque: L 0 moton. ----------------------------------------------------- sn90 MgL /. Mg sn0 0,.e. no rotaton 6
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Example: For the system n fgure (a) wth M.5 kg, m. kg, and R 0. m, fnd at, and. Draw the FBD for the masses, Fgures (b) and (c), and then apply Newton s nd law as follows: - For m: Fgure (b), consder the moton s gong down: ma mg T (), - For M: Fgure (c), consder the rotaton clockwse s postve: I MR () TR (3) CM Equatng () = (3) and usng a R mples T Ma (4) Substtute (4) n (), we can have m. ma mg Ma a g 4.8 m/s M m.5. T Ma 6.0 N a 4.8 4 rad/s Clockwse R 0. --------------------------------------------------------------------------- Example: A mass, m = 5.0 kg, hangs from a strng and descends wth a lnear acceleraton a. The other end s attached to a mass m = 4.0 kg whch sldes on a frctonless horzontal table. The strng goes over a pulley (a unform dsk) of mass M =.0 kg and radus R = 5.0 cm (see Fgure a). Fnd the value of a. a b The equatons of moton for the two masses are gven by: ma mg T (), m m a mg ( T T ) (3) ma T () 7
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 We are takng the clockwse drecton s postve (see fgure b, then ICM MR (4) T-T R (5) Equatng (4) = (5) and usng a R mples Ma ( T T ) (6) Use Eqs. (3) and (6) to solve for a, one fnds: mg m a 4.9 m m M / s -------------------------------------------------------- m Example: In Fg. 0-4, block has mass 0.460 kg, block has mass m 0.500 kg, and the pulley, whch s mounted on a horzontal axle wth neglgble frcton, has radus R = 5.00 cm. When released from rest, block falls 75.0 cm n 5.00 s wthout the cord slppng on the pulley. (a) What s the magntude of the acceleraton of the blocks? What are (b) tenson and (c) tenson acceleraton? (e) What s ts rotatonal nerta? (a) We use constant acceleraton knematcs. If down s taken to be postve and a s the acceleraton of the heaver block m, then ts coordnate s gven by y at, so T? (d) What s the magntude of the pulley s angular y ( 0. 750 m) a 6 00 0. m / s. t ( 500. s) Block has an acceleraton of 6.00 0 m/s upward. (b) Newton s second law for block s m g T ma, where s ts mass and tenson force on the block. Thus, T m ( g a ) (0.500 kg) 9.8 m/s 6.000 m/s 4.87 N. m T T s the (c) Newton s second law for block s m g T m a, where s the tenson force on the block. Thus, T m ( g a ) (0.460 kg) 9.8 m/s 6.000 m/s 4.54 N. (d) Snce the cord does not slp on the pulley, the tangental acceleraton of a pont on the rm of the pulley must be the same as the acceleraton of the blocks, so a 6. 00 0 m / s 0. rad / s. Clockwse R 500. 0 m (e) The net torque actng on the pulley s ( T T) R. Equatng ths to I we solve for the rotatonal nerta: T T R 4.87 N 4.54 N5.000 m.38 0 I kg m..0 rad/s ------------------------------------------------------------------------------ T 8
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 0-8 WORK AND ROTATIONAL KINETIC ENERGY In lnear moton, we knew that W K K K mv mv ; W Fdx, f F s constant f f W P Fv t ----------------------------------------------- Smlarly, for rotatonal moton, we can have: W K K K I I ; f f f W d ( ) f s constant dw P dt ------------------------------------------------- f x f x (power, rotaton about fxed axs) Example: A horzontally-mounted dsk wth moment of nerta I spns about a frctonless axle. At tme. A constant torque s appled to the dsk, causng t to come to stop n tme t. How much Power s requred to dsspate the wheel s energy durng ths tme? Gven that: 0,. The Power requred to dsspate the wheel s ntal t 0, the ntal angular speed of the dsk s f energy s calculated usng P=W/t, where W s the Work requred to change the wheel s knetc energy from ts ntal value to 0: W P, W K K f K I 0 t f I I W I P t t --------------------------------------------------------------- Example: The engne delvers.0 0 5 W to a plane fan at 400 rev/mn 5 rad/s. How much work does the engne do n one revoluton? Snce s vt t; t, then the perodc tme wll be: T s 0.05 s. Consequently, W 5 P W PT. 0 0.05 3000 J t ----------------------------------------------- Example: A grndng wheel of moment of nerta 0f 0.0 kg.m s brought to rest, n 0 revolutons, from an ntal angular velocty of 3000 rpm = 34. rad/s. What s the power dsspated? 0 W K I (0.0) 34. 493.5J, t 0. s; (3000/ 60) K P t 3.47 0 W ------------------------------------------------------------------------------ 9
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Sample Problem 0. Work, rotatonal knetc energy, torque, dsk ----------------------------------------------- Analogous Lnear and Angular Quanttes Lnear Angular Relaton Lnear dsplacement s Angular dsplacement Lnear speed v Angular speed Lnear acceleraton Mass (Inerta) a m F Angular acceleraton Moment of nerta Torque Force Lnear momentum mv Angular momentum mvr I Lnear mpulse Ft Angular mpulse t I s r v r a r I mr rf Lnear Angular F ma I K. E. K. E. work = Fs mv work = I power = Fv power = Secton Summary We now have some understandng of why objects rotate the way they do. We bult the laws of rotatonal moton n analogy to Newton s Laws of Moton for translaton, Newton s Frst Law for Rotaton Every object wll move wth a constant angular velocty unless a torque acts on t. Newton s Second Law for Rotaton Angular acceleraton of an object s drectly proportonal to the net torque actng on t and nversely proportonal to ts rotatonal nerta. ----------------------------------------- 0
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Extra problems Q. A strng s wrapped around a sold dsk of mass m, radus R. The strng s stretched n the vertcal drecton and the dsk s released as shown n the Fgure. Fnd the tenson (T) n the strng. ma mg T (), a a Icm TR; mr TR ma T () R R () () a g T g 3 3 -------------------------------------------------- Q: A strng (one end attached to the celng) s wound around a unform sold cylnder of mass M =.0 kg and radus R = 0 cm (see Fgure). The cylnder starts fallng from rest as the strng unwnds. The lnear acceleraton of the cylnder s: ma mg T (), a a I cm TR ; mr () R TR ma T R () a g 6.53 m/s 3 --------------------------------------------------------------------------- Q: A 6 kg block s attached to a cord that s wound around the rm of a flywheel of radus 0.0 m and hangs vertcally, as shown n Fg 4. The rotatonal nerta of the flywheel s 0.50 kg m. When the block s released and the cord unwnds, the acceleraton of the block s: ma mg T (), a a I TR ; 0.5 T 0. T.5 a () R 0. 6 () 6a 6g.5 a a g 5.5 m/s 8.5 ---------------------------------------------------------------------- H.W.: A torque of 0.80 N.m appled to a pulley ncreases ts angular speed from 45.0 rpm to 80 rpm n 3 seconds. Fnd the moment of nerta of the pulley. 0.7 kg.m -----------------------------------------------------------------------------------------------------
Prof. Dr. I. Nasser Chapter 0-III November 6, 07 Q: A unform.0 kg cylnder of radus 0.5 m s suspended by two strngs wrapped around t, as shown n Fgure 4. The cylnder remans horzontal whle descendng. The acceleraton of the center of mass of the cylnder s: Start wth the equatons of moton: ma mg T (), I tr (), a r Wth the gven nformaton I mr, m kg, r 0.5 m, one has: (3) T ma 4 3 m () ma mg a g 6.53 3 s (3), ------------------------------------------------------------------------ Q: In Fg. 0-4, two blocks, of mass m = 400 g and m = 600 g, are connected by a massless cord that s wrapped around a unform dsk of mass M= 500 g and radus R =.0 cm. The dsk can rotate wthout frcton about a fxed horzontal axs through ts center; the cord cannot slp on the dsk. The system s released from rest. Fnd (a) the magntude of the acceleraton of the blocks, (b) the tenson T n the cord at the left, and (c) the tenson T n the cord at the rght. We choose postve coordnate drectons (dfferent choces for each tem) so that each s acceleratng postvely, whch wll allow us to set a a R (for smplcty, we denote ths as a). Thus, we choose upward postve for m, downward postve for m and (somewhat unconventonally) clockwse for postve sense of dsk rotaton. Applyng Newton s second law to mm and (n the form of Eq. 0-45) to M, respectvely, we arrve at the followng three equatons. T m g m a m g T m a T R T R I (a) The rotatonal nerta of the dsk s I MR (Table 0-(c)), so we dvde the thrd equaton (above) by R, add them all, and use the earler equalty among acceleratons to obtan: F m g m g G I H m m MJ a K whch yelds a 4 g.57 m/s 5. (b) Pluggng back n to the frst equaton, we fnd T 9 mg 4.55 N 5 where t s mportant n ths step to have the mass n SI unts: m = 0.40 kg. (c) Smlarly, wth m = 0.60 kg, we fnd T 5 mg 4.94 N. 6 ------------------------------------------