Rolling, Torque, and Angular Momentum

Transcription

1 C H A P T E R 1 1 Rolling, Toque, and Angula Momentum 11-1 RLLING AS TRANSLATIN AND RTATIN CMBINED Leaning bjectives Afte eading this module, ou should be able to Identif that smooth olling can be consideed as a combination of pue tanslation and pue otation Appl the elationship between the cente-of-mass speed and the angula speed of a bod in smooth olling. Ke Ideas Fo a wheel of adius R olling smoothl, v com vr, whee v com is the linea speed of the wheel s cente of mass and v is the angula speed of the wheel about its cente. The wheel ma also be viewed as otating instantaneousl about the point P of the oad that is in contact with the wheel. The angula speed of the wheel about this point is the same as the angula speed of the wheel about its cente. What Is Phsics? As we discussed in Chapte 10, phsics includes the stud of otation. Aguabl, the most impotant application of that phsics is in the olling motion of wheels and wheel-like objects. This applied phsics has long been used. Fo eample, when the pehistoic people of Easte Island moved thei gigantic stone statues fom the qua and acoss the island, the dagged them ove logs acting as olles. Much late, when settles moved westwad acoss Ameica in the 1800s, the olled thei possessions fist b wagon and then late b tain.toda, like it o not, the wold is filled with cas, tucks, motoccles, biccles, and othe olling vehicles. The phsics and engineeing of olling have been aound fo so long that ou might think no fesh ideas emain to be developed. Howeve, skateboads and inline skates wee invented and engineeed fail ecentl, to become huge financial successes. Steet luge is now catching on, and the self-ighting Segwa (Fig. 11-1) ma change the wa people move aound in lage cities. Appling the phsics of olling can still lead to supises and ewads. u stating point in eploing that phsics is to simplif olling motion. Rolling as Tanslation and Rotation Combined Hee we conside onl objects that oll smoothl along a suface; that is, the objects oll without slipping o bouncing on the suface. Figue 11-2 shows how complicated smooth olling motion can be Although the cente of the object moves in a staight line paallel to the suface, a point on the im cetainl does not. Howeve, we can stud this motion b teating it as a combination of tanslation of the cente of mass and otation of the est of the object aound that cente. Justin Sullivan/Gett Images, Inc. Figue 11-1 The self-ighting Segwa Human Tanspote. 295

2 296 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM Figue 11-2 A time-eposue photogaph of a olling disk. Small lights have been attached to the disk, one at its cente and one at its edge. The latte taces out a cuve called a ccloid. Richad Megna/Fundamental Photogaphs v com P s Figue 11-3 The cente of mass of a olling wheel moves a distance s at velocit v com while the wheel otates though angle u. The point P at which the wheel makes contact with the suface ove which the wheel olls also moves a distance s. s θ P v com To see how we do this, petend ou ae standing on a sidewalk watching the biccle wheel of Fig as it olls along a steet.as shown, ou see the cente of mass of the wheel move fowad at constant speed v com. The point P on the steet whee the wheel makes contact with the steet suface also moves fowad at speed v com, so that P alwas emains diectl below. Duing a time inteval t, ou see both and P move fowad b a distance s. The biccle ide sees the wheel otate though an angle u about the cente of the wheel, with the point of the wheel that was touching the steet at the beginning of t moving though ac length s. Equation elates the ac length s to the otation angle u s ur, (11-1) whee R is the adius of the wheel. The linea speed v com of the cente of the wheel (the cente of mass of this unifom wheel) is ds/. The angula speed v of the wheel about its cente is du/. Thus, diffeentiating Eq with espect to time (with R held constant) gives us v com vr (smooth olling motion). (11-2) A Combination. Figue 11-4 shows that the olling motion of a wheel is a combination of puel tanslational and puel otational motions. Figue 11-4a shows the puel otational motion (as if the otation ais though the cente wee stationa) Eve point on the wheel otates about the cente with angula speed v. (This is the tpe of motion we consideed in Chapte 10.) Eve point on the outside edge of the wheel has linea speed v com given b Eq Figue 11-4b shows the puel tanslational motion (as if the wheel did not otate at all) Eve point on the wheel moves to the ight with speed v com. The combination of Figs. 11-4a and 11-4b ields the actual olling motion of the wheel, Fig. 11-4c. Note that in this combination of motions, the potion of the wheel at the bottom (at point P) is stationa and the potion at the top (a) Pue otation v = v com T + (b) Pue tanslation = (c) Rolling motion v = v com T T v = 2v com ω v com ω v com P v = v com P v = v com P v = v com + v com = 0 Figue 11-4 Rolling motion of a wheel as a combination of puel otational motion and puel tanslational motion. (a) The puel otational motion All points on the wheel move with the same angula speed v. Points on the outside edge of the wheel all move with the same linea speed v v com.the linea velocities v of two such points, at top (T) and bottom (P) of the wheel, ae shown. (b) The puel tanslational motionall points on the wheel move to the ight with the same linea velocit v com. (c) The olling motion of the wheel is the combination of (a) and (b).

3 11-1 RLLING AS TRANSLATIN AND RTATIN CMBINED 297 Figue 11-5 A photogaph of a olling biccle wheel. The spokes nea the wheel s top ae moe blued than those nea the bottom because the top ones ae moving faste, as Fig. 11-4c shows. Coutes Alice Hallida (at point T) is moving at speed 2v com, faste than an othe potion of the wheel. These esults ae demonstated in Fig. 11-5, which is a time eposue of a olling biccle wheel. You can tell that the wheel is moving faste nea its top than nea its bottom because the spokes ae moe blued at the top than at the bottom. The motion of an ound bod olling smoothl ove a suface can be sepaated into puel otational and puel tanslational motions, as in Figs. 11-4a and 11-4b. Rolling as Pue Rotation Figue 11-6 suggests anothe wa to look at the olling motion of a wheel namel, as pue otation about an ais that alwas etends though the point whee the wheel contacts the steet as the wheel moves. We conside the olling motion to be pue otation about an ais passing though point P in Fig. 11-4c and pependicula to the plane of the figue. The vectos in Fig then epesent the instantaneous velocities of points on the olling wheel. T Question What angula speed about this new ais will a stationa obseve assign to a olling biccle wheel? Answe The same v that the ide assigns to the wheel as she o he obseves it in pue otation about an ais though its cente of mass. To veif this answe, let us use it to calculate the linea speed of the top of the olling wheel fom the point of view of a stationa obseve. If we call the wheel s adius R, the top is a distance 2R fom the ais though P in Fig. 11-6, so the linea speed at the top should be (using Eq. 11-2) v top (v)(2r) 2(vR) 2v com, in eact ageement with Fig. 11-4c. You can similal veif the linea speeds shown fo the potions of the wheel at points and P in Fig. 11-4c. Rotation ais at P Figue 11-6 Rolling can be viewed as pue otation, with angula speed v, about an ais that alwas etends though P.The vectos show the instantaneous linea velocities of selected points on the olling wheel.you can obtain the vectos b combining the tanslational and otational motions as in Fig Checkpoint 1 The ea wheel on a clown s biccle has twice the adius of the font wheel. (a) When the biccle is moving, is the linea speed at the ve top of the ea wheel geate than, less than, o the same as that of the ve top of the font wheel? (b) Is the angula speed of the ea wheel geate than, less than, o the same as that of the font wheel?

4 298 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM 11-2 FRCES AND KINETIC ENERGY F RLLING Leaning bjectives Afte eading this module, ou should be able to Calculate the kinetic eneg of a bod in smooth olling as the sum of the tanslational kinetic eneg of the cente of mass and the otational kinetic eneg aound the cente of mass Appl the elationship between the wok done on a smoothl olling object and the change in its kinetic eneg Fo smooth olling (and thus no sliding), conseve mechanical eneg to elate initial eneg values to the values at a late point. Ke Ideas A smoothl olling wheel has kinetic eneg K 1 2 I comv 2 1 2v 2 com, whee I com is the otational inetia of the wheel about its cente of mass and M is the mass of the wheel. If the wheel is being acceleated but is still olling smoothl, the acceleation of the cente of mass a is elated to the com Daw a fee-bod diagam of an acceleating bod that is smoothl olling on a hoiontal suface o up o down a amp Appl the elationship between the cente-of-mass acceleation and the angula acceleation Fo smooth olling of an object up o down a amp, appl the elationship between the object s acceleation, its otational inetia, and the angle of the amp. angula acceleation a about the cente with a com ar. If the wheel olls smoothl down a amp of angle u, its acceleation along an ais etending up the amp is a com, g sin u 1 I com /MR 2. The Kinetic Eneg of Rolling Let us now calculate the kinetic eneg of the olling wheel as measued b the stationa obseve. If we view the olling as pue otation about an ais though P in Fig. 11-6, then fom Eq we have K 1 2 I Pv 2, (11-3) in which v is the angula speed of the wheel and I P is the otational inetia of the wheel about the ais though P. Fom the paallel-ais theoem of Eq (I I com Mh 2 ), we have I P I com MR 2, (11-4) in which M is the mass of the wheel, I com is its otational inetia about an ais though its cente of mass, and R (the wheel s adius) is the pependicula distance h. Substituting Eq into Eq. 11-3, we obtain K 1 2 I comv 2 1 2R 2 v 2, and using the elation v com vr (Eq. 11-2) ields K 1 2 I comv 2 1 2v 2 com. (11-5) 1 2 I comv 2 We can intepet the tem as the kinetic eneg associated with the otation of the wheel about an ais though its cente of mass (Fig. 11-4a), and the 1 tem Mv 2 2 com as the kinetic eneg associated with the tanslational motion of the wheel s cente of mass (Fig. 11-4b). Thus, we have the following ule A olling object has two tpes of kinetic eneg a otational kinetic eneg ( 1 2 I comv 2 ) due to its otation about its cente of mass and a tanslational kinetic eneg ( 1 2 Mv 2 com) due to tanslation of its cente of mass.

5 11-2 FRCES AND KINETIC ENERGY F RLLING 299 The Foces of Rolling Fiction and Rolling If a wheel olls at constant speed, as in Fig. 11-3, it has no tendenc to slide at the point of contact P, and thus no fictional foce acts thee. Howeve, if a net foce acts on the olling wheel to speed it up o to slow it, then that net foce causes acceleation a com of the cente of mass along the diection of tavel. It also causes the wheel to otate faste o slowe, which means it causes an angula acceleation a.these acceleations tend to make the wheel slide at P.Thus, a fictional foce must act on the wheel at P to oppose that tendenc. If the wheel does not slide, the foce is a static fictional foce f and the motion is smooth olling. We can then elate the magnitudes of the linea acceleation a com and the angula acceleation a b diffeentiating Eq with espect to time (with R held constant). n the left side, dv com / is a com, and on the ight side dv/ is a. So, fo smooth olling we have a com ar (smooth olling motion). (11-6) If the wheel does slide when the net foce acts on it, the fictional foce that acts at P in Fig is a kinetic fictional foce fk. The motion then is not smooth olling, and Eq does not appl to the motion. In this chapte we discuss onl smooth olling motion. Figue 11-7 shows an eample in which a wheel is being made to otate faste while olling to the ight along a flat suface, as on a biccle at the stat of a ace. The faste otation tends to make the bottom of the wheel slide to the left at point P. A fictional foce at P, diected to the ight, opposes this tendenc to slide. If the wheel does not slide, that fictional foce is a static fictional foce fs (as shown), the motion is smooth olling, and Eq applies to the motion. (Without fiction, biccle aces would be stationa and ve boing.) If the wheel in Fig wee made to otate slowe, as on a slowing biccle, we would change the figue in two was The diections of the cente-ofmass acceleation a com and the fictional foce fs at point P would now be to the left. s P a com Figue 11-7 A wheel olls hoiontall without sliding while acceleating with linea acceleation a com, as on a biccle at the stat of a ace. A static fictional foce fs acts on the wheel at P, opposing its tendenc to slide. f s Rolling Down a Ramp Figue 11-8 shows a ound unifom bod of mass M and adius R olling smoothl down a amp at angle u, along an ais.we want to find an epession fo the bod s F N Foces F N and F g cosθ meel balance. Foces F g sin θ and f s detemine the linea acceleation down the amp. F g sin θ θ θ R P f s The toque due to f s detemines the angula acceleation aound the com. F g cos θ F g Figue 11-8 A ound unifom bod of adius R olls down a amp.the foces that act on it ae the gavitational foce F g, a nomal foce F N, and a fictional foce f s pointing up the amp. (Fo clait, vecto F N has been shifted in the diection it points until its tail is at the cente of the bod.)

6 300 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM acceleation a com, down the amp.we do this b using Newton s second law in both its linea vesion (F net Ma) and its angula vesion (t net Ia). We stat b dawing the foces on the bod as shown in Fig The gavitational foce F g on the bod is diected downwad. The tail of the vecto is placed at the cente of mass of the bod. The component along the amp is F g sin u, which is equal to Mg sin u. 2. A nomal foce F N is pependicula to the amp. It acts at the point of contact P, but in Fig the vecto has been shifted along its diection until its tail is at the bod s cente of mass. 3. A static fictional foce fs acts at the point of contact P and is diected up the amp. (Do ou see wh? If the bod wee to slide at P, it would slide down the amp.thus, the fictional foce opposing the sliding must be up the amp.) We can wite Newton s second law fo components along the ais in Fig (F net, ma ) as f s Mg sin u Ma com,. (11-7) This equation contains two unknowns, f s and a com,. (We should not assume that f s is at its maimum value f s,ma. All we know is that the value of f s is just ight fo the bod to oll smoothl down the amp, without sliding.) We now wish to appl Newton s second law in angula fom to the bod s otation about its cente of mass. Fist, we shall use Eq (t F ) to wite the toques on the bod about that point. The fictional foce fs has moment am R and thus poduces a toque Rf s, which is positive because it tends to otate the bod counteclockwise in Fig Foces F and F g N have eo moment ams about the cente of mass and thus poduce eo toques. So we can wite the angula fom of Newton s second law (t net Ia) about an ais though the bod s cente of mass as Rf s I com a. (11-8) This equation contains two unknowns, f s and a. Because the bod is olling smoothl, we can use Eq (a com ar) to elate the unknowns a com, and a. But we must be cautious because hee a com, is negative (in the negative diection of the ais) and a is positive (counteclockwise). Thus we substitute a com, /R fo a in Eq Then, solving fo f s, we obtain f s I com. R 2 Substituting the ight side of Eq fo f s in Eq. 11-7, we then find a com, a com, g sin u 1 I com /MR 2. (11-9) (11-10) We can use this equation to find the linea acceleation a com, of an bod olling along an incline of angle u with the hoiontal. Note that the pull b the gavitational foce causes the bod to come down the amp, but it is the fictional foce that causes the bod to otate and thus oll. If ou eliminate the fiction (b, sa, making the amp slick with ice o gease) o aange fo Mg sin u to eceed f s,ma, then ou eliminate the smooth olling and the bod slides down the amp. Checkpoint 2 Disks A and B ae identical and oll acoss a floo with equal speeds.then disk A olls up an incline, eaching a maimum height h, and disk B moves up an incline that is identical ecept that it is fictionless. Is the maimum height eached b disk B geate than, less than, o equal to h?

7 11-3 THE Y-Y 301 Sample Poblem Ball olling down a amp A unifom ball, of mass M 6.00 kg and adius R, olls smoothl fom est down a amp at angle u 30.0 (Fig. 11-8). (a) The ball descends a vetical height h 1.20 m to each the bottom of the amp.what is its speed at the bottom? KEY IDEAS The mechanical eneg E of the ball Eath sstem is conseved as the ball olls down the amp.the eason is that the onl foce doing wok on the ball is the gavitational foce, a consevative foce. The nomal foce on the ball fom the amp does eo wok because it is pependicula to the ball s path. The fictional foce on the ball fom the amp does not tansfe an eneg to themal eneg because the ball does not slide (it olls smoothl). Thus, we conseve mechanical eneg (E f E i ) K f U f K i U i, (11-11) whee subscipts f and i efe to the final values (at the bottom) and initial values (at est), espectivel. The gavitational potential eneg is initiall U i Mgh (whee M is the ball s mass) and finall U f 0. The kinetic eneg is initiall K i 0. Fo the final kinetic eneg K f, we need an additional idea Because the ball olls, the kinetic eneg involves both tanslation and otation, so we include them both b using the ight side of Eq Calculations Substituting into Eq gives us ( 1 I 2 comv v com ) 0 0 Mgh, (11-12) whee I com is the ball s otational inetia about an ais though its cente of mass, v com is the equested speed at the bottom, and v is the angula speed thee. Because the ball olls smoothl, we can use Eq to substitute v com /R fo v to educe the unknowns in Eq Doing so, substituting 5 MR2 fo I com (fom Table 10-2f), and then solving fo v com give us 4.10 m/s. (Answe) Note that the answe does not depend on M o R. (b) What ae the magnitude and diection of the fictional foce on the ball as it olls down the amp? KEY IDEA v com 2( 10 7 )gh 2(10 7 )(9.8 m/s2 )(1.20 m) Because the ball olls smoothl, Eq gives the fictional foce on the ball. Calculations Befoe we can use Eq. 11-9, we need the ball s acceleation a com, fom Eq a com, 2 g sin u 1 I com /MR g sin u MR2 /MR 2 (9.8 m/s2 ) sin Note that we needed neithe mass M no adius R to find a com,. Thus, an sie ball with an unifom mass would have this smoothl olling acceleation down a 30.0 amp. We can now solve Eq as a f s I com, com 2 a R 2 5 MR2 com, 2 R Ma 2 5 com, 2 5 (6.00 kg)(3.50 m/s2 ) 8.40 N m/s 2. (Answe) Note that we needed mass M but not adius R. Thus, the fictional foce on an 6.00 kg ball olling smoothl down a 30.0 amp would be 8.40 N egadless of the ball s adius but would be lage fo a lage mass. Additional eamples, video, and pactice available at WilePLUS 11-3 THE Y-Y Leaning bjectives Afte eading this module, ou should be able to Daw a fee-bod diagam of a o-o moving up o down its sting Identif that a o-o is effectivel an object that olls smoothl up o down a amp with an incline angle of 90. Ke Idea Fo a o-o moving up o down its sting, appl the elationship between the o-o s acceleation and its otational inetia Detemine the tension in a o-o s sting as the o-o moves up o down its sting. A o-o, which tavels veticall up o down a sting, can be teated as a wheel olling along an inclined plane at angle u 90.

8 302 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM R (a) R 0 R 0 F g (b) Figue 11-9 (a) A o-o, shown in coss section. The sting, of assumed negligible thickness, is wound aound an ale of adius R 0.(b) A fee-bod diagam fo the falling o-o. nl the ale is shown. T The Yo-Yo A o-o is a phsics lab that ou can fit in ou pocket. If a o-o olls down its sting fo a distance h, it loses potential eneg in amount mgh but gains kinetic eneg in both tanslational ( 1 Mv 2 and otational ( 1 I 2 com) 2 comv 2 ) foms. As it climbs back up, it loses kinetic eneg and egains potential eneg. In a moden o-o, the sting is not tied to the ale but is looped aound it. When the o-o hits the bottom of its sting, an upwad foce on the ale fom the sting stops the descent. The o-o then spins, ale inside loop, with onl otational kinetic eneg. The o-o keeps spinning ( sleeping ) until ou wake it b jeking on the sting, causing the sting to catch on the ale and the o-o to climb back up. The otational kinetic eneg of the o-o at the bottom of its sting (and thus the sleeping time) can be consideabl inceased b thowing the o-o downwad so that it stats down the sting with initial speeds v com and v instead of olling down fom est. To find an epession fo the linea acceleation a com of a o-o olling down a sting, we could use Newton s second law (in linea and angula foms) just as we did fo the bod olling down a amp in Fig The analsis is the same ecept fo the following 1. Instead of olling down a amp at angle u with the hoiontal, the o-o olls down a sting at angle u 90 with the hoiontal. 2. Instead of olling on its oute suface at adius R, the o-o olls on an ale of adius R 0 (Fig. 11-9a). 3. Instead of being slowed b fictional foce f s, the o-o is slowed b the foce on it fom the sting (Fig. 11-9b). T The analsis would again lead us to Eq Theefoe, let us just change the notation in Eq and set u 90 to wite the linea acceleation as g a com, (11-13) 1 I com /MR 2 0 whee I com is the o-o s otational inetia about its cente and M is its mass. A o-o has the same downwad acceleation when it is climbing back up TRQUE REVISITED Leaning bjectives Afte eading this module, ou should be able to Identif that toque is a vecto quantit Identif that the point about which a toque is calculated must alwas be specified Calculate the toque due to a foce on a paticle b taking the coss poduct of the paticle s position vecto and the foce vecto, in eithe unit-vecto notation o magnitude-angle notation Use the ight-hand ule fo coss poducts to find the diection of a toque vecto. Ke Ideas In thee dimensions, toque t is a vecto quantit defined The magnitude of t is given b elative to a fied point (usuall an oigin); it is t F sin f F F, t F, whee f is the angle between F and, F is the component whee F is a foce applied to a paticle and is a of F pependicula to, and is the moment am of F. position vecto locating the paticle elative to the fied The diection of t is given b the ight-hand ule fo coss point. poducts.

9 11-4 TRQUE REVISITED 303 Coss into F. Toque τ is in the positive diection. τ (= F) φ F (edawn, with tail at oigin) τ A (a) φ F φ F A φ F φ A Line of action of F (b) (c) Figue Defining toque. (a) A foce, ling in an plane, acts on a paticle at point A.(b) This foce poduces a toque t ( F ) on the paticle with espect to the oigin. B the ight-hand ule fo vecto (coss) poducts, the toque vecto points in the positive diection of. Its magnitude is given b F in (b) and b F in (c). F F Toque Revisited In Chapte 10 we defined toque t fo a igid bod that can otate aound a fied ais. We now epand the definition of toque to appl it to an individual paticle that moves along an path elative to a fied point (athe than a fied ais). The path need no longe be a cicle, and we must wite the toque as a vecto t that ma have an diection. We can calculate the magnitude of the toque with a fomula and detemine its diection with the ight-hand ule fo coss poducts. Figue 11-10a shows such a paticle at point A in an plane.a single foce F in that plane acts on the paticle, and the paticle s position elative to the oigin is given b position vecto. The toque t acting on the paticle elative to the fied point is a vecto quantit defined as t F (toque defined). (11-14) We can evaluate the vecto (o coss) poduct in this definition of t b using the ules in Module 3-3. To find the diection of t, we slide the vecto F (without changing its diection) until its tail is at the oigin, so that the two vectos in the vecto poduct ae tail to tail as in Fig b. We then use the ight-hand ule in Fig. 3-19a, sweeping the finges of the ight hand fom (the fist vecto in the poduct) into F (the second vecto). The outstetched ight thumb then gives the diection of t. In Fig b, it is in the positive diection of the ais. To detemine the magnitude of t, we appl the geneal esult of Eq (c ab sin f), finding t F sin f, (11-15) whee f is the smalle angle between the diections of and F when the vectos ae tail to tail. Fom Fig b, we see that Eq can be ewitten as t F, (11-16) whee F is the component of F pependicula to (F sin f). Fom Fig c, we see that Eq can also be ewitten as t F, (11-17) whee is the moment am of F ( sin f) (the pependicula distance between and the line of action of F ). Checkpoint 3 The position vecto of a paticle points along the positive diection of a ais. If the toque on the paticle is (a) eo, (b) in the negative diection of, and (c) in the negative diection of, in what diection is the foce causing the toque?

10 304 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM Sample Poblem Toque on a paticle due to a foce In Fig a, thee foces, each of magnitude 2.0 N, act on a paticle. The paticle is in the plane at point A given b position vecto, whee 3.0 m and u 30. What is the toque, about the oigin, due to each foce? KEY IDEA Because the thee foce vectos do not lie in a plane, we must use coss poducts, with magnitudes given b Eq (t F sin f) and diections given b the ight-hand ule. Calculations Because we want the toques with espect to the oigin, the vecto equied fo each coss poduct is the given position vecto. To detemine the angle f between and each foce, we shift the foce vectos of Fig a, each in tun, so that thei tails ae at the oigin. Figues 11-11b, c, and d, which ae diect views of the plane, show the shifted foce vectos F 1, F 2, and F 3, espectivel. (Note how much easie the angles between the foce vectos and the position vecto ae to see.) In Fig d, the angle between the diections of and F 3 is 90 and the smbol means F 3 is diected into the page. (Fo out of the page, we would use.) Now, appling Eq , we find t 1 F 1 sin f 1 (3.0 m)(2.0 N)(sin 150) 3.0 N m, t 2 F 2 sin f 2 (3.0 m)(2.0 N)(sin 120) 5.2 N m, and t 3 F 3 sin f 3 (3.0 m)(2.0 N)(sin 90) 6.0 N m. (Answe) Net, we use the ight-hand ule, placing the finges of the ight hand so as to otate into F though the smalle of the two angles between thei diections. The thumb points in the diection of the toque.thus t 1 is diected into the page in Fig b; t 2 is diected out of the page in Fig c; and t 3 is diected as shown in Fig d. All thee toque vectos ae shown in Fig e. A (a) F 3 A F 1 F 2 θ (b) θ = 30 φ 1 = 150 F 1 F 1 Coss into F 1. Toque τ 1 is into the figue (negative ). F 1 (c) θ = 30 φ 2 = 120 F 2 F 2 Coss into F 2. Toque τ 2 is out of the figue (positive ). F 2 τ 1 τ 2 (d) θ = 30 τ 3 θ F 3 τ 3 Coss into F 3. Toque τ 3 is in the plane. τ 3 (e) θ τ 3 The thee toques. Figue (a) A paticle at point A is acted on b thee foces, each paallel to a coodinate ais.the angle f (used in finding toque) is shown (b) fo and (c) fo. (d) Toque t 3 is pependicula to both and (foce is diected into the plane of the figue). (e) The toques. F 1 F 2 F 3 F 3 Additional eamples, video, and pactice available at WilePLUS

11 11-5 ANGULAR MMENTUM ANGULAR MMENTUM Leaning bjectives Afte eading this module, ou should be able to Identif that angula momentum is a vecto quantit Identif that the fied point about which an angula momentum is calculated must alwas be specified Calculate the angula momentum of a paticle b taking the coss poduct of the paticle s position vecto and its momentum vecto, in eithe unit-vecto notation o magnitude-angle notation Use the ight-hand ule fo coss poducts to find the diection of an angula momentum vecto. and and ae Ke Ideas The angula momentum of a paticle with linea momentum whee f is the angle between p, p v p, mass m, and linea velocit v is a vecto quantit the components of p and v pependicula to, and is defined elative to a fied point (usuall an oigin) as the pependicula distance between the fied point and p m( v ). the etension of p. The magnitude of is given b The diection of is given b the ight-hand ule Position ou ight hand so that the finges ae in the diection of. mv sin f Then otate them aound the palm to be in the diection of p. p mv You outstetched thumb gives the diection of. p mv, Angula Momentum Recall that the concept of linea momentum p and the pinciple of consevation of linea momentum ae etemel poweful tools. The allow us to pedict the outcome of, sa, a collision of two cas without knowing the details of the collision. Hee we begin a discussion of the angula countepat of p, winding up in Module 11-8 with the angula countepat of the consevation pinciple, which can lead to beautiful (almost magical) feats in ballet, fanc diving, ice skating, and man othe activities. Figue shows a paticle of mass m with linea momentum p ( mv ) as it passes though point A in an plane. The angula momentum of this paticle with espect to the oigin is a vecto quantit defined as (= p) p (edawn, with tail at oigin) φ (a) A p φ p φ p m( v ) (angula momentum defined), (11-18) whee is the position vecto of the paticle with espect to. As the paticle moves elative to in the diection of its momentum p ( mv ), position vecto otates aound. Note caefull that to have angula momentum about, the paticle does not itself have to otate aound. Compaison of Eqs and shows that angula momentum beas the same elation to linea momentum that toque does to foce. The SI unit of angula momentum is the kilogammete-squaed pe second (kg m 2 /s), equivalent to the joule-second (J s). Diection. To find the diection of the angula momentum vecto in Fig , we slide the vecto p until its tail is at the oigin. Then we use the ight-hand ule fo vecto poducts, sweeping the finges fom into p. The outstetched thumb then shows that the diection of is in the positive diection of the ais in Fig This positive diection is consistent with the counteclockwise otation of position vecto about the ais, as the paticle moves. (A negative diection of would be consistent with a clockwise otation of about the ais.) Magnitude. To find the magnitude of, we use the geneal esult of Eq to wite mv sin f, (11-19) whee f is the smalle angle between and p when these two vectos ae tail φ A p Etension of p (b) Figue Defining angula momentum.a paticle passing though point A has linea momentum p ( mv ), with the vecto p ling in an plane.the paticle has angula momentum ( p ) with espect to the oigin. B the ight-hand ule, the angula momentum vecto points in the positive diection of.(a) The magnitude of is given b p mv.(b) The magnitude of is also given b p mv.

12 306 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM to tail. Fom Fig a, we see that Eq can be ewitten as p mv, (11-20) whee p is the component of p pependicula to and v is the component of v pependicula to. Fom Fig b, we see that Eq can also be ewitten as p mv, (11-21) whee is the pependicula distance between and the etension of p. Impotant. Note two featues hee (1) angula momentum has meaning onl with espect to a specified oigin and (2) its diection is alwas pependicula to the plane fomed b the position and linea momentum vectos and p. Checkpoint In pat a of the figue, paticles 1 and 2 move aound point in cicles with adii 2 m and 4 m. In pat b, paticles 3 and 4 tavel along staight lines at pependicula distances of 4 m and 2 m fom point. Paticle 5 moves diectl awa fom.all five paticles have the same mass and the same constant speed. (a) Rank the paticles accoding to the magnitudes 2 of thei angula momentum about point, geatest fist. (b) Which paticles have negative angula momentum about point? (a) (b) 4 5 Sample Poblem Angula momentum of a two-paticle sstem Figue shows an ovehead view of two paticles moving at constant momentum along hoiontal paths. Paticle 1, with momentum magnitude p kgm/s, has position vecto 1 and will pass 2.0 m fom point. Paticle 2, with momentum magnitude p kgm/s, has position vecto 2 and will pass 4.0 m fom point. What ae the magnitude and diection of the net angula momentum L about point of the twopaticle sstem? KEY IDEA To find, we can fist find the individual angula momenta 1 and 2 and then add them. To evaluate thei magnitudes, we can use an one of Eqs though Howeve, Eq is easiest, because we ae given the pependicula distances 1 ( 2.0 m) and 2 ( 4.0 m) and the momentum magnitudes p 1 and p 2. L Calculations Fo paticle 1, Eq ields 1 1 p 1 (2.0 m)(5.0 kgm/s) 10 kgm 2 /s. To find the diection of vecto 1, we use Eq and the ight-hand ule fo vecto poducts. Fo 1 p 1, the vecto poduct is out of the page, pependicula to the plane of Fig This is the positive diection, consistent with the counteclockwise otation of the paticle s position vecto Figue Two paticles pass nea point. 1 aound as paticle 1 moves. Thus, the angula momentum vecto fo paticle 1 is 1 10 kgm 2 /s. Similal, the magnitude of p 2 (4.0 m)(2.0 kgm/s) 8.0 kgm 2 /s, and the vecto poduct 2 p 2 is into the page, which is the negative diection, consistent with the clockwise otation of 2 aound as paticle 2 moves. Thus, the angula momentum vecto fo paticle 2 is kgm 2 /s. The net angula momentum fo the two-paticle sstem is L kgm 2 /s (8.0 kgm 2 /s) 2.0 kgm 2 /s. (Answe) The plus sign means that the sstem s net angula momentum about point is out of the page. is p p 1 Additional eamples, video, and pactice available at WilePLUS

13 11-6 NEWTN S SECND LAW IN ANGULAR FRM NEWTN S SECND LAW IN ANGULAR FRM Leaning bjective Afte eading this module, ou should be able to Appl Newton s second law in angula fom to elate the toque acting on a paticle to the esulting ate of change of the paticle s angula momentum, all elative to a specified point. Ke Idea Newton s second law fo a paticle can be witten in angula fom as t net d, whee t is the net toque acting on the paticle and net is the angula momentum of the paticle. Newton s Second Law in Angula Fom Newton s second law witten in the fom F (single paticle) (11-22) net dp epesses the close elation between foce and linea momentum fo a single paticle. We have seen enough of the paallelism between linea and angula quantities to be pett sue that thee is also a close elation between toque and angula momentum. Guided b Eq , we can even guess that it must be t net d (single paticle). (11-23) Equation is indeed an angula fom of Newton s second law fo a single paticle The (vecto) sum of all the toques acting on a paticle is equal to the time ate of change of the angula momentum of that paticle. Equation has no meaning unless the toques t and the angula momentum ae defined with espect to the same point, usuall the oigin of the coodinate sstem being used. Poof of Equation We stat with Eq , the definition of the angula momentum of a paticle whee is the position vecto of the paticle and v is the velocit of the paticle. Diffeentiating* each side with espect to time t ields d m( v ), m dv d v. (11-24) Howeve, dv / is the acceleation a of the paticle, and d / is its velocit v. Thus, we can ewite Eq as d m( a v v ). *In diffeentiating a vecto poduct, be sue not to change the ode of the two quantities (hee and v) that fom that poduct. (See Eq )

14 308 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM Now v v 0 (the vecto poduct of an vecto with itself is eo because the angle between the two vectos is necessail eo). Thus, the last tem of this epession is eliminated and we then have d m( a ) ma. We now use Newton s second law (F net ma ) to eplace ma with its equal, the vecto sum of the foces that act on the paticle, obtaining d F net (11-25) ( F ). Hee the smbol indicates that we must sum the vecto poducts F fo all the foces. Howeve, fom Eq , we know that each one of those vecto poducts is the toque associated with one of the foces.theefoe, Eq tells us that t net d. This is Eq , the elation that we set out to pove. Checkpoint 5 The figue shows the position vecto of a paticle F 3 F at a cetain instant, and fou choices fo the diection 2 of a foce that is to acceleate the paticle.all fou choices lie in the plane. (a) Rank the choices accoding to the magnitude of the time ate of change (d /) the poduce in the angula momentum of the paticle about point, geatest F 1 F 4 fist. (b) Which choice esults in a negative ate of change about? Sample Poblem Toque and the time deivative of angula momentum Figue 11-14a shows a feee-fame of a kg paticle moving along a staight line with a position vecto given b (2.00t 2 t)î 5.00ĵ, with in metes and t in seconds, stating at t 0. The position vecto points fom the oigin to the paticle. In unit-vecto notation, find epessions fo the angula momentum of the paticle and the toque t acting on the paticle, both with espect to (o about) the oigin. Justif thei algebaic signs in tems of the paticle s motion. KEY IDEAS (1) The point about which an angula momentum of a paticle is to be calculated must alwas be specified. Hee it is the oigin. (2) The angula momentum of a paticle is given b Eq ( p m( v )). (3) The sign associated with a paticle s angula momentum is set b the sense of otation of the paticle s position vecto (aound the otation ais) as the paticle moves clockwise is negative and counteclockwise is positive. (4) If the toque acting on a paticle and the angula momentum of the paticle ae calculated aound the same point, then the toque is elated to angula momentum b Eq ( t d /). Calculations In ode to use Eq to find the angula momentum about the oigin, we fist must find an epession fo the paticle s velocit b taking a time deivative of its position vecto. Following Eq ( v d /), we wite v d ((2.00t2 t)î 5.00ĵ) (4.00t 1.00)î, with v in metes pe second. Net, let s take the coss poduct of and v using the template fo coss poducts displaed in Eq a b (a b b a )î (a b b a )ĵ (a b b a )kˆ. Hee the geneic a is and the geneic is v. Howeve, because we eall don t want to do moe wok than needed, let s fist just think about ou substitutions into b

15 11-6 NEWTN S SECND LAW IN ANGULAR FRM 309 v (m) 5 (m) t 2 s t 1 s t (a) (m) 10 3 (b) (m) v Both angula momentum and toque point out of figue, in the positive diection. (c) (d) (e) t Figue (a) A paticle moving in a staight line, shown at time t 0. (b) The position vecto at t 0, 1.00 s, and 2.00 s. (c) The fist step in appling the ight-hand ule fo coss poducts. (d) The second step. (e) The angula momentum vecto and the toque vecto ae along the ais, which etends out of the plane of the figue. the geneic coss poduct. Because lacks an component and because v lacks an o component, the onl noneo tem in the geneic coss poduct is the ve last one (b a )kˆ. So, let s cut to the (mathematical) chase b witing v (4.00t 1.00)(5.00)kˆ (20.0t 5.00)kˆ m 2 /s. Note that, as alwas, the coss poduct poduces a vecto that is pependicula to the oiginal vectos. To finish up Eq , we multipl b the mass, finding (0.500 kg)[(20.0t 5.00)kˆ m 2 /s] (10.0t 2.50)kˆ kg m 2 /s. (Answe) The toque about the oigin then immediatel follows fom Eq t d (10.0t 2.50)kˆ kgm 2 /s 10.0kˆ kgm 2 /s kˆ Nm, (Answe) which is in the positive diection of the ais. u esult fo tells us that the angula momentum is in the positive diection of the ais. To make sense of that positive esult in tems of the otation of the position vecto, let s evaluate that vecto fo seveal times t 0, t 1.00 s, t 2.00 s, ĵ m 3.00î 5.00ĵ m 10.0î 5.00ĵ m B dawing these esults as in Fig b, we see that otates counteclockwise in ode to keep up with the paticle. That is the positive diection of otation. Thus, even though the paticle is moving in a staight line, it is still moving counteclockwise aound the oigin and thus has a positive angula momentum. We can also make sense of the diection of b appling the ight-hand ule fo coss poducts (hee v, o, if ou like, m v, which gives the same diection). Fo an moment duing the paticle s motion, the finges of the ight hand ae fist etended in the diection of the fist vecto in the coss poduct ( ) as indicated in Fig c. The oientation of the hand (on the page o viewing sceen) is then adjusted so that the finges can be comfotabl otated about the palm to be in the diection of the second vecto in the coss poduct ( v ) as indicated in Fig d. The outstetched thumb then points in the diection of the esult of the coss poduct. As indicated in Fig e, the vecto is in the positive diection of the ais (which is diectl out of the plane of the figue), consistent with ou pevious esult. Figue 11-14e also indicates the diection of t, which is also in the positive diection of the ais because the angula momentum is in that diection and is inceasing in magnitude. Additional eamples, video, and pactice available at WilePLUS

16 310 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM 11-7 ANGULAR MMENTUM F A RIGID BDY Leaning bjectives Afte eading this module, ou should be able to Fo a sstem of paticles, appl Newton s second law in angula fom to elate the net toque acting on the sstem to the ate of the esulting change in the sstem s angula momentum Appl the elationship between the angula momentum of a igid bod otating aound a fied ais and the bod s otational inetia and angula speed aound that ais If two igid bodies otate about the same ais, calculate thei total angula momentum. Ke Ideas The angula momentum L of a sstem of paticles is the vecto sum of the angula momenta of the individual paticles L 1 2 n n The time ate of change of this angula momentum is equal to the net etenal toque on the sstem (the vecto sum of i1 i. the toques due to inteactions of the paticles of the sstem with paticles etenal to the sstem) t net dl (sstem of paticles). Fo a igid bod otating about a fied ais, the component of its angula momentum paallel to the otation ais is L Iv (igid bod, fied ais). The Angula Momentum of a Sstem of Paticles Now we tun ou attention to the angula momentum of a sstem of paticles with espect to an oigin. The total angula momentum L of the sstem is the (vecto) sum of the angula momenta of the individual paticles (hee with label i) L n n i1 i. (11-26) With time, the angula momenta of individual paticles ma change because of inteactions between the paticles o with the outside. We can find the esulting change in b taking the time deivative of Eq Thus, L (11-27) Fom Eq , we see that d i/ is equal to the net toque net,i on the ith paticle. We can ewite Eq as dl (11-28) n net,i. i1 That is, the ate of change of the sstem s angula momentum L is equal to the vecto sum of the toques on its individual paticles. Those toques include intenal toques (due to foces between the paticles) and etenal toques (due to foces on the paticles fom bodies etenal to the sstem). Howeve, the foces between the paticles alwas come in thid-law foce pais so thei toques sum to eo. Thus, the onl toques that can change the total angula momentum L of the sstem ae the etenal toques acting on the sstem. Net Etenal Toque. Let net epesent the net etenal toque, the vecto sum of all etenal toques on all paticles in the sstem. Then we can wite Eq as dl n i1 d i. net dl (sstem of paticles), (11-29)

17 11-7 ANGULAR MMENTUM F A RIGID BDY 311 which is Newton s second law in angula fom. It sas The net etenal toque net acting on a sstem of paticles is equal to the time ate of change of the sstem s total angula momentum L. Equation is analogous to F net dp / (Eq. 9-27) but equies eta caution Toques and the sstem s angula momentum must be measued elative to the same oigin. If the cente of mass of the sstem is not acceleating elative to an inetial fame, that oigin can be an point. Howeve, if it is acceleating, then it must be the oigin. Fo eample, conside a wheel as the sstem of paticles. If it is otating about an ais that is fied elative to the gound, then the oigin fo appling Eq can be an point that is stationa elative to the gound. Howeve, if it is otating about an ais that is acceleating (such as when it olls down a amp), then the oigin can be onl at its cente of mass. The Angula Momentum of a Rigid Bod Rotating About a Fied Ais We net evaluate the angula momentum of a sstem of paticles that fom a igid bod that otates about a fied ais. Figue 11-15a shows such a bod.the fied ais of otation is a ais, and the bod otates about it with constant angula speed v. We wish to find the angula momentum of the bod about that ais. We can find the angula momentum b summing the components of the angula momenta of the mass elements in the bod. In Fig a, a tpical mass element, of mass m i, moves aound the ais in a cicula path.the position of the mass element is located elative to the oigin b position vecto i. The adius of the mass element s cicula path is i, the pependicula distance between the element and the ais. The magnitude of the angula momentum espect to, is given b Eq of this mass element, with whee p i and v i ae the linea momentum and linea speed of the mass element, and 90 is the angle between and p i i. The angula momentum vecto i fo the mass element in Fig a is shown in Fig b; its diection must be pependicula to those of and p i i. The Components. We ae inteested in the component of i that is paallel to the otation ais, hee the ais.that component is i i sin u ( i sin u)(m i v i ) i m i v i. The component of the angula momentum fo the otating igid bod as a whole is found b adding up the contibutions of all the mass elements that make up the bod. Thus, because v v, we ma wite L n i ( i )( p i )(sin 90) ( i )(m i v i ), i1 v n i n i1 i1 2 m i i. m i v i i n (11-30) We can emove v fom the summation hee because it has the same value fo all points of the otating igid bod. 2 The quantit m i i in Eq is the otational inetia I of the bod about the fied ais (see Eq ).Thus Eq educes to i1 i m i (v i ) i Δ m i i Figue (a) A igid bod otates about a ais with angula speed v.a mass element of mass m i within the bod moves about the ais in a cicle with adius i. The mass element has linea momentum p i, and it is located elative to the oigin b position vecto i. Hee the mass element is shown when i is paallel to the ais. (b) The angula momentum i, with espect to, of the mass element in (a).the component is also shown. i i i θ (a) θ (b) p i θ i L Iv (igid bod, fied ais). (11-31)

18 312 CHAPTER 11 RLLING, TRQUE, AND ANGULAR MMENTUM Table 11-1 Moe Coesponding Vaiables and Relations fo Tanslational and Rotational Motion a Tanslational Rotational F Foce Toque ( F ) Linea momentum p Angula momentum ( p ) Linea momentum b P ( p i) Angula momentum b L ( i) Linea momentum b P Mv com Angula momentum c L Iv Newton s second law b F Newton s second law b net dl net dp Consevation law d P a constant Consevation law d L a constant a See also Table b Fo sstems of paticles, including igid bodies. c Fo a igid bod about a fied ais, with L being the component along that ais. d Fo a closed, isolated sstem. We have dopped the subscipt, but ou must emembe that the angula momentum defined b Eq is the angula momentum about the otation ais.also, I in that equation is the otational inetia about that same ais. Table 11-1, which supplements Table 10-3, etends ou list of coesponding linea and angula elations. Checkpoint 6 In the figue, a disk, a hoop, and a solid sphee ae made to spin about fied cental aes (like a Disk Hoop Sphee top) b means of stings F F F wapped aound them, with the stings poducing the same constant tangential foce F on all thee objects.the thee objects have the same mass and adius, and the ae initiall stationa. Rank the objects accoding to (a) thei angula momentum about thei cental aes and (b) thei angula speed, geatest fist, when the stings have been pulled fo a cetain time t CNSERVATIN F ANGULAR MMENTUM Leaning bjective Afte eading this module, ou should be able to When no etenal net toque acts on a sstem along a specified ais, appl the consevation of angula momentum to elate the initial angula momentum value along that ais to the value at a late instant. Ke Idea The angula momentum L o L i L f This is the law of consevation of angula momentum. of a sstem emains constant if the net etenal toque acting on the sstem is eo L a constant (isolated sstem) (isolated sstem). Consevation of Angula Momentum So fa we have discussed two poweful consevation laws, the consevation of eneg and the consevation of linea momentum. Now we meet a thid law of this tpe, involving the consevation of angula momentum. We stat fom

19 11-8 CNSERVATIN F ANGULAR MMENTUM 313 Eq (t net dl /), which is Newton s second law in angula fom. If no net etenal toque acts on the sstem, this equation becomes dl / 0, o L a constant (isolated sstem). (11-32) This esult, called the law of consevation of angula momentum, can also be witten as net angula momentum at some initial time t i o L i L f (isolated sstem). (11-33) Equations and tell us net angula momentum at some late time t f, If the net etenal toque acting on a sstem is eo, the angula momentum of the sstem emains constant, no matte what changes take place within the sstem. L Equations and ae vecto equations; as such, the ae equivalent to thee component equations coesponding to the consevation of angula momentum in thee mutuall pependicula diections. Depending on the toques acting on a sstem, the angula momentum of the sstem might be conseved in onl one o two diections but not in all diections L ω i If the component of the net etenal toque on a sstem along a cetain ais is eo, then the component of the angula momentum of the sstem along that ais cannot change, no matte what changes take place within the sstem. I i This is a poweful statement In this situation we ae concened with onl the initial and final states of the sstem; we do not need to conside an intemediate state. We can appl this law to the isolated bod in Fig , which otates aound the ais. Suppose that the initiall igid bod somehow edistibutes its mass elative to that otation ais, changing its otational inetia about that ais. Equations and state that the angula momentum of the bod cannot change. Substituting Eq (fo the angula momentum along the otational ais) into Eq , we wite this consevation law as I i v i I f v f. (11-34) Hee the subscipts efe to the values of the otational inetia I and angula speed v befoe and afte the edistibution of mass. Like the othe two consevation laws that we have discussed, Eqs and hold beond the limitations of Newtonian mechanics. The hold fo paticles whose speeds appoach that of light (whee the theo of special elativit eigns), and the emain tue in the wold of subatomic paticles (whee quantum phsics eigns). No eceptions to the law of consevation of angula momentum have eve been found. We now discuss fou eamples involving this law. 1. The spinning voluntee Figue shows a student seated on a stool that can otate feel about a vetical ais. The student, who has been set into otation at a modest initial angula speed v i, holds two dumbbells in his outstetched hands. His angula momentum vecto L lies along the vetical otation ais, pointing upwad. The instucto now asks the student to pull in his ams; this action educes his otational inetia fom its initial value I i to a smalle value I f because he moves mass close to the otation ais. His ate of otation inceases makedl, I f L Rotation ais (a) (b) Figue (a) The student has a elativel lage otational inetia about the otation ais and a elativel small angula speed. (b) B deceasing his otational inetia, the student automaticall inceases his angula speed. The angula momentum L of the otating sstem emains unchanged. ω f