5 The Physics of Rotating Bodies

Transcription

1 Prnceton Unversty 1996 Ph101 Laboratory The Physcs of Rotatng Bodes Introducton Newton s law F = m a descrbes the moton of the center of mass of an object. But n general the moton of a rgd body conssts of a rotaton about the center of mass as well as movement of the c.m. In ths Lab you wll explore the physcs of rotaton n four stuatons: 1. Measurement of the moment of nerta of a massve pulley. 2. Measurement of energy balance n a block-and-tackle system. 3. Verfcaton of the vector character of the laws of angular momentum. 4. Hands-on experence wth physcs toys n whch you are part of the rotatng system. Because ths lab offers the opportunty of rapd verfcaton of the mpressve phenomena of rotatonal moton, we wll be somewhat more ambtous than the course text n parts 3 and 4 of the Lab. An Appendx gves vector dervatons of the laws of rotatng objects for those who enjoy such detals. For parts 1 and 2 of the Lab we need only those smplfcatons of the theory that hold for rotary moton about a symmetry axs (axle) wth fxed drecton. 5.1 Moment of Inerta of a Massve Pulley In ths part you should verfy that the moment of nerta of a unform dsk s I calc = 1 2 MR2 = π 2 ρr4 H (1) where R s the radus, H s the thckness and ρ s the densty. The dsk to be studed s a pulley that actually conssts of an alumnum dsk and a small alumnum cylnder bolted together so as to rotate about ther common axs. The total moment of nerta of the pulley s the sum of the moments of ts two components Calculate the moment of nerta of the pulley by measurng the rad and thcknesses of the dsk and cylnder, and note that densty ρ =2.70 g/cm 3 for alumnum. Neglgble error wll occur n the calculaton f you gnore the varous holes and grooves n the dsks. An error analyss of eq. (1) tells us that σ I,calc I { max 4 σ } R R, σ H. (2) H Hence s t especally mportant to make a good measurement of the radus of the larger dsk. The goal of ths part s 1% accuracy. To measure the moment of nerta of the pulley you wll use a varaton of Atwood s machne, sketched n Fg. 1. A mass m s suspended from a strng that wraps around the smaller dsk of the pulley at radus r. The tenson T n the strng exerts a torque of magntude τ = Tr (3)

2 Prnceton Unversty 1996 Ph101 Laboratory 5 2 about the axle of the pulley. Ths torque leads to an angular acceleraton α of the pulley related by τ = Iα (4) From a measurement of the angular acceleraton you could then deduce the moment of nerta as I = Tr α, (5) f you knew the tenson T. Fgure 1: Sketch of the apparatus to measure the moment of nerta of a pulley. The tenson T cannot be measured drectly, but can be nferred by ts effect on the moton of the fallng mass m: mg T = ma, or T = m(g a), (6) where a s the vertcal acceleraton of mass m. To complete the analyss we note that snce the strng wraps around the pulley at radus r, a = αr, (7)

3 Prnceton Unversty 1996 Ph101 Laboratory 5 3 and hence eq. (5) becomes I meas = mgr α mr2. (8) To measure the angular acceleraton use the computer tmer as sketched n Fg. 1. Poston a LED-photodode par (photogate) so that the lght beam ntercepts the holes near the outer edge of the pulley. To check ths, use the Photogate Status Check of program pt n drectory c:\tmer. When you spn the pulley the computer should ndcate Photogate # 1 Unblocked brefly whenever one of the holes n the pulley passes by the photogate. The strng should be attached to a small screw n the hub of the pulley to prevent the hangng weght from httng the floor. For precson results, wrap the strng around the smaller hub of the pulley n a sngle layer. The radus r used n your analyss must nclude the radus of the strng! Do not wrap the strng n the groove of the pulley. To take data use opton Moton Tmer of the Man Menu. The computer starts takng data when the photogate frst becomes unblocked, and stops only when you press Enter/Return. Therefore you must ht the Return key before the weght completely unwnds the strng from the pulley. Use opton Dsplay Table of Data to check the qualty of your data. The tmes should be steadly decreasng. If not, take a new data set. When satsfed, Prnt Table of Data and also use Fle Optons and Save Data to a Fle. To analyze your data, Return to Data Analyss Optons Menu, select Graph Data, and Velocty vs. Tme. We wsh to analyze the angular velocty of the pulley. The computer has recorded the tme at the end of each half turn of the pulley, correspondng to π radans. Use Enter length (n meters) Then the graph wll be labelled Velocty (m/s) but t wll have the meanng of angular velocty n rad/s. Among the Select Graph Style optons, actvate Pont Protectors, Regresson Lne and Statstcs. Accept the automatc scalng optons and dsplay your graph. If satsfactory, Prnt Graph. The ftted slope, M, of the graph s just the angular acceleraton α n rad/s 2 (provded your data s vald: the fallng hanger ddn t ht anythng; the velocty graph fts well to a straght lne...). Measure mass m and radus r (ncludng the thckness of the strng) and then calculate the moment of nerta I meas usng eq. (8). If your values of I calc and I meas dffer by more than 1% you should try to resolve the dscrepancy. If the dfference s large, look for a gross error n value of R or H used n I calc, etc. If the dfference s only a few percent, you should also consder the sources of error n I meas, for whch an error analyss of eq. (8) tells us σ I,meas I { } σm max m, σ r r, σ α. (9) α

4 Prnceton Unversty 1996 Ph101 Laboratory Statcs and Dynamcs of a Block and Tackle [Analyze the data from part 1 before proceedng.] In ths part you wll examne the work-energy relaton for a system that ncludes rotatng objects. Construct a block-and-tackle system as sketched n Fg. 2. The pulley you used n part 1 wll be called the fxed pulley n that ts axle s fxed. A strng s attached to the axle of the fxed pulley, passes around the movng pulley, back up over the fxed pulley and down to the hangng mass. The strng should be long enough that when the hangng mass (m 1 ) s near the fxed pulley the the movng pulley s close the the floor. The strng should loop (twce f necessary to avod slppng) around the groove n the hub of the fxed pulley (and, of course, n the groove of the movng pulley). Fgure 2: The block-and-tackle arrangement for part 2. You may wsh to wnd the strng used n part 1 around the hub of the fxed pulley and secure t out of the way of the groove wth a pece of tape. Do not remove ths strng! Wegh the movng pulley frst. It should nclude a 100-g weght as a load. Adjust the hangng weght untl the system s n statc equlbrum. Keep addng weghts untl the hangng mass would fall f only 1-2 g more were added. What s the relaton between the hangng weght, the weght of the movng pulley and the tenson n the strng? Next, add about 200 g to the hangng mass so that t falls down, rasng the movng pulley n the process.

5 Prnceton Unversty 1996 Ph101 Laboratory 5 5 Use the computer to record the tmes at the end of successve half turns of the moton. Stop the data collecton when the hangng mass nears the movng pulley f the hangng mass hts the movng pulley durng ts fall the data thereafter wll be bad. Plot (angular) velocty vs. tme to see f the data fall along a straght lne. Repeat the data collecton f necessary. In the data analyss you wll compare the fnal knetc energy of the system to the change n potental energy. For ths you need to know the fnal angular velocty, ω f, and the dstance h 1 through whch the hangng weght fell. The angular acceleraton α s agan the slope M of your graph of angular velocty vs. tme whch has been ft to the form ω = B + Mt. We nfer that the weght started fallng at tme t 0 = B/M. Lett f be the tme of the last velocty pont on your graph. Then ω f = αδt = MΔt, and θ f = 1 2 αδt2 = 1 2 MΔt2, (10) where Δt = t f t 0 = t f + B M, (11) and the hangng weght fell by h 1 = rθ f, (12) where r s the radus on the fxed pulley around whch the strng was wrapped. Agan, remember to nclude the radus of the strng n the value of r. For better accuracy, use eq. (10) to calculate ω f rather than readng t off the graph. Do you agree that whle the hangng weght fell by h 1 the movng pulley rose by dstance h 2 = h 1 /2? Hence the change n potental energy was ( ΔPE = m 1 m ) 2 gh 1, (13) 2 where m 1 s the hangng mass and m 2 s the mass of the movng pulley. The fnal knetc energy ncludes that due to the center-of-mass moton of the hangng mass and of the movng pulley, as well as that due to the rotaton of the two pulleys (see Appendx 5.5.5). Do you agree that v 1 =2v 2 = ω f r,andω 2 = v 2 /r 2,whereω 2 s the angular velocty of the movng pulley and r 2 s the radus on the movng pulley at whch the strng s wrapped? The fnal knetc energy s then KE f = 1 2 m 1v m 2v Iω2 f I 2ω 2 2, (14) where I s the moment of nerta of the fxed pulley and I 2 s the moment of nerta of the wheel of the movng pulley. However, the moment of nerta I 2 of the movng pulley s so small t can be gnored. Calculate KE f 1 (( m 1 + m ) ) 2 r 2 + I ωf 2 2 4, (15) usng the value for I that you found n part 1. Compare your results for ΔPE and KE f.you may omt an error analyss n part 2, but f energy s not conserved to wthn 5-10% look for a gross error somewhere.

6 Prnceton Unversty 1996 Ph101 Laboratory Precesson of a Gyroscope [Analyze parts 1 and 2 before proceedng to parts 3 and 4.] A gyroscope s a rapdly spnnng object whose axle has freedom to change drecton. You wll use a dsk very smlar to the pulley of part 1 as a gyroscope, spnnng t up to hgh angular velocty wth one of the electrc spnners located at specal tables n the lab. Most people fnd t dsconcertng to hold a gyroscope, but ths s exactly what we encourage you to do. We hope t s left to the magnaton as to what happens f you drop a gyroscope. If you feel bold, try (gently) twstng the axle of a spnnng gyroscope before proceedng. Gyroscopes are governed by the general law of rotaton τ = d L dt, (16) whch s derved n Appendx We must take careful note of drectons as well as magntudes n understandng ths equaton. In words, an appled torque vector causes a change n the angular momentum vector of the system n the drecton of the appled torque. In addton, the magntude of the rate of change of angular momentum s just the magntude of the torque. Frst, where s the angular momentum vector of a gyroscope? If the angular velocty ω of the spn of gyroscope about ts axle s large we can gnore other components of the angular velocty that arse when the drecton of the axle s changng. Then the angular momentum vector ponts along the axle of the gyroscope and we can wrte L = I ω, (17) followng eq. (39). Use the rght-hand rule (Appendx 5.5.1) to determne whch way L and ω pont along the axle: f the fngers of your rght hand curl n the sense of rotaton of the gyroscope, then the thumb of your rght hand ponts along L and ω. Now suppose the forces on the gyroscope are suffcent to hold the support pont on the axle fxed, but a torque s appled. Equaton (16) tells us that that f the drectons of the torque vector τ and of the angular momentum vector L are not the same the drecton of L wll change. Then, snce L has the drecton of the axle, the latter drecton must change also. As the drecton of the torque s at rght angles to the drecton of the force that causes t, the drecton toward whch the axle moves s surprsng (at frst). We now gve a more quanttatve argument. Suppose the axle (and hence the angular momentum vector L) of the gyroscope s horzontal, and a vertcal force F s appled to the axle at dstance d from the pont of support of the gyroscope. Then the resultng torque has magntude τ = Fdand drecton perpendcular to the axle but n the horzontal plane. Ths torque wll precess the angular momentum vector n the horzontal plane, always pullng t a rght angles to ts current drecton. The angular momentum vector (and hence the axle) wll slowly move n horzontal crcle at a unform rate. We wll descrbe the rate of change of drecton of the axle as angular velocty Ω. Just as the magntude of the velocty of a partcle at radus r n unform crcular moton wth angular velocty Ω s dr/dt =Ωr, the rate of change of the rotatng angular momentum s dl =ΩL. (18) dt

7 Prnceton Unversty 1996 Ph101 Laboratory 5 7 Combnng the equatons, τ = Fd= dl =ΩL =ΩIω, (19) dt and hence we predct Ω= Fd (20) Iω for the precessonal angular velocty of the gyroscope. Confrm the concept of gyroscopc precesson n the followng experment. Frst, suspend a non-spnnng gyroscope by ts strng and handle from one of the small cranes attached to the lab tables. Add a hanger and weghts to the loop of strng at free end of the axle untl the axle s balanced horzontally. Determne ths balance to wthn about 10 g, and screw down the retanng nut on the hanger to hold the weghts fxed. Measure the dstance d along the axle between the two support strngs. Remove the hanger from the gyroscope, and the gyroscope from the crane. Put a sngle pece of maskng tape on the brass dsk of the gyroscope (f none s already present) to assst n later measurement of the spn rate. Spn up the gyroscope and restore t to the crane and add the hangng weght. At ths tme the gyroscope should have no net forces and no net torques on t, and so should just spn wth ts axle n a fxed drecton. Use the rght-hand rule to predct the drecton of the angular momentum vector. For ths, you must have pad attenton to whch way the gyroscope s spnnng. Now add another 500 g to the hanger, whch wll cause a torque about the pont where the strng from the crane attaches to the gyroscope. You may need to secure the extra weght wth a pece of tape. Can you correctly predct the drecton of the torque, and hence of the precesson of the axle? Measure the perod T of precesson of the gyroscope wth a stopwatch (or use the Keyboard Tmng Modes of program pt). Measure the rotatonal frequency f of the spnnng gyroscope wth a stroboscope. The electrcal spnners operate at about 3600 RPM, whch s the maxmum frequency of your gyroscope. If you work quckly you can complete the measurement of the perod of precesson before the gyroscope spn has dropped below 1800 RPM. It may be easer to measure the spn frequency wth the stroboscope just before and after measurng the precesson perod T, and average the two values of f. Practce takng measurements wth the stroboscope frst. When the stroboscope flashes at frequency f the pece of tape on the spnnng dsk appears frozen. But as a check that you have found the rght value of f, adjust the frequency to 2f. You should now see two mages of the tape. If you see only one, or cannot freeze the mage at all near 2f, thevalueoff you started wth was too low. The perod s related to the precesson rate by Ω = 2π/T, the spn angular velocty s related to the spn frequency f n revolutons per second by ω =2πf, and the vertcal force F =Δmg where Δm s the extra weght you added to the balanced gyroscope. Hence eq. (20) becomes T = 4π2 If Δmgd. (21) The moment of nerta of the brass gyroscope s about kg-m 2. Compare the predcton (21) for the perod T wth your measured value.

8 Prnceton Unversty 1996 Ph101 Laboratory 5 8 Thus far, the axle of the gyroscope should have remaned horzontal (unless t has lost a lot of ts spn). If you want to change the angle of the axle to the vertcal, you must gve dl/dt a vertcal component. Ths requres a vertcal torque, whch n turn requres a horzontal force. Push (but don t grab) the axle horzontally to see f the gyroscope responds as predcted. Feel free to try these experments whle holdng the axle of the gyroscope n your hand. Part offers another chance to explore closely related phenomena. You can also operate the gyroscope wthout the hanger. If you release the axle when t s horzontal the gyroscope agan precesses wth ts axle always n a horzontal plane. More generally, the vertcal angle of the axle wll reman constant whle the precessng axle sweeps out a cone. Ths volates the common sense expectaton that the vertcal angle of the axle should change due to the pull of gravty. We hope you now apprecate how the gyroscope ndeed responds to the torque nduced by gravty, but accordng to the vector laws of rotatng objects. 5.4 Angular-Momentum Toys [You don t need to report on ths part n your lab wrteup.] Rotatng Stool and Hand Weghts A specal stool ( Slver ) s provded that can rotate about a vertcal axs. St on the stool and hold the hand weghts at arms length. Have your partner set you spnnng. Pull the hand weghts n, and your angular velocty should ncrease. That s, the angular momentum, Iω, about the vertcal axs s conserved, and snce I MR 2 f you decrease R you decrease I and so ω must ncrease Rotatng Stool and Spnnng Bcycle Wheel Use the small hand spnner to spn up the bcycle wheel. The wheel has lead weghts around the rm to ncrease ts moment of nerta. St on the rotatng stool holdng the axs of the bcycle wheel horzontal ntally. Then twst the axle of the wheel towards the vertcal. Can you understand the subsequent moton on the bass of conservaton of angular momentum? Water Flow n a Tube Pour water nto the funnel attached to a copper tube that bends n a loop wth ts outlet drectly below the funnel. What s the resultng moton of the system? Is angular momentum conserved n ths devce?

9 Prnceton Unversty 1996 Ph101 Laboratory Appendx: Vector Formulaton of the Laws of Rotatng Objects The Rght-Hand Rule The laws of rotatonal moton can be expressed compactly f we use the vector cross product. Thus the vector expresson for torque τ s τ = r F, (22) where vector r ponts from some pvot pont (chosen by you!) to the pont of applcaton of the force F. Recall that the force of gravty of an extended object can be regarded as beng appled at the center of mass. The magntude, τ, of the torque obeys τ = rf sn θ, (23) where r and F are the magntudes of the radus and forces vectors, and angle θ ( 180 by defnton) s the angle between vectors r and F. Especally n parts 3 and 4 of ths Lab we must also keep track of the drectons of vectors defned by cross products. The torque vector τ of eq. (22) s defned as beng perpendcular to both r and F (or equvalently, perpendcular to the plane contanng vectors r and F ). Ths defnton does not yet completely specfy the drecton of τ. It could be pontng ether way along the axs perpendcular to the plane of r and F. The rght-hand rule s needed to complete the specfcaton: poston your rght hand so that your fngers ntally pont along the drecton of + r, and so that they can be curled nto the drecton of F by a rotaton of 180 ; your thumb then ponts n the drecton of vector τ. Lteral applcaton of the rght-hand rule sometmes requre dramatc contortons of your rght arm. You must resst the temptaton to use your left arm even f t feels better, because ths would reverse the sgn of the drecton of τ. (Try t once to verfy ths.) The Cross Product of F = m a The law of rotatonal moton can be elegantly derved by takng the cross product of F = m a: τ = r F = r m a = r m d v dt = d dt ( r m v) d L dt, (24) wherewehaveusedthedentty ( ) ( d d r ( r v) = dt dt v + r d v ) ( =( v v)+ r d v ) = r d v dt dt dt, (25) snce the cross product of any vector wth tself s zero. The law of rotatonal moton, eq. (24), s thus τ = d L dt, where L r m v angular momentum. (26)

10 Prnceton Unversty 1996 Ph101 Laboratory 5 10 An extended body can be thought of as a collecton of pont masses m wth postons and veloctes r and v and angular momentum L = r m v. (27) Rotaton About a Symmetry Axs There s an mportant smplfcaton to eq. (27) for objects that are rotatng about a symmetry axs (= axle) whose drecton s constant. Defne ω to be the magntude of the angular velocty of rotaton about the axle. We can also form a vector called ω wth magntude ω by assgnng t the drecton of the axle accordng to the rght-hand rule. Namely, f the fngers of your rght hand curl n the sense of rotaton, then the thumb of your rght hand ponts along the drecton of ω. The utlty of the vector defnton of ω s that the velocty of any pont r n the rotatng object can be wrtten v = ω r. (28) In ths equaton we have assumed that the orgn of the coordnate system for the poston vectors r les on the axle. We can verfy that eq. (28) matches the more famlar relaton for the magntude of the velocty: v = ωr, (29) where R s the perpendcular dstance between the pont r and the axle. Namely, the defnton of the vector cross product tells us that for eq. (28) we have v = ωr sn θ, (30) where θ s the angle between the axle and the vector r. But the perpendcular dstance R s just r sn θ, and eq. (29) follows. Usng eq. (28) we can rewrte eq. (27) as L = r m ( ω r )= m [r 2 ω ( r ω) r ], (31) usng the vector dentty a ( b c) =( a c) b ( a b) c. (32) To go further, we note that poston vector r can be broken up nto components parallel and perpendcular to the axle: r = R + r cos θ ˆω, (33) where R has magntude R = r sn θ and ponts perpendcular to the axle towards pont, andˆω s a unt vector along the axle. Snce vectors R and ˆω are perpendcular we have that R ˆω =0,so ( r ω) r = r ω cos θ ( R + r cos θ ˆω) =r ω cos θ R + r 2 cos 2 θ ω. (34)

11 Prnceton Unversty 1996 Ph101 Laboratory 5 11 Insertng ths nto eq. (30) we have L = m r 2 (1 cos 2 θ ) ω + ω m r cos θ R = I ω + ω m r cos θ R, (35) notng that m r 2 (1 cos 2 θ )= m r 2 sn 2 θ = m R 2 ω = I. (36) The moment of nerta I can also be wrtten I = m R 2 = ρr 2 dvol, (37) wth ρ =massdensty. If the mass dstrbuton s symmetrc about the axle a marvelous smplfcaton occurs n eq. (35). The symmetry mples that each mass m at r = R + r cos θ ˆω has a partner of equal mass at poston R + r cos θ ˆω, and hence m r cos θ R =0. (38) Thus for objects rotatng about a symmetry axs we have L = I ω, (39) Fnally, usng eq. (39) for the angular momentum n the basc torque equaton (26) we arrve at the famlar form τ = I dω = Iα, (40) dt where α s the angular acceleraton Moments of Inerta of Some Smple Objects An mmedate applcaton of eq. (37) s that a compact mass M at radus R from an axle has moment of nerta I = MR 2 (pont mass on a slng). (41) Smlarly, a hoop of mass M all at radus R from an axle has I = MR 2 (hoop). (42) For a unform dsk of mass M, outer radus R and thckness H the volume s πr 2 H,so the densty s ρ = M/(πR 2 H). The volume element s dvol = dr Rdφ dh, and I = M R 2π H RdR dφ dh R 2 = 1 πr 2 H MR2 (unform dsk). (43) For a unform sphere of mass M, and radus R the volume s (4/3)πR 3, so the densty s ρ =3M/(4πR 3 ). The volume element s dvol = dr R sn θdθ Rdφ, and the dstance from apont(r, θ, φ) to the axs s R sn θ, so I = 3M R π 2π R 2 dr sn θdθ dφr 2 sn 2 θ = 2 4πR MR2 (unform sphere). (44)

12 Prnceton Unversty 1996 Ph101 Laboratory Parallel-Axs Theorem If the momentum of nerta of an object of mass M about a symmetry axs s I 0, then the moment of nerta about a parallel axs at dstance h from the symmetry axs s I = I 0 + Mh 2. (45) To see ths we use relaton (37), rewrtten as I = M R 2,where R s a vector perpendcular to the new axs from that axs to pont. If we defne h as the vector perpendcular to the symmetry axs from the symmetry axs to the new axs and R as the vector perpendcular to the symmetry axs from the symmtery axs to pont then R = R h. Furthermore, the center of mass of the object les on the symmetry axs, so M R =0. Then I = M ( R h) 2 = M R 2 + h 2 M 2 h M R = I 0 + Mh 2, (46) as clamed Knetc Energy of Rotaton A movng object has knetc energy KE = 1 m v 2 2 = 1 m v v. (47) 2 In general the moton of a rgd body can be descrbed by the velocty v cm of the center of mass and a rotaton about some axs through the center of mass wth angular velocty ω. If r s the poston vector from the center of mass to pont then that pont has velocty recallng eq. (28). On squarng ths we fnd usng the vector dentty v = v cm + ω r, (48) v v = v 2 cm +2 v cm ω r + ω 2 r 2 ( ω r ) 2, (49) ( a b) ( c d)=( a c)( b d) ( a d)( b c). (50) Also note that ω r = ωr cos θ,andthatω 2 r 2(1 cos2 θ )=ω 2 R 2 where as before R s the perpendcular dstance from the axs of rotaton to pont. The knetc energy can now be wrtten KE = 1 m vcm v cm ω m r + 1 m R 2 2 ω2 = 1 2 Mv2 cm Iω2, (51) usng the fact that m r = M r cm =0sncewehavetakenthecenterofmasstoatthe orgn. Equaton (51) s true whether or not the axs of rotaton s a symmetry axs, but the nterestng topc of rotatons about non-symmetry axes wll have to be pursued elsewhere. Equaton (51) contans the memorable result that the total knetc energy of a movng object s the sum of the knetc energy of the center of mass moton plus the knetc energy of rotaton about the center of mass.