1 Hz = one cycle per second

Transcription

1 Rotatonal Moton Mchael Fowler, UVa Physcs, 14E Sprng 009 Mar 5 Prelmnares: Unts for Angular Velocty The tachometer on your car dashboard tells you your car engne s angular speed n rpm, revolutons per mnute, and probably redlnes around 6 or 7,000 rpm. A common unt for electromagnetc oscllatons (to be dscussed nest semester) s cycles per second: there s even an offcal name, the Hertz, abbrevated Hz: 1 Hz = one cycle per second But angular moton sn t restrcted to complete cycles we need to measure fractons of a cycle as well, for example, n reorentng an arcraft carrer, the Hertz s not gong to be a handy unt. A complete crcle s dvded nto 360 degrees, wrtten 360. (Ths was done by the Babylonans, 4,000 years ago, probably because the Sun moves very close to 1 a day through the sky, relatve to the fxed stars.) So we could talk about angular velocty n degrees per second. However, from a practcal pont of vew, we often need to relate angular velocty to the speed of some part of the wheel. For example, f I whrl a stone around n a slng at some number of revolutons per second, or degrees per second, then let go, how fast s the stone movng? R v? What lnear speed corresponds to n revolutons per second? n revs per sec Let s say the angular velocty s n degrees per second, or n/360 Hz. If the slng s r meters long, so the stone s gong n a crcle of radus r (gnorng the slght movement of my hand), the stone travels πr meters per cycle, so ts speed v s gven n terms of ts angular speed n by: v nr. 360

2 Ths rather clumsy formula s a consequence of followng the Babylonans n defnng the unt of angle as 1/360 of a full crcle. We d have a much ncer formula f nstead we defned our unt of angle as 1/π of a full crcle! So we defne: the radan = 1/π of a full crcle. Ths works out to be about 57.3, and snce the complete crcumference of a crcle s πr, the dstance around the crcle correspondng to an angle of one radan s one radus. Duh. Angle of one radan the red sold arc has length equal to the radus of the crcle. snθ θ for θ small for r = 1, the arc length s θ, the short black lne = snθ. It s easy to see that radans are the natural angle unt for trg: look at the sne of a small angle. Our default notaton for an angle s θ, measured from the x-axs, wth counterclockwse as postve. Angular velocty n radans per second wll usually be denoted by ω: d/ dt. So f a wheel s rotatng at postve angular velocty ω radans per second about a fxed axle, a pont on the wheel at dstance r from the axle s movng at speed a formula we ll be usng a lot. v r The standard notaton for angular acceleraton s d / dt d / dt. Notce that f we have constant angular acceleraton, formulas for angular velocty and angular dsplacement (meanng angle turned through) are just lke those for lnear moton under constant force:

3 3 t 0 t t ( ). 0 0 Just be sure your system has constant angular acceleraton before usng these formulas! Torque: Seesaw Physcs Torque s a measure of the effectveness of a force n gettng somethng rotatng: actually we ve come across ths already n the dscusson of the seesaw. If a chld sts at the end of a level empty seesaw, t begns to rotate. O.K., t doesn t get far, but there s angular acceleraton. And we know the f the chld sts close to the pvot pont, the seesaw turns more slowly. We ve also seen that two chldren, one twce as heavy sttng at half the dstance, balance we sad they had the same leverage. Ths leverage s actually what s called torque. It s also called the moment of a force about an axs of rotaton. It s the product of the weght of the chld and the dstance from the pvot or axle. So, for a 0 kg chld at a dstance of 1.5m, the torque about the pvot pont s 300 N.m, settng g = 10. d d mg mg Balancng a seesaw We can also have two forces on the same sde of the pvot balancng f one of them s upwards. If a 0kg chld s sttng half way to the pvot, wth no-one on the other sde, you could lft the seesaw to a horzontal poston by supplyng an upward force equvalent to 10 kg weght at the end of the seesaw, the same sde as the chld.

4 4 d d T = mg mg Balancng a seesaw the hard way One further pont: what f you suppled the upward force by tyng a (lght) rope to the end of the seesaw, and tuggng at some angle? In that case, the tenson n the rope, where t pulls on the seesaw, s equvalent to a force perpendcular to the seesaw plus a force parallel. The latter component does nothng (unless the seesaw can move sdeways), the upward component s all that counts. d d T = mg angle 30 mg Only the upward rope force component counts mg 3mg mg There s another way to look at ths whch s sometmes useful: nstead of resolvng the force nto components at ts pont of applcaton, as n the dagram above, we can take the whole force, actng along the lne n space through ts pont of applcaton, but calculate ts leverage the torque by takng the lever arm to be the dstance from the pvot pont to the lne of the force:

5 5 Torque = force dstance of lne of force from pvot = 4mgdsn30 = mgd d angle 30 T = mg dsn30 Ether way, the formula for torque strength s: Fr sn where F s the magntude of the force, r the dstance from the pvot of the pont of applcaton, and θ the angle between the vector r from the pvot pont to where the force acts and the force F. Rotatonal Equvalent of Newton s Frst Law By the tme Galleo notced that a smooth ball on a flat surface wth lttle frcton would keep the same velocty a long tme, t was already known that a rotatng grndstone, wth a well-oled axle, would keep spnnng a long tme too and ths, taken to the deal lmt, s Newton s Frst Law for rotaton: n the absence of torque, a body wll spn at constant angular velocty ndefntely. It s worth mentonng that a vertcal grndstone must have the axle through the center of gravty. If ths s not the case, gravty wll exert a torque, and although the wheel mght keep spnnng, t won t be at constant angular velocty. In fact, the torque on such an off-center wheel s gven by Mgr sn Mgx, where r s the dstance of the center of mass from the axle, and θ the angle between r and the vertcal. To prove ths, magne the wheel to be made up of a large number of glued-together small masses m at postons r.

6 6 Torque around axle from small part of an off-center wheel τ = m gx. axle x m m g As dscussed n the precedng secton, the torque about the axle of the weght of the small mass m s equal to m g multpled by the perpendcular dstance from the axle to the lne of acton of the force, that s, m gx, takng the axle as the orgn of coordnates. Evdently the magntude of the total gravtatonal torque from the mbalance from the defnton of the center of mass. m gx Mgx Knds of Equlbrum If there s no frcton at the axle, the wheel can only be at rest f the center of mass les on the vertcal lne through the axle. If the center of mass s drectly above the axle, the wheel mght be at rest, but the slghtest push wll start t turnng: ths s called unstable equlbrum. If the center of mass s below the axle, a slght push to one sde wll cause the wheel to swng back then oscllate, a topc we ll return to later ths s stable equlbrum. If the center of mass s exactly at the axle, the wheel wll rotate at a steady rate ths s neutral equlbrum. These all have analoges n lnear moton: assumng no frcton, unstable equlbrum s a ball perched on a mountantop, stable s a ball sttng at the bottom of a valley, and neutral s a ball on a flat plane. In two dmensons, there are more complcated possbltes: a ball n the mddle of a saddle mght be stable for dsplacements parallel to the horse, but unstable for dsplacement the other way. Stuatons analogous to ths horsy example do n fact occur n physcal systems. Rotatonal Equvalent of Newton s Second Law A Smple Example: Introducng the Moment of Inerta Newton s Second Law, n the form F ma, states that an object undergoes lnear acceleraton n response to an net external force, and the mass determnes the magntude of ths response. It s pretty

7 7 clear by now that the angular verson must be that an object undergoes angular acceleraton n response to an net external torque but t s less clear what plays the role of mass ths tme. To fnd out, we ll look at an example of angular acceleraton whch s almost the same as lnear acceleraton. M ω = dθ/dt R axle F The mass M has speed v = Rω and acceleraton a = Rdω/dt = Rα Suppose a mass M s frmly attached to a horzontal dsk of neglgble mass on a frctonless vertcal axle, so that the mass M must go around n a horzontal crcle. Assume the mass s ntally at rest, then the force F s appled. What happens n the frst few moments? Well, of course, the mass accelerates wth acceleraton a = F/M. As soon as t reaches a sgnfcant speed, the dsk wll begn to exert a force towards the center, to keep t on ts crcular path, but let s just concentrate on those frst few moments before that effect becomes mportant. Let s translate that ntal acceleraton nto angular language. Snce the angular velocty s related to the actual velocty by v = R, the angular acceleraton must be related to the ordnary acceleraton a along a tangental drecton by a = R. So we can see how to connect the two acceleratons. Now we must connect the two forces, or, rather, the lnear force F wth the torque actng on the system mass + wheel. The torque s just equal to the force F multpled by ts leverage arm R, that s, = FR. Puttng ths together wth a = R, we fnd that F = Ma translates to: = FR = MaR = MR Comparng ths equaton, = MR, wth F = Ma, we notce that the quantty MR plays the role of mass, or nerta, n rotatonal dynamcs for ths smple case. It s called the moment of nerta, and labeled I. We show n the next secton that for more complcated rgd bodes, = I s stll correct, and I s a sum over the body of terms lke MR.

8 8 Moments of Inerta for Extended Bodes We ve only found the moment of nerta for a sngle mass attached to a lght dsk. What about somethng more complcated, for example a heavy dsk? The key to fndng the moment of nerta for a sold body s to vsualze t as a lot of small masses all glued together. Thnk specfcally of a heavy wheel of radus R, rotatng about a fxed axle wth neglgble frcton. Imagne the wheel to be ntally at rest, then a force F s appled to the rm tangentally, so that the torque s FR. Of course, the whole wheel begns to rotate, so thnk what happens to one of the small masses m we are magnng the wheel to be made up of. If t s at a dstance r from the axle, and the ntal angular acceleraton of the wheel s, ths small mass, m say, must experence an ntal acceleraton r, so there must be a total force on t of f = m r, and therefore a torque actng on t of f r = m r. Rgd wheel made up of many lttle masses m : when external torque FR s appled, m has ntal lnear acceleraton r α, so experences torque: axle r m F τ = f r = m r But where, exactly does ths torque come from? Ultmately, t must come from the force F beng appled to a dfferent part of the wheel, but the mmedate cause for the small mass m s acceleraton s just the pulls and pushes of ts neghbors. In other words, f you push on one part of the wheel, snce t s a sold body ts rgdty transmts the effect of the force to all the other parts, so the wheel rotates as a whole provded of course the force s not strong enough to tear the wheel apart. Snce the torque felt by one small part of the wheel τ = m r, the total torque felt by all the parts of the wheel s: τ = Στ = m r = dmr wth the ntegral beng over all the tny masses dm makng up the wheel.

9 9 Now comes a crucal pont. How do we relate ths torque to the external force torque FR? What about all the nternal forces the fact that each part of the wheel feels forces from neghborng parts? How do we take account of ths very complcated stuaton? The answer s: we don t have to! By Newton s Thrd Law, all those nternal forces are n equal opposte pars, actons and reactons between neghborng masses. Ths means that when we sum over all parts of the wheel, we count all these forces, and they all cancel each other n pars. Therefore the total torque s just that from the external forces. It follows that the analog of Newton s Second Law for rotaton about an axle s Torque = I where the moment of nerta I s gven by m r = dmr. (And remember that as always the acceleraton must be measures n radans/sec.) Moments of Inerta: Hoops, Dsks, Cylnders, Rods. Fndng the moment of nerta l s now just a matter of dong elementary ntegrals. For the case of a hoop of mass M, where all the mass can be taken at the same dstance R from the axs of rotaton, no ntegral at all s necessary: For a hoop, I = MR. A dsk can be thought of as constructed of successve hoops, lke a two-dmensonal onon. As a practcal matter, the best approach s to use the two-dmensonal densty, call t, (unts: kg/m ) so that the total mass of the dsk s densty x total area, M = R. Then the hoop whch s that part of the dsk between r and r + Δr from the center has mass m = rδr, and that hoop has moment of nerta mr = (rδrr. A dsk can be seen as made of concentrc hoops glued together. R axle r r + Δr

10 10 Then just do an ntegral to add up the moments of nerta of all the nested hoops from 0 to R to fnd the moment of nerta of the dsk: R dsk I r dr R MR. 0 The same expresson works for a cylnder, whch for these purposes s just a stack of dsks. What about a rod about one end? Assume ts (constant) lnear densty s λ kg/m., so a length dx has mass λdx. Then the moment of nerta s L 3 I x dx L / 3 ML / 3. 0 Suppose such a rod s hnged at one end, horzontal, and falls. What s the ntal acceleraton of ts end? MgL / I ( ML / 3) 3 gl /. So the end accelerates at Lα = 3g/. If t s at frst sght surprsng that the end of the rod accelerates downwards faster then g, consder the followng system: a very lght rod, hnged at one end, wth a large weght attached to ts mddle. A very lght rod hnged to a wall at one end has a mass at ts center: how fast wll the rod s free end accelerate downwards on release from a horzontal poston?

11 11 If let go from a horzontal poston, the weght wll completely domnate the stuaton, and accelerated downwards very close to g. Ths means the far end of the lght rod s forced downwards at close to g. Our example above s not that extreme, but the same dea. The Parallel Axs Theorem If we know the moment of nerta of an object about a lne through the center of mass, the moment of nerta about any parallel lne s easly found. We ll prove ths for a two-dmensonal object (really a thn three-dmensonal object): new axs r h m r center of mass The moment of nerta of the two-dmensonal object about any axs perpendcular to the object s smply related to that about the parallel axs through the center of mass. Wth the notaton n the dagram, I mr, and recall mr 0. Now I m r m ( r h) newaxs I h m r m h I Mh. Ths s the Parallel Axs Theorem: I I Mh newaxs. where h s the perpendcular dstance between the new axs and the parallel axs through the center of gravty. It s easy to extend ths proof to a three-dmensonal object: takng the parallel axes to be n the z-drecton, replace r wth (x, y ).

12 1 Varyng Moment of Inerta Newton dd not actually wrte hs Second Law as F = ma = mdv/dt, but as F = dp/dt, where p = mv s called the momentum, and n fact ths s the correct descrpton of nature n cases where the mass of an object vares, such as a rocket expellng fuel, or a fallng randrop growng from condensaton. Smlarly, the rotatonal Second Law = Icould be wrtten more generally as d dt to allow for the possblty of a varyng moment of nerta, and ths s much more common than a varyng mass. Thnk of a spnnng dancer or skater throwng her arms out and brngng them n, or of a dver curlng up and uncurlng hs rotatng body. The external torques are pretty small under these condtons, the varaton n spn speed s largely due to the changng moment of nerta and the constancy of the total angular momentum IThe observed varatons n angular velocty confrm the valdty of the more general law. Rotatonal Knetc Energy A rotatng object has knetc energy even f ts center of mass s at rest. Thnk of t as composed of many small masses m at dstances r from the center of mass. Then f the angular velocty s m has speed v = r, and knetc energy ½ m r, and summng these, If a net torque s actng, ths energy wll change: d dt I KE of rotaton = ½ I. d dt 1 I I I so the rate of workng, the power, of a torque equals the magntude of the torque multpled by the angular velocty, just as a force F movng somethng at speed v s workng at a rate Fv. Ths equaton can also be ntegrated: d d I dt dt d dt 1 The total change n knetc energy equals the ntegral of torque tmes angular dstance, vald even f the torque s varyng. Rollng Down a Ramp For example, we ll take the case of a hoop of radus R rollng down a ramp wthout slppng. Suppose at some nstant the center of the hoop s movng wth speed v. Of course, other ponts on the hoop are

13 13 movng at dfferent speeds. The pont at the bottom n contact wth the ramp sn t movng at all at that nstant. But n the frame of reference n whch the center of mass s momentarly statonary, that pont s movng backwards at R. Ths means that the angular speed and the lnear speed of the center of mass are related by v = R. v v = Rω v ω 0 Veloctes relatve to the ramp for noslp: zero at pont of contact, v at center of hoop, v at farthest pont. Veloctes of parts of hoop relatve to ts center: v = Rω perpendcular to the radus. What s the total energy of the rollng hoop? Imagne as usual that t s made up of a large number of small masses m. If the small mass m has velocty v, the total energy s ½ mv. Snce ts both movng along and rollng, let s wrte the velocty of the small mass m as v v u where u s the velocty of the small mass m relatve to the center of mass. Then the total knetc energy of all the small masses mv m v u Mv v mu mu. The frst term s just the ordnary knetc energy of lnear moton, the last term s the same as the knetc energy of rotaton for the center of mass at rest. The mddle term s zero, because the sum s just the d lnear momentum n the center of mass frame, whch s zero. ( mu mr 0 n the frame dt n whch the center of mass s at rest.)

14 14 Therefore, the total energy s just the sum of the translatonal knetc energy and the rotatonal knetc energy: E Mv I 1 1 where I s the moment of nerta about the center of mass. So what does ths mean for the hoop? If t s rollng at speed v, t s rotatng at angular speed v/r, and snce ts moment of nerta s MR, ts total energy s E Mv I Mv MR ( v / R) Mv. So t has twce the knetc energy of a block of the same mass sldng down at the same speed. Put another way, a block sldng down through a heght h wll have speed gven by v = gh, the hoop wll only reach speed v = gh, because t wll have lost the same amount of potental energy, so must have the same total knetc energy ncludng the rotatonal knetc energy, whch the block ddn t have. The equaton F = ma stll descrbes ts lnear moton down the ramp, and after descendng a dstance x ts velocty s gven by v = ax. Evdently, the rollng hoop has only half the acceleraton of the sldng block, so must only be subject to half the force pullng t downhll. The dfference s the retardng statc frctonal force at the pont of contact: that must be just half of the gravtatonal force component pontng down the hll. If there sn t enough frcton, the hoop wll slde wthout ncreasng ts angular velocty. (Note: the hoop has only half the acceleraton of the block, but reaches a fnal velocty greater than half that of the block, because t accelerates for a longer tme.)