. (7.1.1) This centripetal acceleration is provided by centripetal force. It is directed towards the center of the circle and has a magnitude

Transcription

1 Lecture #7-1 Dynamics f Rtatin, Trque, Static Equilirium We have already studied kinematics f rtatinal mtin We discussed unifrm as well as nnunifrm rtatin Hwever, when we mved n dynamics f rtatin, the nly examples we cnsidered were examples f unifrm circular mtin S, nw we shall talk aut dynamics f nnunifrm circular mtin irst f all recall what yu knw aut dynamics f unifrm circular mtin The simplest case f the unifrm circular mtin is when a pint-like ject mves arund a circle f radius r with cnstant speed v In this case it has centripetal acceleratin a v r (711) This centripetal acceleratin is prvided y centripetal frce It is directed twards the center f the circle and has a manitude m v r (71) The sinificance f this frce is just the net frce actin n the dy We have already discussed, that this centripetal frce is nt a special frce ut the result f the vectr additin f all the frces actin n the dy Nte that this centripetal frce nly prvides centripetal acceleratin, ut it des nt have any cmpnent in the tanential directin f the circle, s it can nt prduce tanential acceleratin and a nnunifrm circular mtin 1 Trque Nw let us cnsider rtatin f a dy arund fixed axis f rtatin We saw that the mst effective way t descrie kinematics f this mtin is y means f the anular variales These variales include: anular displacement, anular velcity and anular acceleratin Nw we have t intrduce similar variales in rder t e ale t descrie dynamics f rtatin The main suject f dynamics is frce Let us see what rtatinal anal f this quantity is In the case f translatinal mtin, it des nt matter at which pint f the dy yu apply the frce as ln as it is a riid dy, which mves as a whle Indeed if yu cnsider a heavy x n the flr, yu can either pull it frm the frnt r push it frm

2 ehind with a same frce The result is the same if these frces had the same manitude and directin Hwever, it is different in the case f rtatin r instance, yu miht ntice that the drkn is always lcated as far frm the dr's hine as pssile If yu push this dr riht near the hine, yu praly will nt e ale t pen it On the ther hand even very heavy drs can e pened withut t much f the effrt, if yu push them fr the drkns S, in the case f rtatin it mattes nt nly what frce yu apply, ut hw far frm rtatinal axis yu apply it Directin als matters If yu push r pull this dr in the directin aln its surface it will nt d any d fr penin f the dr Yu can pen it nly if yur frce has a cmpnent in the directin perpendicular t the dr In eneral, if yu cnsider a riid dy rtatin arund fixed rtatinal axis sme frce can e applied in aritrary directin at the pint, which is lcated at distance r frm the dy's rtatinal axis The psitin f this pint is defined y the radius vectr r cnnectin it with the pivt pint (rtatinal axis) At this pint we shall intrduce the new physical quantity, trque, which has the manitude r sin, (713) where is the smaller anle etween vectrs r and (Nte: If yu are familiar with a cncept f the crss prduct f tw vectrs, then yu can see that trque f frce is a vectr r If vectr r and frce frm a plane perpendicular t rtatinal axis, trque will e parallel r antiparallel t rtatinal axis) Since frce is actin in aritrary directin it may have a radial cmpnent r aln the r -vectr as well as tanential cmpnent sin perpendicular t the r - vectr This means that r r, (714) where r r sin is the distance in perpendicular directin etween rtatinal axis and the extended line runnin thruh vectr This extended line is called line f actin and r is called the mment arm f frce We shall use the same sin cnventin fr trque as we used fr any ther anular quantity The trque which causes

3 cunterclckwise rtatin is psitive, the trque which causes clckwise rtatin is neative In the example aut the dr, we saw that if the frce acts perpendicular t the dr, it will e the mst efficient way t pen it Accrdin t equatins 713, 714 trque has its maximum manitude in this case If the frce acts aln the dr it des nt mve the dr and trque f this frce is equal t zer It is als equal t zer, if the frce is applied directly t the hine, ecause r=0 and the dr des nt mve S, trque plays the same rle fr rtatin as frce des fr translatin The SI unit fr trque is Newtn-meter (Nm) Trque eys superpsitin principle If there are several frces actin n the dy, ne can find trques fr each f thse frces and calculate the net trque with respect t the chsen rtatinal axis The dy will rtate arund this axis with nn-zer anular acceleratin, if the net trque is nt zer It is in same way as the dy can nly mve with nnzer acceleratin, if the net frce actin n it is nt zer On the ther hand rtatin with cnstant anular speed may nly ccur at zer net trque, as it was in the case f Newtn's first law fr translatinal mtin Hwever, it still requires the existence f nnzer centripetal frce even thuh it has the zer trque Static equilirium A dy is in static equilirium if it is at rest, nt mvin and nt rtatin This means that the dy has zer acceleratin a 0 and zer anular acceleratin 0 We saw frm the Newtn's first law that zer acceleratin still des nt uarantee that the dy is at rest It can mve with cnstant linear velcity r rtate with cnstant anular velcity But if the ttal acceleratin f the dy is zer, we can always find a reference frame in which this dy has zer velcity When talkin aut static equilirium, we will use this special reference frame in which dy is at rest T e in static equilirium is f extreme imprtance fr many enineerin structures, such as uildins and rides The state f equilirium can e descried as either stale r unstale The dy is in stale equilirium, if it returns t this state after ein displaced y a small frce r instance, a marle placed n the ttm f the hemispherical wl will return t its psitin if slihtly displaced On the ther hand if I put the same marle n the tp f a sphere and displace it slihtly, it will rll ut This is

4 the example f unstale equilirium Different enineerin structures nt nly have t e in equilirium, they als have t e in stale equilirium t survive the influence f the envirnment Mtin f the dy is verned y Newtn's secnd law The asence f mtin means that this law takes the frm net net 0, 0 (715) The first f these tw equatins shws that the dy is nt mvin, ecause the vectr sum f all external frces actin n it is zer The secnd equatin shws that the dy is nt rtatin, ecause the sum f all the trques due t external frces arund any axis is zer Since equatins 715 are vectr equatins, they can e rewritten in terms f vectr cmpnents In all the prlems which we will cnsider here, the crdinate system can e chsen in such a way, that all the frces are in x-y plane, which reduces the numer f equatins t three net, x net, y net x y 0 0, 0, (716) When cnsiderin the prlems, it is imprtant t knw nt nly what frces are actin n a dy, ut als at which pint each f the frces is applied If the dy has sme cmplicated shape, it is difficult t understand, where the effective pint f applicatin fr ravitatinal frce is Talkin aut effective pint, we mean that ravitatinal frce is in fact actin n different parts f the dy, ut we can replace all these frces y ne effective ravitatinal frce, which acts in such a way, that the net frce and the net trque actin n the dy will nt chane We have already discussed the center f mass The effective frce f ravity acts at the center f mass This statement is true, if ravitatinal acceleratin is the same fr all the parts f the dy, which is usually true fr dies small cmpared t the size f the Earth S, we can always draw ravitatinal frce vectr as ein applied at the center f mass

5 Example 711 A hrizntal eam f mass m is supprted at each end A lck f mass M rests a quarter f the way frm ne end What is the vertical frce n each f the supprts? The frces actin n the eam are shwn in the picture They all are in the same plane, which we will cnsider as x-y plane The vertical frce n each f the supprts is the same ut ppsite t frces 1, actin frm the supprts t the eam Let us use summatin f trques first, t find ne f thse frces We will pick a pivt pint at the first supprt We shall call the lenth f the eam t e L S the trque equatin will e L 3 m LM L 4 0 In this equatin the ravitatinal frce actin n the eam is m (it acts at the center f the eam) and M is the frce actin frm the lck n the eam, which is 3/4 L frm the pivt pint I can divide the equatin y lenth f the eam, s ives 1 3 m M 0, 4 m 3M m 3M 4 4 T find the secnd frce, let us use summatin f frces in vertical directin It

6 y 1 m M 0, s m M m M m 3 M m M m M Example 71 In the fiure shwn elw assume that the anle is knwn t e The mass f the eam is m and suspended lck has the mass M Determine frce f the wall and tensin T in the cale All the frces actin n the eam are shwn in the picture They all are in the same plane, which I called x-y plane Crdinate axes are als shwn in the picture I will chse the lenth f the eam t e L In fact we will see that the answer des nt depend n this lenth, exactly as it was in the previus prlem We will chse the pivt pint at the pint, where this eam is attached t the wall Let us find trques and cmpnents fr all the frces actin n the eam The first trque L Lm due t weiht f the eam is 1 r 1 1 sin 1 m sin 70 Here I have used the half f the eam's lenth as distance r 1 frm the pivt pint t the pint where the frce acts, since the weiht f the eam acts at its center f mass The secnd trque is due t the weiht f the suspended lck, s r sin LM sin 70 LM, since the lck is suspended at the end f the eam The third trque is due t the frce f tensin T in the cale It is r sin LT sin The trque due t frce f the wall is

7 in t e zer, ecause this frce acts at the pivt pint The equatin fr the trque summatin ecmes: Lm LM LT sin 0, m M T sin 0, T m M / sin S we have fund the manitude f the frce f tensin in the cale It has tw cmpnents: Ty T sin m M, Tx T cs m / M / tan T find the frce f the wall, let us first determine its cmpnents The equatin x 0 takes the frm x Tx 0, x Tx m M tan The secnd equatin y m M T 0, y HG y 0 will e I K J m m y m M M The manitude f the frce y x H G I K J HG m m M tan can e fund thruh Pytharean therem as I K J Exercise: What can yu say aut frce if M=0? Example 713 See the picture elw Assume that the ladder has mass f 50k and makes an anle 30 derees with the rund and that 100k man is 3/4 f the way up the ladder Assume that the wall is frictinless Calculate the minimum cefficient f frictin etween the rund and the ladder fr the ladder nt t slip

8 As efre all the frces actin n the ladder are in the same x-y plane (see the picture) T slve this prlem let us first use the equatin f trque summatin I will chse the ttm f the ladder t e the pivt pint I shall call the weiht f the ladder t e m and the weiht f the persn M Let the lenth f the ladder e L, the hrizntal (n frictin) frce cmin frm the wall W and the frce f the rund G The frce f the rund has tw cmpnents: the hrizntal cmpnent plays the same rle as frictin frce, which keeps the ladder frm slippin and the vertical cmpnent f rund's frce is the nrmal frce L The first trque due ladder's ravitatinal frce is 1 r1 1 sin1 m sin 60, 3L the secnd trque f the man is r sin M sin 60 The third trque is due 4 0 t hrizntal frce f the wall, it is3 r3 3 sin3 L W sin 30 The trque f the rund frce is zer since it acts at the pivt pint, s L 3L z m sin60 M sin60 L W sin

9 Gx m 3M sin60 W sin 30, 4 m 3M sin60 m 3MI W K J tan sin I HG K J I HG K J HG Nw let us discuss frce G The summatin f frces in x directin ives us 0, r r y directin ne has Gy m M 0, s W M m Gy Gx W T find minimum cefficient f frictin we have t rememer that ur case this frmula ecmes,, r s 50k 3100k 4 50k 100k Gx Gy Gx Gy 100k tan 60 tan 60 tan k 3 fr max m 3M 4 m M N In tan 60,