Newton's second law in action

Transcription

1 Newon's second law in acion In many cases, he naure of he force acing on a body is known I migh depend on ime, posiion, velociy, or some combinaion of hese, bu is dependence is known from experimen In such cases, Newon's law becomes an equaion of moion which we can solve The soluion allows us o predic he posiion of he body a any ime, as long as we know is iniial posiion and velociy This predicive qualiy is he main power of Newon s law In his secion, we will use Newon s second law in his way We will consider he case of onedimensional moion under a consan force, boh wih and wihou damping This has many applicaions, one of which is verical moion near earh s surface Moion under a consan force When we say ha he force acing on a body is consan, we mean ha i does no vary over ime, is he same no maer where he body is, and does no depend on he body's velociy In oher words, F is jus a number The following graph shows a case in which he consan force happens o be in he posiive direcion: F We are going o use Newon's law o show ha he graph of he posiion of he body versus ime, x(), is a parabola We begin wih Newon s law, F=ma Using he definiion of he acceleraion as he second derivaive of x(), we find he following equaion of moion: m d 2 x ( ) = F d 2 This is a differenial equaion for he funcion x() Tha is, i is an equaion whose soluion is a whole funcion of ime, no jus an algebraic number I says ha x() is a funcion whose second derivaive wih respec o ime is a consan, F/m Therefore, we know wha is firs derivaive mus be; ha is, we can inegrae once and find

2 dx( ) d = F m + consan, where he consan does no depend on ime Now, he lef-hand side is he velociy So he consan mus be fixed by he value of he velociy a some paricular ime For simpliciy, le's suppose ha his ime is =, and ha he velociy a ha ime is v Subsiuing = ino boh sides, we find ha he value of he consan is v The velociy is herefore given by v ( ) = F m + v We hen inegrae again, and find x ( ) = 1 2 F m 2 + v + anoher consan The new consan is fixed by he value of he posiion a anoher paricular ime, which we may again ake o be = If he posiion hen is x, hen he value of he new consan is x The final answer for he posiion as a funcion of ime is hus given by posiion and velociy The reason we have o know wo quaniies is because Newon's law gives rise o a second-order differenial equaion Tha is, he highes derivaive which appears is he second derivaive Le's ry o picure such a moion physically We will do his by considering some special cases Special case: zero iniial velociy, posiive force Suppose he body is iniially a res a x =, and he force acs in he posiive x direcion Then we know wha will happen - he body will jus speed up in he direcion of he force The posiion as a funcion of ime is given by x ( ) = F 2 m 2 This is (one half of) a parabola poining upwards: x x() x ( ) = F 2 m 2 + v + x This formula allows us o find he posiion a any ime, as long as we know he values of he iniial

3 The slope a = is zero, corresponding o zero iniial velociy Special case: posiive v, negaive force Suppose now ha he iniial velociy is posiive, bu he force is negaive Tha means ha he body will slow down, evenually sop, and hen speed up in he opposie direcion Since he force is consan, he body coninues o speed up indefiniely (unil Newonian mechanics iself breaks down - a opic for laer discussion) Here is a plo of x() in he case jus described: x x slope = v x() Because we have he soluion for x(), we may answer virually any quesion we like abou he properies of he moion For example, suppose we wan o know he maximum value of x reached by he body, and a wha ime i is reached This may be handled by sandard mehods of calculus, bu we will insead do i here using he mehods of analyic geomery We complee he square in he soluion for x(): x ( ) = F 2 m ä å + mv ë ì 2 + x ã F í mv 2 2 F This makes i clear ha he pah in he -x plane is a parabola The parabola opens downward in our case, because he facor F/2m is negaive The maximum value of x occurs when he firs erm is zero, ie when mv =, F which is posiive) The maximum value of x is x max = x mv 2, 2 F which is greaer han x Here s a MAPLE inpu line which solves he presen differenial equaion symbolically: soln:=op(2,dsolve({m*diff(x(),$2)=f, x()=x[],d(x)()=v[]},x())); The nex line plos a soluion: plo(subs({m=1,f=-1,x[]=1,v[]=1}, soln),=3,2); (Here s how o copy hese lines from he presen

4 documen and pase hem ino MAPLE) You can easily change he values of he parameers before execuing he inpu If you wish, you can also plo he velociy by changing soln o diff(soln,) in he above line Applicaion: verical moion near earh A very imporan physical siuaion o which he above applies direcly is he case of a body moving verically near he surface of he earh Alhough we will no sudy he graviaional ineracion unil laer, you may be familiar wih he relevan fac: near he surface of he earh, all bodies (whaever heir inerial mass) have an acceleraion of approximaely 981 m s 2 in he downward direcion This is rue as long as he effecs of air resisance are negligible This value is given a special symbol, g, and is called he graviaional acceleraion a earh s surface: g 981 m s 2 Le's orien our coordinaes verically, wih posiive values poining upwards Then he force due o graviy is F = mg The minus sign indicaes ha he force is direced downwards (If we had oriened he coordinaes wih posiive values poining downwards, he minus sign would no be presen in he equaion for F) f This force is consan - does no depend on ime, posiion, or velociy We can herefore ake over he above formulas direcly, wih mg subsiued for F In paricular, we find ha a body fired upwards wih velociy v reaches a maximum heigh of v 2 2 g

5 above is saring heigh, a ime v g The pah in he -x plane is given by he parabola shown wo figures ago I is imporan o realize ha he pah in space is a sraigh line, no a parabola We are considering verical moion only, a presen (Laer, when we come o consider moion in more han one dimension, we will find ha a projecile can move on a parabolic pah in space Tha parabolic pah is no he same as he presen one in he -x plane - don' ge hem confused!) The resources for his secion conain a movie he body is he sum of he consan force F ha showing his verical moion along wih graphs ofwe have jus been considering, and he resisive posiion, velociy and acceleraion For force F res : comparison, here is also a movie showing he same quaniies for a body dropped from res F ne = F + F res Resisive forces In many physical cases, here is some resisance o moion For example, a body could be sliding along a rack wih fricion presen Or, a body could be moving verically near he earh, wih air resisance In many cases, he force resising he moion is proporional o he velociy of he body Mahemaically, his is wrien F res = bv The quaniy b is a posiive consan, whose value depends on he properies of he maerial providing he resisance I is no a fundamenal consan of naure The minus sign indicaes ha he force resiss he moion, so is direced opposie o he velociy We would like o illusrae he procedure of solving Newon s law when such a force is involved We will suppose ha he oal force on I is always a good idea o use physical inuiion o ge an idea of he naure of he soluion, before beginning he mahemaics In he presen case, i is easy o see one aspec of he soluion As ime goes on, he exernal consan force will jus balance he resisive force, giving zero ne force The body will hen move wih a consan velociy called he erminal velociy v

6 b v The above figure shows he equal and opposie forces in red The ne force is v F ne = F bv =, which gives for he erminal velociy v = F b Le s solve he equaion of moion and see how his is refleced in he soluion Newon's law reads m dv( ) = F bv( ), d which we have wrien enirely in erms of he velociy v() and is firs derivaive How o solve his? Well, if F were zero, we would have dv/d= (b/m)v, which has is soluion some consan imes exp(-b/m) By inspecion, we find ha we can accoun for nonzero F by simply adding a consan, F/b Tha is, v ( ) = F b + consan H exp ä å b ã m ë ì í The value of he consan mus be relaed o he F value of v a some paricular ime For convenience, le s say ha a = he value of v is v Then he consan is v F/b The velociy is hus given by v( ) = F b + ä å v F ë ì exp ä å b ã b í ã m ë ì í As becomes large, he second erm vanishes and v() approaches F/b, as we know i should Noice ha he second erm never acually becomes zero a any finie ime - i jus ges closer and closer Le's now see how he inroducion of he resisive force changes he resuls we found earlier in he case where he iniial velociy is posiive and he force is negaive 1) A wha ime max does he posiion reach is maximum value? The answer is he ime a which he velociy is zero Solving, we find ha his is given by max = m b ln ä å bv 1 ë ì ã F í I is posiive because bv / F is negaive 2) Wha is he maximum value x max of he posiion? To answer his, we need o solve for

7 x() We inegrae our expression for v() once, obaining x ( ) = F b m ä å v b F ë ì exp ä å b ã b í ã m ë ì + consan í Le s say ha x=x a = Subsiuing ino he above expression yields consan = x + m ä å v b F ë ì ã b í Hence, he full soluion for x() is x( ) = x + F b + m b ä å v F ë ì ä å 1 exp ä å b ã b í ã ã m ë ì ë ì í í To answer our quesion, we subsiue max ino our expression for x(), obaining x max = x + mv b + Fm ln ä å bv 1 ë ì b 2 ã F í Here is a plo of x() showing he effec of resisance; a plo wih he same iniial condiions bu no resisance is shown in red for comparison We see ha he curve wih resisance deviaes noiceably from he ideal parabolic form x() slope = v wih resisance no resisance Here is a MAPLE inpu line which solves he presen differenial equaion symbolically: soln:=op(2,dsolve({m*diff(x(),$2)=fb*diff(x(),),x()=x[],d(x)()=v[]}, x())); The nex line plos a soluion: plo(subs({m=1,f=-1,b=1,x[]=,v[]=1}, soln),=2); (Here s how o copy hese lines from he presen documen and pase hem ino MAPLE) You can easily change he values of he parameers before execuing If you wish, you can also plo he velociy by changing soln o diff(soln,) in he above line Checking our answers: In he nex secion, we will consider a couple of ways o check he answers we obained in his

8 secion In paricular, we will see how o make sure ha he unis are correc (dimensional analysis), and also how o make sure ha he case of zero damping is recovered when we se he damping o zero in he more general case