Thomas Whitham Sith om Mechanics in Mathematics Unit M Rectilinea Motion Dynamics of a paticle Pojectiles Vectos Cicula motion
. Rectilinea Motion omation and solution of simple diffeential equations in which elocity, o acceleation is gien as a function of time. t=0 a d d a Eample A paticle leaes O with a speed of m/s in the diection of 0 and at any time t s late its acceleation is t m/s. Show that subsequent to stat of motion, the paticle changes diection twice and find the distance fom O whee it occus. t t=0 =m/s d Acceleation t t t c t 0, 0 0 c t t d Velocity t t a t t t D
t 0, 0 D 0 t t t Possible diection change when 0 i.e. t t 0 t 0 t i.e. afte second and afte seconds. when t, when t, a m/s acceleates towads O fom est. Change of diection at m a m/s acceleates away fom O fom est. Change of diection at 0m. Dynamics of a paticle o Wok is done by a foce when it moes its point of application. o a constant foce, a few eamples follow: (i) (iii) WD.cos WD (ii) WD 0
(i) WD NBthis might be epessed by saying that wok is done against Eample A hoizontal foce moes a paticle of mass kg in a staight line fom est to est acoss a hoizontal suface with 7 though 0m. ind the wok done in moing the paticle, in the absence of any ai esistance. N P g 0m Let P be the applied foce. Although it won t be constant, at any time, all it has to do is to oecome, the fictional foce. WD by P = WD against = 0 Vetical Sliding N g N WD 08. 68J 7 g 8. o a foce which is aiable in magnitude in the diection in which it is applied N 0 A B
WD by in the displacement AB = d whee is epessed as a function of. Eample ind the wok done in stetching an elastic sting of modulus and natual length a to an etension e T NL= a Hee (again) the wok done by the applied foce will be equal to the wok done in oecoming the Tension T in the sting. HL o o o T a WD = = e e d 0 a a 0 a Enegy is that possessed by a body giing it capacity fo wok. Kinetic enegy is that possessed by a body in itue of its motion and is gien by the amount of wok it can do in being bought to est. o a paticle of mass m moing with speed, its KE = e m Positional Potential enegy of a body is that which it possesses in itue of its position and is gien by the amount of wok it can do in moing to some base position h This paticle of mass m has PPE = mgh Base position (PPE = 0)
o o 5 Stoed Potential enegy in a stetched sting o sping is that which it possesses in its capacity to do wok in being estoed to its natual length. (fo a sping it also applies to compession). o a sping, WD in estoation to Natual length is equal to the wok done in stetching it fom Natual length to the gien etension. etension natual length SPE (fom fomula aboe) Change in enegy ist note that if a paticle of mass m falls though a height h, the wok done by gaity = mgh = loss in PPE. (egadless of whatee othe foces ae acting). Similaly fo a paticle of mass m aised though a height h the wok done against gaity = mgh = gain in PPE A wok enegy equation can be witten down in one of the following ways, eithe (i) o (ii) (i) Oeall gain in enegy = wok done by foces poducing it. Oeall loss in enegy = wok done against esistances Includes PPE doesn t include weight (ii) Gain in Kinetic enegy = wok done by all foces poducing it. Loss in Kinetic enegy = wok done against all esistances. Includes weight
Eample 0 6 A 5kg pacel is placed at the top of a chute which is 0cm long in the fom of a spial with a fall of 0m fom top to bottom. The aeage fictional esistance to the motion of the pacel is N thoughout. ind the speed of the pacel when it aies at the bottom of the chute. Due to the fictional foce thee will be an oeall loss in enegy = 0 60J 0m Loss in PPE = 5 9.8 0 980J Gain in KE = 5 5 J o Wok/Enegy equation 980 60 5, 8, 8 The altenatie is along the lines Gain in PPE = WD by gaity wok done 5 against esistance 59.8 0 0 Conseation of enegy Mechanical enegy is conseed fo a system of foces which ae conseatie. These ae foces which don t inole conesion into othe foms of enegy (heat, sound, light fo eample). Weights, nomal eactions at smooth contacts, tensions in stings ae eamples of conseatie foces. iction, and impulsie foces ae not conseatie foces, a wok enegy equation can be witten down (i) as in the last section o (ii) by consideing change fom one fom of mechanical enegy to anothe o m/s
7 (iii) by equating total enegy (PPE+SPE+KE) in one position to that of anothe, Eample A light elastic sting of length l and modulus mg is attached at one end to a fied point A. A paticle of mass m is attached to the othe end of the sting. (i) If the paticle hangs in equilibium calculate the etended length of the sting. (ii) If the paticle is held at A and is allowed to fall fom est find, assuming no ai esistance, a) the maimum speed of the paticle in ensuing motion b) the length of the sting when the paticle is instantaneously at est. (i) In equilibium position Vetical T mg l l HL T mge l T e e l Equilibium position ma etended length = 5 l mg (ii) a) Ma speed occus when it passes though the equilibium position, since at that instant, the acceleation will be zeo.
8 An enegy equation will be along the lines of Gain in KE + Gain in SPE = Loss in PPE Enegy m mg. l l 5 mg. l gl 5 9 gl gl ma gl b) In falling fom est at A to instantaneous est, the loss in PPE will be equal to the gain in SPE. l So, if the maimum etension is C say Enegy mgl C C.mg l C l C l lc l 0 C C lc l 0 C l At instantaneous est C length of sting in this position = l o Powe is a measue of wok ate. As an engine woks at a paticula ate we efe to this as the powe of the engine. Eample A pump ejects 000kg of wate pe minute fom its souce with a speed of m/s. ind the powe which the pump will need to deelop if it is 80% efficient.
000 kg 00kg 60 9 of wate is delieed pe second at a speed of m/s KE gien to wate pe second = 80% of equied powe =.6 kw 006 600J 00 80 equied powe =.6 kw If the pump was.5m aboe the souce of the wate then o Eample PE gien to wate pe second = 00 9.8.5 900J Total enegy gien to wate pe second = 600 + 900 = 6500J 80% of equied powe = 6.5 kw 00 80 equied powe = 6.5 8.5kW Vehicles in motion The foce which popels a ca (fo eample) fowad is supplied by the engine. If at a gien instant this foce is Newtons when the ca is moing at m/s, the powe delieed by the engine will be gien by H Whee H watts It is essential that you emembe that, in the powe equation H, is the foce supplied by the engine and in the equation of motion (N) which is acceleation. ma, is the esultant foce poducing the The engine of a ca is woking at a steady ate of 0kw. The ca, of mass 00kg is being dien along a hoizontal oad against constant esistance to motion of 6N. ind (a) the acceleation of the ca when its speed is 8 m/s (b) the maimum speed of the ca.
a (a) Powe 0000 8 8m/s 6 50 N N 6 00a a 0. 77 m/s 0 (b) Constant speed P 6. Pojectiles 6 ma 0. ma 7 m/s Powe 0000 6 The assumption will be that motion is fee unde gaity i.e. thee will be no esistie foces, and a paticle model will apply. You will be epected to deie epessions fo ange, time of flight, maimum height, equation of path, etc.. Quotation of fomulae will gain no cedit. Eample of with O. y P A paticle is pojected fom the oigin with speed u at an angle u sin u u cos ma Always do this fist. P(,y) u sin 0 u cos
OP Hoizontal s ut (the only equation fo hoizontal motion) Vetical ucos. t y usin. t gt * Eliminating t between the equations establishes the equation of the y tajectoy fo a gien elocity of pojection tan g sec u ; ** being in quadatic fom the paabolic natue of the tajectoy follows, along with symmeties. Putting and 0 u sin though O u sin T g y in ** gies the ange on a hoizontal plane R Now 0 g Maimum height aboe plane can be found using u sin H g u sin om R, g R maimum ange is quotable}. u g y in * gies the time of flight 5 u as ; ma fo. {the angle 5 fo The diection of motion fo any gien position can be found fom ** by diffeentiation. The diection of motion at a gien time t can be found fom elocity components. usin gt u cos usin gt tan u cos
The etical component will be gien by u at To find the angle of pojection fo possible accessibility to a gien point (a, b) a quadatic equation in tan can be found fom * ga b a tan tan tan sec Two eal oots u two alues of, one lowe and one highe tajectoy Equal oots point only just accessible No eal oots point is not accessible.. Vectos The basic pinciples undelying the use of ectos, in application to mechanics, can be found in Pue. (i) o a paticle moing in D o D space, elatie to an oigin O. O = position ecto at time t. = elocity at time t, and into this is built the diection of motion at any instant. a = acceleation. Diffeentiation will be by component, as will integation. d a d a d Eample At time t a paticle has position ecto ti sin t j k cos. Show that it will always be moing in a diection pependicula to its position ecto. cos ti sin t j k
d sin ti cos t j cos t sin t. sin t. cos t 0 sin t cos t sin t cos t 0 Eample A paticle moes with constant acceleation a i. ind the elocity and position ectos at time t gien that its initial elocity and position ectos wee d i ti c when t 0, j k 0 j k, 0 k 0 c j k ti j k d when t 0, k ti j k t i t j tk d 0 d k t i t j t k
t=0 (ii) 0 motion with constant elocity will be linea P The diagam shows the motion of a t paticle P with constant elocity. The elocity is the diection ecto and 0 t 0 o Collisions and closest appoach poblems Let A and B be two paticles with position ectos A and B at time t, each moing with constant elocity. The displacement of B fom A will be gien by AB. A collision will occu if thee is a alue of t fo which AB 0 AB. If not, then the closest appoach can be found by minimising A A B Eample At 0 j t, paticles A and B hae position ectos i j i and elocities 6i j and 8i 5 j espectiely 5 and (with usual SI units) Show that the paticles will collide if they maintain thei initial elocities. 5i j t6i j i j t8i j A B 5 B
5 AB i j ti t j t i t j B A Clealy, when t AB 0, hence collision afte seconds. Eample At 0 i j and i j espectiely. The elocities of A and B ae t, paticles A and B hae position ectos i j and i j of closest appoach. A espectiely. ind the time and distance i j t i j i j ti j B 5i 5 j ti j 5 ti 5 tj AB B A AB 5 t 5 t 5t 5 0t 50 5 t 5 5 5 0t t 5 0t t t t 50 5t t 50 5 Time of closest appoach is t, when AB 5 ds s and put 0, {fo calculus appoach, let 5t 0t 50 etc}. (iii) Motion in a moing medium eg a boat in a flowing ie, an aicaft in an ai cuent. Hee the still wate o still ai elocity will be compounded with that of the cuent to gie a esultant elocity. A ecto appoach might be equied, although it isn t othewise necessay.
Eample 6 An aicaft has to fly a distance of 00km due east in a wind fom the noth west blowing at 0 km/h. The aicaft opeates with a still ai speed of 500km/h. ind the couse which the pilot should set and the time fo the jouney. Take units i, j in the diections due east and due noth. j stat 500 cos 500sin 500cos 0 500cos 0 500sin 0 500sin 0 but to tael due east, 500sin 0 0 couse to be set at 085. appo. then 58.97... 0 i 0 sin destination 0 500 5 time taken t = 00 58.97.. hous. min s Eample A ie with paallel banks flows at.5m/s. A stong swimme has a still wate speed of.8m/s. He entes the wate at a point on one bank and swims so that he is always facing the opposite bank. ind how long it takes him to coss if the ie is 5m wide, and find how fa down steam he is caied. ind also the time it would take him to swim staight acoss.
7 j 5m.5i i.8j.5i. 8 j.5ti. 8t j c 0.5ti. 8t j c Reaches the othe side when j ] 5.8t 5 t 5s Distance caied downsteam i].5 5 7. 5m u.5i aj a j u. 5i u. 5i a j Now u. 8.5 a. 8 (i) Eample Paticle Statics Time taken to coss =.5 a. a 5 0.99 0.99 a 0.99 5.s ind the esultant of a foce of 5N acting at O in the diection of i j k and a foce of N acting at O in the diection 6i j k.
8 k j i 0 5 0 5. 5 k j i 9 6 8 6. 9 6 6 N 8 k j i R o Components, esolutes, esoled pats of ectos The component o esolute of in the diection of a is gien by cos which is equal to a ˆ. The (ecto) esoled pat of in the diection of a is theefoe a a ˆ ˆ.. The (ecto) esoled pat of in the diection pependicula to a is gien by a a ˆ.ˆ. Eample Resole the foce k j i 5 into ecto components in the diections of k j i a and pependicula to a 6 6 0. 5 ˆ. a a
Components ae 9 6 i j k 5 i j k i j k i k and o Equilibium of foces acting on a paticle 0 () Wok, Powe, Enegy Wok done by the constant foce in the displacement is gien by. The powe H due to a foce at the instant when its point of application is moing with elocity is gien by H. The Kinetic enegy of mass m at the instant when its elocity is is gien by K m. {o (i) m of couse} Newton s Second Law If a paticle of mass m is subject to a esultant foce poducing in it an acceleation a then ma. Eample The position ecto at time t of a paticle P of mass m is gien by t sin ti t cos tj ind the elocity and the acceleation a of the paticle at time t and show that the acceleation is of constant magnitude.
d ind also (i) 0 the foce acting on the paticle at time t, and its powe (ii) the Kinetic enegy oe the inteal t cos ti sin tj d a sin ti cos t j sin t cos t (i) ma msin ti cos t j a QED. H sin t cos t. m. msin t sin t cos t cos t sin t cos t cos t sin t m sin t cos t m cos t sin t (ii) KE = m m cos t sin t Gain in KE in the inteal 5. Cicula Motion (no ai esistance) (i) t = m.5 m = m m m o a paticle moing in a hoizontal cicle with constant angula speed its linea speed at any instant is gien by the adius of the cicle. a, a being Thee will be an acceleation diected towads the cente of the cicle gien by a o a Poblem soling esole etically (the only diection fo esolution)
equate to ma o m a the cental foce poide. Eample component of the tension. The conical pendulum the cental foce is poided by the Vetical T cos mg Cental foce: N T sin m T Also l sin Cente of cicle mg Eample Coneing ca (model ca as a paticle) moing with speed. The cental foce is poided by fiction. N mg 0 m Vetical Cental oce N iction N The inequality hee gies limiting speed if sliding is not to occu. N mg
Eample Banking N Banked tack o cone O mg The diagam shows an optimum speed situation in which no fiction is equied; the hoizontal component of the nomal eaction is the cental foce poide. Vetical N cos mg Cental foce N N sin m equied O N mg Too fast! iction now Vetical N cos mg sin Cental foce N iction N m cos N sin
(ii) o a paticle moing in a etical cicle the speed will be aiable and thee ae two acceleation components to conside Radial component of acceleation a a d a a t Tangential component of acceleation a t d d a Poblem soling No questions will be asked egading a t Enegy equations usually fom initial to a geneal position Cental foce N in the geneal position Eample A commonly occuing poblem is fo a paticle to be pojected hoizontally, with speed U say and to stat to moe in a etical cicle. The geneal position shows it moing with speed when the angula displacement is a cos Rad a m u mg mg a acos mu Enegy
u ga cos * Cental oce m mg cos a m mg cos u ga cos a mu mgcos ** a (a) if the paticle is theaded on a wie { = eaction due to wie} o is inside a tube { = eaction due to wall of tube} the motion is constained to a cicula path. The condition then fo complete cicles will be that 0 when ; thus you will be able to show leads to u ga. Note that the eaction could anish at some point in the top half of the cicle and subsequently act outwads. o eample if be able to show that 0 when cos. u ga then you will (b) If the paticle is at the end of a light inetensible sting { = tension in sting} o inside a smooth sphee { = eaction due to sphee} then complete cicles ae possible only if 0 when. You will be able to show then that u 5ga fo complete cicles.