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MAT1112_ch11_v2007

Course: MAT 2070, Spring 2011
School: Université du Québec à...
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11 Applications CHAPITRE de lintgrale multiple. e Ce chapitre sera tr`s bref. Il existe un grand nombre dapplications de lintgrale multiple. Il sut de e e penser aux notions desprance et de variance en probabilits ou encore des quations intgrales. Beaucoup e e e e de ces applications seront discutes dans dautres cours. Ici nous nnumrerons que quelques-unes, surtout e e e relies ` la physique. Plusieurs quantits...

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11 Applications CHAPITRE de lintgrale multiple. e Ce chapitre sera tr`s bref. Il existe un grand nombre dapplications de lintgrale multiple. Il sut de e e penser aux notions desprance et de variance en probabilits ou encore des quations intgrales. Beaucoup e e e e de ces applications seront discutes dans dautres cours. Ici nous nnumrerons que quelques-unes, surtout e e e relies ` la physique. Plusieurs quantits physiques peuvent tre exprimes comme des intgrales multiples. ea e e e e De tels expressions sont fondes sur la dnition de lintgrale comme la limite dune somme. e e e Si une quantit de mati`re est contenue dans une rgion R de R3 et (x, y, z ) est la densit par unit e e e e e e e de volume au point (x, y, z ), alors le centre de masse (x, y, z ) de celle-ci est dni au moyen dintgrales. En a physique, pour un syst`me de n particules, alors la composante x de ce centre par rapport ` laxe des x est e dnie par lquation e e m1 x1 + m2 x2 + . . . + mn xn m1 + m2 + . . . + mn dans laquelle mi est la masse et xi est la coordonne par rapport ` laxe des x de la position de la i-i`me e a e e c e e e u particule. On dnit y et z de la mme faon. Si la quantit de mati`re est distribue continment dans e la rgion R et (x, y, z ) est la densit au point (x, y, z ), alors nous sommes amens ` dnir son centre de e e eae masse (x, y, z ) par x= x (x, y, z ) dx dy dz , (x, y, z ) dx dy dz R R y (x, y, z ) dx dy dz (x, y, z ) dx dy dz R R y= et z = z (x, y, z ) dxdydz . (x, y, z ) dx dy dz R R Il est aussi possible dans la situation prcdente de dcrire le moment dinertie par rapport ` un axe. ee e a Nous nous limiterons ` dcrire ce moment par rapport ` laxe des z . En physique, le moment dinertie dun ae a syst`me de n particules par rapport ` un axe de rotation est dni par lquation e a e e 2 2 2 m1 r1 + m2 r2 + . . . + mn rn dans laquelle mi est la masse et ri est la distance laxe ` donn de la i-i`me particule. Si la quantit de a e e e mati`re est distribue continment dans la rgion R et (x, y, z ) est la densit au point (x, y, z ), alors le e e u e e moment dinertie I par rapport ` laxe des z comme axe de rotation sera a (x2 + y 2 ) (x, y, z ) dx dy dz. I= R Ce moment dinertie permet de dcrire lnergie cintique dun corps rigide qui tourne autour dun axe avec e e e une vitesse angulaire comme I 2 /2. Il y a une version 2-dimensionnelle du moment dinertie. Si R est une rgion de R2 et (x, y ) est la densit e e par unit daire au point (x, y ). Alors son moment dinertie par rapport ` laxe des x est R y 2 (x, y ) dx dy . e a Exercice 11.1: Soient la rgion R ` lintrieur du ttra`dre dans R3 dont les sommets sont (0, 0, 0), (1, 0, 0), (0, 2, 0) et e a e ee (0, 0, 1) et la fonction de densit (x, y, z ) = x + y + z . Dterminer le centre de masse de cette rgion. e e e 97 Exercice 11.2: Dterminer le moment dinertie par rapport ` laxe des z de la rgion R constitue des points de R3 tels que e a e e z 0, (x2 + y 2 + z 2 ) 1 et 3z 2 (x2 + y 2 ). Exercice 11.3: a) Dterminer le moment dinertie par rapport ` laxe des z de la rgion R constitue des points de R3 tels e a e e que 0 z H , (x2 + y 2 ) R2 . b) Dterminer le moment dinertie par rapport ` laxe des z de la rgion R constitue des points de R3 tels e a e e que H/2 y H/2, (x2 + z 2 ) R2 . Exercice 11.4: Soient a et b, deux nombres rels tels que 0 < a < b. Considrons le tore T engendr par la rotation du e e e cercle dquation (y b)2 + z 2 = a2 autour de laxe des z . Calculer la masse T (x, y, z ) dx dy dz de ce tore e en sachant que la densit est constante sur les cercles contenus dans des plans parall`les au plan des x, y et e e dont les centres sont situs sur laxe des z et quen un point A, elle est proportionnelle ` la distance de A au e a centre du mridien qui porte A. e 98
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Université du Québec à Montréal - MAT - 2070
CHAPITRE 12Intgrales impropres, fonctions gamma et bta et transforme de Laplace.eeeDans ce chapitre, nous revenons aux intgrales simples, mais cette fois soit lintervalle dintgration, soiteela fonction ` intgrer, soit les deux ne sont pas borns. T
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Kettering - MECH - 115
00 introductionmy expectationsyour expectationsgradingon-line quizzeswhat I expect from you: active learningbefore classread chapter before we beginwarm-up quiz (on-line) on reading*during classthink/pair/share quizzes*be prepared with question
Kettering - MECH - 115
01 Motiondiagramsuse movie to studymotionfixed time betweenframesdon't pan the camera!usually 1/30th of a secstrobe pictures arefasterfaster objectshave greater positionchange between framesWhich car is going faster, A or B? Assume there are
Kettering - MECH - 115
02 Velocity &amp;accelerationrv(average) velocity: tvector displacement rtime interval tdraw the velocity vector vvthe same size as the displacement vectordon't forget the difference in meaning, unitsvelocity:how fast it's goingAND in what dir
Kettering - MECH - 115
03 Motion in1 dimensioncoordinate systemoriginpositive direction: conventionx: right y: upposition-time graphsposition on vertical axis(dependent variable)even if position ishorizontaltime on horizontal axis(independent variable)can pick off
Kettering - MECH - 115
04 position, velocity and accelerationif v is constant s=v s tdisplacementchange in positionarea under v(t)workbook:2.2b page 2-3displacement when v isn't constanttfs f = si v s t dttiintegral = areaunder curvebetween curve andaxisintegrat
Kettering - MECH - 115
05 problem solving: constant amodel: particle modelvisualize: shift between these as neededmotion diagrampictorial representation: labelv fs = v is a s tsolve12s f = si v is t a s t 2assess22v fs = v is 2 a s s units? sensible? believable?
Kettering - MECH - 115
06 Instantaneousaccelerationa s=d vsdt v s = a s dtfollow up:workbook 2.28page 2-16Rank in order, from largest to smallest,the accelerations aA aC at points A C.A) aA &gt; aB &gt; aCB) aC &gt; aA &gt; aBC) aC &gt; aB &gt; aAD) aB &gt; aA &gt; aCWhich velocity-vers
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07 Vectors and componentsvector: magnitude &amp; directionscalar: magnitude onlyreview:follow up:Workbook 3.1-3Page 3-1Which figure showsA1 A 2 ?A1 A 2 A 3 ?multiplying a vector by a scalar. by a positive scalar. by a negative scalarfollow up: 3.
Kettering - MECH - 115
08 Identifying forcesA force is. a push or a pullI push/pull on your handdo I exert a force? do you?does it matter if our hands move?I push/pull on a doorknobdo I exert a force? does the door?does it matter if the door moves?does a spring exert a
Kettering - MECH - 115
stnd09 Newton's 1 and 2 lawsIssues:Is a force needed to produce uniform motion?What happens to motion if a constant force is applied?DemonstrationndNewton's 2 lawFnet1a= Fi=mmiobservations from experiment:more mass lesser accelerationmo
Kettering - MECH - 115
10 Applying Newton's LawsThe basic equations areNewton's 1st law0F = F i =netior Newton's 2nd lawF = F i = m anetiEquation hunting won't solve these problemsNeeded: Problem solving strategyapply the same patternvary the approach to match th
Kettering - MECH - 115
11 Mass and weightmass:a F i=m iweight: force of gravityw=m gsame mass m in both formulasdirection: downwardscoincidence? General Relativity?apparent weight: normal force of scale on feettry bouncing on a scale!Inclined plane#1#2#3#1: (A)
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12 frictioncontact forces at an interface between surfacesnormal force acts perpendicular to interfacefriction force acts parallel to interfacefriction equationsMODEL observed behaviorfriction caused bysurface bondingtypes of frictionstatic: no m
Kettering - MECH - 115
13motion in 2drva , , are vectors, with componentstrajectory (path) graph: y(x)rv avg =trecall: vector subtraction infollow up: workbook 6.2 (page 6-1)instantaneousvelocity &amp;accelerationdrv=dtdva=dtv is tangent to trajectorycompo
Kettering - MECH - 115
14 projectile motionno horizontal acceleration: constant horizontal velocitydemo: knock 2 balls off a table (different v)mga==g=mmacceleration is vertically downwardsF netlisten for the impacts. does one hit first?follow ups:workbook probl
Kettering - MECH - 115
15 UniformCircular Motionangular position qpositive =counterclockwise (ccw)from +x axisalways measured in radians2p rad = 1 rev = 360arc length s = r which looks bigger?a dime at arm's length s 2 cm, r 0.8 mthe moon s = 3.5 x 106 m, r = 3.8 x 1
Kettering - MECH - 115
n16 Circular dynamicsTdemo: object moves on string in horizontal circle wnet force is towards center, constantthis produces constant acceleration towards centerthis results in uniform circular motiontension can also create linear acceleration!circ
Kettering - MECH - 115
17 Non-uniform circular motiondemo: loop-the-loopat the bottom: FBDnormal force is upweight is downnet force is towards centernwat the top: FBDnormal force is downnet force is towards centernweight is downwvcritical issues2mat the bottom
Kettering - MECH - 115
18 Newton'srd3 lawobjects exert forces on each otherroles of object/agent switchinteraction pairs never act on the same object!problems now have several systemsforces includeexternal forcesinteraction forcesdifferent FBDs: draw dotted line to co
Kettering - MECH - 115
19 Ropes &amp; Pulleystension holds a rope togetherend of rope exerts this force onobject it's touchingacceleration constraint:cutting the rope severs bonds,removes forces: ends fly apartmagnitude of accelerations ofanything tied to the same rope ist
Kettering - MECH - 115
20 Momentum and ImpulseConservation laws common: e.g. mass:pour 100 g of vinegar into 100 g of oil.what's the mass of the vinaigrette?Many other physical quantities often conservedmomentumangular momentumenergychargemomentumpv m vector: direc
Kettering - MECH - 115
21 Conservationof Momentum#1 = mosquito#2 = truckduring the collision, whichhas the greater magnitude.force?acceleration?impulse?change in momentum?change in velocity?(A) 1(B) 2(C) samesuppose:collision lasts tm1 &lt; m2|(vix)1| = |(vix)2|c
Kettering - MECH - 115
22 Kinetic and Potential EnergyTwo types of energy (many variants)Kinetic (involves motion)translational kinetic energyfollow up: WB 10.6,7 page 10-2potential energy (involves position)gravitational potential energy12K= mv2U g=m g yfollow up:
Kettering - MECH - 115
23 SpringsHooke's (force) law: F sp s =k s = k s s e each end exerts this force!spring constant kstiffness: demoForce : stretch ratiomass mForce : acceleration ratiofollow up:WB 10.15 (p. 10-5)Spring FBDsdraw a FBD for the object on each endA
Kettering - MECH - 115
24 elastic collisionsall collisions conserve pm1 v fx 1 m 2 v fx 2 = m1 v ix 1perfectly elastic collisions also conserve K111222m1 v fx 1 m 2 v fx 2 = m1 v ix 1222tedious algebra (p 287-8) yields v fx 1 = v fx 2=m1 m 2m1 m 22 m1m1 m 2
Kettering - MECH - 115
25 WorksfW = F s dssiWork force * distancenot generallyOnly the component of theforce in the direction of thedisplacement does workIF the force is constantW = F s cos Only if also q = 0 willW =F sW and DK: demoswork is an energy transfercar
Kettering - MECH - 115
26 variableforcesW = F s dssiWork force * distancesfnot generallyIF the force and the displacementare parallelsfW = F s dssiOnly if the force is also constantwillW =F sA 1 kg particle starts at x = 0 mwith vx = + 2 m/s.Which is true?(A)
Kettering - MECH - 115
27 Conservation of EnergyK f U f E th = K i U i W extscenarios:what's the system?(A) + (B) - (C) 0push a block across atable at constant speedDK=?DU=?a block slides to a haltWdiss = ?pick up a block, andplace it on a ledgeWext = ?toss a bloc
Kettering - MECH - 115
28 Rotationalkinematicsangular velocityddt = dtsign conventionCCW = +CW = -linear velocitydsvtdt s = v t dtrelationshipv t = r28 Rotationalkinematicsangular accelerationddt = dtsign conventionif a, w havesame sign: speeding upoppo
Kettering - MECH - 115
29 Torque, Rotational Dynamicstorque: rotational analog of forcesign convention: CCW = +, CW = - =r F sin follow up:WB 13.8,9 (p13-3)two complementary viewstangential forcemoment arm =d F =r F tfollow up: WB 13.11 (p13-4)follow up: WB 13.10 (p
Kettering - MECH - 115
30 Rotational kinetic energyAnother variety of kinetic energyFollow up:WB 13.28 p 13-11WB 13.29 p 13-111K rot = I 22Ex:RMwhich is moving fastestSame M, m, R, hhwhen the block hits?mmodelDisc, wheel are rigid bodiesBlock is a particleStr
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