(3.2.3) r x x x y y y. 2. Average Velocity and Instantaneous Velocity 2 1, (3.2.2)

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1 Lecture 3- Kinemtics in Two Dimensions Durin our preious discussions we he been tlkin bout objects moin lon the striht line. In relity, howeer, it rrely hppens when somethin moes lon the striht pth. For instnce, if we re tlkin bout trffic, only few hihwys re relly striht. If we consider motion of the bll durin footbll me, we cn lmost neer tret it s n object moin lon the striht line. This is why we he to study more complicted cse of two dimensionl motion, like cr moin on cured hihwy or the bll s motion durin ny sport me. 1. Position nd Displcement Think wht we need to chne in our pproch in order to be ble to describe position of the object in two or three dimensionl cse. To describe position of the object in two-dimensionl spce, we he to introduce coordinte system, which now hs not only one is, but two es nd y. It hs lredy been mentioned tht the choice of the system ery much depends on the problem. You he to pick the oriin nd directions of the es in order to obtin the simplest possible description of the problem. Position of the object cn now be defined by mens of the position ector (rdius ector) r which connects the oriin of the coordinte system with the point where the object is locted. As ny other ector the position ector cn be resoled into components r ˆ yyˆ (3..1) The coefficients, nd y re lso known s the object's rectnulr coordintes. For some problems with specific symmetry it is more conenient to use not rectnulr but sphericl, cylindricl or some other coordintes. As the object moes, its position is chnin from the initil position r 1 to the finl position r. We shll define displcement ector, r, s the chne of position, in sme wy s we did in one dimension, so r r r 1, (3..) 1 1 ˆ ˆ, r ˆ y yˆ ˆ y yˆ, r y y y 1 1 r ˆ yyˆ. (3..3). Aere Velocity nd Instntneous Velocity

2 Our second step in description of one-dimensionl motion, ws the nswer to the question: How fst the object is moin? In sme wy, now we he to discuss this question for the motion in seerl dimensions. We still my use the sme definition of the ere elocity displcement r time interl t. (3..4) Let us notice tht displcement nd elocity re ectors. Displcement only depends on the oriinl nd finl positions of the object but not on the ctul pth of the object from the oriinl position to its finl position. Aere elocity lso depends on the choice of the time interl for which it is tken. Followin eqution 3..3 we cn rewrite eqution for the ere elocity s ˆ yyˆ y ˆ ˆ y. (3..5) t t t Think bout rphicl interprettion of ere elocity in three-dimensionl cse. The instntneous elocity t some moment in time is defined s r lim t t. (3..6) This mens tht instntneous elocity hs tnentil direction to the object's pth t eery point of its trjectory. So we he y lim ˆ yyˆ lim ˆ lim yˆ ˆ ˆ yy, (3..7) t t t t t t where elocity ector components re instntneous direction of trel. 3. Aere Accelertion nd Instntneous Accelertion lim t t, y y lim. This elocity ector shows us the t t Howeer, most probbly the object under considertion is not moin with constnt elocity. So, we lso he to tke into ccount how fst its elocity is chnin. This mens tht we he to introduce ccelertion s we did in the one-dimensionl cse. If the elocity is chnin s the object trels from point 1 to point durin the time interl t, one cn define the ere ccelertion s chne in elocity time interl 1 t t. (3..8) Accelertion is lso ector nd ere ccelertion depends on the time interl for which it is tken. If this time interl shrinks to zero, we he instntneous ccelertion lim t t (3..9)

3 Think wht is the min difference between ccelertion in -dimensionl cse compred to1- dimensionl cse. Since elocity nd ccelertion re ectors, the object hs ccelertion not only if it chnes speed (mnitude of elocity), but lso if it only chnes the direction of motion. One cn lso rewrite eqution 3..9 s lim ˆ ˆ lim ˆ lim y ˆ ˆ ˆ y y y y y. (3..1) t t t t t t 4. Two-dimensionl motion with constnt ccelertion We he lredy studied one-dimensionl motion with constnt ccelertion. All of the equtions which we he obtined in tht cse re still lid but he to be etended into dimensions, so const, const, y const. t, t, t. y y y 1 r r t t, 1 t t, 1 y y yt yt. 5. Projectile Motion (3..11) (3..1) (3..13) As we he lredy discussed, ll the objects he the sme bsolute lue of ccelertion, 9. 8m s, when they re fllin down ner the erth s surfce nd this ccelertion is pointin downwrds to the erth's surfce. To simplify the problem we shll inore ir resistnce, which is quite ccurte pproimtion s fr s speed of the object is not too lre nd there is no wind. We shll only consider the problem in two dimensions. So, we will only study the object which moes in erticl plne. We choose the -is to be horizontl is in the sme plne s the object moes nd is y to be erticl is in the sme plne with positie direction oin upwrds. The object hs n oriinl elocity,, in the sme plne (see the picture). Such n object is clled projectile nd its motion projectile motion.

4 We cn resoled the oriinl elocity into two components which is ˆ yˆ, y cos, sin. y (3..14) Here,, is the nle between the direction of is nd the oriinl elocity. We cn then resole the ccelertion ector, but it is een esier. The only component tht ccelertion hs is erticl component y,, (3..15) yˆ Since rittionl ccelertion is constnt we cn consider this problem usin the set of equtions It cn be reduced to two one-dimensionl problems, becuse for projectile erticl motion nd horizontl motion re completely independent. This fct cn be confirmed eperimentlly by wtchin two blls, one of which is fllin striht down, nother is shot horizontlly by sprin. Their motion in erticl direction will be ectly the sme. They will coer the sme distnces durin the sme time interls nd it tkes the sme time for both of them to rech the round. For horizontl motion we he, cos const, t cos t. (3..16)

5 For erticl motion, the similr set of equtions ies const, y t sin t, y y y t y t y y yt y sin t. (3..17) The lst set of equtions illustrtes behior of the erticl component of elocity depicted in my picture. At first it is equl y then it ets smller due to decelertion of the object by rittionl field. At the hihest point of the trjectory the erticl component of elocity becomes zero, so it only moes horizontlly, then it is ccelerted in the downwrd direction by the sme rittionl field. Finlly it reches the round with the sme speed s it hd oriinlly, nd it mkes the sme nle below the horizon s the oriinl nle boe the horizon ws. The trjectory my look like complicted cure. Howeer, it cn be shown by elimintin time from equtions nd 3..17, tht it is just prbol (for simplicity, we set ytn, (3..18) ( cos ), y ). Eercise: Proe eqution Eercise: How eqution will look like if nd y? In the picture you cn lso see the horizontl distnce, R, which projectile hs treled before it returns to its initil lunchin leel. Let us find this distnce, clled the horizontl rne of the projectile. For the coordintes t the finl point one hs R, y y. With ccount of equtions 3..16, it becomes R cos t, t sin t. Elimintin time from these two equtions one hs R sin cos sin. (3..19) Eercise: Proe eqution 3..19, by elimintin time from the preious eqution. This lst eqution cn help us to nswer the interestin question: At which nle one needs to shoot from the un, in order to et the lrest horizontl rne R? If R hs the lrest lue for the

6 ien oriinl speed of projectile, this mens tht sine of the nle in eqution 3..19, should rech its mimum lue, which is 1, so we he sin 1, 9 o o 45., This mens tht horizontl rne will rech its mimum, if you lunch projectile t 45 derees nle nd it is equl R m o sin 9. Emple At wht nle reltie to the horizon one hs to shoot projectile in order to he the projectile's heiht t the hihest point equl projectile's horizontl rne? We just proed tht projectile's horizontl rne is R sin cos. Now let us see wht projectile's heiht is. Since t the hihest point of the trjectory the erticl component of elocity is zero, we he ( y y ), y y y where, sin, nd y y H, so y y y (sin ) H, H (sin ), (sin ) H. In this problem we he H=R, so (sin ) sin cos, sin 4, cos tn 4, 1 o tn (4) 76. sin cos,