Lecture 3.1 Scalars and Vectors, Kinematics in Two and Three Dimensions

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1 1. Sclrs n Vectors Lecture 3.1 Sclrs n Vectors, Kinemtics in Two n Three Dimensions Phsics is quntittive science, where everthing cn be escribe in mthemticl terms. As soon s the sstem of units hs been estblishe, phsicl quntities cn be presente s numbers, reflecting the mgnitue of the quntit. However, mn phsicl quntities hve not just mgnitues but irections s well. Think bout emples of such phsicl quntities. Other phsicl quntities m onl hve the mgnitue but not irection. Think bout emples of those quntities. For instnce, if we re tlking bout motion of cr, it is not just moving, but it is moving in certin irection, such s to the North or to the West or in somewhere else. So, in orer to escribe its motion b mens of isplcement or velocit, we nee not onl the mgnitue of these quntities, but lso we hve to specif their irection. Such quntities re known s vectors. On the other hn this cr lso hs mss which onl hs the mgnitue but hs no irection. This tpe of quntit is clle sclr. Even though sclr hs no irection but it still hs sign. This m not be true for mss which, s fr s we know, is lws positive. But there re lot of other sclrs, such s chrge or temperture tht cn be negtive s well s positive. We shll represent vector quntit s letter with n rrow, for instnce A, n we will use the sme letter but without the rrow for the mgnitue of this vector A A. To represent vectors in the picture, we will lso use rrows with lengths proportionl to the mgnitues of these vectors. There re severl vectors shown in the picture below. Vector A hs mgnitue lrger thn vector B n it is longer in the picture.

2 Let vectors A n B show how cr moves from one plce to nother plce. If this cr first moves long vector A n then long vector B, the finl position of the cr is the sme s if it just goes long vector C. This is wh we will cll this vector C to be the sum of vectors A n B C A B As ou cn see, vector summtion is not the sme s lgebric summtion. The mgnitue of vector C oes not equl to the lgebric sum of the mgnitues of vectors A n B. This is becuse we re ing not just bsolute vlues but lso irections. The picture suggests prcticl w of how one cn two vectors geometricll. You hve to rw both vectors in the sme scle with til of the secon vector t the he of the first vector. Then the vector-sum of those two vectors is vector which etens from the til of the first vector to the he of the lst vector. Since vectors cn be shifte without chnging their bsolute vlue n irection, we cn them in n orer. This gives us the sme commuttive n ssocitive lws s in the cse of the lgebric ition A B B A, A B D A B D i i. (3.1.1) One cn lso introuce the A vector, which hs the sme mgnitue A but the opposite irection compre to vector A. This mens tht A A cn lso subtrct them. For instnce in the picture B C A. i 0, n now not onl we cn vectors but we. Vectors n their components. The geometricl metho of ing vectors, escribe bove, helps us to visulize the picture, but it is not too much of the use, if we wnt to fin the ect mgnitue of vector C n specif its irection. First of ll we nee to clrif, wht we men b irection. In sme w s we cn obtin vector's mgnitue b compring its length to stnr unit, we cn efine its irection b fining the ngle which it mkes with some stnr irection. The choice of the stnr irection, s well s the choice of units, is somewht rbitrr n epens on ech prticulr problem. Sometimes to choose the right stnr irections mens to solve the problem. For instnce, if ll the vectors in the problem re oriente long the sme irection, this irection will be the perfect choice for the

3 stnr irection. In generl, when we hve two vectors, s in the picture, the form plne surfce. So, this problem becomes two-imensionl problem. To solve it, we hve to choose two stnr irections on this surfce. Usull the re chosen to be perpeniculr to ech other n form the - coorinte sstem. In this coorinte sstem we choose i n j (other nottion n is lso in use) to be bsis (unit) vectors in the irection of es n. Vector i is imensionless vector of the unit length pointing in the positive -irection. Vector j is imensionless vector of the unit length pointing in the positive -irection. Then we cn resolve n vector into components in such w tht A A i A j. (3.1.) If is the ngle between is n irection of vector A then A A Acos, Asin. Since n re perpeniculr, the mgnitue of vector Pthgoren theorem, n its irection using trigonometr s (3.1.3) A cn be etermine bse on, A A A A tn. A Since we live in 3-imensionl spce, if we more thn vectors, in generl, the re not necessril in the sme - plne. This is wh the thir is z must be introuce perpeniculr to - plne. The unit vector in the positive irection of is z is usull clle k ( z nottion is lso in use). It is customr to use right-hne coorinte sstem, such s if ou plce our right hn roun z-es in w tht our fingers woul sweep vector i into vector j through the smller ngle between them, our outstretche thumb points in the irection of k. An vector cn then be resolve in 3 components. 3. Kinemtics in two n three imensions During our previous meetings we hve been tlking bout objects moving long the stright line. In relit, however, it rrel hppens when something moves long the stright pth. For instnce, if we re tlking bout trffic, onl few highws re rell stright. If we consier

4 motion of the bll uring footbll gme, we cn lmost never tret it s n object moving long the stright line. This is wh we hve to stu more complicte cses: two n three imensionl motion, like cr moving on curve highw or the bll s motion uring n sport gme. 4. Position n Displcement Think wht we nee to chnge in our pproch in orer to be ble to escribe position of the object in two or three imensionl cse. To escribe position of the object in three-imensionl spce, we hve to introuce coorinte sstem, which now hs not onl one is, but three es, n z. It will be the righthne coorinte sstem, s it ws escribe before in m lecture bout vectors. I hve lre mentione tht the choice of the sstem ver much epens on the problem. You hve to pick the origin n irections of the es in orer to obtin the simplest possible escription of the problem. Position of the object cn now be efine b mens of the position vector (rius vector) r which connects the origin of the coorinte sstem with the point where the object is locte. As n other vector the position vector cn be resolve into components r iˆ j ˆ zkˆ (3.1.4) The coefficients, n z re lso known s the object's rectngulr coorintes. For some problems with specific smmetr it is more convenient to use not rectngulr but sphericl, clinricl or some other coorintes. As the object moves, its position is chnging from the initil position r 1 to the finl position r. We shll efine isplcement vector, r, s chnge the of position, in sme w s we i in one imension, so r r r 1, (3.1.5) ˆ ˆ r iˆ ˆj z k iˆ ˆ j z k, r iˆ ˆ j z z kˆ, r iˆ j ˆ zkˆ. (3.1.6) 5. Averge Velocit n Instntneous Velocit Our secon step in escription of one-imensionl motion, ws the nswer to the question: How fst the object is moving? In sme w, now we hve to iscuss this question for the motion in severl imensions. We still m use the sme efinition of the verge velocit

5 isplcement r v vg time intervl t. (3.1.7) Let us notice tht isplcement n velocit re vectors. Displcement onl epens on the originl n finl positions of the object but not on the ctul pth of the object from the originl position to its finl position. Averge velocit lso epens on the choice of the time intervl for which it is tken. Following eqution we cn rewrite eqution for the verge velocit s iˆ j ˆ zkˆ ˆ ˆ z v ˆ vg i j k. (3.1.8) t t t t Think bout grphicl interprettion of verge velocit in three-imensionl cse. The instntneous velocit t some moment in time is efine s r v lim t 0 t r. (3.1.9) This mens tht instntneous velocit hs tngentil irection to the object's pth t ever point of its trjector. So we hve ˆ ˆ ˆ ˆ ˆ z v i j zk i j kˆ v ˆ ˆ ˆ i v j vzk, (3.1.10) where velocit vector components re v lim, v t 0 t This velocit vector shows us the instntneous irection of trvel. lim t 0 t n v z z z lim. t 0 t 6. Averge Accelertion n Instntneous Accelertion However, most probbl the object uner consiertion is not moving with constnt velocit. So, we lso hve to tke into ccount how fst its velocit is chnging. This mens tht we hve to introuce ccelertion s we i in the one-imensionl cse. If the velocit is chnging s the object trvels from point 1 to point uring the time intervl t, one cn efine the verge ccelertion s vg chnge in velocit time intervl v v1 v t t. (3.1.11) Accelertion is lso vector n verge ccelertion epens on the time intervl for which it is tken. If this time intervl shrinks to zero, we hve instntneous ccelertion v v F r r t t H G I lim K J 0. (3.1.1)

6 Think wht is the min ifference between ccelertion in 3-imensionl cse compre to the 1-imensionl cse. Since velocit n ccelertion re vectors, the object hs ccelertion not onl if it chnges spee (mgnitue of velocit), but lso if it onl chnges the irection of motion. One cn lso rewrite eqution s e j, (3.1.13) v v i v j v k i v vz z j k i j z k where components of ccelertion re v, v n v z z z. Emple (Smple problem from the book) A rbbit runs cross prking lot, on which set of coorinte es, strngel enough been rwn. The rbbit's position re given b 0. 31t 7. t 8, 0. t 91. t 30, (m) (m) where coorintes re given in meters n time in secons. The picture shows trjector of this rbbit. () At time t=15s, wht is rbbit's position vector? We cn represent this vector in the unit vector nottion, which is r i j, where b g b b g b gb g b gb g gb g b gb g m, m,

7 b g b g so r 66m i 57m. j The mgnitue of this vector will be 1 r b66mg b57mg F 87m n it mkes ngle H G I 1 K J F 57I o tn tn K J with respect to is. "Minus" sign mens clockwise ngle. (b) Wht is rbbit's velocit t the sme time t=15s? Let us use efinition of velocit to fin the components of velocit vector t this time. v v 0. 31t 7. t t 7., 0. t 91. t 30i 0. 44t 91.. So for t=15s, we hve v v m s, v v m s. b g b b g b gb g gb g In unit vector nottion the velocit will be v 1. m s i. 5 m s j b g b g, i b g b g n it mkes ngle which hs the mgnitue of v v v 1. m s. 5m s 33. m s tn F H G I KJ F HG 1 v 1. 5 tn v 1. I K J 130 o with is. This ngle is negtive, becuse it is in the clockwise irection with respect to is. It shows the irection of motion t time t=15s, which is tngentil to the trjector t tht moment. Eercise Wht is the verge velocit of the rbbit for the first 15s of his trip? (c) Wht is rbbit's ccelertion t time t=15s? Let us use our efinition of ccelertion to fin the components of the ccelertion vector t this time. From the velocit eqution, which we hve erive lst time, one hs b v 0. 6t m s, v b0. 44t 91. g 0. 44m s. g So ccelertion sts constnt ll the time uring the rbbit's trip n for t=15s, we hve tht in unit vector nottion ccelertion will be 0. 6m s i 0. 44m s j i i, HG

8 e j e j n it mkes ngle it hs the mgnitue 0. 6m s 0. 44m s 0. 76m s tn F H G I KJ F HG tn 0. 6 I K J 145 o with positive -irection. 7. Two-imensionl motion with constnt ccelertion We hve lre stuie one-imensionl motion with constnt ccelertion. All of the equtions which we hve obtine in tht cse re still vli but hve to be etene into imensions, so const, const, const. v v t, 0 v v t, 0 v v t. 0 (3.1.14) (3.1.15) 1 r r0 v0t t, 1 0 v0t t, 1 0 v0t t. (3.1.16)