MTIN IN TW R THREE DIMENIN 3 LEARNING GAL studing this chpter, ou will lern:?if cr is going round cure t constnt speed, is it ccelerting? If so, in wht direction is it ccelerting? Wht determines where btted bsebll lnds? How do ou describe the motion of roller coster cr long cured trck or the flight of circling hwk? If ou throw wter blloon horizontll from our window, will it tke the sme mount of time to hit the ground s blloon tht ou simpl drop? We cn t nswer these kinds of questions using the techniques of Chpter 2, in which prticles moed onl long stright line. Insted, we need to etend our descriptions of motion to two- nd three-dimensionl situtions. We ll still use the ector quntities displcement, elocit, nd ccelertion, but now these quntities will no longer lie long single line. We ll find tht seerl importnt kinds of motion tke plce in two dimensions onl tht is, in plne. These motions cn be described with two components of position, elocit, nd ccelertion. We lso need to consider how the motion of prticle is described b different obserers who re moing reltie to ech other. The concept of reltie elocit will pl n importnt role lter in the book when we stud collisions, when we eplore electromgnetic phenomen, nd when we introduce Einstein s specil theor of reltiit. This chpter merges the ector mthemtics of Chpter 1 with the kinemtic lnguge of Chpter 2. As before, we re concerned with describing motion, not with nlzing its cuses. ut the lnguge ou lern here will be n essentil tool in lter chpters when we stud the reltionship between force nd motion. How to represent the position of bod in two or three dimensions using ectors. How to determine the ector elocit of bod from knowledge of its pth. How to find the ector ccelertion of bod, nd wh bod cn he n ccelertion een if its speed is constnt. How to interpret the components of bod s ccelertion prllel to nd perpendiculr to its pth. How to describe the cured pth followed b projectile. The ke ides behind motion in circulr pth, with either constnt speed or ring speed. How to relte the elocit of moing bod s seen from two different frmes of reference. 71
72 CHAPTER 3 Motion in Two or Three Dimensions 3.1 Position nd Velocit Vectors 73 3.1 The position ector r from the origin to point P hs components,, nd z. The pth tht the prticle follows through spce is in generl cure (Fig. 3.2). z z z Position P of prticle t gien time hs coordintes,, z. r r i^ zk^ P Position ector of point P hs components,, z: r 5 i ^ 1 j ^ 1 zk. ^ j^ 3.2 The erge elocit between points P 1 nd P 2 hs the sme direction s the displcement D r. z Prticle s position t time t 2 Prticle s pth P 2 r 5 2 Dr r 1 Prticle s pth P 1 P 2 The instntneous elocit ector is tngent to the pth t ech point. Dr Prticle s position t time t 1 3.3 The ectors 1 nd 2 re the instntneous elocities t the points P 1 nd P 2 shown in Fig. 3.2. P 1 3.1 Position nd Velocit Vectors To describe the motion of prticle in spce, we must first be ble to describe the prticle s position. Consider prticle tht is t point P t certin instnt. The position ector r of the prticle t this instnt is ector tht goes from the origin of the coordinte sstem to the point P (Fig. 3.1). The Crtesin coordintes,, nd z of point P re the -, -, nd z-components of ector r. Using the unit ectors we introduced in ection 1.9, we cn write (position ector) (3.1) During time interl the prticle moes from P 1, where its position ector is r 1, to P 2, where its position ector is r The chnge in position (the displcement) during this interl is D r 5 2. r 2 2 r 1 5 1 2 2 1 2d^ 1 1 2 2 1 2e^ 1 1 z 2 2 z 1 2 k^. We define the erge elocit during this interl in the sme w we did in Chpter 2 for stright-line motion, s the displcement diided b the time interl: (erge elocit ector) (3.2) Diiding ector b sclr is rell specil cse of multipling ector b sclr, described in ection 1.7; the erge elocit is equl to the displcement ector D r multiplied b 1/, the reciprocl of the time interl. Note tht the -component of Eq. (3.2) is - 5 1 2 2 1 2 / 1 t 2 2 t 1 2 5D/. This is just Eq. (2.2), the epression for erge -elocit tht we found in ection 2.1 for one-dimensionl motion. We now define instntneous elocit just s we did in Chpter 2: It is the limit of the erge elocit s the time interl pproches zero, nd it equls the instntneous rte of chnge of position with time. The ke difference is tht position r nd instntneous elocit re now both ectors: D r 5 lim 0 5 d r (instntneous elocit ector) (3.3) The mgnitude of the ector t n instnt is the speed of the prticle t tht instnt. The direction of t n instnt is the sme s the direction in which the prticle is moing t tht instnt. Note tht s 0, points P 1 nd P 2 in Fig. 3.2 moe closer nd closer together. In this limit, the ector D r becomes tngent to the pth. The direction of D r in the limit is lso the direction of the instntneous elocit. This leds to n importnt conclusion: At eer point long the pth, the instntneous elocit ector is tngent to the pth t tht point (Fig. 3.3). It s often esiest to clculte the instntneous elocit ector using components. During n displcement D r, the chnges D, D, nd Dz in the three coordintes of the prticle re the components of D r. It follows tht the components,, nd z of the instntneous elocit re simpl the time derities of the coordintes,, nd z. Tht is, 5 d r 5 d^ 1 e^ 1 zk^ 5 r 2 2 r 1 5 D r t 2 2 t 1 5 d z 5 dz (components of instntneous elocit) The -component of is 5 d/, which is the sme s Eq. (2.3) the epression for instntneous elocit for stright-line motion tht we obtined in ec- (3.4) tion 2.2. Hence Eq. (3.4) is direct etension of the ide of instntneous elocit to motion in three dimensions. We cn lso get this result b tking the deritie of Eq. (3.1). The unit ectors d^, e^, nd k^ re constnt in mgnitude nd direction, so their derities re zero, nd we find This shows gin tht the components of re d/, d/, nd dz/. The mgnitude of the instntneous elocit ector tht is, the speed is gien in terms of the components,, nd b the Pthgoren reltion 0 0 5 5 " 1 2 1 z (3.6) Figure 3.4 shows the sitution when the prticle moes in the -plne. In this cse, z nd re zero. Then the speed (the mgnitude of ) is z 5 " 2 1 nd the direction of the instntneous elocit is gien b the ngle in the figure. We see tht (We lws use Greek letters for ngles. We use for the direction of the instntneous elocit ector to oid confusion with the direction u of the position ector of the prticle.) The instntneous elocit ector is usull more interesting nd useful thn the erge elocit ector. From now on, when we use the word elocit, we will lws men the instntneous elocit ector (rther thn the erge elocit ector). Usull, we won t een bother to cll ector; it s up to ou to remember tht elocit is ector quntit with both mgnitude nd direction. Emple 3.1 5 d r 5 d d dz d^ 1 e^ 1 k^ z tn 5 Clculting erge nd instntneous elocit A robotic ehicle, or roer, is eploring the surfce of Mrs. The lnding crft is the origin of coordintes, nd the surrounding Mrtin surfce lies in the -plne. The roer, which we represent s point, hs - nd -coordintes tht r with time: 5 2.0 m 2 1 0.25 m/s 2 2 t 2 5 1 1.0 m/s 2 t 1 1 0.025 m/s 3 2 t 3 () Find the roer s coordintes nd its distnce from the lnder t t 5 2.0 s. (b) Find the roer s displcement nd erge elocit ectors during the interl from t 5 0.0 s to t 5 2.0 s. (c) Derie generl epression for the roer s instntneous elocit ector. Epress the instntneous elocit t t 5 2.0 s in component form nd lso in terms of mgnitude nd direction. LUTIN IDENTIFY: This problem inoles motion in two dimensions tht is, in plne. Hence we must use the epressions for the displcement, erge elocit, nd instntneous elocit ectors obtined in this section. (The simpler epressions in ections 2.1 nd 2.2 don t inole ectors; the ppl onl to motion long stright line.) ET UP: Figure 3.5 shows the roer s pth. We ll use Eq. (3.1) for position r, the epression D r 5 r 2 2 r 1 for displcement, 2.5 2.0 1.5 1.0 0.5 (m) 3.4 The two elocit components for motion in the -plne. t 5 2.0 s r 2 r 1 The instntneous elocit ector is lws tngent to the pth. 0.5 1.0 1.5 nd re the - nd - components of. 5 128 t 5 1.0 s r 0 Roer s pth 2.0 0 t 5 0.0 s Prticle s pth in the -plne 3.5 At t 5 0 the roer hs position ector r 0 nd instntneous elocit ector 0. Likewise, r 1 nd 1 re the ectors t t 5 1.0 s; r nd re the ectors t t 5 2.0 s. 2 2 (3.5) (3.7) Eq. (3.2) for erge elocit, nd Eqs. (3.5) nd (3.6) for instntneous elocit nd its direction. The trget ribles re stted in the problem. (m) Continued
74 CHAPTER 3 Motion in Two or Three Dimensions 3.2 The Accelertion Vector 75 EXECUTE: () At time t 5 2.0 s the roer s coordintes re Then we cn write the instntneous elocit ector s 5 2.0 m 2 1 0.25 m/s 2 212.0 s 2 2 5 1.0 m 5 1 1.0 m/s 212.0 s 2 1 1 0.025 m/s 3 212.0 s 2 3 5 2.2 m The roer s distnce from the origin t this time is r 5 " 2 1 2 5 " 1 1.0 m 2 2 1 1 2.2 m 2 2 5 2.4 m (b) To find the displcement nd erge elocit, we epress the position ector r s function of time t. From Eq. (3.1), this is r 5 d^ 1 e^ 5 32.0 m 2 1 0.25 m/s 2 2 t 2 4 d^ 1 311.0 m/s 2 t 1 1 0.025 m/s 3 2 t 3 4 e^ At time t 5 0.0 s the position ector r is r 0 5 1 2.0 m 2d^ 1 1 0.0 m 2e^ 2 From prt () the position ector r t time t 5 2.0 s is r 2 5 1 1.0 m 2d^ 1 1 2.2 m 2e^ Therefore the displcement from t 5 0.0 s to t 5 2.0 s is D r 5 r 2 2 r 0 5 1 1.0 m 2d^ 1 1 2.2 m 2e^ 2 1 2.0 m 2d^ 5 1 21.0 m 2d^ 1 1 2.2 m 2e^ During the time interl from t 5 0.0 s to t 5 2.0 s, the roer moes 1.0 m in the negtie -direction nd 2.2 m in the positie -direction. From Eq. (3.2), the erge elocit during this interl is the displcement diided b the elpsed time: 5 D r 1 21.0 m 2d^ 1 1 2.2 m 2e^ 5 2.0 s 2 0.0 s 5 1 20.50 m/s 2d^ 1 1 1.1 m/s 2 e^ The components of this erge elocit re - 520.50 m/s - 5 1.1 m/s (c) From Eq. (3.4), the components of instntneous elocit re the time derities of the coordintes: 0 5 d 5 1 20.25 m /s 2 212t 2 5 d 5 1.0 m /s 1 1 0.025 m/s 3 213t 2 2 At time t 5 2.0 s, the components of instntneous elocit re 5 1 20.50 m/s 2 212.0 s 2 521.0 m/s The mgnitude of the instntneous elocit (tht is, the speed) t t 5 2.0 s is The direction of with respect to the positie -is is gien b the ngle, where, from Eq. (3.7), Your clcultor will tell ou tht the inerse tngent of 21.3 is 252. ut s we lerned in ection 1.8, ou he to emine sketch of ector to decide on its direction. Figure 3.5 shows tht the correct nswer for is 252 1 180 5 128. EVALUATE: Tke moment to compre the components of erge elocit tht we found in prt (b) for the interl from t 5 0.0 s to t 5 2.0 s 1-520.50 m/s, - 5 1.1 m/s 2 with the components of instntneous elocit t t 5 2.0 s tht we found in prt (c) 1 521.0 m/s, 5 1.3 m/s 2. The comprison shows tht, just s in one dimension, the erge elocit ector oer n interl is in generl not equl to the instntneous elocit t the end of the interl (see Emple 2.1). You should clculte the position ector, instntneous elocit ector, speed, nd direction of motion t t 5 0.0 s nd t 5 1.0 s. Figure 3.5 shows the position ectors r nd instntneous elocit ectors t t 5 0.0 s, 1.0 s, nd 2.0 s. Notice tht t eer point, is tngent to the pth. The mgnitude of increses s the roer moes, which shows tht its speed is incresing. Test Your Understnding of ection 3.1 In which of these situtions would the erge elocit ector oer n interl be equl to the instntneous elocit t the end of the interl? (i) bod moing long cured pth t constnt speed; (ii) bod moing long cured pth nd speeding up; (iii) bod moing long stright line t constnt speed; (i) bod moing long stright line nd speeding up. 3.2 The Accelertion Vector 5 d^ 1 e^ 5 1 20.50 m/s 2 2 t d^ 1 31.0 m/s 1 1 0.075 m/s 3 2 t 2 4 e^ 5 1.0 m/s 1 1 0.075 m/s 3 212.0 s 2 2 5 1.3 m/s 5 " 1 5 " 1 21.0 m/s 2 2 1 1 1.3 m/s 2 2 5 1.6 m/s tn 5 5 1.3 m /s 521.3 so 5128 21.0 m/s Now let s consider the ccelertion of prticle moing in spce. Just s for motion in stright line, ccelertion describes how the elocit of the prticle chnges. ut since we now tret elocit s ector, ccelertion will describe chnges in the elocit mgnitude (tht is, the speed) nd chnges in the direction of elocit (tht is, the direction in which the prticle is moing). In Fig. 3.6, cr (treted s prticle) is moing long cured rod. The ectors nd represent the cr s instntneous elocities t time t 1, when the 1 2 3.6 () A cr moing long cured rod from P to (b) btining D 5 b ector subtrction. (c) The ector 5 D 2 2 1 P 2. 1 / represents the erge ccelertion between P 1 nd P 2. () P 1 P 2 This cr ccelertes b slowing while rounding cure. (Its instntneous elocit chnges in both mgnitude nd direction.) cr is t point P 1, nd t time t 2, when the cr is t point P 2. The two elocities m differ in both mgnitude nd direction. During the time interl from t 1 to t 2, the ector chnge in elocit is 2 2 1 5 D (Fig. 3.6b). We define the erge ccelertion of the cr during this time interl s the elocit chnge diided b the time interl t 2 2 t 1 5: 5 2 1 5 D t 2 2 t 1 (erge ccelertion ector) (3.8) Aerge ccelertion is ector quntit in the sme direction s the ector D (Fig. 3.6c). Note tht is the ector sum of the originl elocit nd the chnge D 2 1 (Fig. 3.6b). The -component of Eq. (3.8) is - 5 1 2 2 / 1 t 2 2 t 1 2 5D /, which is just Eq. (2.4) for the erge ccelertion in stright-line motion. As in Chpter 2, we define the instntneous ccelertion t point P 1 s the limit of the erge ccelertion when point P pproches point nd D 2 P 1 nd both pproch zero. The instntneous ccelertion is lso equl to the instntneous rte of chnge of elocit with time. ecuse we re not restricted to stright-line motion, instntneous ccelertion is now ector (Fig. 3.7): D 5 lim 0 5 d (instntneous ccelertion ector) (3.9) The elocit ector, s we he seen, is tngent to the pth of the prticle. ut Figs. 3.6c nd 3.7 show tht if the pth is cured, the instntneous ccelertion ector lws points towrd the conce side of the pth tht is, towrd the inside of n turn tht the prticle is mking. CAUTIN An prticle following cured pth is ccelerting When prticle is moing in cured pth, it lws hs nonzero ccelertion, een when it moes with constnt speed. This conclusion m seem contrr to our intuition, but it s rell just contrr to the eerd use of the word ccelertion to men tht speed is incresing. The more precise definition gien in Eq. (3.9) shows tht there is nonzero ccelertion wheneer the elocit ector chnges in n w, whether there is chnge of speed, direction, or both. To conince ourself tht prticle hs nonzero ccelertion when moing on cured pth with constnt speed, think of our senstions when ou ride in cr. When the cr ccelertes, ou tend to moe inside the cr in (b) P 1 P 2 1 D 5 2 To find the cr s erge ccelertion between P 1 nd P 2, we first find the chnge in elocit D b subtrcting from. (Notice tht 1 D 5.)? (c) 3.7 Instntneous ccelertion t point in Fig. 3.6. P 1 P 1 To find the instntneous ccelertion t P 1... P 2 P 1 D P 2 D 5 The erge ccelertion hs the sme direction s the chnge in elocit, D.... we tke the limit of s P 2 pproches P 1... P 1 5 lim 0 D Instntneous ccelertion points towrd conce side of pth.... mening tht D nd pproch 0.
76 CHAPTER 3 Motion in Two or Three Dimensions 3.2 The Accelertion Vector 77 3.8 When the rcher shoots the rrow, it ccelertes both forwrd nd upwrd. Thus its ccelertion ector hs both horizontl component 1 2 nd erticl component 1 2. Emple 3.2 Clculting erge nd instntneous ccelertion Let s return to the motions of the robotic roer in Emple 3.1. We found tht the components of instntneous elocit t n time t re 5 d 5 1 20.25 m /s 2 212t 2 5 d 5 1.0 m /s 1 1 0.025 m/s 3 213t 2 2 nd tht the elocit ector is 5 d^ 1 e^ 5 1 20.50 m/s 2 2 t d^ 1 31.0 m/s 1 1 0.075 m/s 3 2 t 2 4 e^ () Find the components of the erge ccelertion in the interl from t 5 0.0 s to t 5 2.0 s. (b) Find the instntneous ccelertion t t 5 2.0 s. LUTIN IDENTIFY: This emple uses the ector reltionships mong elocit, erge ccelertion, nd instntneous ccelertion. ET UP: In prt () we first determine the lues of nd t the beginning nd end of the interl, nd then use Eq. (3.8) to clculte the components of the erge ccelertion. In prt (b) we direction opposite to the cr s ccelertion. (We ll discoer the reson for this behior in Chpter 4.) Thus ou tend to slide towrd the bck of the cr when it ccelertes forwrd (speeds up) nd towrd the front of the cr when it ccelertes bckwrd (slows down). If the cr mkes turn on leel rod, ou tend to slide towrd the outside of the turn; hence the cr hs n ccelertion towrd the inside of the turn. We will usull be interested in the instntneous ccelertion, not the erge ccelertion. From now on, we will use the term ccelertion to men the instntneous ccelertion ector. Ech component of the ccelertion ector is the deritie of the corresponding component of elocit: 5 d 5 d In terms of unit ectors, determine the instntneous ccelertion components t n time t b tking the time derities of the elocit components s in Eq. (3.10). EXECUTE: () If we substitute t 5 0.0 s or t 5 2.0 s into the epressions for nd, we find tht t the beginning of the interl 1 t 5 0.0 s 2 the elocit components re nd tht t the end of the interl 1 t 5 2.0 s 2 the components re (The lues t t 5 2.0 s re the sme s we found in Emple 3.1.) Thus the components of erge ccelertion in this interl re (b) Using Eq. (3.10), we find 5 d - 5 D - 5 D (components of instntneous ccelertion) 5 0.0 m/s 5 1.0 m/s 521.0 m/s 5 1.3 m/s 5 21.0 m /s 2 0.0 m/s 520.5 m/s 2 2.0 s 2 0.0 s 5 1.3 m /s 2 1.0 m/s 5 0.15 m/s 2 2.0 s 2 0.0 s 520.50 m/s 2 5 d (3.10) (3.11) The -component of Eqs. (3.10) nd (3.11), 5 d /, is the epression from ection 2.3 for instntneous ccelertion in one dimension, Eq. (2.5). Figure 3.8 shows n emple of n ccelertion ector tht hs both - nd -components. ince ech component of elocit is the deritie of the corresponding coordinte, we cn epress the components,, nd of the ccelertion ector s The ccelertion ector 5 d 2 2 5 d 2 2 z 5 d 2 z 2 5 d d^ 1 d e^ 1 d z k^ itself is z 5 d z z 5 d 2 d^ 1 d 2 2 e^ 1 d 2 z 2 k^ 2 (3.12) (3.13) 5 1 0.075 m/s 3 212t 2 We cn write the instntneous ccelertion ector s 5 d^ 1 e^ 5 1 20.50 m/s 2 2d^ 1 1 0.15 m/s 3 2 t e^ At time t 5 2.0 s, the components of instntneous ccelertion re 520.50 m/s 2 5 1 0.15 m/s 3 212.0 s 2 5 0.30 m/s 2 The ccelertion ector t this time is 5 1 20.50 m/s 2 2d^ 1 1 0.30 m/s 2 2 e^ The mgnitude of ccelertion t this time is 5 " 2 2 1 5 " 1 20.50 m/s 2 2 2 1 1 0.30 m/s 2 2 2 5 0.58 m/s 2 The direction of with respect to the positie -is is gien b the ngle b, where tn b 5 5 0.30 m /s 2 20.50 m/s 2 520.60 b5180 2 31 5 149 EVALUATE: You should use the results of prt (b) to clculte the instntneous ccelertion t t 5 0.0 s nd t 5 1.0 s. Figure 3.9 shows the roer s pth nd the elocit nd ccelertion ectors t Prllel nd Perpendiculr Components of Accelertion The ccelertion ector for prticle cn describe chnges in the prticle s speed, its direction of motion, or both. It s useful to note tht the component of ccelertion prllel to prticle s pth tht is, prllel to the elocit tells us bout chnges in the prticle s speed, while the ccelertion component perpendiculr to the pth nd hence perpendiculr to the elocit tells us bout chnges in the prticle s direction of motion. Figure 3.10 shows these components, which we lbel i nd '. To see wh the prllel nd perpendiculr components of he these properties, let s consider two specil cses. In Fig. 3.11 the ccelertion ector is in the sme direction s the elocit 1, so hs onl prllel component (tht is, The elocit chnge D i ' 5 0). during smll time interl is in the sme direction s nd hence in the sme direction s The elocit t the end of gien b 2 5 1 1 D 1. 2,, is in the sme direction s 1 but hs greter mgnitude. Hence during the time interl the prticle in Fig. 3.11 moed in stright line with incresing speed. In Fig. 3.11b the ccelertion is perpendiculr to the elocit, so hs onl perpendiculr component ' (tht is, i 5 0). In smll time interl, the elocit chnge D is er nerl perpendiculr to Agin 2 5 1 1 D 1., but in this cse nd he different directions. As the time interl 1 2 3.11 The effect of ccelertion directed () prllel to nd (b) perpendiculr to prticle s elocit. () Accelertion prllel to prticle s elocit: Chnges mgnitude but not direction of elocit. Prticle moes in stright line with chnging speed. 1 D (b) Accelertion perpendiculr to prticle s elocit: Chnges direction but not mgnitude of elocit. Prticle follows cured pth t constnt speed. t 5 0.0 s, 1.0 s, nd 2.0 s. Note tht nd re not in the sme direction t n of these times. The elocit ector is tngent to the pth t ech point, nd the ccelertion ector points towrd the conce side of the pth. 3.9 The pth of the robotic roer, showing the elocit nd ccelertion t t 5 0.0 s 1 nd 1 nd nd t 5 2.0 s 1 nd 0 0 2, t 5 1.0 s 1 1 2, 2 2 2. 2.5 2.0 1.5 1.0 0.5 (m) D f 2 t 2.0 s b = 149 1 = 128 t 1.0 s 0 0.5 1.0 1.5 Component of prllel to the pth P 2.0 Roer s pth 0 t 0.0 s (m) 3.10 The ccelertion cn be resoled into component i prllel to the pth (tht is, long the tngent to the pth) nd component ' perpendiculr to the pth (tht is, long the norml to the pth). Component of perpendiculr to the pth Tngent to pth t P Prticle s pth Norml to pth t P
78 CHAPTER 3 Motion in Two or Three Dimensions 3.3 Projectile Motion 79 3.12 Velocit nd ccelertion ectors for prticle moing through point P on cured pth with () constnt speed, (b) incresing speed, nd (c) decresing speed. () When speed is constnt long cured pth... P... ccelertion is norml to the pth. Norml t P Emple 3.3 Clculting prllel nd perpendiculr components of ccelertion For the roer of Emples 3.1 nd 3.2, find the prllel nd perpendiculr components of the ccelertion t t 5 2.0 s. LUTIN IDENTIFY: We wnt to find the components of the ccelertion ector tht re prllel nd perpendiculr to the elocit ector. ET UP: We found the directions of nd in Emples 3.2 nd 3.1, respectiel. This will llow us to find the ngle between the two ectors nd hence the components of. EXECUTE: In Emple 3.2 we found tht t t 5 2.0 s the prticle hs n ccelertion of mgnitude 0.58 m/s 2 t n ngle of 149 with respect to the positie -is. From Emple 3.1, t this sme time the elocit ector is t n ngle of 128 with respect to the positie -is. o Fig. 3.9 shows tht the ngle between nd is 149 2 128 5 21 (Fig. 3.13). The prllel nd perpendiculr components of ccelertion re then i 5 cos 21 5 1 0.58 m/s 2 2 cos 21 5 0.54 m/s 2 ' 5 sin 21 5 1 0.58 m/s 2 2 sin 21 5 0.21 m/s 2 pproches zero, the ngle f in the figure lso pproches zero, D becomes perpendiculr to both 1 nd 2, nd 1 nd 2 he the sme mgnitude. In other words, the speed of the prticle sts the sme, but the direction of motion chnges nd the pth of the prticle cures. In the most generl cse, the ccelertion hs components both prllel nd perpendiculr to the elocit, s in Fig. 3.10. Then the prticle s speed will chnge (described b the prllel component i) nd its direction of motion will chnge (described b the perpendiculr component ' ) so tht it follows cured pth. Figure 3.12 shows prticle moing long cured pth for three different situtions: constnt speed, incresing speed, nd decresing speed. If the speed is constnt, is perpendiculr, or norml, to the pth nd to nd points towrd the conce side of the pth (Fig. 3.12). If the speed is incresing, there is still perpendiculr component of, but there is lso prllel component hing the sme direction s (Fig. 3.12b). Then points hed of the norml to the pth. (This ws the cse in Emple 3.2.) If the speed is decresing, the prllel component hs the direction opposite to, nd points behind the norml to the pth (Fig. 3.12c). We will use these ides gin in ection 3.4 when we stud the specil cse of motion in circle. (b) When speed is incresing long cured pth... P Norml t P... ccelertion points hed of the norml. 3.13 The prllel nd perpendiculr components of the ccelertion of the roer t t 5 2.0 s. 21 Perpendiculr component of ccelertion (c) When speed is decresing long cured pth... P Prllel component of ccelertion Position of roer t t 5 2.0 s Pth of roer EVALUATE: The prllel component i is in the sme direction s, which mens tht the speed is incresing t this instnt; the lue of i 5 0.54 m/s 2 mens tht the speed is incresing t rte of 0.54 m/s per second. The perpendiculr component ' is not zero, which mens tht t this instnt the roer is chnging direction nd following cured pth; in other words, the roer is turning.... ccelertion points behind the norml. Norml t P Conceptul Emple 3.4 Accelertion of skier A skier moes long ski-jump rmp s shown in Fig. 3.14. The rmp is stright from point A to point C nd cured from point C onwrd. The skier picks up speed s she moes downhill from point A to point E, where her speed is mimum. he slows down fter pssing point E. Drw the direction of the ccelertion ector t points, D, E, nd F. LUTIN Figure 3.14b shows our solution. At point the skier is moing in stright line with incresing speed, so her ccelertion points downhill, in the sme direction s her elocit. At point D the skier is moing long cured pth, so her ccelertion hs component perpendiculr to the pth. There is lso component in the direction of her motion becuse she is still speeding up t this point. o the ccelertion ector points hed of the norml to her pth t point D. The skier s speed is instntneousl not chnging t point E; the speed is mimum t this point, so its deritie is zero. There is no prllel component of, nd the ccelertion is perpendiculr to her motion. Finll, t point F the ccelertion hs perpendiculr component (becuse her pth is cured t this point) nd prllel component opposite to the direction of her motion (becuse she s slowing down). o t this point, the ccelertion ector points behind the norml to her pth. In the net section we ll emine the skier s ccelertion fter she flies off the rmp. Test Your Understnding of ection 3.2 A sled trels oer the crest of snow-coered hill. The sled slows down s it climbs up one side of the hill nd gins speed s it descends on the other side. Which of the ectors (1 through 9) in the figure correctl shows the direction of the sled s ccelertion t the crest? (Choice 9 is tht the ccelertion is zero.) 3.3 Projectile Motion A projectile is n bod tht is gien n initil elocit nd then follows pth determined entirel b the effects of grittionl ccelertion nd ir resistnce. A btted bsebll, thrown footbll, pckge dropped from n irplne, nd bullet shot from rifle re ll projectiles. The pth followed b projectile is clled its trjector. To nlze this common tpe of motion, we ll strt with n idelized model, representing the projectile s single prticle with n ccelertion (due to grit) tht is constnt in both mgnitude nd direction. We ll neglect the effects of ir resistnce nd the curture nd rottion of the erth. Like ll models, this one hs limittions. Curture of the erth hs to be considered in the flight of long-rnge missiles, nd ir resistnce is of crucil importnce to sk dier. Neertheless, we cn lern lot from nlsis of this simple model. For the reminder of this chpter the phrse projectile motion will impl tht we re ignoring ir resistnce. In Chpter 5 we will see wht hppens when ir resistnce cnnot be ignored. Projectile motion is lws confined to erticl plne determined b the direction of the initil elocit (Fig. 3.15). This is becuse the ccelertion due to 3.14 () The skier s pth. (b) ur solution. () A (b) Direction of motion C 0 D E 3 2 4 1 5 8 6 7 or 9: ccelertion 5 0 3.15 The trjector of projectile. 5 0, 5 2g F led s pth A projectile moes in erticl plne tht contins the initil elocit ector 0. Its trjector depends onl on 0 nd on the downwrd ccelertion due to grit. Trjector
80 CHAPTER 3 Motion in Two or Three Dimensions 3.3 Projectile Motion 81 3.16 The red bll is dropped from rest, nd the ellow bll is simultneousl projected horizontll; successie imges in this stroboscopic photogrph re seprted b equl time interls. At n gien time, both blls he the sme -position, -elocit, nd -ccelertion, despite hing different -positions nd -elocities. NLINE 3.1 oling Projectile Motion Problems 3.2 Two lls Flling 3.3 Chnging the -elocit 3.4 Projectile --Accelertions grit is purel erticl; grit cn t moe the projectile sidews. Thus projectile motion is two-dimensionl. We will cll the plne of motion the -coordinte plne, with the -is horizontl nd the -is erticll upwrd. The ke to nlzing projectile motion is tht we cn tret the - nd -coordintes seprtel. The -component of ccelertion is zero, nd the -component is constnt nd equl to 2g. ( definition, g is lws positie; with our choice of coordinte directions, is negtie.) o we cn nlze projectile motion s combintion of horizontl motion with constnt elocit nd erticl motion with constnt ccelertion. Figure 3.16 shows two projectiles with different -motion but identicl -motion; one is dropped from rest nd the other is projected horizontll, but both projectiles fll the sme distnce in the sme time. We cn then epress ll the ector reltionships for the projectile s position, elocit, nd ccelertion b seprte equtions for the horizontl nd erticl components. The components of re 5 0 52g (projectile motion, no ir resistnce) (3.14) ince the -ccelertion nd -ccelertion re both constnt, we cn use Eqs. (2.8), (2.12), (2.13), nd (2.14) directl. For emple, suppose tht t time t 5 0 our prticle is t the point 1 0, 0 2 nd tht t this time its elocit components he the initil lues 0 nd 0. The components of ccelertion re 5 0, 52g. Considering the -motion first, we substitute 0 for in Eqs. (2.8) nd (2.12). We find 5 0 5 0 1 0 t (3.15) (3.16) For the -motion we substitute for, for, 0 for 0, nd 52gfor : 5 0 2 gt 5 0 1 0 t 2 1 2 gt 2 (3.17) (3.18) It s usull simplest to tke the initil position (t t 5 0) s the origin; then 0 5 0 5 0. This might be the position of bll t the instnt it lees the thrower s hnd or the position of bullet t the instnt it lees the gun brrel. Figure 3.17 shows the pth of projectile tht strts t (or psses through) the origin t time t 5 0. The position, elocit, nd elocit components re shown t equl time interls. The -component of ccelertion is zero, so is constnt. The -component of ccelertion is constnt nd not zero, so chnges b equl mounts in equl times, just the sme s if the projectile were lunched erticll with the sme initil -elocit. At the highest point in the trjector, 5 0. We cn lso represent the initil elocit 0 b its mgnitude 0 (the initil speed) nd its ngle 0 with the positie -is (Fig. 3.18). In terms of these quntities, the components nd of the initil elocit re 0 5 1 0 cos 0 2 t 5 1 0 sin 0 2 t 2 1 2 gt 2 5 0 cos 0 0 5 0 cos 0 0 5 0 sin 0 5 0 sin 0 2 gt (3.19) Using these reltionships in Eqs. (3.15) through (3.18) nd setting 0 5 0 5 0, we find (projectile motion) (3.20) (projectile motion) (3.21) (projectile motion) (3.22) (projectile motion) (3.23) These equtions describe the position nd elocit of the projectile in Fig. 3.17 t n time t. We cn get lot of informtion from these equtions. For emple, t n time the distnce r of the projectile from the origin (the mgnitude of the position ector r ) is gien b r 5 " 2 1 2 The projectile s speed (the mgnitude of its elocit) t n time is 0 5 " 2 1 (3.24) (3.25) The direction of the elocit, in terms of the ngle it mkes with the positie - direction (see Fig. 3.17), is gien b 3.18 The initil elocit components 0 nd 0 of projectile (such s kicked soccer bll) re relted to the initil speed 0 nd initil ngle 0. 0 0 5 0 sin 0 () NLINE 0 0 0 5 0 cos 0 3.5 Initil Velocit Components 3.6 Trget Prctice I 3.7 Trget Prctice II 3.19 The nerl prbolic trjectories of () bouncing bll nd (b) blobs of molten rock ejected from olcno. uccessie imges of bll re seprted b equl time interls. uccessie peks decrese in height becuse bll loses energ with ech bounce. 3.17 If ir resistnce is negligible, the trjector of projectile is combintion of horizontl motion with constnt elocit nd erticl motion with constnt ccelertion. tn 5 (3.26) 0 0 0 0 At the top of the trjector, the projectile hs zero erticl elocit ( 5 0), but its erticl ccelertion is still 2g. 5 2g 3 3 3 0 3 Verticll, the projectile ehibits constnt-ccelertion motion in response to the erth s grittionl pull. Thus, its erticl elocit chnges b equl mounts during equl time interls. The elocit ector is tngent to the trjector t ech point. We cn derie n eqution for the trjector s shpe in terms of nd b eliminting t. From Eqs. (3.20) nd (3.21), which ssume 0 5 0 5 0, we find nd t 5 / 1 0 cos 0 2 g 5 1 tn 0 2 2 2 2 0 cos 2 0 (3.27) Don t worr bout the detils of this eqution; the importnt point is its generl form. The quntities 0, tn 0, cos 0, nd g re constnts, so the eqution hs the form 5 b 2 c 2 (b) Trjectories re nerl prbolic. 0 Horizontll, the projectile ehibits constnt-elocit motion: Its horizontl ccelertion is zero, so it moes equl -distnces in equl time interls. 3 where b nd c re constnts. This is the eqution of prbol. In projectile motion, with our simple model, the trjector is lws prbol (Fig. 3.19). When ir resistnce isn t lws negligible nd hs to be included, clculting the trjector becomes lot more complicted; the effects of ir resistnce
82 CHAPTER 3 Motion in Two or Three Dimensions 3.3 Projectile Motion 83 3.20 Air resistnce hs lrge cumultie effect on the motion of bsebll. In this simultion we llow the bsebll to fll below the height from which it ws thrown (for emple, the bsebll could he been thrown from cliff). (m) 100 50 250 2100 sebll s initil elocit: 0 5 50 m/s, 0 5 53.1 (m) 100 200 300 With ir resistnce No ir resistnce depend on elocit, so the ccelertion is no longer constnt. Figure 3.20 shows computer simultion of the trjector of bsebll both without ir resistnce nd with ir resistnce proportionl to the squre of the bsebll s speed. We see tht ir resistnce hs er lrge effect; the mimum height nd rnge both decrese, nd the trjector is no longer prbol. (If ou look closel t Fig. 3.19b, ou ll see tht the trjectories of the olcnic blobs deite in similr w from prbolic shpe.) 2. List the unknown nd known quntities, nd decide which unknowns re our trget ribles. For emple, ou might be gien the initil elocit (either the components or the mgnitude nd direction) nd sked to find the coordintes nd elocit components t some lter time. In n cse, ou ll be using Eqs. (3.20) through (3.23). (Certin other equtions gien in ection 3.3 m be useful s well.) Mke sure tht ou he s mn equtions s there re trget ribles to be found. 3. tte the problem in words nd then trnslte those words into smbols. For emple, when does the prticle rrie t certin point? (Tht is, t wht lue of t?) Where is the prticle when its elocit hs certin lue? (Tht is, wht re the lues of nd when or hs the specified lue?) ince 5 0 t the highest point in trjector, the question When does the projectile rech its highest point? trnsltes into Wht is the lue of t when 5 0? imilrl, When does the projectile return to its initil eletion? trnsltes into Wht is the lue of t when 5 0? EXECUTE the solution: Use Eqs. (3.20) through (3.23) to find the trget ribles. Resist the tempttion to brek the trjector into segments nd nlze ech segment seprtel. You don t he to strt ll oer when the projectile reches its highest point! It s lmost lws esier to use the sme es nd time scle throughout the problem. Use the lue g 5 9.8 m/s 2. EVALUATE our nswer: As lws, look t our results to see whether the mke sense nd whether the numericl lues seem resonble. Conceptul Emple 3.5 Accelertion of skier, continued Let s consider gin the skier in Conceptul Emple 3.4. Wht is her ccelertion t points G, H, nd I in Fig. 3.21 fter she flies off the rmp? Neglect ir resistnce. LUTIN Figure 3.21b shows our nswer. The skier s ccelertion chnged from point to point while she ws on the rmp. ut s 3.21 () The skier s pth during the jump. (b) ur solution. Problem oling trteg 3.1 () F Projectile Motion NTE: The strtegies we used in ections 2.4 nd 2.5 for strightline, constnt-ccelertion problems re lso useful here. IDENTIFY the relent concepts: The ke concept to remember is tht throughout projectile motion, the ccelertion is downwrd nd hs constnt mgnitude g. Note tht the projectile-motion equtions don t ppl to throwing bll, becuse during the throw the bll is cted on b both the thrower s hnd nd grit. These equtions come into pl onl fter the bll lees the thrower s hnd. (b) soon s she lees the rmp, she becomes projectile. o t points G, H, nd I, nd indeed t ll points fter she lees the rmp, the skier s ccelertion points erticll downwrd nd hs mgnitude g. No mtter how complicted the ccelertion of prticle before it becomes projectile, its ccelertion s projectile is gien b 5 0, 52g. G H I ET UP the problem using the following steps: 1. Define our coordinte sstem nd mke sketch showing our es. Usull it s esiest to tke the -is s being horizontl nd the -is s being upwrd nd to plce the origin t the initil 1 t 5 0 2 position where the bod first becomes projectile (such s where bll lees the thrower s hnd). Then the components of the (constnt) ccelertion re 5 0, 52g, nd the initil position is 0 5 0, 0 5 0. Emple 3.6 A bod projected horizontll A motorccle stunt rider rides off the edge of cliff. Just t the edge his elocit is horizontl, with mgnitude 9.0 m/s. Find the motorccle s position, distnce from the edge of the cliff, nd elocit fter 0.50 s. LUTIN IDENTIFY: nce the rider lees the cliff, he is in projectile motion. His elocit t the edge of the cliff is therefore his initil elocit. ET UP: Figure 3.22 shows our sketch. We plce the origin of our coordinte sstem t the edge of the cliff, where the motorccle first becomes projectile, so 0 5 0 nd 0 5 0. The initil elocit is purel horizontl (tht is, 0 5 0), so the initil elocit components re 0 5 0 cos 0 5 9.0 m/s nd 0 5 0 sin 0 5 0. To find the motorccle s position t time t 5 0.50 s, we use Eqs. (3.20) nd (3.21), which gie nd s functions of time. We then find the distnce from the origin using Eq. (3.24). Finll, we use Eqs. (3.22) nd (3.23) to find the elocit components nd t t 5 0.50 s. EXECUTE: Where is the motorccle t t 5 0.50 s? From Eqs. (3.20) nd (3.21), the - nd -coordintes re 5 0 t 5 1 9.0 m/s 210.50 s 2 5 4.5 m 1 52 2 gt 2 1 52 2 1 9.8 m /s 2 210.50 s 2 2 521.2 m The negtie lue of shows tht t this time the motorccle is below its strting point. Wht is the motorccle s distnce from the origin t this time? From Eq. (3.24), r 5 " 2 1 2 5 " 1 4.5 m 2 2 1 1 21.2 m 2 2 5 4.7 m Wht is the elocit t time t 5 0.50 s? From Eqs. (3.22) nd (3.23), the components of elocit t this time re 5 0 5 9.0 m/s 52gt 5 1 29.8 m/s 2 210.50 s 2 524.9 m/s 3.22 ur sketch for this problem. At this point, the bike nd rider become projectile. The motorccle hs the sme horizontl elocit s when it left the cliff t t 5 0, but in ddition there is downwrd (negtie) erticl elocit. If we use unit ectors, the elocit t t 5 0.50 s is 5 d^ 1 e^ 5 1 9.0 m/s 2d^ 1 1 24.9 m/s 2 e^ We cn lso epress the elocit in terms of mgnitude nd direction. From Eq. (3.25), the speed (mgnitude of the elocit) t this time is 5 " 2 1 5 " 1 9.0 m/s 2 2 1 1 24.9 m/s 2 2 5 10.2 m/s From Eq. (3.26), the ngle of the elocit ector is 5rctn 5 rctn1 24.9 m /s 9.0 m/s 2 5229 At this time the elocit is 29 below the horizontl. EVALUATE: Just s shown in Fig. 3.17, the horizontl spect of the motion is unchnged b grit; the motorccle continues to moe horizontll t 9.0 m/s, coering 4.5 m in 0.50 s. The motorccle initill hs zero erticl elocit, so it flls erticll just like bod 1 relesed from rest nd descends distnce 2 gt 2 5 1.2 m in 0.50 s.
84 CHAPTER 3 Motion in Two or Three Dimensions Emple 3.7 Height nd rnge of projectile I: A btted bsebll A btter hits bsebll so tht it lees the bt t speed 0 5 37.0 m/s t n ngle 0 5 53.1, t loction where g 5 9.80 m/s 2. () Find the position of the bll, nd the mgnitude nd direction of its elocit, t t 5 2.00 s. (b) Find the time when the bll reches the highest point of its flight nd find its height h t this point. (c) Find the horizontl rnge R tht is, the horizontl distnce from the strting point to where the bll hits the ground. LUTIN IDENTIFY: As Fig. 3.20 shows, the effects of ir resistnce on the motion of bsebll ren t rell negligible. For the ske of simplicit, howeer, we ll ignore ir resistnce for this emple nd use the projectile-motion equtions to describe the motion. ET UP: Figure 3.23 shows our sketch. We use the sme coordinte sstem s in Fig. 3.17 or 3.18 so we cn use Eqs. (3.20) through (3.23) without n modifictions. ur trget ribles re (1) the position nd elocit of the bll 2.00 s fter it lees the bt, (2) the elpsed time fter leing the bt when the bll is t its mimum height tht is, when 5 0 nd the -coordinte t this time, nd (3) the -coordinte t the time when the -coordinte is equl to the initil lue 0. The bll lees the bt meter or so boe ground leel, but we neglect this distnce nd ssume tht it strts t ground leel 1 0 5 0 2. The initil elocit of the bll hs components 0 5 0 cos 0 5 1 37.0 m/s 2 cos 53.1 5 22.2 m/s 0 5 0 sin 0 5 1 37.0 m/s 2 sin 53.1 5 29.6 m/s EXECUTE: () We wnt to find,,, nd t time t 5 2.00 s. From Eqs. (3.20) through (3.23), 5 0 t 5 1 22.2 m/s 212.00 s 2 5 44.4 m 5 0 t 2 1 2 gt 2 5 1 29.6 m/s 212.00 s 2 2 1 2 1 9.80 m /s 2 212.00 s 2 2 5 39.6 m 5 0 5 22.2 m/s 5 0 2 gt 5 29.6 m/s 2 1 9.80 m/s 2 212.00 s 2 5 10.0 m/s The -component of elocit is positie, which mens tht the bll is still moing upwrd t this time (Fig. 3.23). The mgnitude nd direction of the elocit re found from Eqs. (3.25) nd (3.26): 5 " 1 5 " 1 22.2 m/s 2 2 1 1 10.0 m/s 2 2 5 24.3 m/s 5rctn 1 10.0 m /s 22.2 m/s 2 5 rctn 0.450 5 24.2 The direction of the elocit (tht is, the direction of motion) is 24.2 boe the horizontl. (b) At the highest point, the erticl elocit is zero. When does this hppen? Cll the time t 1 ; then 5 0 2 gt 1 5 0 t 1 5 0 g 5 29.6 m /s 9.80 m/s 2 5 3.02 s 3.23 ur sketch for this problem. The height h t this time is the lue of when t 5 t 1 5 3.02 s: h 5 0 t 1 2 1 2 gt 2 1 5 1 29.6 m/s 213.02 s 2 2 1 2 1 9.80 m /s 2 213.02 s 2 2 5 44.7 m (c) We ll find the horizontl rnge in two steps. First, when does the bll hit the ground? This occurs when 5 0. Cll this time t 2 ; then 5 0 5 0 t 2 2 1 2 gt 2 2 5 t 2 A 0 2 1 2 gt 2 This is qudrtic eqution for t 2. It hs two roots: t 2 5 0 nd t 2 5 2 0 g 5 2 1 29.6 m /s 2 9.80 m/s 2 5 6.04 s There re two times t which 5 0; t 2 5 0 is the time the bll lees the ground, nd t 2 5 2 0/g 5 6.04 s is the time of its return. This is ectl twice the time to rech the highest point tht we found in prt (b), t 1 5 0/g 5 3.02 s, so the time of descent equls the time of scent. This is lws true if the strting nd end points re t the sme eletion nd ir resistnce cn be neglected. The horizontl rnge R is the lue of when the bll returns to the ground tht is, t t 5 6.04 s: R 5 0 t 2 5 1 22.2 m/s 216.04 s 2 5 134 m The erticl component of elocit when the bll hits the ground is 5 0 2 gt 2 5 29.6 m/s 2 1 9.80 m/s 2 216.04 s 2 5229.6 m/s Tht is, hs the sme mgnitude s the initil erticl elocit 0 but the opposite direction (down). ince is constnt, the ngle 5253.1 (below the horizontl) t this point is the negtie of the initil ngle 0 5 53.1. EVALUATE: It s often useful to check results b getting them in different w. For emple, we cn check our nswer for the mimum height in prt (b) b ppling the constnt-ccelertion formul Eq. (2.13) to the -motion: 2 5 0 2 1 2 1 2 0 2 5 0 2 2 2g 1 2 0 2 Emple 3.8 Height nd rnge of projectile II: Mimum height, mimum rnge For projectile lunched with speed t initil ngle 0 (between 0 nd 90 ), derie generl epressions for the mimum height h nd horizontl rnge R (Fig. 3.23). For gien 0, wht lue of 0 gies mimum height? Wht lue gies mimum horizontl rnge? LUTIN IDENTIFY: This is rell the sme eercise s prts (b) nd (c) of Emple 3.7. The difference is tht we re looking for generl epressions for h nd R. We ll lso be looking for the lues of 0 tht gie the mimum lues of h nd R. ET UP: In prt (b) of Emple 3.7 we found tht the projectile reches the high point of its trjector (so tht 5 0) t time t 1 5 0/g, nd in prt (c) of Emple 3.7 we found tht the projectile returns to its strting height (so tht 5 0 ) t time t 2 5 2 0/g. (As we sw in Emple 3.7, t 2 5 2t 1.) To determine the height h t the high point of the trjector, we use Eq. (3.21) to find the -coordinte t t 1. To determine R, we substitute t 2 into Eq. (3.20) to determine the -coordinte t t 2. We ll epress our nswers in terms of the lunch speed 0 nd lunch ngle 0 using Eq. (3.19). EXECUTE: From Eq. (3.19), 0 5 0 cos 0 nd 0 5 0 sin 0. Hence we cn write the time t 1 when 5 0 s Then, from Eq. (3.21), the height t this time is h 5 1 0 sin 0 2 1 0 sin 0 g 2 2 1 2 1 g 0 sin 0 g 2 2 5 0 2 sin 2 0 2g t 1 5 0 g 5 0 sin 0 g For gien lunch speed 0, the mimum lue of h occurs when sin 0 5 1 nd 0 5 90 tht is, when the projectile is lunched stright up. Tht s wht we should epect. If it is lunched horizontll, s in Emple 3.6, 0 5 0 nd the mimum height is zero! The time t 2 when the projectile returns to the ground is 0 t 2 5 2 0 g 5 2 0 sin 0 g 3.3 Projectile Motion 85 At the highest point, 5 0 nd 5 h. ubstituting these, long tl rnge R 5 134 m in prt (c) is greter thn the 99.7-m distnce with 0 5 0, we find from home plte to the right-field fence t feco Field in ettle. 0 5 (The bll s height when it crosses the fence is more thn enough to 0 2 2gh cler it, so this bll is home run.) h 5 0 2g 5 1 29.6 m /s 2 2 2 1 9.80 m/s 2 2 5 44.7 m In rel life, btted bll with the initil speed nd ngle we e used here won t go s high or s fr s we e clculted. (If it did, home runs would be fr more common nd bsebll would be fr which is the sme height we obtined in prt (b). less interesting gme.) The reson is tht ir resistnce, which we It s interesting to note tht h 5 44.7 m in prt (b) is comprble neglected in this emple, is ctull n importnt fctor t the to the 52.4-m height boe the pling field of the roof of the tpicl speeds of pitched nd btted blls (see Fig. 3.20). Hubert H. Humphre Metrodome in Minnepolis, nd the horizon- The horizontl rnge R is the lue of t this time. From Eq. (3.20), R 5 1 0 cos 0 2 t 2 5 1 0 cos 0 2 We cn now use the trigonometric identit to rewrite this s R 5 0 2 sin 2 0 g 2 0 sin 0 g 2 sin 0 cos 0 5 sin 2 0 The mimum lue of sin 2 0 is 1; this occurs when 2 0 5 90, or 0 5 45. This ngle gies the mimum rnge for gien initil speed. EVALUATE: Figure 3.24 is bsed on composite photogrph of three trjectories of bll projected from spring gun t ngles of 30, 45, nd 60. The initil speed 0 is pproimtel the sme in ll three cses. The horizontl rnges re nerl the sme for the 30 nd 60 ngles, nd the rnge for 45 is greter thn either. Cn ou proe tht for gien lue of 0 the rnge is the sme for both n initil ngle 0 nd n initil ngle 90 2 0? CAUTIN Height nd rnge of projectile We don t recommend memorizing the boe epressions for h nd R. The re pplicble onl in the specil circumstnces we he described. In prticulr, the epression for the rnge R cn be used onl when lunch nd lnding heights re equl. There re mn endof-chpter problems to which these equtions do not ppl. 3.24 A lunch ngle of 45 gies the mimum horizontl rnge. The rnge is shorter with lunch ngles of 30 nd 60. A 45 lunch ngle gies the gretest rnge; other ngles fll shorter. Lunch ngle: 0 5 30 0 5 45 0 5 60
86 CHAPTER 3 Motion in Two or Three Dimensions 3.4 Motion in Circle 87 Emple 3.9 Different initil nd finl heights You toss bll from our window 8.0 m boe the ground. When the bll lees our hnd, it is moing t 10.0 m/s t n ngle of 20 below the horizontl. How fr horizontll from our window will the bll hit the ground? Ignore ir resistnce. LUTIN IDENTIFY: As in our clcultion of the horizontl rnge in Emples 3.7 nd 3.8, we re tring to find the horizontl coordinte of projectile when it is t gien lue of. The difference here is tht this lue of is not equl to the initil -coordinte. EXECUTE: To determine t, we rewrite Eq. (3.21) in the stndrd form for qudrtic eqution for t: The roots of this eqution re t 5 1 2 gt 2 2 1 0 sin 0 2 t 1 5 0 0 sin 0 6 Ä 1 2 0 sin 0 2 2 2 41 1 2 g 2 21 1 2 g 2 3.26 The trnquilizer drt hits the flling monke. Without grit The monke remins in its initil position. The drt trels stright to the monke. Therefore, the drt hits the monke. Trjector of drt without grit Dshed rrows show how fr the drt nd monke he fllen t specific times reltie to where the would be without grit. At n time, the he fllen b the sme mount. Drt s fll ET UP: nce gin we choose the -is to be horizontl nd the -is to be upwrd, nd we plce the origin of coordintes t 5 0 sin 0 6 " 0 sin 2 0 2 2g the point where the bll lees our hnd (Fig. 3.25). We he g 0 Drt s fll 0 5 10.0 m/s nd 0 5220 ; the ngle is negtie becuse the 1 10.0 m /s 2 sin 1 220 2 initil elocit is below the horizontl. ur trget rible is the 6" 1 10.0 m/s 2 2 sin 2 1 220 2 2 2 1 9.80 m/s 2 2128.0 m 2 R Trjector of drt 0 with grit lue of t the point where the bll reches the ground tht is, 5 9.80 m/s 2 when 528.0 m. ecuse the initil nd finl heights of the bll d 5 21.7 s or 0.98 s With grit re different, we cn t simpl use the epression for the horizontl The monke flls stright down. rnge found in Emple 3.8. Insted, we first use Eq. (3.21) to find We cn discrd the negtie root, since it refers to time before the At n time t, the drt hs fllen b the sme mount the time t when the bll reches 528.0 m nd then clculte the lue of t this time using Eq. (3.20). bll left our hnd. The positie root tells us tht the bll tkes 0.98 s to rech the ground. From Eq. (3.20), the bll s -coordinte s the monke reltie to where either would be in the bsence of grit: D drt 5 D monke 5 2 gt 2. t tht time is Therefore, the drt lws hits the monke. 3.25 ur sketch for this problem. 5 1 0 cos 0 2 t 5 1 10.0 m/s 23cos 1 220 2410.98 s 2 For the drt we use Eq. (3.21): EVALUATE: We he proed tht t the time the -coordintes re 5 9.2 m drt 5 1 0 sin 0 2 t 2 1 equl, the -coordintes re lso equl; drt imed t the initil The bll hits the ground horizontl distnce of 9.2 m from our 2 gt 2 position of the monke lws hits it, no mtter wht 0 is. This result is lso independent of the lue of g, the ccelertion due to window. o we see tht if d tn 0 5 1 0 sin 0 2 t t the time when the two grit. With no grit 1 g 5 0 2, the monke would remin EVALUATE: The root t 521.7 s is n emple of fictionl -coordintes re equl, then monke 5 drt, nd we he hit. To motionless, nd the drt would trel in stright line to hit him. solution to qudrtic eqution. We discussed these in Emple 2.8 proe tht this hppens, we replce t with d/ 1 0 cos 0 2, the time With grit, both fll the sme distnce A 1 2gt 2 below their in ection 2.5; ou should reiew tht discussion. when monke 5 drt. ure enough, we find tht g 5 0 positions, nd the drt still hits the monke (Fig. 3.26). With our choice of origin we hd initil nd finl heights d 0 5 0 nd 528.0 m. Cn ou use Eqs. (3.16) nd (3.18) to 1 0 sin 0 2 t 5 1 0 sin 0 2 5 d tn show tht ou get the sme nswers for t nd if ou choose the 0 cos 0 0 origin to be t the point on the ground directl below where the bll lees our hnd? 1 2 d tn 0 Drt s fll Monke s fll Emple 3.10 The zookeeper nd the monke A monke escpes from the zoo nd climbs tree. After filing to entice the monke down, the zookeeper fires trnquilizer drt directl t the monke (Fig. 3.26). The cleer monke lets go t the sme instnt the drt lees the gun brrel, intending to lnd on the ground nd escpe. how tht the drt lws hits the monke, regrdless of the drt s muzzle elocit (proided tht it gets to the monke before he hits the ground). LUTIN IDENTIFY: In this emple we he two bodies in projectile motion: the trnquilizer drt nd the monke. The drt nd the monke he different initil positions nd initil elocities, but the go into projectile motion t the sme time. To show tht the drt hits the monke, we he to proe tht t some time the monke nd the drt he the sme -coordinte nd the sme -coordinte. ET UP: We mke the usul choice for the - nd -directions, nd plce the origin of coordintes t the end of the brrel of the trnquilizer gun (Fig. 3.26). We ll first use Eq. (3.20) to find the time t when the -coordintes monke nd drt re the sme. Then we ll use Eq. (3.21) to check whether monke nd drt re lso equl t this time; if the re, the drt hits the monke. EXECUTE: The monke drops stright down, so monke 5 d t ll times. For the drt, Eq. (3.20) tells us tht drt 5 1 0 cos 0 2 t. When these -coordintes re equl, d 5 1 0 cos 0 2 t, or d t 5 0 cos 0 To he the drt hit the monke, it must be true tht monke 5 drt t this sme time. The monke is in one-dimensionl free fll; his position t n time is gien b Eq. (2.12), with pproprite smbol chnges. Figure 3.26 shows tht the monke s initil height is d tn 0 (the opposite side of right tringle with ngle 0 nd djcent side d), nd we find monke 5 d tn 0 2 1 2 gt 2 Test Your Understnding of ection 3.3 In Emple 3.10, suppose the trnquilizer drt hs reltiel low muzzle elocit so tht the drt reches mimum height t point P before striking the monke, s shown in the figure. When the drt is t point P, will the monke be (i) t point A (higher thn P), (ii) t point (t the sme height s P), or (iii) t point C (lower thn P)? Ignore ir resistnce. 3.4 Motion in Circle When prticle moes long cured pth, the direction of its elocit chnges. As we sw in ection 3.2, this mens tht the prticle must he component of ccelertion perpendiculr to the pth, een if its speed is constnt (see Fig. 3.11b). In this section we ll clculte the ccelertion for the importnt specil cse of motion in circle. NLINE P A C 4.1 Mgnitude of Centripetl Accelertion
88 CHAPTER 3 Motion in Two or Three Dimensions 3.27 A cr in uniform circulr motion. The speed is constnt nd the ccelertion is directed towrd the center of the circulr pth. 3.28 Finding the elocit chnge D, erge ccelertion, nd instntneous ccelertion rd for prticle moing in circle with constnt speed. () A point moes distnce Ds t constnt speed long circulr pth. P 1 Cr speeding up long circulr pth Component of ccelertion prllel to elocit: Chnges cr s speed Ds P 2 R Df R (b) The corresponding chnge in elocit nd erge ccelertion Df D (c) The instntneous ccelertion rd Component of ccelertion perpendiculr to elocit: Chnges cr s direction These two tringles re similr. The instntneous ccelertion in uniform circulr motion R lws points towrd the center of the circle. Cr slowing down long circulr pth Component of ccelertion perpendiculr to elocit: Chnges cr s direction Component of ccelertion prllel to elocit: Chnges cr s speed Uniform Circulr Motion When prticle moes in circle with constnt speed, the motion is clled uniform circulr motion. A cr rounding cure with constnt rdius t constnt speed, stellite moing in circulr orbit, nd n ice skter skting in circle with constnt speed re ll emples of uniform circulr motion (Fig. 3.27; compre Fig. 3.12). There is no component of ccelertion prllel (tngent) to the pth; otherwise, the speed would chnge. The ccelertion ector is perpendiculr (norml) to the pth nd hence directed inwrd (neer outwrd!) towrd the center of the circulr pth. This cuses the direction of the elocit to chnge without chnging the speed. ur net project is to show tht the mgnitude of the ccelertion in uniform circulr motion is relted in simple w to the speed of the prticle nd the rdius of the circle. Figure 3.28 shows prticle moing with constnt speed in circulr pth of rdius R with center t. The prticle moes from P 1 to P 2 in time. The ector chnge in elocit D during this time is shown in Fig. 3.28b. The ngles lbeled Df in Figs. 3.28 nd 3.28b re the sme becuse is perpendiculr to the line nd 1 P 1 2 is perpendiculr to the line P 2. Hence the tringles in Figs. 3.28 nd 3.28b re similr. The rtios of corresponding sides of similr tringles re equl, so The mgnitude of the erge ccelertion during is therefore 0 D 0 The mgnitude of the instntneous ccelertion t point P 1 is the limit of this epression s we tke point P 2 closer nd closer to point P 1 : ut the limit of Ds/ is the speed t point P 1. Also, P 1 cn be n point on the pth, so we cn drop the subscript nd let represent the speed t n point. Then rd 5 2 R 5 Ds R 5 0 D 0 or 0 D 0 5 R Ds 5 R Ds 5 lim 0 R 5 R lim Ds 0 Uniform circulr motion: Constnt speed long circulr pth (uniform circulr motion) (3.28) We he dded the subscript rd s reminder tht the direction of the instntneous ccelertion t ech point is lws long rdius of the circle, towrd Ds Accelertion is ectl perpendiculr to elocit; no prllel component To center of circle its center. ecuse the speed is constnt, the ccelertion is lws perpendiculr to the instntneous elocit. This is shown in Fig. 3.28c; compre with the right-hnd illustrtion in Fig. 3.27. We he found tht in uniform circulr motion, the mgnitude of the instntneous ccelertion is equl to the squre of the speed diided b the rdius R of the circle. Its direction is perpendiculr to nd inwrd long the rdius. ecuse the ccelertion is lws directed towrd the center of the circle, it is sometimes clled centripetl ccelertion. The word centripetl is deried from two Greek words mening seeking the center. Figure 3.29 shows the directions of the elocit nd ccelertion ectors t seerl points for prticle moing with uniform circulr motion. CAUTIN Uniform circulr motion s. projectile motion The ccelertion in uniform circulr motion hs some similrities to the ccelertion in projectile motion without ir resistnce, but there re lso some importnt differences. In both uniform circulr motion (Fig. 3.29) nd projectile motion (Fig. 3.29b) the mgnitude of ccelertion is the sme t ll times. Howeer, in uniform circulr motion the direction of chnges continuousl so tht it lws points towrd the center of the circle. (At the top of the circle the ccelertion points down; t the bottom of the circle the ccelertion points up.) In projectile motion, b contrst, the direction of remins the sme t ll times. We cn lso epress the mgnitude of the ccelertion in uniform circulr motion in terms of the period T of the motion, the time for one reolution (one complete trip round the circle). In time T the prticle trels distnce equl to the circumference 2pR of the circle, so its speed is When we substitute this into Eq. (3.28), we obtin the lterntie epression Emple 3.11 rd 5 4p2 R T 2 5 2pR T (3.29) (uniform circulr motion) (3.30) Centripetl ccelertion on cured rod An Aston Mrtin V8 Vntge sports cr hs lterl ccelertion of 0.96g, which is 1 0.96 219.8 m/s 2 2 5 9.4 m/s 2. This represents the mimum centripetl ccelertion tht the cr cn ttin without skidding out of the circulr pth. If the cr is treling t constnt 40 m/s (bout 89 mi/h, or 144 km/h), wht is the minimum rdius of cure it cn negotite? (Assume tht the cure is unbnked.) LUTIN IDENTIFY: ecuse the cr is moing t constnt speed long cure tht is segment of circle, we cn ppl the ides of uniform circulr motion. ET UP: We use Eq. (3.28) to find the trget rible R (the rdius of the cure) in terms of the gien centripetl ccelertion rd nd speed. () Uniform circulr motion r rd rd r rd r 3.4 Motion in Circle 89 3.29 Accelertion nd elocit () for prticle in uniform circulr motion nd (b) for projectile with no ir resistnce. rd (b) Projectile motion r rd rd r r Accelertion is constnt in mgnitude nd direction. Accelertion hs constnt mgnitude but ring direction. Velocit nd ccelertion re lws perpendiculr. Velocit nd ccelertion re perpendiculr onl t the pek of the trjector. EXECUTE: We re gien nd, so we sole Eq. (3.28) for R: rd R 5 2 5 1 40 m /s 2 2 5 170 m (bout 560 ft) rd 9.4 m/s 2 EVALUATE: ur result shows tht the required turning rdius R is proportionl to the squre of the speed. Hence een smll reduction in speed cn mke R substntill smller. For emple, reducing b 20% (from 40 m/s to 32 m/s) would decrese R b 36% (from 170 m to 109 m). Another w to mke the required turning rdius smller is to bnk the cure. We will inestigte this option in Chpter 5. r r r r
90 CHAPTER 3 Motion in Two or Three Dimensions 3.5 Reltie Velocit 91 Emple 3.12 Centripetl ccelertion on crnil ride In crnil ride, the pssengers trel t constnt speed in circle of rdius 5.0 m. The mke one complete circle in 4.0 s. Wht is their ccelertion? LUTIN IDENTIFY: The speed is constnt, so this is problem inoling uniform circulr motion. ET UP: We re gien the rdius R 5 5.0 m nd the period T 5 4.0 s, so we cn use Eq. (3.30) to clculte the ccelertion. Alterntiel, we cn first clculte the speed using Eq. (3.29) nd then find the ccelertion using Eq. (3.28). EXECUTE: From Eq. (3.30), rd 5 4p2 1 5.0 m 2 1 4.0 s 2 2 5 12 m/s 2 We ll check this nswer b using Eq. (3.28) fter first determining the speed. From Eq. (3.29), the speed is the circumference of the circle diided b the period T: 5 2pR T 5 2p 1 5.0 m 2 4.0 s The centripetl ccelertion is then Hppil, we get the sme nswer for with both pproches. EVALUATE: As in Emple 3.11, the direction of is lws towrd the center of the circle. The mgnitude of is greter thn g, the ccelertion due to grit, so this is not ride for the fintherted. (ome roller costers subject their pssengers to ccelertions s gret s 4g.) rd 5 7.9 m/s rd 5 2 R 5 1 7.9 m /s 2 2 5.0 m 5 12 m /s 2 Test Your Understnding of ection 3.4 uppose tht the prticle in Fig. 3.30 eperiences four times the ccelertion t the bottom of the loop s it does t the top of the loop. Compred to its speed t the top of the loop, is its speed t the bottom of the loop (i) "2 times s gret; (ii) 2 times s gret; (iii) 2 "2 times s gret; (i) 4 times s gret; or () 16 times s gret. 3.5 Reltie Velocit You e no doubt obsered how cr tht is moing slowl forwrd ppers to be moing bckwrd when ou pss it. In generl, when two obserers mesure the elocit of moing bod, the get different results if one obserer is moing reltie to the other. The elocit seen b prticulr obserer is clled the elocit reltie to tht obserer, or simpl reltie elocit. Figure 3.31 shows sitution in which understnding reltie elocit is etremel importnt. We ll first consider reltie elocit long stright line, then generlize to reltie elocit in plne. 3.31 Airshow pilots fce complicted problem inoling reltie elocities. The must keep trck of their motion reltie to the ir (to mintin enough irflow oer the wings to sustin lift), reltie to ech other (to keep tight formtion without colliding), nd reltie to their udience (to remin in sight of the specttors). 3.30 A prticle moing in erticl loop with ring speed, like roller coster cr. lowest speed: Lest rdil ccelertion, zero tngentil ccelertion peeding up: Tngentil ccelertion in sme direction s tn tn rd 5 rd rd rd r 5 rd lowing down: Tngentil ccelertion rd opposite to tn tn Fstest speed: Gretest rdil ccelertion, zero tngentil ccelertion Nonuniform Circulr Motion We he ssumed throughout this section tht the prticle s speed is constnt. If the speed ries, we cll the motion nonuniform circulr motion. An emple is roller coster cr tht slows down nd speeds up s it moes round erticl loop. In nonuniform circulr motion, Eq. (3.28) still gies the rdil component of ccelertion rd 5 /R, which is lws perpendiculr to the instntneous elocit nd directed towrd the center of the circle. ut since the speed hs different lues t different points in the motion, the lue of rd is not constnt. The rdil (centripetl) ccelertion is gretest t the point in the circle where the speed is gretest. In nonuniform circulr motion there is lso component of ccelertion tht is prllel to the instntneous elocit. This is the component i tht we discussed in ection 3.2; here we cll this component tn to emphsize tht it is tngent to the circle. From the discussion t the end of ection 3.2 we see tht the tngentil component of ccelertion is equl to the rte of chnge of speed. Thus rd 5 2 R (nonuniform circulr motion) (3.31) The ector ccelertion of prticle moing in circle with ring speed is the ector sum of the rdil nd tngentil components of ccelertions. The tngentil component is in the sme direction s the elocit if the prticle is speeding up, nd in the opposite direction if the prticle is slowing down (Fig. 3.30). In uniform circulr motion there is no tngentil component of ccelertion, but the rdil component is the mgnitude of d /. CAUTIN tn nd tn 5 d 0 0 Uniform s. nonuniform circulr motion Note tht the two quntities d 0 0 nd d P P re not the sme. The first, equl to the tngentil ccelertion, is the rte of chnge of speed; it is zero wheneer prticle moes with constnt speed, een when its direction of motion chnges (such s in uniform circulr motion). The second is the mgnitude of the ector ccelertion; it is zero onl when the prticle s ccelertion ector is zero tht is, when the prticle moes in stright line with constnt speed. In uniform circulr motion 0 d / 0 5 rd 5 /r; in nonuniform circulr motion there is lso tngentil component of ccelertion, so 0 d / 0 5 " 2 rd 1 2 tn. Reltie Velocit in ne Dimension A pssenger wlks with elocit of 1.0 m/s long the isle of trin tht is moing with elocit of 3.0 m/s (Fig. 3.32). Wht is the pssenger s elocit? It s simple enough question, but it hs no single nswer. As seen b second pssenger sitting in the trin, she is moing t 1.0 m/s. A person on biccle stnding beside the trin sees the wlking pssenger moing t 1.0 m/s 1 3.0 m/s 5 4.0 m/s. An obserer in nother trin going in the opposite direction would gie still nother nswer. We he to specif which obserer we men, nd we spek of the elocit reltie to prticulr obserer. The wlking pssenger s elocit reltie to the trin is 1.0 m/s, her elocit reltie to the cclist is 4.0 m/s, nd so on. Ech obserer, equipped in principle with meter stick nd stopwtch, forms wht we cll frme of reference. Thus frme of reference is coordinte sstem plus time scle. Let s use the smbol A for the cclist s frme of reference (t rest with respect to the ground) nd the smbol for the frme of reference of the moing trin. In stright-line motion the position of point P reltie to frme A is gien b P / (the position of P with respect to A), nd the position of P reltie to frme is A gien b P / (see Fig. 3.32b). The position of the origin of A with respect to the origin of is / A. Figure 3.32b shows tht (3.32) In words, the totl distnce from the origin of A to point P equls the distnce from the origin of to point P plus the distnce from the origin of A to the origin of. The -elocit of P reltie to frme A, denoted b P / is the deritie of with respect to time. The other elocities re similrl A-, P / obtined. o the time deritie A of Eq. (3.32) gies us reltionship mong the rious elocities: d P / A P / A- 5 P / - 1 / A- P / A 5 P / 1 / A 5 d P/ 1 d / A (reltie elocit long line) (3.33) Getting bck to the pssenger on the trin in Fig. 3.32, we see tht A is the cclist s frme of reference, is the frme of reference of the trin, nd point P represents the pssenger. Using the boe nottion, we he or P / - 511.0 m/s / A- 513.0 m/s 3.32 () A pssenger wlking in trin. (b) The position of the pssenger reltie to the cclist s frme of reference nd the trin s frme of reference. () (b) A Cclist's frme P (pssenger) A (cclist) Trin s frme /A (trin) Velocit of trin reltie to cclist Position of pssenger in both frmes P A, A /A P/ P/A