Motion in Two Dimensions Sections Covered in the Text: Chapters 6 & 7, except 7.5 & 7.6

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1 Motion in Two Dimensions Sections Covered in the Tet: Chapters 6 & 7, ecept 7.5 & 7.6 It is time to etend the definitions we developed in Note 03 to describe motion in 2D space. In doin so we shall find that the vector nature of the definitions displacement, velocit and acceleration take on more importance than before. Havin developed these tools we then appl them to describe an important, special eample of motion in 2D space, namel the motion of a projectile. As before, we shall model the object as a point particle. Displacement, Velocit and Acceleration Consider an object modeled as a point particle movin alon an arbitrar path in the -plane (Fiure 6-1). We assume that we are able to detect the particle s position at an point and to measure the corres-pondin clock time. Two positions i and f in the particle s path are shown. Let the vectors that locate these positions be ri and r f, respectivel. The time that elapses between the two positions is t = t f t i. r i i Δr = r f r i r f d f path of object Fiure 6-1. An object modelled as a point particle moves alon an arbitrar path in 2D space. Distance Traveled The distance d the particle travels between the two positions is indicated approimatel in the fiure. This would be the number obtained if it were possible to la a fleible tape measure alon the particle s path. d is difficult to measure or calculate in all but the simplest of paths. B its nature, d is alwas a positive number. Displacement The displacement of the particle between the two positions is different from the distance travelled; it is the difference between the correspondin position vectors: Δr = r f ri. Obviousl, in contrast to motion in 1D space, position vectors in 2D space do not, in eneral, alwas point in the same direction. Moreover, in eneral, the displacement vector points in a direction different from the position vectors that define its limits. Velocit Two tpes of velocit are defined in 2D space as in 1D space: averae velocit and instantaneous velocit. Averae Velocit The averae velocit of the particle is defined as the displacement divided b the correspondin elapsed time t: v = Δr Δt. [6-1] Since averae velocit is proportional to displacement (the proportionalit factor bein 1/ t, a scalar), it is a vector pointin in the same direction as the displacement vector. In an elapsed time the averae velocit depends on the end positions of the interval (as does the displacement), and is therefore independent of the path taken b the particle. Instantaneous Velocit The instantaneous velocit of the particle is defined as the limit of the averae velocit as the elapsed time t becomes vanishinl small, i.e., as t 0: Δr v = lim v = lim Δt 0 Δt 0 Δt = dr dt. [6-2] Thus the instantaneous velocit vector equals the first derivative of the position vector with respect to time (or the slope of the tanent to the r(t) function). The direction of the instantaneous velocit vector at an position in a particle s path is alon a tanent line to the path at that position and in the direction of motion. The instantaneous speed is the manitude of the instantaneous velocit. 06-1

2 Fiure 6-1 depicts the motion of the particle in terms of position or displacement vectors; but the motion can also be described in terms of velocit vectors. Let the instantaneous velocit vectors at positions i and f be as drawn in Fiure 6-2. In eneral, the chane in velocit vector (small diaram in the fiure) points in a direction different from the instantaneous velocit vectors. Since, in eneral, the instantaneous velocit vectors at positions i and f are unequal, the particle is, b definition, underoin an acceleration. i a = Δv Δt v f f path of object v f Δv = v f i v f Δv = v f Fiure 6-3. The averae acceleration vector of the particle has the direction of the chane in velocit vector. Acceleration is a different vector tpe than is velocit or displacement. f path of object Δv a = lim a = lim Δt 0 Δt 0 Δt = dv dt. [6-4] Fiure 6-2. The motion of a particle described in terms of velocit vectors. The velocit and position vectors (compare this fiure with Fiure 6-1) are drawn with different head stles to emphasize their different natures. Acceleration Two tpes of acceleration are defined in 2D space: averae acceleration and instantaneous acceleration. Averae Acceleration The averae acceleration of a particle over an elapsed time t is defined as the chane in velocit divided b the elapsed time: a = v f t f t i v f = Δv Δt. [6-3] The averae acceleration is a vector divided b a scalar, and is therefore a vector. The averae acceleration vector points in the same direction as the chane in velocit vector (see Fiure 6-3). Instantaneous Acceleration The instantaneous acceleration of a particle is defined as the limit of the averae acceleration as the elapsed time becomes vanishinl small, i.e., as t 0: Thus the instantaneous acceleration vector is the first derivative of the instantaneous velocit vector with respect to time. The direction of the instantaneous acceleration vector at an position in a particle s path is alon a tanent line to the velocit function at that position. Two-Dimensional Motion with Constant Acceleration The kinematic equations of motion we developed in Note 03 (eqs[3-7], [3-11] and [3-12])) can be eneralized b writin them in full vector notation. We shall simpl state the results here. For convenience we assume the particle is movin with constant acceleration in 2D space. The instantaneous position vector of a particle at a point (,) is r = ) i + ) j. [6-5] If, over an elapsed time t, the velocit of the particle chanes from vi to v f then v f = vi + at. [6-6] If, over the elapsed time t the position of the particle chanes from ri to r f then r f = ri + vit at2. [6-7] 06-2

3 In 2D space eq[6-6] ives rise to an equation for each of the - and -components v f = v i + a t and v f = v i + a t. Eq[6-7] ives rise to the equations [6-8] Note 06 tion is to derive this kind of epression for projectile motion from kinematics. To bein our description, we model the problem in -space with the -direction positive upward (Fiure 6-4). We assume that the particle is projected from the oriin (0,0) with an initial velocit vi at an anle θ i with respect to the horizontal. The position of the particle at an subsequent clock time t is iven b the position vector r. f = i + v i t a t 2 and f = i + v i t a t 2. [6-9] (R/2,h) v i These equations emphasize that motion with constant acceleration in 2D space is equivalent to two independent motions in the - and -directions havin constant accelerations a and a, respectivel. The component of motion in the -direction does not affect the component of motion in the -direction and viceversa. With these tools we are read to describe the motion of a projectile. v i r i θ i vi v f v i Projectile Motion One of the earl successes of Natural Science in the renaissance was the correct description of the motion of a projectile. This problem was pursued b the best minds of the da, includin Da Vinci, Galileo and others, as it had important usae in the art of war. The burnin question was if a cannonball is fired with a certain velocit at a certain anle above the horizon, how hih will the ball o and how far will it travel? In other words, what will be the cannonball s maimum heiht and rane? This was important in puttin a cannonball where ou wanted it to o. We have all seen a similar motion in baseball. When a plaer hits or throws the ball, the ball travels in a curved path of a certain maimum heiht and rane. One can demonstrate photoraphicall that if the resistance of the air is neliible then the path of the projectile is a parabola. 1 In other words the functional form of the path is Fiure 6-4. The path of a projectile in 2D space. The position and velocit vectors are drawn with different head stles to emphasize their different natures. We make the followin assumptions: 1 the acceleration of the particle is a constant of value a = directed downwards 2 the effect of air resistance is neliible. 2 This means that the acceleration of the particle has a non-zero component in the -direction onl; the - component of the acceleration is zero. The initial velocit vector can be resolved into its - and -components v i = and v i =. Substitutin these epressions into eqs[6-8] and [6-9] and takin a = 0 and a = ields () = a + b 2, [6-10] v f = v i = = constant [6-11] where a and b are constants. One objective of this sec- v f = v i t = t [6-12] 1 As the astronomers are quick to point out the correct path is an ellipse. However, if the particle moves close to the Earth so that its acceleration remains constant then to a ood approimation the path can be described b a parabola. 2 If we do not make this assumption the problem is more complicated. Anone who has watched a baseball ame has seen baseballs veer downwards, to the left or riht in response to air friction or the wind. 06-3

4 f = i + v i t = ( )t, [6-13] f = i + v i t 1 2 t2 = ( )t 1 2 t2 [6-14] If we solve for t in eq[6-13] and substitute it into eq[6-14] we et 2 f = ( tanθ ) f 2v 2 i cos 2 θ i f [6-15] This epression has the form of eq[6-10] and proves that the path of the projectile is, indeed, a parabola accordin to the assumptions we have made. Horizontal Rane and Maimum Heiht We are now in a position to derive epressions for the maimum heiht h and horizontal rane R. These occur, respectivel, at the special positions (R/2, h) and (R, 0) in the particle s path. At the hihest point of the particle s path, v f = 0. Thus from eq[6-12] we can calculate the time the particle takes to reach this position: t h = Substitutin this epression for t h into eq[6-14] and replacin f with h, we have h = ( ) v i. 1 2 v i, 2 which reduces to = 2v 2 i R = v 2 i sin 2θ i,, [6-17] since sin2θ = 2sinθcosθ. It can be seen that, other thins bein equal, a maimum R occurs for θ i = 45. As before, R can be increased b increasin or b operatin in an environment where is small. We now have the equations we need to describe an aspect of the motion of a projectile. Let us consider an eample. Eample Problem 6-1 An Eample of Projectile Motion A ball is thrown from the top of one buildin toward a tall buildin 50.0 m awa (Fiure 6-5). The initial velocit of the ball is 20.0 m.s 1, 40.0 above the horizontal. How far above or below the oriinal level does the ball strike the opposite wall? 40.0 which reduces to h = v 2 i sin 2 θ i 2. [6-16] 50.0 m It can be seen that, other thins bein equal, maimum h occurs for θ i = 90. h can be increased b increasin or b operatin in an environment (on a planet or small astronomical bod) where is ver small. Now the rane R is the horizontal distance traveled b the particle in twice the time it takes to reach the peak, that is, in a time 2t h. Usin t h from above and notin that h = R at t = 2t h, we have R = ( )2t h = ( ) 2v i, Fiure 6-5. A ball is projected at 40.0 above the horizontal. Solution: We have the followin components of the initial velocit: = (20.0m.s 1 )cos40.0 o = 15.3m.s 1 = (20.0m.s 1 )sin 40.0 o = 12.9m.s 1 The horizontal component of the initial velocit remains unchaned at 06-4

= v f = v = 15.3m.s 1 Since the -component of acceleration is zero, we can use the epression = v t to solve for t where t is the time that elapses between launch and impact on the opposite wall.

27s. 15.3m.s For the vertical motion takin up as positive and down as neative we have = v 0 t + 1 2 a t2 Fiure 6-6. Two drivers A and B move past an observer O. = (12.9m.s 1 )(3.27s) + 1 2 ( 9.80m.

5 = v f = v = 15.3m.s 1 Since the -component of acceleration is zero, we can use the epression = v t to solve for t where t is the time that elapses between launch and impact on the opposite wall. Thus t is Note 06 same event with the vector r po'. O and his/her reference frame is movin with velocit vo'o relative to O. The two position vectors are related b r PO = r PO' + vo' Ot. t = 50m 1 = 3.27s. 15.3m.s For the vertical motion takin up as positive and down as neative we have = v 0 t a t2 Fiure 6-6. Two drivers A and B move past an observer O. = (12.9m.s 1 )(3.27s) ( 9.80m.s 2 )(3.27s) 2 = 10.2m. ' P is the vertical distance above the oriinal level. Since is neative here, the ball hits the opposite wall at a point 10.2 m below the oriinal level. r PO r PO' Another, special acceleration is defined for objects movin with uniform speed in a circle. Thouh this topic is presented at this point in the tetbook, we postpone it until a later note. Relative Velocit Thus far we have taken the oriin of our coordinate sstem as a fied point (the point we have taken as 0 in a 1D frame or (0,0) in a 2D frame). There will be times when we need to relate certain measurements made in different sstems, in particular when one sstem is movin with respect to the another. Take for eample the motion illustrated in Fiure 6-6. Two cars with drivers A and B move past a third observer O who is standin beside the road. O measures the speeds of A and B at 60 km/hr. But accordin to observer A what is the speed of B? As we all know from eperience A would sa that B s speed is 0 km/hr. In other words the speed of B relative to O is 60 km/hr, but relative to A is 0 km/hr. We can quantif the phsics of these observations with the help of the position vectors in Fiure 6-7. Observers O and O in their own reference frames (S and S ) observe the same event P. O locates the event with the position vector r po, while O locates the O S v O' O t O' S' v O' O Fiure 6-7. Observers in stationar and movin reference frames describe the same event P with different position and velocit vectors. If we differentiate this epression with respect to time we obtain an epression for the correspondin velocit vectors: d ( dt r ) PO = d ( r PO' + vo'ot) dt ives v po = vpo' + vo' o. The subscripts po, po and O O mean p wrt O, p wrt O and O wrt O. Thus the velocit of P wrt O equals the velocit of P wrt O plus the velocit of O wrt O. This result is made clearer with the help of an eample. 06-5

6 Eample Problem 6-2 A Boat Crossin a River A boat propelled with a speed of m.s 1 in still water moves directl across a river that is 60.0 m wide. The river flows with a speed of m.s 1 relative to the riverbank. (a) At what anle, relative to the straiht-across direction, must the boat be pointed? (b) How lon does it take the boat to cross the river? d = 60.0 m v WO = 0.300m.s 1 v BW = 0.500m.s 1 θ vbo O downriver Solution: Let the velocit of the water wrt an observer on the riverbank (manitude m.s 1 ) be denoted vwo. Let the velocit of the boat relative to the water (manitude m.s 1 ) be denoted vbw. Let the velocit of the boat relative to the riverbank be denoted vbo. The vectors are arraned as shown in Fiure 6-8. (a) The boat must therefore be steered upstream b the anle θ = sin 1 v WO v BW = sin = 37.0 o upstream with respect to the water. It is the subsequent dra of the water flowin downstream that results in the boat movin directl across the river. Fiure 6-8. Velocit diaram for a boat crossin a river. (b) The speed of the boat with respect to the riverbank is the manitude of the vector vbo. This is v BO = (cos37 o )(0.500m.s 1 ) = 0.400m.s 1. This is the resultant speed of the boat wrt the riverbank. The time required for the boat to o directl across the river is the distance travelled divided b this speed: t = d v BO = 60.0m 1 = 150s m.s The boat crosses the river in 150 seconds. To Be Mastered Definitions: distance travelled, displacement, averae velocit, instantaneous velocit, instantaneous speed, averae acceleration, instantaneous acceleration kinematic equations (committed to memor) projectile motion problem relative velocit problems 06-6

7 Tpical Quiz/Test/Eam Questions Note A child throws a ball straiht up into the air and catches the ball on its return. Let the maimum heiht reached b the ball be h and the total elapsed time be t. Assume the ball is at the same, neliible heiht above the Earth when thrown and cauht and nelect the effect of air friction. (a) What is the distance travelled b the ball? (b) What is the speed of the ball at its hihest point? (c) What is the averae velocit of the ball? (d) What is the acceleration of the ball at its hihest point? 2. A pilot in an airplane is movin at the speed of 15.0 m.s 1 parallel to flat round m below as shown in the fiure. He releases a sack of flour intendin to hit a taret. Assume air friction is neliible and answer the followin questions: sack B C m A taret (a) Which of A, B or C most closel approimates the path of the sack of flour? Eplain. (b) How lare must the distance from plane to taret be in order that the sack hits the taret? (c) How lon is the sack in the air? 3. A particle is projected horizontall from a position 10.0 m above round (see the fiure on the net pae). It hits the round a horizontal distance of m awa. It is proposed that the path of the particle in -space can be described b the function () = a + b + c 2 where and are in m and a, b and c are constants. Based on the facts iven find values for a, b and c. 06-7

8 particle 10.0 m m 06-8

REVIEW: Going from ONE to TWO Dimensions with Kinematics. Review of one dimension, constant acceleration kinematics. v x (t) = v x0 + a x t

REVIEW: Going from ONE to TWO Dimensions with Kinematics. Review of one dimension, constant acceleration kinematics. v x (t) = v x0 + a x t Lecture 5: Projectile motion, uniform circular motion 1 REVIEW: Goin from ONE to TWO Dimensions with Kinematics In Lecture 2, we studied the motion of a particle in just one dimension. The concepts of