Motion in Two and Three Dimensions

Transcription

1 PH 1-A Fall 014 Motion in Two and Three Dimensions Lectures 4,5 Chapter 4 (Halliday/Resnick/Walker, Fundamentals of Physics 9 th edition) 1

2 Chapter 4 Motion in Two and Three Dimensions In this chapter we will continue to study the motion of objects without the restriction we put in chapter to moe along a straight line. Instead we will consider motion in a plane (two dimensional motion) and motion in space (three dimensional motion). The following ectors will be defined for twoand three- dimensional motion: Displacement Aerage and instantaneous elocity Aerage and instantaneous acceleration We will consider in detail projectile motion and uniform circular motion as examples of motion in two dimensions. Finally we will consider relatie motion, i.e. the transformation of elocities between two reference systems which moe with respect to each other with constant elocity.

3 Position Vector The position ector r of a particle is defined as a ector whose tail is at a reference point (usually the origin O) and its tip is at the particle at point P. Example : The position ector in the figure is: r xiˆ yj ˆzkˆ P r 3iˆˆj5kˆ m 3

4 Displacement Vector For a particle that changes postion ector from r1 to r we define the displacement ector r as follows: r r r1 The position ectors r1 and r are written in terms of components as: r xiˆ yj ˆ zk r x iˆ y ˆjzk ˆ ˆ The displacement r can then be written as: r x x iˆ y y ˆj z z kˆxiˆyj ˆzkˆ x x x 1 y y y 1 z z z 1 t 1 t 4

5 Problem. A watermelon seed has the following coordinates: x 5.0 m, y 8.0 m, z0 m. Find its position ector (a) in unitector notation and as (b) a magnitude and (c) an angle relatie to the positie direction of the x axis. (d) Sketch the ector on a right handed coordinate system. If the seed is moed to the xyz coordinates (3.00m, 0m, 0m) what is its displacement (e) in unit-ector notation as as (f) a magnitude and (g) an angle relatie to the positie x direction? r xiˆ yj ˆzkˆ 5

6 Aerage and Instantaneous Velocity Following the same approach as in chapter we define the aerage elocity as: displacement aerage elocity = time interal ag r xiˆyj ˆzkˆ xiˆ yj ˆ zkˆ t t t t t t t + Δt We define as the instantaneous elocity (or more simply the elocity) as the limit: lim t 0 r t dr 6

7 If we allow the time interal t to shrink to zero, the following things happen: 1. Vector r moes towards ector r1 and r 0 r. The direction of the ratio (and thus ag )approaches the direction t of the tangent to the path at position 1 3. ag d ˆ ˆ ˆ dx ˆ dy ˆ dz xi yjzk i j kˆ ˆ ˆ ˆ xi y jzk The three elocity components are gien by the equations: t t + Δt dr x y dx dy z dz 7

8 Problem 6. An electron position is gien by r3.00tiˆ4.00t ˆj.00 k, with t in seconds and r in meters. (a) In unit-ector notation, what is the electron's elocity t ( )? At t.00 s, what is (b) in unit-ector notation (a) Eq. leads to ˆ and as (c) a magnitude and (d) an angle relatie to the positie direction of the x axis? dr ˆ t ( ) d (3.00ti 4.00t ˆj +.00k) ˆ (3.00 m/s)i ˆ (8.00 tm/s) ˆj (b) Ealuating this result at t =.00 s produces = (3.00i ˆ 16.0j) ˆ m/s. (c) The speed at t =.00 s is (3.00 m/s) ( 16.0 m/s) 16.3 m/s. (d) The angle of at that moment is m/s tan 79.4 or m/s where we choose the first possibility (79.4 measured clockwise from the +x direction, or 81 counterclockwise from +x) since the signs of the components imply the ector is in the fourth quadrant. 8

9 Aerage and Instantaneous Acceleration The aerage acceleration is defined as: change in elocity aerage acceleration = time interal a ag 1 t t We define as the instantaneous acceleration as the limit: d d lim ˆ ˆ ˆ d d x ˆ y ˆ dz a ˆ ˆ ˆ ˆ xi y jzk i j k axi ay jazk t t 0 Note: Unlike elocity, the acceleration ector does not hae any specific relationship with the path. The three acceleration components are gien by the equations: a x d x a y d y a z d z a d 9

10 Problem 11. The position r of a particle moing in an xy plane is gien by 3 ˆ 4 ( ) ( ) ˆ r t t i t j, with r in meters and t in seconds. In unit-ector notation, calculate (a) r, (b), (c) a for t.00 s. (d) W hat is the angle between the positie direction of the x axis and a line tangent to the particle at t =.00 s? dr dx ax In parts (b) and (c), we use. For part (d), we find the direction of the elocity computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the gien expression, we obtain r [.00(8) 5.00()]i ˆ + [ (16)] ˆj (6.00 ˆi 106 ˆ j) m t.00 (b) Taking the deriatie of the gien expression produces t t ˆ t 3 ( ) = ( ) i 8.0 j ˆ where we hae written (t) to emphasize its dependence on time. This becomes, at t =.00 s, = (19.0 ˆi 4 ˆj) m/s. (c) Differentiating the ( t ) found aboe, with respect to t produces which yields a =(4.0 ˆi 336 ˆj) m/s at t =.00 s. ˆ 1.0t i 84.0t j, ˆ (d) The angle of, measured from +x, is either 1 4 m /s tan 85. or m /s where we settle on the first choice ( 85., which is equialent to 75 measured counterclockwise from the +x axis) since the signs of its components imply that it is in the fourth quadrant. 10

11 Projectile Motion The motion of an object in a ertical plane under the influence of graitational force is known as projectile motion The projectile is launched with an initial elocity The horizontal and ertical elocity components are: iˆ ˆj o ox oy g cos ox o o sin oy o o Projectile motion will be analyzed in a horizontal and a ertical motion along the x- and y-axes, respectiely. These two motions are independent of each other. Motion along the x- axis has zero acceleration. Motion along the y-axis has uniform acceleration a y = -g 11

12 Horizontal Motion: x 0 0 a x 0 The elocity along the x-axis does not change cos (eqs.1) xx t cos t (eqs.) Vertical Motio n: a g Along the y-axis the projectile is in free fall y o ox o o gt gt y 0sin0 gt (eqs.3) y yo oyt 0sin0t (eqs.4) y sin If we eliminate t between equations 3 and 4 we get: g y y 0 0 o g Here x and y are the coordinates o o of the launching point. For many problems the launching point is taken at the origin. In this case x o 0 and y 0 o Note: In this analysis of projectile motion we neglect the effects of air resistance 1

13 The equation of the path: gt x o cos o t (eqs.) y 0sin 0t (eqs.4) If we eliminate t between equations and 4 we get: y tan o x cos o g o x This equation describes the path of the motion The path equations has the form: y axbx This is the equation of a parabola Note: The equation of the path seems too complicated to be useful. Appearances can deceie: Complicated as it is, this equation can be used as a short cut in many projectile motion problems 13

14 cos (eqs.1) x cos t (eqs.) O x 0 0 R sin Acos A sin A A t o gt y 0sin 0 gt (eqs.3) y 0sin 0t (eqs.4) O / Horizontal Range: The distance OA is defined as the horizantal range R At point A we hae: y 0 From equation 4 we hae: gt gt 0sin0t 0 t0sin0 0 This equation has two solutions: Solution 1. t 0 This solution correspond to point O and is of no interest Solution. sin 0 0 gt 0 0 From solution we get: t If we substitute t in eqs. we get: o 0 This solution correspond to point A sin g o o R sinocoso sin o g g R has its maximum alue when 45 R max o g sin o 3/ 14

15 t A H g Maximum height H H o sin o g The y-component of the projectile elocity is: At point A: 0 sin gt t H y 0 0 sin gt sin g gt 0sin0 g 0sin0 () 0sin0 0sin0 H y t t g sin g o o 15 y g

16 t A Maximum height H (encore) H g H o sin o g We can calculate the maximum height using the third equation of kinematics for motion along the y-axis: y yo a y yo In our problem: y 0, y H, sin, 0, and ag gh H yo o yo o o y g g yo osin o 16

17 Problem : A car dries straight off the edge of a cliff that is 54m high. The police at the scene of the accident note that the point of impact is 130m from the base of a cliff. How fast was the car traeling when it went oer the cliff? During this time the car traels a horizontal distance of 130m

18 Problem : A dier springs upward from a board that is three meters aboe the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75.0 degrees with respect to the horizontal surface of the water. Determine her initial elocity, both magnitude and direction

19 The motion of a ballistic missile can be regarded as the motion of a projectile because along the greatest part of its trajectory the missile is in free fall. Suppose that a missile is to strike a target 1000 km away. What minimum speed must the missile hae at the beginning of its trajectory? What maximum height does it reach when launched with this minimum speed? How long does it take to reach the target? For these calculations assume that g = 9.8 m/s eerywhere along the trajectory. ~7.5 min

20 Problem: A projectile is fired at an initial elocity of 35.0 m/s at an angle of 30.0 degrees aboe the horizontal from the roof of a building 30.0 m high, as shown. Find a) The maximum height of the projectile b) The time to rise to the top of the trajectory c) The total time of the projectile in the air d) The elocity of the projectile at the ground e) The range of the projectile

21 Problem 43. A ball is shot from the ground into the air. At a height of 9.1 m, its elocity is (7.6iˆ 6.1 ˆj) m.s, with iˆ horizontal and ˆj upward. (a) To what maximum height does the ball rise? (b) What total horizontal distance does the ball trael? What are the (c) magnitude and (d) angle (below the horizontal) of the ball's elocity just before it hits the ground? We designate the gien elocity ( 7.6 m / s ) ˆi ( 6.1 m / s ) ˆj as 1 as opposed to the elocity when it reaches the max height or the elocity when it returns to the ground 3 and take 0 as the launch elocity, as usual. The origin is at its launch point on the ground. (a) Different approaches are aailable, but since it will be useful (for the rest of the problem) to first find the initial y elocity, that is how we will proceed. g y (6.1 m /s) (9.8 m /s )(9.1 m ) 1 y 0 y 0 y which yields 0 y = 14.7 m/s. Knowing that y must equal 0, we use y = h for the maximum height: y 0 y g h 0 (14.7 m /s) (9.8 m /s ) h which yields h = 11 m. (b) Recalling the deriation of Eq. 4-6, but using 0 y for 0 sin 0 and 0x for 0 cos 0, we hae y t g t, R 0 x t which leads to R 0 x 0 y / g. Noting that 0x = 1x = 7.6 m/s, we plug in alues and obtain R = (7.6 m/s)(14.7 m/s)/(9.8 m/s ) = 3 m. (c) Since 3x = 1x = 7.6 m/s and 3y = 0 y = 14.7 m/s, we hae 3 3 x 3 y (7.6 m /s) ( 14.7 m /s) 17 m /s. (d) The angle (measured from horizontal) for 3 is one of these possibilities: m ta n 6 3 o r m where we settle on the first choice ( 63, which is equialent to 97 ) since the signs of its components imply that it is in the fourth quadrant. 1

22 Uniform circular Motion A particles is in uniform circular motion it moes on a circular path of radius r with constant speed. Een though the speed is constant, the elocity is not. The reason is that the direction of the elocity ector changes from point to point along the path. The fact that the elocity changes means that the acceleration is not zero. The acceleration in uniform circular motion has the following characteristics: 1. Its ector points towards the center C of the circular path, thus the name centripetal. Its magnitude a is gien by the equation: a r r Q r C r P The time T it takes to complete a full reolution is known as the period. It is gien by the equation: R T r

23 ˆ ˆ sin ˆ cos ˆ yp xp xi y j i j sin cos r r Here xp and yp are the coordinates of the rotating particle yp ˆ xp ˆ d dyp Acceleration = ˆ dxp i j a i ˆj r r r r dyp dxp We note that: y cos and x sin a i j a a a r r r r cos ˆ sin ˆ x y cos sin ay ax r P / r sin tan tan a points towards C / cos x sin cos y C A C cos sin 1 3

24 Problem 56. An earth satellite moes in a circular orbit 640 km aboe Earth's surface with a period of 98.0 min. What are the (a) speed and (b) magnitude of the centripital acceleration of the satellite? T r We apply to sole for speed and to find acceleration a. (a) Since the radius of Earth is m, the radius of the satellite orbit is r = ( ) m = m. a r Therefore, the speed of the satellite is c h b gb g 6 r m T min 60 s /min m/s. (b) The magnitude of the acceleration is a r c h m/s m m/s. 4

25 Relatie Motion in One Dimension: The elocity of a particle P determined by two different obserers A and B aries from obserer to obserer. Below we derie what is known as the transformation equation of elocities. This equation gies us the exact relationship between the elocities each obserer perceies. Here we assume that obserer B moes with a known constant elocity BA with respect to obserer A. Obserer A and B determine the coordinates of particle P to be x PA and x PB, respectiely. xpa xpb xba Here xba is the coordinate of B with respect to A d d d We take deriaties of the aboe equation: xpa xpb xba PA PB BA If we take deriaties of the last equation and take d BA into account that 0 apa apb Note: Een though obserers A and B measure different elocities for P, they measure the same acceleration 5

26 Relatie Motion in Two Dimensions: Here we assume that obserer B moes with a known constant elocity BA with respect to obserer A in the xy-plane. Obserers A and B determine the position ector of particle P to be rpa and rpb, respectiely. r r r We take the time deriatie of both sides of the equation PA PB BA d r PA d r d PB r BA PA PB BA PA PB BA If we take the time deriatie of both sides of the last equation we hae: d d d dba P B B If we take into account that 0 a PA a PA A PB Note: As in the one dimensional case, een though obserers A and B measure different elocities for P, they measure the same acceleration 6

27 Problem 76. A light plane attain an airspeed of 500 km/h. The pilot sets out for a destination 800 km due north but discoers that the plane must be headed 0.0 east of due north to fly there directly. The plane arries in.00 h. What were the (a) magnitude and (b) direction of the wind elocity? The destination is D = 800 km j^ where we orient axes so that +y points north and +x points east. This takes two hours, so the (constant) elocity of the plane (relatie to the ground) is pg = (400 km/h) j^. This must be the ector sum of the plane s elocity with respect to the air which has (x,y) components (500cos70º, 500sin70º) and the elocity of the air (wind) relatie to the ground ag. Thus, which yields (400 km/h) j^ = (500 km/h) cos70º i^ + (500 km/h) sin70º j^ + ag (a) The magnitude of ag (b) The direction of ag is is ag =( 171 km/h)i^ ( 70.0 km/h)j^. ag ( 171 km/h) ( 70.0 km/h) 185 km/h. pg ag pa km/h tan.3 (south of west). 171 km/h 7