Modeling Ballistics and Planetary Motion

Transcription

1 Discipline Couses-I Semeste-I Pape: Calculus-I Lesson: Lesson Develope: Chaitanya Kuma College/Depatment: Depatment of Mathematics, Delhi College of Ats and Commece, Univesity of Delhi Institute of Lifelong Leaning, Univesity of Delhi pg. 1

2 Table of Contents: Chapte: 1: Leaning Outcomes : Intoduction 3: Modeling the Motion of a Pojectile in a vacuum o 3.1: Height, Flight Time and Range fo Ideal Pojectile Motion 4: Ideal Tajectoy of Pojectile is Paabolic Execise-I 5: Keple's Lews o 5.1: Keple s Second Law Execise-II Summay Refeences 1. Leaning Outcomes Afte you have ead this chapte, you should be able to Modeling the Motion of a Pojectile in a vacuum Height, Flight Time and Range fo Ideal Pojectile Motion Ideal Tajectoy of Pojectile is Paabolic Keple's Laws fo Planetay Motion Poof of Keple's second Law Applications of Keple's Laws Institute of Lifelong Leaning, Univesity of Delhi pg.

3 . Intoduction: In this section, we shall discuss modeling the motion of a pojectile in a vacuum and Keple s second law. 3. Modeling the Motion of a Pojectile in a vacuum: In ode to deive equations fo pojectile motion, we assume that pojectile behaves like a paticle moving in a vetical co-odinate plane and that the only foce acting on the pojectile duing its flight is the constant foce of gavity, which always points staight down. Let the pojectile is launched fom the oigin at time t = into the fist quadant with an initial velocity v. Let v makes an angle with hoizontal, then v v cos i v sin j If we use the simple notion v fo the initial speed v, then v v i v ˆj (1) cos ˆ sin The pojectile s initial position is iˆ ˆj () Newton s second law of motion states that the foce acting on the pojectile is equal the pojectile s mass m times its acceleation. Let is the pojectile s position vecto and t is time. If the foce is only the gavitation foce mgˆ j, then Institute of Lifelong Leaning, Univesity of Delhi pg. 3

4 m d m g ˆj d gj ˆ We have to find value poblem as a function of t by solving the following initial d gj d Initial conditions: and v, when t =. Integating above equation, we have d g t ˆj v Again integating, we get 1 t gt ˆj v t Substituting the values of v and fom equations (1) and () gives 1 t gt ˆj vcos t iˆ vsint ˆj Institute of Lifelong Leaning, Univesity of Delhi pg. 4

5 Figue 1 The vecto equation fo ideal pojectile motion is ˆ 1 t vcos t i vsint gt ˆj (3) Whee is the pojectile s angle of elevation and v is the pojectile s initial speed Figue The components of t give the paametic equations x(t) = (v cos )t and y (t) = (v sin ) t - 1 gt. (4) Institute of Lifelong Leaning, Univesity of Delhi pg. 5

6 whee x(t) is distance downange and y(t) is the height of the pojectile at time t > Value Addition: Motion of a pojectile in a vacuum Conside a pojectile that tavels in a vacuum in a co-odinate plane, with x axis along level gound. Let the pojectile is fied fom height of s with initial speed v and angle of elevation, then a time t (t > ), it will at the point (x (t), y (t)) whee 1 y t gt v t s. and sin x t v cos t Example 1: A pojectile is fied fom the oigin ove hoizontal gound at an initial speed of 4 m/sec and a launch angle 3. Whee will the pojectile be 8 second late? Solution : given, v = 4, = 3, g = 9.8 and t = 8 we have to find the pojectile s components 8 seconds afte fiing. ˆ 1 t vcos t i vsint gt ˆj ˆ 1 8 4cos 3 8i 4 sin 3 t ˆj 3 ˆ i ˆ j iˆ ˆj 16 3 iˆ ˆj Institute of Lifelong Leaning, Univesity of Delhi pg. 6

7 Theefoe, 8 seconds afte fiing, the pojectile is about 186 m in the ai and 16 3 m downange Height, Flight Time and Range: Equation (3) enables us to answe most questions about the motion fo a pojectile fied fom the oigin. The pojectile eaches its highest point, when its vetical velocity component is zeo. dy v sin gt t v sin g Fo this value of t, the value of y is vsin 1 vsin ymax v sin g g g v g sin To find when the pojectile lands when fied ove hoizontal gound, we set the vetical component equal to zeo in equation (3) and the time of flight T satisfies v T gt 1 sin 1 T v sin g T v sin T, T g Institute of Lifelong Leaning, Univesity of Delhi pg. 7

8 Since is the time, when the pojectile is fied. T v sin g must be time, when the pojectile stikes the gound. To find the pojectile s ange R, we find the value of the hoizontal component, when T v sin g x ( v cos ) t R v sin g v cos v g v sin cos sin g The ange is lagest when sin = 1 = 9 = 45 Value Addition: Height, Flight Time and Range fo Ideal Pojectile Motion Fo ideal pojectile motion, when an object is launched fom the oigin ove a hoizontal suface with initial speed v and launch angle Institute of Lifelong Leaning, Univesity of Delhi pg. 8

9 Maximum height y max v sin g v sin Flight time T g Range R v sin g Example : Find the maximum height, flight time and ange of a pojectile fied fom the oigin ove hoizontal gound at an initial speed 5 m/sec and a launch angle of 6 Solution: Maximum height y max v sin g 3 5 5sin = 9566 m (appoximately) v sin Flight time = T g 3 5sin = 88.4 second (Appoximately) Institute of Lifelong Leaning, Univesity of Delhi pg. 9

10 Range = R v sin g sin =,98 m (Appoximately) The position vecto of the pojectile is ˆ 1 t vcos t i vsint gt ˆj ˆ 5cos 6 5sin t i t t ˆj 5 ˆ t i t t j ˆ 4. Ideal Tajectoy of Pojectile is Paabolic: The paametic equations fo the motion of a pojectile povide useful geneal infomation about pojectile s motion. If, we can x eliminate t by substituting t v cos in equation (4), we obtain the Catesian equation. g y x tan x v cos The equation is of the fom y ax bx Institute of Lifelong Leaning, Univesity of Delhi pg. 1

11 Figue 3 Hence its gaph is a paabola Figue 3 is the path of a pojectile fied fom (x, y ) with an initial velocity v at an angle of with hoizontal. Example 3: A boy standing at the edge of a cliff thows a ball upwad at 3 angle with an initial speed of 64 ft/s. Suppose that when the ball leaves the boy s hand, it is 48 ft above the gound at the base of the cliff. (i) (ii) (iii) What ae the time of flight of the ball and its ange? What ae the velocity of the ball and its speed at impact? What is the highest point eached by the ball duing its flight? Solution: given, g = 3 ft/s, s = 48 ft. v 64 ft s and = 3 The paametic equations of ball s tajectoy ae x t v cos t 64cos3 t 3 3 t 1 yt g t v sin t s Institute of Lifelong Leaning, Univesity of Delhi pg. 11

12 1 3 t 64sin3 t t 3 t 48 (i) The ball hits the gound, when y = 16t + 3t + 48 =, fo t > t -4t 6 = t t 3 = t 3t + t 3 = t (t 3) + 1 (t 3) = (t 3) (t + 1) = t = 3 and t = -1 So time of flight is T = 3 seconds. The ange is R = x(3) = 3 3 (3) = (Appoximately) The ball hits the gound about 166 ft fom the base of the cliff. (ii) We find that x' t 3 3 y ' t 3t 3 Hence the velocity at time t is 3 3ˆ 3 3 v t i t ˆj Thus, at impact, the velocity is Institute of Lifelong Leaning, Univesity of Delhi pg. 1

13 3 3 3ˆ 64 ˆ v i j v Its speed is = (Appoximately) (iii) The speed at impact is about 85 ft/s The ball attains its maximum height, when the upwad v t component of its velocity y (t) = 3t + 3 = t = 1 Theefoe, the maximum height attained by the ball is ymax y xmax x (Appoximately) Hence the highest point eached by the ball has co-odinates (ounded to the neaest foot) 55, 64 Example 4: A pojectile is fied fom the gound level at an angle of 4 with muzzle speed of 11 ft/s. Find the time of flight and the ange. Solution. Given, v = 11 ft/s, = 4 and g = 3 ft/s Institute of Lifelong Leaning, Univesity of Delhi pg. 13

14 The time of flight is given by T v sin 11 sin 4 g (Appoximately) The ange R is given by R v 11 sin 8 sin g (appoximately) Hence, the pojectile tavels about 37 ft hoizontally befoe it hits the gound and the flight takes a little moe than 4 seconds. Execise-I 1. A pojectile is fied at a speed of 84 m/sec at an angle of 6. How long will it take to get 1 km downange?. A shell fied fom gound level at an angle of 45 hits the gound m away. What is the muzzle speed of the shell? 3. A base ball is hit when it is 3 ft above the gound. It leaves the bat with initial speed of 15 ft/sec, making an angle of with the hoizontal. At the instant the ball is hit, an instaneous gust of wind blows in the hoizontal diection diectly opposite the diection the ball is taking towad the outfield, adding a component of -8.8 î (ft/sec) to the ball s initial velocity (8.8 ft/sec = 6 mph) (i) Find the vecto equation fo the path of the baseball. Institute of Lifelong Leaning, Univesity of Delhi pg. 14

15 (ii) How high does the baseball go, and when does it each maximum height? (iii) Assuming that the ball is not caught, find its ange and flight time. 4. At what angle (to the neaest tenth of a degee) should a pojectile be fied fom gound level if its muzzle speed is ft/s and the desied ange is 6 ft? 5. A shell is fied at gound level with a muzzle speed of 8 ft/s and at an elevation of 45 fom gound level. (i) Find the maximum height attained by the shell. (ii) Find the time of flight and the ange of the shell. (iii) Find the velocity and speed of the shell at impact. 6. If a shot putte thows a shot fom a height of 5 ft with an angle of 46 and initial speed of 5 ft/s, what is the hoizontal distance of the thow? 5. Keple s Laws: In the seventeenth centuy, the Geman astonome Johannes Keple fomulated thee laws fo descibing planetay motion which ae given below. 1. The planets move about the sun in elliptical obits, with the sun at one focus.. The adius vecto joining a planet to the sun sweeps out equal aeas in equal intevals of time. Institute of Lifelong Leaning, Univesity of Delhi pg. 15

16 3. The squae of the time of one complete evolution of a planet about its obit is popotional to the cube of the length of the semi majo axis of its obit. We shall pove the second law using vecto methods. The othe two laws can be poved similaly. Let û and û denote unit vectos along the adial axis and othogonal to that axis espectively. Figue 4: Desciption of the motion of a paticle along a cuve. Then, in tems of unit vectos î and ĵ, We have ˆ uˆ cos i sin ˆj ˆ uˆ sin i cos ˆ j On diffeentiating above equations w..t., we have duˆ siniˆ cos ˆj uˆ d duˆ cosiˆ sin ˆj uˆ d Institute of Lifelong Leaning, Univesity of Delhi pg. 16

17 Now, suppose the sun S is at the oigin (pole) of a pola co-odinate system and conside the motion of a body B about S. The adial vecto SB can be witten as uˆ cos i sin ˆj ˆ Whee = and the velocity v satisfies d d duˆ v uˆ d duˆ d uˆ. d d d uˆ ˆ u We summaize these fomulas in the following box. Value Addition: Note Pola fomulas fo velocity and Acceleation d d v t uˆ ˆ u a t v t d d d ˆ d u d d ˆ u Institute of Lifelong Leaning, Univesity of Delhi pg. 17

18 t t iˆ t ˆj Example 5: The position vecto of a moving body is fo t >. Expess and the velocity vecto vt in tems of u and u. t tiˆ t ˆj Solution: 4 t t t 4t t t t 4 Hee, x = t, y = t We know that ˆ u t t 4u ˆ Since, d d v t u u ˆ ˆ, d we need d and We find that d 1 1 t 4 t t 4 t t 4 t 4 1 y 1 t 1 t tan tan tan x t Institute of Lifelong Leaning, Univesity of Delhi pg. 18

19 d 1 1 t t 4 1 Thus, t 4 ˆ 4 4 t t v t u t t uˆ 4 t uˆ tuˆ 4 t Keple s Second Law: The adius vecto fom the sun to a planet in its obit sweeps out equal aeas in equal intevals of time. Poof. We will assume that the only foce acting on a planet is the gavitational attaction of the sun. Figue 5: The adial line sweeps out equal aea in equal time. Accoding to the univesal law of gavitation, the foce of attaction is given by GmM F u ˆ (1) Whee G is gavitational constant and m and M ae the masses of the planet and the sun espectively. Institute of Lifelong Leaning, Univesity of Delhi pg. 19

20 By Newton s second law of motion F m a () Whee a is the acceleation of the planet in its obit fom equations (1) and (), we have mm ˆ m a G u GM a u ˆ component The acceleation of a planet in its obit has only a adial This means that the û component of the planet s acceleation is. By examining the pola fomula fo acceleation, we see that this condition is equivalent to the diffeential equation. d d d (i) We know that (ii) d d d Fom equations (i) and (ii), we have d d d 1 1 d 1 1 d d Institute of Lifelong Leaning, Univesity of Delhi pg.

21 log log log c logc c d c d c Let [t 1, t ] and [t 3, t 4 ] be two time intevals of equal length. t t 1 = t 4 t 3 The aea swept out in the time peiod [t 1, t ] is t t 1 1 d S1 d t1 t1 t t1 1 d c sn i ce c 1 1 c t c t t t t1 1 Similaly, the aea swept out in time peiod [t 3, t 4 ] is 1 S ct t S c t t c t t S Hence equal aea is swept out in equal time as claimed by Keple. Institute of Lifelong Leaning, Univesity of Delhi pg. 1

22 Example 6: If the cental acceleation is equal to, find the obit. Solution: We know that h dp p d 3 Integating both side, we have h p c Compaing with b p a l which is an ellipse o hypebola efeed to the focus, we get h c b a l h b l, whee l is the semi-latus ectum a c a Hence, the obit is an ellipse, paabola o hypebola accoding as c is negative, zeo o positive. Also v h 1 c p a 1 Hence, fo elliptic obit, v a 1 Fo hypebolic obit, v a Institute of Lifelong Leaning, Univesity of Delhi pg.

23 Fo paabolic obit, v Value Addition : Deduction Since 1 h is the aeal velocity. 1 h. T = Aea of ellipse = ab peiodic time T ab h ab a b a 3 Whee is constant. Example 7: If v 1 and v ae the velocities of a planet, when it is espectively neaest and fathest fom the sun, pove that 1 ev 1 e v 1 Solution: Fo an elliptic obit, we have 1 v a v e a 1e a a 1 e v 1 1 e a 1e a a 1 e Institute of Lifelong Leaning, Univesity of Delhi pg. 3

24 Hence, v v 1 1 e 1 e v v e e 1 ev 1 e v 1 Example 8: If be the angula velocity of a planet at the neae end of the majo axis, pove that its peiod is 1 e 1 e 3. Solution: By Keple s second law of motion h a e w h 1 and h l a 1 e a 4 1 e 4 h a1 e 3 e 1 3 a 1 e Hence, T 3 1 e a 1 e 3 Execise-II 1. A paticle moves along the pola path (, ), moe 3 sin t, (t) = t 3. Find v t and t t û. a in tems of û and Institute of Lifelong Leaning, Univesity of Delhi pg. 4

25 . If the velocity of the eath at any point of its obit, assumed cicula, wee inceased by about one-half, pove that it would descibe a paabola about the sun as focus. Show about that if a body wee pojected fom the eath with a velocity exceeding 7 miles pe second, it will not etun to the eath. 3. If a planet was suddenly stopped in its obit supposed cicula, show that it would fall into the sun in a time which is times the peiod of the planet s evolution Using the fomula d d v t uˆ ˆ u, Pove that pola acceleation d d ˆ d a t u d. d ˆ u 5. A paticle is pojected fom eath s suface with velocity v. Show that if the dimension of gavity is taken into account, but the esistance of the ai is neglected, the path is an ellipse of majo axis. ga ga v, whee a is the eath s adius. Summay: In this lesson, we have emphasize on the following topics: Modeling the Motion of a Pojectile in a vacuum Height, Flight Time and Range fo Ideal Pojectile Motion Ideal Tajectoy of Pojectile is Paabolic Institute of Lifelong Leaning, Univesity of Delhi pg. 5

26 Keple's Laws fo Planetay Motion Poof of Keple's second Law Applications of Keple's Laws Refeences: 1. H. Anton, I. Bivens and S. Davis, Calculus, John Wiley and Sons, Inc.,.. G.B. Thomas and R.L. Finney, Calculus, Peason Education, Mauice D. Wei, Joel Hass, Fank R. Giodano, Thomas Calculus (Eleventh Edition), Peason Education. 4. M.J. Stauss, G.L. Badley, K.J. Smith, Calculus (3 d Edition), Peason Education. Institute of Lifelong Leaning, Univesity of Delhi pg. 6