Lecture4- Projectile Motion Chapter 4

Transcription

1 1 / 32 Lecture4- Projectile Motion Chapter 4 Instructor: Prof. Noronha-Hostler Course Administrator: Prof. Roy Montalvo PHY-123 ANALYTICAL PHYSICS IA Phys- 123 Sep. 28 th, 2018

2 2 / 32 Objectives Vector form of velocity and acceleration Motion in 2D and 3D Projectile Motion Frame of reference

3 3 / 32 When is 2D or 3D motion needed?

4 4 / 32 Motion vectors Position r = xî + yĵ + zˆk Velocity v = v x î + v y ĵ + v z ˆk Acceleration a = a x î + a y ĵ + a z ˆk

5 5 / 32 Displacement vector

6 6 / 32 Displacement Position r = xî + yĵ + zˆk So for the displacement r = r 2 r ( 1 ) ( ) = x 2 î + y 2 ĵ + z 2ˆk x 1 î + y 1 ĵ + z 1 2ˆk = (x 2 x 1 ) î + (y 2 y 1 ) ĵ + (z 2 z 1 ) ˆk = xî + yĵ + zˆk

7 7 / 32 Velocity Vector Recall average velocity average velocity=displacement/time interval Thus, v avg = r t xî + yĵ + zˆk v avg = t = x t î + y t ĵ + z t ˆk = (v x ) avg î + (v y ) avg ĵ + (v z ) avg ˆk

8 8 / 32 Instantaneous Velocity Vector Instantaneous Velocity limit as dt 0 v = d r dt Thus, v = d ( ) xî + yĵ + zˆk dt = dx dt î + dy dt ĵ + dz dt ˆk = v x î + v y ĵ + v z ˆk individual components v x = dx dt, v y = dy dt, v z = dz dt

9 Acceleration Vector Average accerlation average velocity=change in velocity/time interval Instantenous acceleration a avg = v t average velocity=change in velocity/time interval limit as dt 0 a x = dv dt, a y = dvy dt, a z = dvz dt a = d v dt v = a x î + a y ĵ + a z ˆk 9 / 32

10 10 / 32 Example What is the velocity vector? A spaceship flies with the position vector x = (1. + t 2 m)î (t3m)ĵ, what is it s velocity after 60 seconds? v x = dx dt = d dt = 2tm/s (1 + t 2) = 120m/s 100m/s (1) Notice Significant figures v y = dy dt = d ( t 3) dt = 3t 2 m/s = 10800m/s v = (2t m/s)î (3t2 m/s)ĵ 10000m/s

11 11 / 32 Example What is the velocity vector? A spaceship flies with the position vector x = (1. + t 2 m)î (t3m)ĵ, what is it s velocity after 60 seconds? Magnitude v = = vx 2 + vy 2 (100m/s) 2 + ( 10000m/s) 2 = 10, 000m/s = 10 4 m/s

12 12 / 32 When is 2D or 3D motion needed?

13 13 / 32 Initial Velocity components The initial velocity: v 0 = v 0,x î + v 0,y ĵ With the components of v 0,x = v 0 cos θ and v 0,y = v 0 sin θ

14 14 / 32 Horizontal Motion Only Recall the equation of motion: x x 0 = v 0,x t Substituting in v 0,x = v 0 cos θ x x 0 = v 0 cos θ

15 15 / 32 Vertical Motion Only y y 0 = (v 0 sin θ)t 1 2 gt2 v y = v 0 sin θ 0 gt vy 2 = (v 0 sin θ 0 ) 2 2g(y y 0 ) Derivation in book..

16 In two-dimensional motion in the x-y plane, what is the relationship between the x part of the motion to the y part of the motion? a) The x part of the motion is independent of the y part of the motion. b) The y part of the motion goes as the square of the x part of the motion. c) The x part of the motion is linearly dependent on the y part of the motion. d) The x part of the motion goes as the square of the y part of the motion. e) If the y part of the motion is in the vertical direction, then x part of the motion is dependent on the y part.

17 A ball is rolling down one hill and up another as shown. Points A and B are at the same height. How do the velocity and acceleration change as the ball rolls from point A to point B? a) The velocity and acceleration are the same at both points. b) The velocity and the magnitude of the acceleration are the same at both points, but the direction of the acceleration is opposite at B to the direction it had at A. c) The acceleration and the magnitude of the velocity are the same at both points, but the direction of the velocity is opposite at B to the direction it had at A. d) The horizontal component of the velocity is the same at points A and B, but the vertical component of the velocity has the same magnitude, but the opposite sign at B. The acceleration at points A and B is the same. e) The vertical component of the velocity is the same at points A and B, but the horizontal component of the velocity has the same magnitude, but the opposite sign at B. The acceleration at points A and B has the same magnitude, but opposite direction.

18 16 / 32 Range: horizontal distance the project traveled

19 Range: horizontal distance the project traveled Point when x x 0 = R and y y 0 = 0 Using R = vt = (v 0 cos θ 0 )t and 0 = y y 0 = (v 0 sin θ)t 1 2 gt2 one can find the range of the project: R = v 2 0 g sin 2θ 0 17 / 32

20 What describes the actual path (trajectory) of the projectile? Assume x 0 = 0 and y 0 = 0. gx 2 y = (tan θ 0 ) x 2 (v 0 cos θ 0 ) 2 Note that is has the form of y = ax + bx 2 i.e. a parabola 18 / 32

21 When does it reach its maximum height? Point C is the maximum height where v y = 0. What equation should we use? A x x 0 = v 0 cos θ B y y 0 = (v 0 sin θ)t 1 2 gt2 C v y = v 0 sin θ 0 gt D v 2 y = (v 0 sin θ 0 ) 2 2g(y y 0 ) E a = dv dt 19 / 32

22 20 / 32 What is the maximum height it reaches? Point C is the maximum height where v y = 0. Then, 0 = v 0 sin θ 0 t c 0.5gt 2 C t C = v 0 sin θ 0 g

23 What is the maximum height it reaches? What equation should we use to obtain the height (h C )? Hint θ = 90 degrees A x x 0 = v 0 cos θ B y y 0 = (v 0 sin θ)t 1 2 gt2 C v y = v 0 sin θ 0 gt D v 2 y = (v 0 sin θ 0 ) 2 2g(y y 0 ) E a = dv dt 21 / 32

24 22 / 32 What is the maximum height it reaches? Using t C = v 0 sin θ 0 g, h C = y y 0 = (v 0 sin θ)t 1 2 gt2 h C = (v 0 sin θ 0 ) 2 2g

25 23 / 32 Air resistance

26 24 / 32 Secret Agent Problems

27 A bicyclist is riding at a constant speed along a horizontal, straight-line path. The rider throws a ball straight up to a height a few meters above her head. Ignoring air resistance, where will the ball land? a) in front of the rider b) behind the rider c) in the opposite hand to the one that threw it d) in the same hand that threw the ball e) This cannot be determined without knowing the speed of the rider and the maximum height of the ball.

28 Frame of reference 1D Previous slides used: Motion described in FIXED reference frame (coordinate system) However: Nothing prevents us from choosing another frame moving at constant velocity v = const 25 / 32

29 26 / 32 Frame of reference- position Coordinate in v = const frame x PA = x PB + x BA

30 27 / 32 Frame of reference- velocity Velocity in v = const frame d dt (x PA) = d dt (x PB) + d dt (x BA)

31 28 / 32 Frame of reference- velocity Velocity in v = const frame v PA = v PB + v BA

32 Frame of reference- acceleration Acceleration in v = const frame d dt (a PA) = d dt (a PB) + d dt (a BA) }{{} =0 29 / 32

33 30 / 32 Frame of reference- acceleration Acceleration in v = const frame a PA = a PB

34 31 / 32 Examples

35 e) 25 m/s at 27 below the horizontal direction An eagle takes off from a tree branch on the side of a mountain and flies due west for 225 m in 19 s. Spying a mouse on the ground to the west, the eagle dives 441 m at an angle of 65 relative to the horizontal direction for 11 s to catch the mouse. Determine the eagle s average velocity for the thirty second interval. a) 19 m/s at 44 below the horizontal direction b) 22 m/s at 65 below the horizontal direction c) 19 m/s at 65 below the horizontal direction d) 22 m/s at 44 below the horizontal direction

36 A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance. a) 10.0 m/s b) 20.0 m/s c) 30.0 m/s d) 40.0 m/s e) The height of the cliff must be specified to answer this question.

37 32 / 32 Next Week Newton s Laws