Chapter 3 Kinematics in two and three dimensions. x and y components 1

Transcription

1 Chapter 3 Kinematics in two and three dimensions x and y components 1

2 Start with 1D Motion 3 independent equations Derive these 2 from the other 3 v = v + at 0 v = 1 avg 2 (v + v) 0 x = x 0 + v 0 t at 2 x = x 0 + v av t v 2 = v a(x! x 0 ) 2

3 A projectile at 1.5 m high is shot horizontally with speed v 0 that causes it to land 1.25 m away. y Example v 0 x 0 What is the time it is in the air What is v 0? 1.5 m y 0 What is v x and v y and the speed v when it lands? m x 3

4 x direction 3 independent equations Derive these 2 from the other 3 v x = v 0 x + a x t x = x 0 + v 0 x t a x t 2 v avgx = 1 2 (v 0 x + v x ) x = x 0 + v avgx t v 2 = v 2 + 2a (x! x ) x 0 x x 0 4

5 y direction 3 independent equations Derive these 2 from the other 3 v y = v 0 y + a y t y = y 0 + v 0 y t a yt 2 v 2 = v 2 + 2a (y! y ) y 0 y y 0 v avgy = 1 2 (v 0 y + v y ) y = y 0 + v avgy t 5

6 Now we want to use what we know about x and y components of a vector to help understand motion in the xy plane: We call this Projectile Motion Galileo separated motion into Horizontal and Vertical Horizontal: v is constant because a = 0 Vertical: a is constant because a = g (assume air resistance or friction is 0) 6

7 Combine x and y motion to find resultant. Consider a ball projected horizontally with initial speed v 0 x = x 0 + v 0 x t y = y 0! 1 2 gt 2 x 0 = 0 y x 0. v 0 v x = v 0 x = v 0 v y =!gt v 0 y = 0 y 0 0 x You can eliminate t to find trajectory in terms of x and y. Get a parabola. 7

8 A projectile at 1.5 m high is shot horizontally with speed v 0 that causes it to land 1.25 m away. What is time in the air? y x 0 v 0 t = time in the air: 1.5 m y 0 y = y o! 1 2 gt 2 x 0 = 1.50! 5t 2 t 2 = 0.30 t = 0.55 What is v 0? m 8

9 A projectile at 1.5 m high is shot horizontally with speed v 0 that causes it to land 1.25 m away. y v 0 x 0 What is v 0? v x = v m y 0 v o = x / t = 1.25 / 0.55 = 2.27m / s m x 9

10 A projectile at 1.5 m high is shot horizontally with speed v 0 that causes it to land 1.25 m away? y v 0 Find v y and the speed v when it lands? 1.5 m y 0 x 0 x v y =!gt =!10(0.55) =!5.5m / s m v = v x 2 + v y 2 v = (2.27) 2 + (!5.50) 2 v = 5.05 m /s 10

11 x! x 0 = (v 0 cos" 0 )t y! y 0 = (v 0 sin" 0 )t! 1 2 gt 2 v y = v 0 sin" 0! gt Projectile Motion v y 2 = (v 0 sin" 0 ) 2! 2g(y! y 0 ) v x = v 0 cos" 0 x = (v 0 cos! 0 )t y = (v 0 sin! 0 )t " 1 2 gt 2 v y = v 0 sin! 0 " gt v y 2 = (v 0 sin! 0 ) 2 " 2g(y " y 0 ) v x = v 0 cos! 0 x 0 = 0 y 0 = 0 11

12 x = x 0 + v 0 x t y = y 0 + v 0 y t! 1 2 gt 2 Projectile Motion v y = v 0 y! gt y v x = v 0 x v 0 v 0y x! x 0 = (v 0 cos" 0 )t y! y 0 = (v 0 sin" 0 )t! 1 gt 2 2 v y = v 0 sin" 0! gt v 2 y = (v 0 sin" 0 ) 2! 2g(y! y 0 ) v x = v 0 cos" 0 y 0 x 0 θ 0 v 0x v 0 x = v 0 cos! 0 v 0 y = v 0 sin! 0 x 12

13 Slow Pitch Problem 0.25 s A baseball is released at 3 ft above ground level. A stroboscopic plot shows the position of the ball every 0.25 sec. Answer the following questions. a) How far does it go when it reaches the batter? b) How long does it take the ball to reach the bat? c) What is the initial speed of the ball? R=40 ft t = 1.25 s 13

14 H θ 0 Find v 0x v 0 x = R / t c) What is the initial speed of the ball? v 0 = (v 2 0x +v2 0y )1/2 v 0 v 0 x = 40 ft /1.25s = 32 ft / s θ 0 v 0x v 0y To find v 0y. Consider what happens when the ball reaches the batter. y! y 0 = (v 0 y )t! 1 2 gt 2 0 = v 0y t! 1 2 gt 2 v 0y = 1 2 gt v 0 y = 1 (32)(1.25) = 20 ft / s 2 FINALLY v 0 = = 38 ft / s 14

15 H d) How high does the ball go above the ground? It reaches the maximum when v y = 0. H = v 0 y t! 1 gt 2 2 v y = v 0 y! gt = 0 t = v 0 y / g H = (v 0 y ) 2 / g! 1 g(v / 2 0 y g)2 H = v 2 0 y / 2g H = 202 = 6.25 ft 2! 32 y = = 9.25 ft (Above the ground) 15

16 Slow Pitch Problem 0.25 s H θ 0 e) What is the launch angle? v 0 θ 0 v 0x v 0y sin! 0 = v 0 y v 0! 0 = sin "1 ( v 0 y v 0 ) = sin "1 ( ) = 32o 16

17 Foul Shot problem Find the initial speed and the time it takes for the ball to go through the basket. x! x 0 = (v 0 cos" 0 )t y! y 0 = (v 0 sin" 0 )t! 1 2 gt 2 x! x 0 = 14! 1 = 13 ft y! y 0 = 10! 7 = 3 ft " 0 = 55 o 13 = (v 0 cos55)t 3 = (v 0 sin55)t! 1 2 gt 2 17

18 13 = (v 0 cos55)t Foul Shot Continued 3 = (v 0 sin55)t! 1 gt = 0.574v 0 t, v 0 t = = 0.819v 0 t! 1 gt = 0.819(22.65)! 1 gt 2 2!15.55 =!16t 2 t 2 = t = ±0.986 v 0 = /0.986 v 0 = ft /s Time it takes for the ball to go through the basket Initial speed of ball 18

19 Horizontal Range formula A cannon ball is fired with speed v 0 = 82 m/s at a ship 560 m from shore. What are the two launch angles needed to hit the ship? x component y component R = v 0 cos! 0 t 0 = v 0 sin! 0 t " 1 2 gt 2 0 = v 0 sin! 0 " 1 2 gt Solve for t t = 2v 0 sin! 0 g R = 2v 2 0 sin! 0 cos! 0 g = v 0 2 sin2! 0 g 19

20 Horizontal Range continued R = 2v 2 0 sin! 0 cos! 0 g = v 2 0 sin2! 0 g 2! 0 = sin "1 ( Rg v 0 2 ) 560 # 9.8 2! 0 = sin "1 ( ) = sin "1 (.816) 82 2! 0 = 54.7 o θ 0 = 27.3 degrees second solution 2θ 0 = = degrees θ 0 = 62.6 degrees 20

21 Important Concepts to Know Net Force = vector sum of all external forces acting on a body. Inertial reference frame - Where Newton s Laws hold true. one that is not accelerating- not the earth Mass - a characteristic that relates force to acceleration- leads to Newton's second Law m = F/a Free body diagram (modeling) important in isolating forces on a body in order to determine its motion 21

22 E: Earth M: Man W: Wagon Two Reference frames moving at a RELATIVE Velocity W V W/E E Constant V W/E a W/E = 0 22

23 Two Reference frames moving at a RELATIVE Velocity E: Earth M: Man W: Wagon W V M/W V W/E E Constant V W/E a W/E = 0 V M/E = V M/W + V W/E 23

24 Two Reference frames moving at a RELATIVE Velocity E: Earth M: Man W: Wagon W V M/W V W/E E Constant V W/E a W/E = 0 V M/E = V M/W + V W/E 0 a M /E = dv M /E dt = dv M /W dt + dv W /E dt a M/E = a M/W 24

25 Two Reference frames moving at a RELATIVE SPEED E: Earth M: Man W: Wagon W V M/W V W/E E Constant V W/E a W/E = 0 V M/E = V M/W + V W/E a M/E = a M/W Acceleration and forces are the same in Earth and Wagon reference frames as long as one frame moves at constant speed relative to the other. 25

26 GALILEAN RELATIVITY Person on a moving ship drops a rock. They see it fall straight down to land at their feet. Someone on shore sees the rock continue to move horizontally as it falls, and says the trajectory is parabolic. Another observer on another ship sees it move horizontally with a different speed and sees a different parabola. But they all see it land at the same time. Galileo concluded that the horizontal motion cannot influence the vertical motion, and that what takes place on the ship is independent of the ship s motion. He argued this applies to all physical and biological processes. The same principle applies to what takes place on a moving Earth. 26

27 RELATIVITY CAR v 0 v 27

28 ConcepTest You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: Dropping a Package 1) quickly lag behind the plane while falling 2) remain vertically under the plane while falling 3) move ahead of the plane while falling 4) not fall at all

29 ConcepTest You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: Dropping a Package 1) quickly lag behind the plane while falling 2) remain vertically under the plane while falling 3) move ahead of the plane while falling 4) not fall at all Both the plane and the package have the same horizontal velocity at the moment of release. They will maintain this velocity in the x- direction, so they stay aligned. Follow-up: What would happen if air resistance is present?

30 ConcepTest From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first? Dropping the Ball I (1) the dropped ball (2) the fired ball (3) they both hit at the same time (4) it depends on how hard the ball was fired (5) it depends on the initial height

31 ConcepTest Dropping the Ball I From the same height (and at the same time), one ball is dropped and another ball is fired horizontally. Which one will hit the ground first? (1) the dropped ball (2) the fired ball (3) they both hit at the same time (4) it depends on how hard the ball was fired (5) it depends on the initial height Both of the balls are falling vertically under the influence of gravity. They both fall from the same height. Therefore, they will hit the ground at the same time. The fact that one is moving horizontally is irrelevant remember that the x and y motions are completely independent!! Follow-up: Is that also true if there is air resistance?

32 ConcepTest Dropping the Ball II 1) the dropped ball In the previous problem, which ball has the greater speed at ground level? 2) the fired ball 3) neither they both have the same speed on impact 4) it depends on how hard the ball was thrown

33 ConcepTest Dropping the Ball II 1) the dropped ball In the previous problem, which ball has the greater speed at ground level? 2) the fired ball 3) neither they both have the same velocity on impact 4) it depends on how hard the ball was thrown Both balls have the same vertical velocity when they hit the ground (since they are both acted on by gravity for the same time). However, the fired ball also has a horizontal velocity. When you add the two components vectorially, the fired ball has a larger net velocity when it hits the ground.

34 ConcepTest Projectile A projectile is launched from the ground at an angle of 30 o. At what point in its trajectory does this projectile have the least speed? 1) just after it is launched 2) at the highest point in its flight 3) just before it hits the ground 4) halfway between the ground and the highest point 5) speed is always constant

35 ConcepTest Projectile A projectile is launched from the ground at an angle of 30 o. At what point in its trajectory does this projectile have the least speed? 1) just after it is launched 2) at the highest point in its flight 3) just before it hits the ground 4) halfway between the ground and the highest point 5) speed is always constant The speed is smallest at the highest point of its flight path because the y- component of the velocity is zero.

36 Dynamics: Newtons Laws of Mo3on Objects undergo motion and accelerations. This is caused by an interaction between bodies. Such interactions are called forces. Recall most of our forces are contact forces Newton s laws of motion: Published Newton s Principia (1642) 36

37 NEWTON S FIRST LAW An object con-nues in a state of rest or of mo-on at constant speed in a straight line unless acted upon by a net force. If you push it and let it go, it will move at constant velocity. Example : Hockey puck of mass m on ice Block of Ice 37

38 Examples of Contact Forces: A push or pull can be a force Normal force Tension in a string Friction Two balls colliding. Example of Noncontact forces Gravitation Electric

39 NEWTON S FIRST LAW An object con-nues in a state of rest or of mo-on at constant speed in a straight line unless acted upon by a net force. If you don t push it, it won t move. A body has inertia. Example : Hockey puck of mass m on ice Block of Ice 39

40 Iner3a Another way of understanding the First Law Inertia is a bodies resistance to change due to forces. Inertia is related to mass. Some examples of iner-a Hit a nail in a piece of wood on an anvil signg on your head Mass on string 40

41 ! F NEWTON S SECOND LAW! a =! F m = 1 m! i! F i = sum of all external forces acting on the body = net force System Mass Acceleration Force SI kg m/s 2 Newton (N) CGS g cm/s 2 dyne (dyn) BE slug (sl) ft/s 2 pound (lb) 41

42 This gives us our weight. Weight = F=mg 200 lbs 1 lb = 4.45 N 200 lbs = 890 N mg= 890 N m=890/9.81=90.7 kg 42

43 NEWTON S THIRD LAW When two bodies interact, the forces on the bodies due to each other are always equal in magnitude and opposite in direc3on. N M F ME F EM! F ME =!! F EM Gravita-onal Interac-on E 43

44 NEWTON S THIRD LAW When two bodies interact, the forces on the bodies due to each other are always equal in magnitude and opposite in direc3on. N M F MT! F MT =!! F TM T F TM Contact Interac-on E 44

45 Rules for drawing free body diagrams. Isolates the forces ac-ng on one body Represent the body by a point. Each force ac-ng on the body is represented by a vector with tail at the point and the length of vector indica-ng the approximate magnitude of the force. If the situa-on consist of several bodies which are rigidly connected, you can s-ll represent all the bodies by a point and use the total mass. Internal forces are not included. 45

46 What is the free body diagram of the block at rest on the table? Free body diagram of object with mass M M Table 46

47 Book leaning against a crate on a table at rest. What are the ac3on reac3on pairs? Table T 47

48 Problem: What is the accelera-on of the system of the two blocks and the contact force between the blocks? What is the net force on Block B? A B 24 kg 31 kg 65 N A B F AB = F BA 65 N Student Version 48 48

49 Problem: What is the accelera-on of the system of the two blocks and the contact force between the blocks? What is the net force on Block B? A B 24 kg 31 kg 65 N a = F / m = 65N / (24kg + 31kg) = 65N / 55kg a = 1.18 m/s 2 A B F AB = F BA 65 N F BA = m A a A Net Force on Block B=65.0 N N= N F BA = 24! 1.18 = 28.3 N Student Version 49 49

50 Now lets look at tension in a string Tension in the string is equal to the weight = 10 N The scale reads the tension in the string 50

51 Is the tension in the string any different when I have weights pulling it down on both sides? 10N 10 N 51

52 Problem 12 kg 24 kg 31 kg T 3 =65 N kg 12 kg 24 kg 31 kg 65 N a) What is the accelera-on of the system? b) Find T 1 c) Find T 2 a) b) c) T 3 = m sys a T 1 = m 1 a T 2 = (m 1 + m 2 )a a = T 3 m sys = 65N ( )kg = 0.97m / s2 T 1 = (12kg)(0.97m / s 2 ) = 11.6N T 2 = ( kg)(0.97m / s 2 ) = 34.9N T 2 = T 3! m 3 a = 65! (31)(0.97) = 65! 30.1 = 34.9N 52