1 / 61

# Chapter 3

Chapter 3. Motion in Two or Three Dimensions. Goals for Chapter 3. Vectors to represent the position of a body Velocity vector using the path of a body Acceleration vector of a body Describe the curved path of projectile Investigate circular motion

Télécharger la présentation

## Chapter 3

E N D

### Presentation Transcript

1. Chapter 3 Motion in Two or Three Dimensions

2. Goals for Chapter 3 • Vectors to represent the position of a body • Velocity vector using the path of a body • Acceleration vector of a body • Describe the curved path of projectile • Investigate circular motion • (Describe the velocity of a body as seen from different frames of reference)

3. Projectile Motion Examples of projectile motion. Notice the effects of air resistance

4. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

5. Projectile motion—Figure 3.15 • Projectile is any body with initial velocity that follows path determined by gravity (& air resistance). • Begin by assuming “g” is constant (near ground), & neglecting air resistance, Earth’s curvature, rotation, & motion.

6. Projectile Motion Understood by analyzing the horizontal and vertical motions separately.

7. Projectile Motion • Photo shows 2 balls starting to fall at the same time. • Yellow ball on right has an initial speed in the x-direction. • Red ball on left has vx0 = 0 • Vertical positions of the balls are identical at identical times • Horizontal position of yellow ball increases linearly in time.

8. Projectile Motion The speed in the x-direction is constant; in the y-direction the object moves with constant accelerationg.

9. The x and y motion are separable—Figure 3.16 We can analyze projectile motion as horizontal motion with constantvelocity and vertical motion with constantacceleration: ax = 0and ay = g.

10. Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, Usually where vertical acceleration is g and is down.

11. Solving Problems Involving Projectile Motion • Read the problem carefully, and choose the object(s) you are going to analyze. • Draw a diagram. • Choose an origin and a coordinate system. • Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. • Examine the x and y motions separately.

12. Solving Problems Involving Projectile Motion • 6. List known and unknown quantities. • Remember that vx never changes, and that vy = 0 at the highest point. • 7. Plan how you will proceed. • Use the appropriate equations; you may have to combine some of them, or find time from one set and use in the other

13. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

14. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We WANT: vx (m/s)

15. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: vy0 = 0 m/s!

16. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: Dy = -50 m!

17. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: Dx = +90 m!

18. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We KNOW: ax = 0! ay = -9.8!

19. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. For Horizontal: Dx = +90 m = vxt vx =90/t How to get t ??

20. Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. Get t from Vertical: Dy = -50 m! Dy = v0yt – ½ gt2 -50 = - ½ gt2

21. Projectile Motion If object is launched at initial angle θ0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.

22. Solving Problems Involving Projectile Motion Example: A kicked football. A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown. It travels through the air (assume no resistance) and lands at the ground.

23. Example: A kicked football. Calculate the maximum height, the time of travel before the football hits the ground, how far away it hits the ground. the velocity vector at the maximum height, the acceleration vector at maximum height.

24. Solving Problems Involving Projectile Motion • STEP 1: Coordinates! (and signs!) • +x horizontally to the right • +y vertically upwards • acceleration of gravity = -9.8 m/s/s

25. Solving Problems Involving Projectile Motion • STEP 2: COMPONENTS! (and signs!) • +vx0 horizontally to the right = v0 cosq • +vy0 vertically upwards = v0 sinq • ay acceleration of gravity = -9.8 m/s/s only in y

26. Solving Problems Involving Projectile Motion • STEP 3: LOOK FOR 3 THINGS! • vf = vi + at • Dx = ½ (vi + vf)*t • Dx = vi*t + ½ at2 • Dx = vf*t – ½ at2 • Vf2 = Vi2 +2aDx

27. Solving Problems Involving Projectile Motion • In x direction, ax = 0! • vfx = vix • Dx = vix*t = v cosq * t

28. Solving Problems Involving Projectile Motion • In y direction, ay = -9.8! • vfy = viy – 9.8t = v0 sin q - gt • Dy = ½ (viy + vfy)*t • Dy = viy*t – 4.9t2 = (v0 sin q) t - ½ gt2 • Dx = vfy*t + 4.9t2 • vfy2 = viy2 – 19.6Dy = (v0 sin q)2 - 2gDy

29. Solving Problems Involving Projectile Motion

30. Solving Problems Involving Projectile Motion • Solve it! • vx = 16 m/s • vy = 12 m/s • Max height = 7.3 m • Max distance in x (“range”) = 39 m • Time = 2.5 seconds

31. Example: Level horizontal range. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y0. (a) Derive a formula for the horizontal range R of a projectile in terms of its initial speed v0 and angle θ0.

32. Solving Problems Involving Projectile Motion Example: Level horizontal range. (b) Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?

33. Solving Problems Involving Projectile Motion Suppose the football was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground?

34. Solving Problems Involving Projectile Motion The wrong strategy! A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.

35. Rescue helicopter drops supplies. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped?

36. Rescue helicopter drops supplies. (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?

37. Solving Problems Involving Projectile Motion Projectile motion is parabolic: Taking the equations for x and y as a function of time, and combining them to eliminate t, we find y as a function of x: This is the equation for a parabola.

38. The equations for projectile motion • If we set x0 = y0 = 0, the equations describing projectile motion are shown at the right. • The trajectory is a parabola.

39. A body projected horizontally • A motorcycle leaves a cliff horizontally. • Follow Example 3.6.

40. Height and range of a projectile • A baseball is batted at an angle. • Vo = 37.0 m/s at 53.1 degrees; @ 2 seconds where is it?

41. Maximum height and range of a projectile • What initial angle will give the maximum height and the maximum range of a projectile? • Follow Example 3.8.

42. Tranquilizing a falling monkey • Where should the zookeeper aim? • Follow Example 3.10.

43. The effects of air resistance—Figure 3.20 • Calculations become more complicated. • Acceleration is not constant. • Effects can be very large. • Maximum height and range decrease. • Trajectory is no longer a parabola.

44. Position vector • The position vector from the origin to point P has components x, y, and z.

45. Average velocity—Figure 3.2 • Average velocity between two points = displacement divided by time • Same direction as the displacement.

46. Instantaneous velocity • Instantaneous velocityis instantaneous rate of change of position vector with respect to time. • Components of instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt. • Instantaneous velocity of a particle is always tangent to its path.

47. Average acceleration • The average acceleration during a time interval t is defined as the velocity change during t divided by t.

48. Average acceleration • The average acceleration during a time interval t is defined as the velocity change during t divided by t.

49. Instantaneous acceleration • Instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time. • Any particle following a curved path is accelerating, even if it has constant speed. • Components are ax = dvx/dt, ay = dvy/dt, and az = dvz/dt.

50. Direction of the acceleration vector • The direction of the acceleration vector depends on whether the speed is constant, increasing, or decreasing, as shown in Figure 3.12.

More Related