1 / 73

Chapter 3

Chapter 3. Motion in Two or Three Dimensions. Goals for Chapter 3. To study position, velocity, and acceleration vectors To apply position, velocity, and acceleration insights to projectile motion To extend our linear investigations to uniform and non-uniform circular motion

debra-spence
Download Presentation

Chapter 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 Motion in Two or Three Dimensions

  2. Goals for Chapter 3 • To study position, velocity, and acceleration vectors • To apply position, velocity, and acceleration insights to projectile motion • To extend our linear investigations to uniform and non-uniform circular motion • To investigate relative velocity

  3. Position relative to the origin—Figure 3.1 • An overall position relative to the origin can have components in x, y, and z dimensions. • The path for a particle is generally a curve.

  4. Average velocity—Figure 3.2 • The average velocity between two points will have the same direction as the displacement.

  5. (x2 – x1)i (y2 – y1)j (z2 – z1)k vav = + + ∆t ∆t ∆t Average velocity vector • During a time interval t the particle moves from P1 , where its position vector is r1, to P2, where its position vector is r2. One dimension

  6. (2D) Instantaneous velocity

  7. dr v = dt

  8. Example 3.1: calculating average and instantaneous velocity

  9. Test your understanding • In which of these situations would the average velocity vector vav over an interval be equal to the instantaneous velocity v at the end of interval: • A body moving along a curved path at constant speed • A body moving along a curved path and speeding up • A body moving along a straight line at constant speed • A body moving along a straight line and speeding up.

  10. 2b 3c t = example If r = bt2i+ ct3j Where b and c are positive constants, when does the velocity vector make an angle of 45.0o with the x- and y-axes? v = dr/dt = 2bt i + 3ct2 j = 1

  11. The acceleration vector—Figure 3.6 • The acceleration vector can result in a change in either the magnitude OR the direction of the velocity. • Consider the race car in Figure 3.6.

  12. (v2x – v1x)i (v2y – y1y)j (v2z – v1z)k aav = + + ∆t ∆t ∆t One dimension

  13. d2x dvx dvy dvz d2y d2z a = i + j + k a = i + j + k dt dt dt dt dt dt ay tanθ = ax |a| = √ax2 + ay2 + az2 Instantaneous Acceleration

  14. All about acceleration • Equal to time rate of change of velocity • ≠ 0 if velocity changes in magnitude or direction. • It does not necessarily have same direction as velocity vector • Acceleration vector lies on concave side of curved path.

  15. Example 3.2 • Given: • Find the components of the average acceleration in the interval from t = 0.0 s to t = 2.0 s. • Find the instantaneous acceleration at t = 2.0 s, its magnitude and its direction.

  16. Parallel and perpendicular components of acceleration • The acceleration vector a for a particle can describe changes in the particle’s speed, its direction of motion, or both. • The component of acceleration parallel to a particle’s path (parallel to the velocity) tells us about changes in the particle's speed. • The acceleration component perpendicular to the path (perpendicular to the velocity) tells us about changes in the particle’s direction of motion.

  17. Consider the different vectors—Figure 3.12 • Notice the acceleration vector change as velocity decreases, remains the same, or increases.

  18. Parallel and perpendicular components of acceleration • Refer to the worked Example 3.3. • Figure 3.13 illustrates the calculation in the example.

  19. Example 3.3 • Given: • Find the parallel and perpendicular components, relative to velocity, of the instantaneous acceleration at t = 2.0 s

  20. Example 3.4 • A skier moves along a ski-jump ramp as shown in the figure. The ramp is straight from point A to point C onward. The skier picks up speed as she moves downhill from point A to point E. Draw the direction of the acceleration vector at points B, D, E and F.

  21. Test your understanding 3.2 • A sled travels over the crest of a snow-covered hill. The sled slows down as it climbs up one side of the hill and gains speed as it descends on the other side. Which of the vectors (1 though 9) in the figure correctly shows the direction of the sled’s acceleration at the crest? Choice 9 is that the acceleration is zero.)

  22. 3.3 Projectile motion • A projectile is any body given an initial velocity that then follows a path determined by the effects of gravity and air resistance. • Begin neglecting resistance.

  23. X and Y motion are separable—Figure 3.16 • The red ball is dropped, and the yellow ball is fired horizontally as it is dropped. • The strobe marks equal time intervals.

  24. Projectiles move in TWO dimensions • Horizontal and Vertical The path of a projectile is called a trajectory

  25. θ Horizontal Component • Velocity is constant: vx0 = vi0cosθ • Acceleration: ax = 0 • Displacement = vx0∙t In other words, the horizontal velocity is CONSTANT. BUT WHY? Gravity DOES NOT work horizontally to increase or decrease the velocity.

  26. θ Vertical Component • Acceleration: ay = -g • Velocity: vy = vy0 – gt = vi0sinθ - gt • Displacement: y = y0 + vy0∙t - ½ gt2

  27. If air resistance is negligible, the trajectory of a projectile is a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. The path of the projectile is parabolic.

  28. The effects of wind resistance—Figure 3.20 • Cumulative effects can be large. • Peak heights and distance fall. • Trajectories cease to be parabolic.

  29. Horizontally Launched Projectiles Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land? t = 10 seconds x = 1000 m

  30. Ground Launched Projectiles NO Vertical Velocity at the top of the trajectory. Vertical Velocity decreases on the way upward Vertical Velocity increases on the way down, Horizontal Velocity is constant velocity

  31. Since the projectile was launched at a angle, the velocity MUST be broken into components!!! voy vo q vox

  32. There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0

  33. You will still use kinematic equations, but YOU MUST use COMPONENTS in the equation. voy vo q vox

  34. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? (b) How far away does it land? (c) How high does it travel? vo=20.0 m/s q = 53

  35. Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (c) How high does it travel? CUT YOUR TIME IN HALF! 13.01 m

  36. A special case… • What if the projectile was launched from the ground at an angle and did not land at the same level height from where it started? In other words, what if you have a situation where the “y-displacement” DOES NOT equal zero? Assuming it is shot from the ground. We see we have one squared term variable, one regular term variable, and a constant number with no variable. What is this? A QUADRATIC EQUATION!

  37. A special case example • An object is thrown from the top of a 100m high cliff at an initial speed of 20 m/s. How long does it take to reach the ground? Well, here is what you have to realize. As it goes up its speed decreases, reaches zero, the increasing back to the ground. When the ball reaches the cliff face it is now traveling at 20 m/s again , but in the opposite direction. DOWNWARD. Thus the speed is considered to be -20 m/s.

  38. Example GOTO F2 ( which is the Algebra screen) CHOOSE SOLVE. Type in the equation. In this example our equation and set the it equal to zero. Then add a comma after the equation and tell it that you want it to solve for, in this case "t".

  39. Example It will respond by showing you exactly what you typed and the TWO ROOTS! Since we are solving for time we choose the positive root as TIME cannot be negative. Thus it took 2.92 seconds for the ball to hit the ground.

  40. If at t = 0, x0 = y0 = 0 then we can find the x, y coordinates and the x, y velocity at time t: The magnitude of the position: The magnitude of the velocity: The direction of the velocity:

  41. Example 3.5 Let’s consider again the skier in example 3.4. What is her acceleration at points G, H, and I after she flies off the ramp? Neglect air resistance. The acceleration at points G, H, I are the same: ax = 0; ay = -9.8 m/s2

  42. Example 3.6:a body projected horizontally • A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. find the motorcycle’s position,distance and velocity from the edge of the cliff, and velocity at 0.50 s.

  43. Example: 3.7 – height and range of a projectile I – a batted baseball

  44. Example 3.8 – height and range of a projectile II: maximum height, maximum range • For a projectile launched with speed v0 at initial angle α0 (between 0o and 90o), derive general expressions for the maximum height h and horizontal range R. For a given v0, what value of α0 gives maximum height? What value gives maximum horizontal range?

  45. Maximum height: vy = 0 0 = v0sinα0 - gt t = v0sinα0 / g h = (v0sinα0)(v0sinα0 /g) – ½ g(v0sinα0 /g)2 h = (v0sinα0)2 / 2g When α0 is 90o, sin α0 is 1, h has maximum value.

  46. Maximum range: The time for the ball to return to the ground is twice the time for the ball to reach the maximum height. t = 2v0sinα0 / g R = (v0cosα0)(2v0sinα0/g) R = v02sin2α0/g When 2α0 is 90o, or when α0 is 45o, R has maximum value.

  47. You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o below the horizontal. How far horizontally from your window will the ball hit the ground? Ignore air resistance.

  48. A monkey escapes from the zoo and climbs a tree. After failing to entice the monkey down, the zookeeper fires a tranquilizer dart directly at the monkey. The clever monkey lets go at the same instant the dart leaves the gun barrel, intending to land on the ground and escape. Show that the dart always hits the monkey, regardless of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).

More Related