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© 2013 Pearson Education, Inc. Chapter Goal: To learn how to solve problems about motion in a plane. Chapter 4 Kinematics in Two Dimensions Slide 4-2.

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Presentation on theme: "© 2013 Pearson Education, Inc. Chapter Goal: To learn how to solve problems about motion in a plane. Chapter 4 Kinematics in Two Dimensions Slide 4-2."— Presentation transcript:

1 © 2013 Pearson Education, Inc. Chapter Goal: To learn how to solve problems about motion in a plane. Chapter 4 Kinematics in Two Dimensions Slide 4-2

2 © 2013 Pearson Education, Inc. Chapter 4 Preview Slide 4-3

3 © 2013 Pearson Education, Inc. Chapter 4 Preview Slide 4-4

4 © 2013 Pearson Education, Inc. Chapter 4 Preview Slide 4-5

5 © 2013 Pearson Education, Inc. The average acceleration of a moving object is defined as the vector: The acceleration points in the same direction as, the change in velocity. As an object moves, its velocity vector can change in two possible ways: 1.The magnitude of the velocity can change, indicating a change in speed, or 2. The direction of the velocity can change, indicating that the object has changed direction. Acceleration Slide 4-20

6 © 2013 Pearson Education, Inc. The figure to the right shows a motion diagram of Maria riding a Ferris wheel. Maria has constant speed but not constant velocity, so she is accelerating. For every pair of adjacent velocity vectors, we can subtract them to find the average acceleration near that point. Acceleration Slide 4-25

7 © 2013 Pearson Education, Inc. At every point Marias acceleration points toward the center of the circle. This is an acceleration due to changing direction, not to changing speed. Acceleration Slide 4-26

8 © 2013 Pearson Education, Inc. A car is traveling around a curve at a steady 45 mph. Is the car accelerating? A.Yes B.No QuickCheck 4.2 Slide 4-27

9 © 2013 Pearson Education, Inc. A car is traveling around a curve at a steady 45 mph. Is the car accelerating? A.Yes B.No QuickCheck 4.2 Slide 4-28

10 © 2013 Pearson Education, Inc. A car is traveling around a curve at a steady 45 mph. Which vector shows the direction of the cars acceleration? QuickCheck 4.3 Slide 4-29 E. The acceleration is zero.

11 © 2013 Pearson Education, Inc. A car is traveling around a curve at a steady 45 mph. Which vector shows the direction of the cars acceleration? QuickCheck 4.3 Slide 4-30 E. The acceleration is zero.

12 © 2013 Pearson Education, Inc. An objects acceleration can be decomposed into components parallel and perpendicular to the velocity. is the piece of the acceleration that causes the object to change speed. is the piece of the acceleration that causes the object to change direction. An object changing direction always has a component of acceleration perpendicular to the direction of motion. Analyzing the Acceleration Vector Slide 4-33

13 © 2013 Pearson Education, Inc. A car is slowing down as it drives over a circular hill. Which of these is the acceleration vector at the highest point? QuickCheck 4.4 Slide 4-34

14 © 2013 Pearson Education, Inc. A car is slowing down as it drives over a circular hill. Which of these is the acceleration vector at the highest point? QuickCheck 4.4 Slide 4-35 Acceleration of changing speed Acceleration of changing direction

15 © 2013 Pearson Education, Inc. The figure to the right shows the trajectory of a particle moving in the x-y plane. The particle moves from position at time t 1 to position at a later time t 2. The average velocity points in the direction of the displacement and is Two-Dimensional Kinematics Slide 4-36

16 © 2013 Pearson Education, Inc. The instantaneous velocity is the limit of avg as t 0. As shown the instantaneous velocity vector is tangent to the trajectory. Mathematically: Two-Dimensional Kinematics Slide 4-37 which can be written: where:

17 © 2013 Pearson Education, Inc. If the velocity vectors angle is measured from the positive x-direction, the velocity components are: Two-Dimensional Kinematics Slide 4-38 where the particles speed is Conversely, if we know the velocity components, we can determine the direction of motion:

18 © 2013 Pearson Education, Inc. The figure to the right shows the trajectory of a particle moving in the x-y plane. The acceleration is decomposed into components and. is associated with a change in speed. is associated with a change of direction. always points toward the inside of the curve because that is the direction in which is changing. Decomposing Two-Dimensional Acceleration Slide 4-41

19 © 2013 Pearson Education, Inc. The figure to the right shows the trajectory of a particle moving in the x-y plane. The acceleration is decomposed into components a x and a y. If v x and v y are the x- and y- components of velocity, then Decomposing Two-Dimensional Acceleration Slide 4-42

20 © 2013 Pearson Education, Inc. If the acceleration is constant, then the two components a x and a y are both constant. In this case, everything from Chapter 2 about constant- acceleration kinematics applies to the components. The x-components and y-components of the motion can be treated independently. They remain connected through the fact that t must be the same for both. Constant Acceleration Slide 4-43

21 © 2013 Pearson Education, Inc. Baseballs, tennis balls, Olympic divers, etc. all exhibit projectile motion. A projectile is an object that moves in two dimensions under the influence of only gravity. Projectile motion extends the idea of free-fall motion to include a horizontal component of velocity. Air resistance is neglected. Projectiles in two dimensions follow a parabolic trajectory as shown in the photo. Projectile Motion Slide 4-44

22 © 2013 Pearson Education, Inc. The start of a projectiles motion is called the launch. The angle of the initial velocity v 0 above the x-axis is called the launch angle. The initial velocity vector can be broken into components. where v 0 is the initial speed. Projectile Motion Slide 4-45

23 © 2013 Pearson Education, Inc. Gravity acts downward. Therefore, a projectile has no horizontal acceleration. Thus: The vertical component of acceleration a y is g of free fall. The horizontal component of a x is zero. Projectiles are in free fall. Projectile Motion Slide 4-46

24 © 2013 Pearson Education, Inc. The figure shows a projectile launched from the origin with initial velocity: The value of v x never changes because theres no horizontal acceleration. v y decreases by 9.8 m/s every second. Projectile Motion Slide 4-47

25 © 2013 Pearson Education, Inc. Example 4.4 Dont Try This at Home! Slide 4-48

26 © 2013 Pearson Education, Inc. Example 4.4 Dont Try This at Home! Slide 4-49

27 © 2013 Pearson Education, Inc. Example 4.4 Dont Try This at Home! Slide 4-50

28 © 2013 Pearson Education, Inc. Example 4.4 Dont Try This at Home! Slide 4-51

29 © 2013 Pearson Education, Inc. A heavy ball is launched exactly horizontally at height h above a horizontal field. At the exact instant that the ball is launched, a second ball is simply dropped from height h. Which ball hits the ground first? If air resistance is neglected, the balls hit the ground simultaneously. The initial horizontal velocity of the first ball has no influence over its vertical motion. Neither ball has any initial vertical motion, so both fall distance h in the same amount of time. Reasoning About Projectile Motion Slide 4-52

30 © 2013 Pearson Education, Inc. Without gravity, the arrow would follow a straight line. Because of gravity, the arrow at time t has fallen a distance ½gt 2 below this line. The separation grows as ½gt 2, giving the trajectory its parabolic shape. Reasoning About Projectile Motion Slide 4-57 A hunter in the jungle wants to shoot down a coconut that is hanging from the branch of a tree. He points his arrow directly at the coconut, but the coconut falls from the branch at the exact instant the hunter shoots the arrow. Does the arrow hit the coconut?

31 © 2013 Pearson Education, Inc. Had the coconut stayed on the tree, the arrow would have curved under its target as gravity causes it to fall a distance ½gt 2 below the straight line. But ½gt 2 is also the distance the coconut falls while the arrow is in flight. So yes, the arrow hits the coconut! Reasoning About Projectile Motion Slide 4-58 A hunter in the jungle wants to shoot down a coconut that is hanging from the branch of a tree. He points his arrow directly at the coconut, but the coconut falls from the branch at the exact instant the hunter shoots the arrow. Does the arrow hit the coconut?

32 © 2013 Pearson Education, Inc. This distance is sometimes called the range of a projectile. Example 4.5 from your textbook shows: The maximum distance occurs for 45. Range of a Projectile Slide 4-59 A projectile with initial speed v 0 has a launch angle of above the horizontal. How far does it travel over level ground before it returns to the same elevation from which it was launched? Trajectories of a projectile launched at different angles with a speed of 99 m/s.

33 © 2013 Pearson Education, Inc. Consider a ball on a roulette wheel. It moves along a circular path of radius r. Other examples of circular motion are a satellite in an orbit or a ball on the end of a string. Circular motion is an example of two-dimensional motion in a plane. Circular Motion Slide 4-79

34 © 2013 Pearson Education, Inc. To begin the study of circular motion, consider a particle that moves at constant speed around a circle of radius r. This is called uniform circular motion. The time interval to complete one revolution is called the period, T. The period T is related to the speed v : Uniform Circular Motion Slide 4-80

35 © 2013 Pearson Education, Inc. Example 4.9 A Rotating Crankshaft Slide 4-81

36 © 2013 Pearson Education, Inc. Consider a particle at a distance r from the origin, at an angle from the positive x axis. The angle may be measured in degrees, revolutions (rev) or radians (rad), that are related by: If the angle is measured in radians, then there is a simple relation between and the arc length s that the particle travels along the edge of a circle of radius r : Angular Position Slide rev = 360 = 2 rad

37 © 2013 Pearson Education, Inc. A particle on a circular path moves through an angular displacement f – i in a time interval t = t f – t i. In analogy with linear motion, we define: As the time interval t becomes very small, we arrive at the definition of instantaneous angular velocity. Angular Velocity Slide 4-83

38 © 2013 Pearson Education, Inc. Angular velocity ω is the rate at which a particles angular position is changing. As shown in the figure, ω can be positive or negative, and this follows from our definition of. A particle moves with uniform circular motion if ω is constant. ω and are related graphically: Angular Velocity Slide 4-84

39 © 2013 Pearson Education, Inc. When angular velocity ω is constant, this is uniform circular motion. In this case, as the particle goes around a circle one time, its angular displacement is 2 during one period t T. The absolute value of the constant angular velocity is related to the period of the motion by: Angular Velocity in Uniform Circular Motion Slide 4-87

40 © 2013 Pearson Education, Inc. Example 4.11 At the Roulette Wheel Slide 4-90

41 © 2013 Pearson Education, Inc. Example 4.11 At the Roulette Wheel Slide 4-91

42 © 2013 Pearson Education, Inc. The tangential velocity component v t is the rate ds/dt at which the particle moves around the circle, where s is the arc length. The tangential velocity and the angular velocity are related by: In this equation, the units of v t are m/s, the units of ω are rad/s, and the units of r are m. Tangential Velocity Slide 4-92

43 © 2013 Pearson Education, Inc. In uniform circular motion, although the speed is constant, there is an acceleration because the direction of the velocity vector is always changing. The acceleration of uniform circular motion is called centripetal acceleration. The direction of the centripetal acceleration is toward the center of the circle. The magnitude of the centripetal acceleration is constant for uniform circular motion. Centripetal Acceleration Slide 4-93

44 © 2013 Pearson Education, Inc. The figure shows the velocity at one instant and the velocity an infinitesimal amount of time dt later. By definition,. By analyzing the isosceles triangle of velocity vectors, we can show that: which can be written in terms of angular velocity as: a ω 2 r. Centripetal Acceleration Slide 4-94

45 © 2013 Pearson Education, Inc. Example 4.12 The Acceleration of a Ferris Wheel Slide 4-101

46 © 2013 Pearson Education, Inc. Suppose a wheels rotation is speeding up or slowing down. This is called nonuniform circular motion. We can define the angular acceleration as The units of are rad/s 2. The figure to the right shows a wheel with angular acceleration 2 rad/s 2. Angular Acceleration Slide 4-103

47 © 2013 Pearson Education, Inc. is positive if | ω | is increasing and ω is counter-clockwise. is positive if | ω | is decreasing and ω is clockwise. is negative if | ω | is increasing and ω is clockwise. is negative if | ω | is decreasing and ω is counter-clockwise. Pg 104 (*****) The Sign of Angular Acceleration Slide 4-104

48 © 2013 Pearson Education, Inc. The fan blade is slowing down. What are the signs of ω and ? A. ω is positive and is positive. B. ω is positive and is negative. C. ω is negative and is positive. D. ω is negative and is negative. E. ω is positive and is zero. QuickCheck 4.15 Slide 4-105

49 © 2013 Pearson Education, Inc. The fan blade is slowing down. What are the signs of ω and ? A. ω is positive and is positive. B. ω is positive and is negative. C. ω is negative and is positive. D. ω is negative and is negative. E. ω is positive and is zero. QuickCheck 4.15 Slide Slowing down means that and have opposite signs, not that is negative

50 © 2013 Pearson Education, Inc. Example 4.14 Back to the Roulette Wheel Slide 4-113

51 © 2013 Pearson Education, Inc. Example 4.14 Back to the Roulette Wheel Slide 4-114

52 © 2013 Pearson Education, Inc. The particle in the figure is moving along a circle and is speeding up. The centripetal acceleration is a r v t 2 /r, where v t is the tangential speed. There is also a tangential acceleration a t, which is always tangent to the circle. The magnitude of the total acceleration is Acceleration in Nonuniform Circular Motion Slide 4-115

53 © 2013 Pearson Education, Inc. A particle moves along a circle and may be changing speed. The distance traveled along the circle is related to : The tangential velocity is related to the angular velocity: The tangential acceleration is related to the angular acceleration: Nonuniform Circular Motion Slide 4-116

54 © 2013 Pearson Education, Inc. Chapter 4 Summary Slides Slide 4-117

55 © 2013 Pearson Education, Inc. General Principles Slide 4-118

56 © 2013 Pearson Education, Inc. General Principles Slide 4-119

57 © 2013 Pearson Education, Inc. Important Concepts Slide 4-120

58 © 2013 Pearson Education, Inc. Important Concepts Slide 4-121


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