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制作 张昆实 Yangtze University

Bilingual Mechanics. Chapter 2 Motion. 制作 张昆实 Yangtze University. 制作 张昆实 Yangtze University. Chapter 2 Motion. 2-1 What Is Physics 2-2 Vectors and Scalars 2-3 Multiplying Vectors 2-4 Motion 2-5 Position and Displacement 2-6 Average Velocity and Instantaneous

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制作 张昆实 Yangtze University

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  1. BilingualMechanics Chapter 2 Motion 制作 张昆实 Yangtze University 制作 张昆实 Yangtze University

  2. Chapter 2 Motion 2-1 What Is Physics 2-2 Vectors and Scalars 2-3 Multiplying Vectors 2-4 Motion 2-5 Position and Displacement 2-6 Average Velocity and Instantaneous Velocity 2-7 Acceleration

  3. Chapter 2 Motion 2-8 Constant Acceleration: A Special Case 2-9 Graphical Integration of Motion Analysis 2-10 Projectile Motion 2-11 Projectile Motion Analyzed 2-12 Uniform Circular Motion 2-13 Relative Motion

  4. 2-1 What Is Physics ★All objects in natureare in constant motion, nothing stands still absolutely. Deals with the concepts discribing motion, without any reference to force. Kinematics (chapter 2) Mechanics Dynamics Deals with the effectthatforces haveon motion . (chapter 3) ★ In this chapter we learnwhat is physics through the motionof particles

  5. 2-2 Vectors and Scalars . ★ Scalars are quantities that havemagnitude only. ★Vectors are quantities that have both magnitudeanddirectionandobeythe vector law of addition. length, time, mass , density, energy , temperature etc. are all scalars. Displacement, velocity, acceleration, force, torque, etc. are all vectors.

  6. 2-2 Vectors and Scalars . ★Three notations of a vector a) unit-vector notation b) component notation c) magnitude-angle notation

  7. magnitude (new vector) 1. direction as as 2. The Scalar Product 3. The Vector Product 2-3 Multiplying Vectors ★ There are three ways in which vectors can be multiplied. Multiplying a Vector by a Scalar Multiplying a Vector by a Vector

  8. 2. The Scalar Product ( dot product ) 3. The Vector Product ( cross product ) 2-3 Multiplying Vectors Multiplying a Vector by a Vector “a dot b”: (scalar) “a cross b”: (vector) magnitude Direction right hand screw rule

  9. 2-4 Motion ★ Objects move in one, two or three dimension(s). ★The moving object is either a particle or an object that moves like a particle. ★To describ the motion of an object a reference frame(参考系) must be chosen and a coordinate system(坐标系) must be constructed on it.

  10. * magnitude： 2-5 Position and Displacement ★Position vector a vector extends from the origin of a coordinate system tothe particle ● unit vector ●scalar components (2-25)

  11. magnitude： direction： 2-5Position and Displacement ★Position vector *

  12. 1 P20 Sample Problem 2-4 2-5Position and Displacement 2 ★Displacement (2-26) (2-27) (2-28)

  13. Solution 2-5Position and Displacement P20 Sample Problem 2-4 A rabbit runs across a parking plot on which a set of coor-dinate axes has been drawn. The coordinates of the rabbit’s position as function of time are given by (2-29) (2-30) With in seconds and and in meters. (a) At , What is the rabbit’s position vector in unit-vector notation and in magnitude-angle notation? (Answer) unit-vector notation

  14. 2-5Position and Displacement P20 Sample Problem 2-4 (a) At , What is the rabbit’s position vector in magnitude-angle notation? (Answer) magnitude-angle notation

  15. 2-5Position and Displacement P20 Sample Problem 2-4 (b)Graph the rabbit’s path for to magnitude-angle notation

  16. 2-6 Average Velocity and • Instantaneous Velocity Average displacement Velocity time interval ★ Average Velocity If a particle moves through a displacement in a time Interval ,Then its (2-32) (2-33)

  17. 2-6 Average Velocity and • Instantaneous Velocity The direction of is always tangent to the particle’s path at the particle’s position ★ Instantaneous Velocity (2-34) ● ● ● (2-35)

  18. 2-6 Average Velocity and • Instantaneous Velocity P22 Sample Problem 2-5 ★ Instantaneous Velocity and its components (2-36) Fig.2-18The velocity of a particle, along with the scalar components of

  19. 2-6 Average Velocity and • Instantaneous Velocity taking derivetives of the components of to get the components of Solution P22 Sample Problem 2-5 For the rabbit in Sample Problem 2-4, Find the velocity attime , in unit-vector notation and in magnitude-angle notation. (2-37) (2-38) (Answer) unit-vector notation

  20. 2-6 Average Velocity and • Instantaneous Velocity The magnitude of : Solution (in the third quadrant) P22 Sample Problem 2-5 Finding the velocity at time , in magnitude-angle notation. (Answer) (2-38) magnitude-anglenotation

  21. t2 t1 2-7 Acceleration ★ Average Acceleration When a particle’s velocity changes from to in a time interval ,Then its average change in velocity accelerationtime interval (2-39)

  22. (2-42) P23 Sample Problem 2-6 2-7 Acceleration ★ Instantaneous acceleration (2-40) (2-41)

  23. taking derivetives of the components of to get the components of Solution (Answer) unit-vector notation 2-7 Acceleration P23 Sample Problem 2-6 For the rabbit in Sample Problem 2-4 and 2-5,Find the acceleration at time, in unit-vector notation and in magnitude-angle notation.

  24. The magnitude of : Solution 2-7 Acceleration P23 Sample Problem 2-6 in magnitude-angle notation. (Answer) (in the second quadrant) magnitude-anglenotation

  25. (2-43) (2-47) • 2-8 Constant Acceleration: A Special Case The average velocity ★ In many cases theacceleration is constant (2-45) Substituting2-43 for : (2-46) Recast this equation: Substituting 2-46into 2-44: Similarly Eq. 2-43 and 2-47: are the basic equations for constant acceleration. (2-44)

  26. (2-48) Eliminate : (2-43) Eliminate : (2-49) (2-47) Eliminate : (2-50) P26 Table 2-1 • 2-8 Constant Acceleration: A Special Case the basic equations2-43 and 2-47for constant acceleration can be combined in three ways to yield additional equations:

  27. Indefinite integral To determine constant C, let (2-47) • 2-8 Constant Acceleration: A Special Case Eq. 2-43 and 2-47 can be ontained by integration Indefinite integral To determine constant C’, let (2-43)

  28. (1) The directions of motion are now along a verticle axis with the positive direction of upward. (2) The free-fall acceleration, being downward, is now negative! • 2-8 Constant Acceleration: A Special Case Toss an object either up or down (eliminate the air resistance)the object accelerates downward at a certain constant rate:free-fall acceleration The value of varies slightly with latitude and with elevation. *** Note: For free fall ***

  29. (2-43) (2-43) (2-47) (2-47) (2-48) (2-48) • 2-8 Constant Acceleration: A Special Case Horizontal motion Vertical motion

  30. 2-9 Graphical Integration of • Motion Analysis Using a graph of an object, the object’s canbe found by integrating on the graph: Area between acceleration curveandtime axis, from to (definite integral) When the acceleration curve is abovethe time axis, the area is positive; When the acceleration curve is belowthe time axis, the area is negative;

  31. 2-9 Graphical Integration of • Motion Analysis Using a graph of an object, the object’s canbe found by integrating on the graph: Area between velocity curve andtime axis, from to (definite integral) When the velocity curve is above the time axis, the area is positive; When the velocity curve is below the time axis, the area is negative;

  32. 2-10 Projectile Motion ★Projectile Motion : A particle moves in a vertical plane with some initial velocity and an angle with respect to the horizontal axis but its acceleration is always the free-fall acceleration , which is downward. (2-58) (2-59)

  33. 2-10 Projectile Motion Fig. 2-29 The projectile bullet always hits the falling coconut

  34. 2-11 Projectile Motion Analyzed ★ In projectile motion , the horizontal motion and the vertical motion are independent of each other. ★the horizontal motion: (2-60)

  35. 2-11 Projectile Motion Analyzed ★the vertical motion: (2-61) (2-62) (2-63)

  36. (2-60) (2-61) (2-64) (trajectory) 2-11 Projectile Motion Analyzed ★ The Equation of the Path Let Solving Eq.2-60 for t and Substituting into Eq.2-61: This is the equation of a parabola, so the path (trajectory) is parabolic.

  37. (2-60) (2-61) (2-65) The horizontal distanceR is maximum for a launch angle of 2-11 Projectile Motion Analyzed The Horizontal Range: the horizontal distancethe Projectile has traveled when it returns to its initial (launch) height. Eliminating t between these two equations yields

  38. In vacuum 2-11 Projectile Motion Analyzed The Effects of the Air (Air resistance): In vacuum: thepath(trajectory) is parabolic. In air: thehorizontal range, the maximum heightof the path are muchless. Air resistance force: density of air, the cross section area of the projectile, mainly the velocity of the body. In Air

  39. Figure 2-32 shows a pirate ship 560m from a fort defending the harbor entrance of an island. A defense canno, located at sea level, fires balls at initial speed . • at what angle from • the horizontal must a • ball be fired to hit the • ship? 2-11 Projectile Motion Analyzed Sample Problem 2-8 (P32) (b) How far should the pirate ship be from the canno if it is to be beyond the maximum range of the cannoballs?

  40. at what angle from the horizontal must a • ball be fired to hit the ship? We know: (2-65) Which gives us 2-11 Projectile Motion Analyzed Sample Problem 2-8 (P32) Solution The firedcannoball is a projectile (2-65)

  41. (2-65) 2-11 Projectile Motion Analyzed Solution: The maximum range corresponds to an elevation angle of Sample Problem 2-8 (P32) (b) How far should the pirate ship be from the canno if it is to be beyond the maximum range of the cannoballs?

  42. (2-70) Period: the time for going around a circle exactly once (circumference of the circle) (2-71) 2-12 Uniform Circular Motion ★Uniform circularMotion : A particle travel around a circle or a circular arc at constant (uniform) speed. The velocity changes only in direction, there are still an acceleration– centripetal acceleration. Fig.2-34

  43. 2-13Relative Motion y y v BA Inthree dimensions: Two observers are watching a moving particle P from the origins of frames A and B, while B moves at , The corresponding axes of frame A and B remain parallel position vecter: B toA : P to A : P toB : P v r BA o r o PB o o o PA o x o Frame B o o o o o o x r Frame A o BA r BA r PA (4-41) r r r + = PA BA PB r PB

  44. (4-41) r r + r = PA PB BA + v v = v PB BA PA constant, (4-43) a a = = v PA PB BA 2-13Relative Motion y y v BA P Take the time derivative + r = r r PB BA PA r o r Get : o PB o o o PA o x o Frame B (4-42) o o o o o o x r Frame A Take the time derivative o BA Since: The acceleration of the particle measured from frames A and B are the same!

  45. (4-41) r (4-38) r + r x + x = x = PA PB BA PA PB BA + r r r = PB BA PA (4-39) + v v = v (4-42) PB BA PA Since: + constant, v v = v v = PB BA BA PA Since: constant, (4-43) a a = = v PA PB BA 2-13Relative Motion in Two or Three Dimensionsin One Dimension + x x x = PB BA PA (4-40) a a = PA PB The acceleration of the particle measured from frames A and B are the same!

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