lecture based on 2016 Pearson Education, Ltd.

Transcription

1 KINEMATICS D R M A R T A S T A S I A K D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S lecture based on 2016 Pearson Education, Ltd.

2 Vectors Reference Frames Displacement Average Velocity Instantaneous Velocity Acceleration Motion at Constant Acceleration Solving Problems Falling Objects Graphical Analysis of Linear Motion Projectile Motion Linear Momentum

3 VECTORS A vector has magnitude as well as direction. Examples: displacement, velocity, acceleration, force, momentum A scalar has only magnitude Examples: time, mass, temperature, energy

4 VECTORS Magnitude Direction the line of action (line segment) and sense (orientation). Origin (tail) of the vector - point of application, initial point sense line segment Origin A D B

5 VECTOR ADDITION ONE DIMENSION A person walks 8 km East and then 6 km East. Displacement =14 km East A person walks 8 km East and then 6 km West. Displacement = 2 km East

6 VECTOR ADDITION A person walks 10 km East and 5.0 km North D R D 1 D 2 D R 2 2 D 1 D2 2 2 D R (10km) (5km) 11. 2km sin 1 sin D ( D 2 R ) sin D 2 D R 1 5km ( ) 11.2 km Order doesn t matter

7 GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD V1 V V 2 3 V R

8 GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD V 1 V V 3 2 V1 V2 V R V 3

9 PARALLELOGRAM METHOD

10 SUBTRACTION OF VECTORS Negative of vector has same magnitude but points in the opposite direction For subtraction, we add the negative vector.

11 MULTIPLICATION BY A SCALAR A vector V can be multiplied by a scalar c; the result is a vector c V that has the same direction but a magnitude c V. If c is negative, the resultant vector points in the opposite direction.

12 ADDING VECTORS BY COMPONENTS Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.

13 Opposite TRIGONOMETRY REVIEW sin Opposite Hypotenuse Hypotenuse cos Adjacent Hypotenuse tan Opposite Adjacent sin cos Adjacent

14 ADDING VECTORS BY COMPONENTS If the components are perpendicular, they can be found using trigonometric functions. sin Opposite Hypotenuse V y V sin V y V cos Adjacent Hypotenuse V x V tan Opp Adj sin cos V x V cos

15 ADDING VECTORS BY COMPONENTS Adding vectors: 1. Draw a diagram; add the vectors graphically. 2. Choose x and y axes. 3. Resolve each vector into x and y components. 4. Calculate each component using sines and cosines. 5. Add the components in each direction. 6. To find the length and direction of the vector, use: sin V y V

16 REFERENCE FRAMES Any measurement of position, distance, or speed must be made with respect to a reference frame.

17 REFERENCE FRAMES Coordinate axes

18 COORDINATE SYSTEM REFERENCE FRAMES

19 DISPLACEMENT distance displacement

20 DISPLACEMENT The displacement is written: Displacement is positive Δx=30m-10m=20m Displacement is negative Δx=10m-30m=-20m

21 VELOCITY Speed: how far an object travels in a given time interval Velocity includes directional information:

22 VELOCITY AS A VECTOR v = x t

23 AVERAGE VELOCITY v = x x 2 1 = Δx t 2 t 1 Δt

24 INSTANTANEOUS VELOCITY constant velocity Δx v = lim t 0 Δt = δx δt varying velocity

25 ACCELERATION

26 ACCELERATION VECTOR

27 ACCELERATION AS A VECTOR a = v t

28 ACCELERATION There is a difference between negative acceleration and deceleration

29 ACCELERATION Negative acceleration is acceleration in the negative direction as defined by the coordinate system.

30 ACCELERATION Deceleration occurs when the acceleration is opposite in direction to the velocity.

31 ACCELERATION The instantaneous acceleration is the average acceleration, in the limit as the time interval becomes infinitesimally short. Δv a = lim t 0 Δt = δv δt = δ2 x δt

32 MOTION AT CONSTANT ACCELERATION The average velocity of an object during a time interval t is The acceleration, assumed constant, is

33 MOTION AT CONSTANT ACCELERATION In addition, as the velocity is increasing at a constant rate, we know that Combining these last three equations, we find:

34 MOTION AT CONSTANT ACCELERATION We can also combine these equations so as to eliminate t: We now have all the equations we need to solve constant-acceleration problems.

35 FALLING OBJECTS The same acceleration

36 FALLING OBJECTS

37 FALLING OBJECTS 9.80 m/s 2

38 FREE FALL HOW FAST? The velocity acquired by an object starting from rest is Velocity = acceleration time So, under free fall, when acceleration is 10 m/s 2, the speed is 10 m/s after 1 s. 20 m/s after 2 s. 30 m/s after 3 s. And so on Pearson Education, Inc.

39 GRAPHICAL ANALYSIS OF LINEAR MOTION constant velocity

40 GRAPHICAL ANALYSIS OF LINEAR MOTION VARYING VELOCITY

41 GRAPHICAL ANALYSIS OF LINEAR MOTION Displacement Displacement

42 SUMMARY Kinematics is the description of how objects move with respect to a defined reference frame. Displacement is the change in position of an object. Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. Instantaneous velocity is the limit as the time becomes infinitesimally short.

43 SUMMARY Average acceleration is the change in velocity divided by the time. Instantaneous acceleration is the limit as the time interval becomes infinitesimally small. The equations of motion for constant acceleration are given in the text; there are four, each one of which requires a different set of quantities. Objects falling (or having been projected) near the surface of the Earth experience a gravitational acceleration of 9.80 m/s 2.

44 KINEMATICS IN TWO DIMENSION D R M A R T A S T A S I A K D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S lecture based on 2016 Pearson Education, Ltd.

45 PROJECTILE MOTION two dimensions parabola

46 PROJECTILE MOTION a y =o t grounded =t dropped vertically

47 EQUATIONS FOR PROJECTILE MOTION Horizontal X a x =0 v x = constant Vertical Y a y = - g v v v g t 0 vx0 y y 0 x x 0 v x0 t y y v y t g t 2 2 v 2 y v 2 y0 2 g ( y y0)

48 INITIAL VELOCITY v y sin v y 0 v 0 v y 0 v 0 sin v x 0 v 0 cos For y=0 v 0 = v v = v x0 = constans

49

50 PROBLEM SOLVING A GENERAL APPROACH Read the problem carefully; then read it again. Draw a sketch, and then a free-body diagram. Choose a convenient coordinate system. List the known and unknown quantities Find relationships between the knowns and the unknowns. Estimate the answer. Solve the problem without putting in any numbers (algebraically); once you are satisfied, put the numbers in. Keep track of dimensions. Make sure your answer is reasonable.

51 EXAMPLE A football is kicked at an angle of 50 above the horizontal with a velocity of 18 m/s. Θ=50 V 0 =18m/s Calculate the maximum height and the range as well as how long it is in the air H max, R, t=? Assume that the ball was kicked at ground level and lands at ground level.

52 EXAMPLE A football is kicked at an angle of 50 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level. 0 v cos m ( 18 )(cos50 ) 11.6m s s / sin m ( 18 )(sin 50 ) 13.8m s s / v v 0 g t 0 v x v y 0 v at top: y y t up vy g 1 H 2 max y max y 0 v yo t gt 2 2 v sin 1 v sin H max 0 vy0 g g 2 g 1 0 (13.8m H )(1.41s) (9.8m 2)(1.41s s 2 s 0 v sin g 2 max ) 13.8m 9.80m s s s H max 9. 7m

53 LEVEL HORIZONTAL RANGE Range is determined by time it takes for ball to return to ground level or perhaps some other vertical value. If ball hits something a fixed distance away, then time is determined by x motion If the motion is on a level field, when it hits: y = y y0 vy0 t g t 0 0 vyo t g t vy0 Solving we find t g We can substitute this in the x equation to find the range R 2 2v y0 2vx0 v yo 2v0 sin 0 cos 0 R x vx0 t vxo ( ) g g g

54 LEVEL HORIZONTAL RANGE We can use a trig identity 2sin cos sin 2 R v 2 0 sin 2 g ( ) Greatest range: θ= 45 0 θ = 30 0 and 60 0 have same range. Caution the range formula has limited usefulness. It is only valid when the projectile returns to the same vertical position.

55 EXAMPLE A football is kicked at an angle of 50 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level. Assume time down = time up For Range: t ( 2)(1.41s) 2.82s R x x 0 v x0 t 0 (11.6 m ) s (2.82s) 33m Could also use range formula R v 2 0 sin 2 g (18m / 2 s ) sin 9.8m / (2)(50 2 s 33m 0 )

56 EXAMPLE-VERTICAL PROJECTION A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s) how far in advance of the recipients (horizontal distance) must the goods be dropped? y v 235m vx m / s x? v y0 0 t Coordinate system is 235 m below plane y 235m 0 g t g t 2 2 x ( 2)( y ) ( 2)(235m) 6.93s 2 g 9.8m / s x 0 v 0 ( xo x 481m t 69.4 m / s )(6.93 s

57 PROJECTILE MOTION IS PARABOLIC In order to demonstrate that projectile motion is parabolic, the book derives y as a function of x. When we do, we find that it has the form: This is the equation for a parabola

58 LINEAR MOMENTUM D R M A R T A S T A S I A K D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S lecture based on 2016 Pearson Education, Ltd.

59 Momentum and Its Relation to Force Conservation of Momentum Collisions and Impulse Conservation of Energy and Momentum in Collisions Elastic Collisions in One Dimension Inelastic Collisions Collisions in Two or Three Dimensions Center of Mass (CM) CM for the Human Body Center of Mass and Translational Motion

60 MOMENTUM AND ITS RELATION TO FORCE Momentum is a vector symbolized by the symbol p, and is defined as The rate of change of momentum is equal to the net force: This can be shown using Newton s second law.

61 CONSERVATION OF MOMENTUM During a collision, measurements show that the TOTAL MOMENTUM DOES NOT CHANGE:

62 CONSERVATION OF MOMENTUM More formally, the law of conservation of momentum states: The total momentum of an isolated system of objects remains constant.

63 CONSERVATION OF MOMENTUM EXPERIMENT Momentum conservation works for a rocket as long as we consider the rocket and its fuel to be one system, and account for the mass loss of the rocket.

64 CONSERVATION OF MOMENTUM EXPERIMENT A Wad of Clay Hits Unsuspecting Sled 1 kg clay ball strikes 5 kg sled at 12 m/s and sticks Momentum before collision: (1 kg)(12 m/s) + (5 kg)(0 m/s) Momentum after = 12 kg m/s (6 kg) (2 m/s)

65 ELASTIC COLLISION: BILLIARD BALLS Whack stationary ball with identical ball moving at velocity v cue 8 To conserve both energy and momentum, cue ball stops dead, and 8-ball takes off with v cue 8 Momentum conservation: mv cue = mv cue, after + mv 8-ball Energy conservation: ½mv 2 cue = ½mv 2 cue, after + ½mv 2 8-ball The only way v 0 = v 1 + v 2 and v 2 0 = v v 2 2 is if either v 1 or v 2 is 0. Since cue ball can t move through 8-ball, cue ball gets stopped.

66 DESK TOY PHYSICS The same principle applies to the suspended-ball desk toy, which eerily knows how many balls you let go Only way to simultaneously satisfy energy and momentum conservation Relies on balls to all have same mass

67 INELASTIC COLLISION Energy not conserved (absorbed into other paths) Non-bouncy: hacky sack, velcro ball, ball of clay Momentum before = m 1 v initial Momentum after = (m 1 + m 2 )v final = m 1 v initial (because conserved) Energy before = ½m 1 v 2 initial Energy after = ½ (m 1 + m 2 )v 2 final + heat energy

68 COLLISIONS AND IMPULSE During a collision, objects are deformed due to the large forces involved. Since, we can Write The definition of impulse:

69 COLLISIONS AND IMPULSE The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

70 CONSERVATION OF ENERGY AND MOMENTUM IN COLLISIONS Momentum is conserved in all collisions. Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic.

71 ELASTIC COLLISIONS IN ONE DIMENSION Here we have two objects colliding elastically. We know the masses and the initial speeds. Since both momentum and kinetic energy are conserved, we can write two equations. This allows us to solve for the two unknown final speeds.

72 INELASTIC COLLISIONS With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy. A completely inelastic collision is one where the objects stick together afterwards, so there is only one final velocity.

73 COLLISIONS IN TWO OR THREE DIMENSIONS Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities

74 COLLISIONS IN TWO OR THREE DIMENSIONS PROBLEM SOLVING Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply. Is there an external force? If so, is the collision time short enough that you can ignore it? Draw diagrams of the initial and final situations, with momentum vectors labeled. Choose a coordinate system. Apply momentum conservation; there will be one equation for each dimension. If the collision is elastic, apply conservation of kinetic energy as well. Solve. Check units and magnitudes of result.

75 SUMMARY Momentum of an object: Newton s second law: Total momentum of an isolated system of objects is conserved. During a collision, the colliding objects can be considered to be an isolated system even if external forces exist, as long as they are not too large. Momentum will therefore be conserved during collisions.

76 SUMMARY In an elastic collision, total kinetic energy is also conserved. In an inelastic collision, some kinetic energy is lost. In a completely inelastic collision, the two objects stick together after the collision.