PS 11 GeneralPhysics I for the Life Sciences

Transcription

1 PS 11 GeneralPhysics I for the Life Sciences M E C H A N I C S I D R. B E N J A M I N C H A N A S S O C I A T E P R O F E S S O R P H Y S I C S D E P A R T M E N T N O V E M B E R 0 1 3

2 Definition Mechanics is the study of motion and its causes

3 Before discussing the cause of motion, we need to know how to describe motion (kinematics) first.

4 Kinematics How would you describe a moving object?

5 Locate the Object Establish Reference Frame Point of reference needed (origin) Reference direction needed Construct the position vector (magnitude and direction) from the origin to the object We can also call this the displacement of the object from the origin It consists of a magnitude (in meters) and direction

6 1-D Position Vector Only two directions available Can be represented by a signed scalar (magnitude) 1 m O P

7 -D Position Vector Direction goes through a 360 angle 0 angle reference needed Position vector = magnitude, angle Polar Coordinates (r, ) are natural r = magnitude, = direction Polar coordinates to cartesian coordinates and back: P x r cos r = r r y r sin = 0 r x y tan 1 y / x

8 Vector Components Similar to polar coordinate transformation A A A x y tan A Acos Asin x 1 A y A / y A x x coordinate yields the x-component A x of vector A y coordinate yields the y-component A y of vector A

9 Second and Third Quadrant Adjustment The direction is always measured from the +x axis 180 tan 1 A y / A x tan -1 (B y /B x ) < 0 for quadrant II tan -1 (C y /C x ) > 0 for quadrant III

10 3-D Position Vector Direction consists of two angles Choice for two angles Geographer s coordinates Polar angle (longitude) Angle from the horizon (latitude) 0 = horizontal view Spherical coordinates Polar angle Azimuthal Angle 0 = view at the top (azimuth)

11 Geographer s Coordinates

12 Spherical Coordinates

13 Exercise: Position Vector -D Specify the position of the back door. Reference point: Reference direction: Magnitude: Direction: 3-D Specify the position of the projector in the classroom. Reference point: Reference direction: Magnitude: Direction:

14 How fast is the object moving? (Velocity Vector) Average velocity v ave displacement time elapsed Instantaneous velocity Velocity at any instant in time Average velocity over a very short time interval Slope of the position vs. time graph Speed = magnitude of velocity v lim t 0 x x t f t x i dx dt x t

15 Example: Runner s Average Velocity During a 3.00 s time interval, a runner s position changes from x 1 = 50.0 m to x = 30.5 m towards you along a straight track. What is the runner s average velocity? Solution v ave x t 30.5m 50.0m 3.00s 19.5m 3.00s 6.50m / s

16 Is it slowing down or picking up speed? (Acceleration Vector) Average acceleration Instantaneous acceleration a ave v f The acceleration at a particular point in time The average acceleration over a very small time interval The slope of the velocity vs. time graph v dv a lim t 0 t dt t v i v t

17 Example: Accelerating Car A car accelerates along a straight road from from rest to 75 kph in 5.0 s. What is the magnitude of the average acceleration? Solution a ave v f t v i 75kph 5.0s 15 km hs 1000m s 4.m / s

18 When we know the acceleration of an object we can figure out how it is moving! Zero acceleration At rest or moving with a constant velocity Constant acceleration Speed varies linearly (direction remains constant) Position varies parabolically (in time) Variable acceleration Non-linear change in speed or changing direction

19 Motion Graphs Position vs. time Velocity vs. time Velocity = time rate of change of position Slope of position vs. time graph Acceleration vs. time Acceleration = time rate of change of velocity Slope of velocity vs. time graph

20 Constant Velocity Graphs x v t t a = 0

21 Constant Acceleration Graphs

22 Example: Graphical Analysis The velocity of a motorcycle driven by a PNP officer is given by the graph below. How far does the officer go after 5 s? 9 s? 13 s? Solution Consider the area under the curve! t=5 s x=(0 m/s)(5 s)=100m t=9 s x=(0m/s)(9s)+(1/ )(4s)(5m/s) =180 m+50 m=30 m t=13 s x=30 m+(1/)(4s)(45m/s)=30m

23 Variable Acceleration Uniform Circular Motion Launching spaceships/satellites Planetary Orbits Motion in the very general sense

24 Questions and Problems for Contemplation Giancoli (6 th edition) Chapter Questions:, 4, 6, 8, 14, 17, 18, 1 Problems: 3, 7, 8, 0, 4, 6, 35, 39, 46, 50, 56 General Problems: 57, 58, 59, 60, 68, 76, 81 First Long Exam Wednesday, Dec. 4, 013 Submit blue book by Dec. Chapters 1, and 3 including notes, assigned questions and problems

25 Free Fall When an object falls through the air, how fast will it fall down? With no air friction Constant acceleration g = -9.8 m/s g = 9.8 m/s With air friction Variable acceleration

26 Free Fall Demo Falling Objects Galileo s experiment

27 Equations of Motion (No Friction) a(t) = a o = -g = -9.8 m/s v(t) = -gt + v o v o = initial velocity x(t) = -(½)gt + v o t + x o x o = initial displacement

28 Reaction Time Calculation Catch the falling meter stick! v o = 0, x o =? g = 9.8 m/s v(t) = -gt x(t) = -(½)gt + x o

29 Contest! teams (3 persons each) As quickly as you can, catch the falling meter stick between your thumb and pointing finger Only one trial!

30 Throwing Objects Upwards Equations with an initial velocity component x max, time of flight, or v o usually to be determined x o = 0 point where object leaves hand g = 9.8 m/s v(t) = -gt + v o x(t) = -(½)gt + v o t

31 Example Calculate the initial velocity of an object thrown upward to a height of.0 m. Solution Find v o. v(t)=0 at the highest point 0 = -(9.8 m/s )t + v o need to find t! x max =.0 m x(t) = -(½)gt + v o t.0 m = -(4.9m/s )t + v o t From the first equation, t = v o /(9.8 m/s ). Thus.0 m = -(4.9m/s )(v o /(9.8 m/s )) + v o (v o /(9.8 m/s )) Solve for v o.0 m = -v o /(19.6 m/s ) + v o /(9.8 m/s )

32 Example continued.0 m = -v o /(19.6 m/s ) + v o /(9.8 m/s ).0 m = v o /(19.6m/s ) v o = 39. m /s v o = m/s v o = 6. 3 m/s

33 Questions and Problems for Contemplation Giancoli (6 th edition) Chapter 3 Questions: 5, 6, 9, 11, 14, 16, 18 Problems:, 8, 18, 8, 35, 38, 47, 48, 49 General Problems: 53, 57, 63, 69, 75

34 Seatwork: Plot the motion graphs for a bouncing ball. You may work in pairs

35 Motion Graphs for Bouncing Ball

36 -D Motion Projectile Motion

37 With No Air Resistance Horizontal direction Constant velocity Vertical direction Constant acceleration Projectile launched with initial velocity v o at an angle from the horizontal v ox = v o cos v oy = v o sin

38 Equations of Motion Horizontal v(t) = v ox v ox = initial velocity horizontal component x(t) = v ox t + x o x o = initial horizontal displacement Vertical a(t) = a o = -g = -9.8 m/s v y (t) = -gt + v oy v oy = initial velocity vertical component y(t) = -½gt + v oy t + y o y o = initial vertical displacement

39 Horizontal v o Will you be able to jump across to the other building if your initial horizontal velocity is 10 m/s? Solution x o = 0, y o = 0 v oy = 0, v ox = 10 m/s x(t) = (10 m/s)t v y (t) = -(9.8 m/s )t y(t) = -(4.9 m/s )t y(t)= -5.0 m = -(4.9 m/s )t 5.0 t s 1. 0s 4.9 x(1.0s) = (10 m/s)(1.0s) = 10 m 10m 1m You will fall short of the building!! v o 1m 5.0 m

40 Exercise What take-off velocity would you need to jump successfully to the other building? If you cannot go any faster, will you succeed by just varying your take-off angle?

41 -D Trajectory of the Projectile Combine horizontal and vertical motion

42 -D Trajectory x(t) = v ox t (1) y(t) = -½gt + v oy t () From (1), t = x/v ox Substitute into (), y v v oy ox x g v ox x (tan ) x v o g cos x y Ax Bx Isn t this an equation for a parabola?!

43 Plotting the Parabola Roots: y = 0 0 = x(a - Bx) Two roots A tan B g v o cos x = 0 (take-off point) v x = A/B (landing point, range R) o sin R g Vertex: x = A/B x coordinate of y max y max = A /B A /4B = A /4B y max v o sin g

44 Time of flight Time from launching point to landing point t x v ox v o R cos v o v sin o cos g v o sin g Time to reach maximum height y max 0 t v oy vo sin g gt This is just half the time of flight!

45 Range of the Projectile Varying the projection angle

46 Azkal s Football Kick v o = 0.0 m/s, 37.0 Calculate Maximum height Time of flight Range Velocity at the maximum height Velocity as it hits the ground

47 Solution Resolve initial velocity into its components v v ox oy v v At maximum height, v y = 0 With y o = 0 y v o o oy cos37.0 (0.0m / s)(0.799) 16.0m / s sin 37.0 (0.0m / s)(0.60) 1.0m / s t t v oy g gt 1.0m/ s 9.80m/ s 1.s (1.0m / s)(1.s) (4.90m / s )(1.s) 7.35m

48 Solution Time of flight Time to go up done Time to go up = time to go down Time of flight = x time to go up =.44s Alternatively, 0 v oy t gt (1.0m / s) t (4.90m / s ) t 0 t[(1.0m / s) (4.90m / s ) t] 1.0m / s t. 45s 4.90m/ s

49 Solution Range = how far will it go horizontally Horizontal displacement at t = t flight x voxt flight (16.0m / s)(.45s) 39. m

50 Solution Velocity at the maximum point v y = 0 v x = 16.0 m/s v = 16.0 m/s, 0 (horizontal) Velocity as the football hits the ground Velocity at t = t flight v x = 16.0 m/s v y =? v y ( 9.8m/ s )(.45s) 1.0m / s 1.0m / s v ( 1.0m / s) (16.0m / s) 0.0m / s tan 1 1.0m/ s 16.0m/ s 36.9

51 Hang-Time Calculate the fraction of time the football spends on the upper half of its flight. Solution From y=(1/) y max to highest point, y max, back to y=(1/) y max y max = 7.35 m y v oy t gt 7.35m (1.0m / s) t (4.9m / s ) t m (1.0m / s) t (4.9m / s ) t t s 1 t. 090s t t.090s 0.359s s 1 t t t 1 flight 1.73s.45s %

52 Jumping From Building A to Building B Jump at an angle of 15 from the horizontal, 10.0 m/s Time to go up to max height Time to go down Determine y max first vo sin (10m / s)sin 15 t 0. 64s g 9.8m / s Now determine t down (free fall from a height of 5.00m +0.34m = 5.34m) y gt Time of flight = t up + t down = 0.64s s = 1.308s Horizontal distance covered x vo sin (10m / s) sin 15 ymax 0. 34m g (9.8m / s ) ( 5.34m) t down s 9.8m / s You ll make voxt flight ( 10.0m / s)cos15 (1.308s) 1. 6m the jump!

53 Basketball Exercise There are two ways to shoot the ball given the same initial velocity High arc Low arc Determine the angles of projection for the two shots mentioned above for your favorite player

54 Reminders LONG TEST 1 on December 4, 013 (Wednesday) Chapters 1, and 3 Submit Blue Book on Dec. Monday

55 Relative Velocity (1-D) The velocity with respect to a particular reference frame The woman The train The road The bike rider Woman s velocity relative to the train is 1.0 m/s Train s velocity relative to bike rider is 3.0 m/s What is the woman s velocity with respect to the bike rider? v P / A vp / B vb/ A

56 Example You are driving north on a straight road at a constant velocity of 88 kph. A truck is traveling at a constant velocity of 104 kph on the opposite lane. Relative velocity of truck with respect to you v v v T / E T / Y Y / E vt / Y vt / E vy / E 104kph 88kph v Y T / 19kph Your relative velocity with respect to the truck vy / T vt / Y 19kph Relative velocities don t change after the truck has passed you!

57 Relative Velocity (-D) Vector addition required Woman is walking at an angle with respect to the train s displacement Train is moving at an angle with respect to the normal to the bike rider s line of sight Position vector r P / A rp / B rb / A Velocity Vector v P / A vp / B vb/ A

58 Example An airplane is headed north at 40 kph. If there is a wind of 100 kph from west to east, determine the resultant velocity of the airplane with respect to the ground. P=plane, A=air, E=earth v P / A 40kph v A / E 100kph v v From the diagram v P / E P/ A A/ E due due West East v P / E 40kph 100kph 60kph 100kph tan 1 40kph 3 E of N

59 Correcting Flight Path In what direction should you fly the plane so that its resultant direction is northwards? v P / A 40kph direction unknown v A / E 100kph due East v v v P / E P/ A A/ E From the diagram, v P / E 40kph 100kph 18kph 100kph sin 1 5 W of N 40kph