2. 4.5Projectile motion 4.6Range of projectile Dynamics 4.8Relative motion in 1D 4.9Relative motion in 3-D 4.7Circular motion (study at home) 5.2Newton 1 5.3-5.5Force, mass, and Newton 2 4.5Projectile motion 4.6Range of projectile Dynamics 4.8Relative motion in 1D 4.9Relative motion in 3-D 4.7Circular motion (study at home) 5.2Newton 1 5.3-5.5Force, mass, and Newton 2 Summary for Lecture 3 Problems Chap. 4 24, 45, 53,

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6. Gallileo Parabola Max range 45 o 2 angles for any range Parabola Max range 45 o 2 angles for any range Running out of “impetus”

7. What do we know from experience? The trajectory depends on: Initial velocity Projection angle Anything else? Air resistance Ignore for now

8. Projectile Motion To specify the trajectory we need to specify every point (x,y) on the curve. That is, we need to specify the displacement vector r at any time (t). x y r(t) r(t) = i x(t) + j y(t) x(t) y(t)

9. v 0 = iv ox + j v oy -g v 0 sin  v 0 cos  Usually we know the initial vector velocity v o. We know acceleration is constant = -g. accel is a vector a = ia x + ja y = i 0 - j g = i v 0 cos  jv 0 sin  What do we know? vovo  x y

10. The vertical component of the projectile motion is the same as for a falling object The horizontal component is motion at constant velocity Finding an expression for Projectile motion

11. To define the trajectory, we need r(t) That is:- we need x(t) and y(t) v x (t) = v 0x + a x t x(t) = v 0 cos  t  x(t) = v 0x t + ½a x t 2 0 since a x =0 v 0 cos  x y r vovo  Consider Horizontal motion velocity = v 0 cos  displacement -g  const Use x - x o = v 0 t + ½ at 2

12. vertical component of displacement a y = -g Use y - y o = ut + ½ at 2 y(t) = v 0y t + ½ a y t 2 Recall v 0 sin  x y r vovo  = v 0 sin  t – ½ gt 2 Re-arranging gives Height (y) as a function of distance (x) for a projectile.

13. y = -k 1 x 2 + k 2 x x y y = kx 2

14. y = -k 1 x 2 + k 2 x x y range vovo

15. Range For what x values does y = 0? i.e. where where Maximum when 2  = 90 0. i.e when  = 45 0 x = 0 or x=0 R sine 0 angle

16. http://www.colorado.edu/physics/phet/web-pages/simulations-base.html

17. R R’  y x Gradient of slope Range up a slope

18. True for x = 0 or

19. y’ If m = 0 (on level ground) y’ = mx’ R’= x’ = R R  R’ y x x’

20. y  vovo For collision, x and y for dart and monkey must be the same at instant t Time to travel distance d isd y for dart y - y 0 = v 0 t + ½ at 2 h -g y o = 0

21. y  h vovo d How far has monkey fallen? y for dart at impact (ie at time ) Therefore the height of the monkey is y = h - dist = v 0 monk t + ½ at 2 g

22. Dynamics Kinematics HOW things move Dynamics WHY things move

23. 40% Q P. R

24. If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO In the absence of a FORCE a body is at rest 1660 AD A body only moves if it is driven. In the absence of a FORCE A body at rest WILL REMAIN AT REST 350 BC

25. Dynamics Aristotle For an object to MOVE we need a force. Newton For an object to CHANGE its motion we need a force Newtons mechanics applies for motion in an inertial frame of reference! ???????

26. He believed that there existed an absolute (not accelerating) reference frame, and an absolute time. The laws of physics are always the same in any inertial reference frame. His laws applied only when measurements were made in this reference frame….. Newton clarified the mechanics of motion in the “real world”. … or in any other reference frame that was at rest or moving at a constant velocity relative to this absolute frame. Inertial reference frame

27. Both Newton’s and Einstein’s mechanics are relativistic! Values of displacement and velocity made in different inertial reference frames are different but can be simply related. The laws of physics (essentially the forces involved) are always the same in any inertial reference frame. Reference frames that are NOT ACCELERATING Values are relative to the reference frame

28. Both Newton’s and Einstein’s mechanics are relativistic! Values of displacement and velocity made in different inertial reference frames are different but can be simply related. The laws of physics (essentially the forces involved) are always the same in any inertial reference frame. Reference frames that are NOT ACCELERATING Values are relative to the reference frame

29. Newton believed that there existed an absolute (not accelerating) reference frame, and an absolute time. The laws of physics are always the same in any inertial reference frame. Einstein recognised that all measurements of position and velocity (and time) are relative. There is no absolute reference frame.

30. Frames of Reference In mechanics we need to specify position, velocity etc. of an object This requires a frame of reference x y z o p The reference frames for the same object may be different

31. P P x y z o Ref. frame A x y z o Ref. frame B

32. Frames of Reference In mechanics we need to specify position, velocity etc. of an object This requires a frame of reference The reference frames for the same object may be different The reference frames may have a constant relative velocity x y z o p

33. x y z o X’ Y’ Z’ o x y z o We need to be able to relate the coordinates and velocity of an object when measured in frames that are moving relative to each other. X Y Z o

34. Frames of Reference The reference frames may have a constant relative velocity We need to be able to relate one set of measurements to the other The connection between inertial reference frames is the “Gallilean transformation”. In mechanics we need to specify position, velocity etc. of an object or event. This requires a frame of reference x y z o p The reference frames for the same object may be different x’ y’ z’ o

35. Ref. Frame P (my seat in Plane) T (Lunch Trolley) x TP Ref. Frame G (ground) x PG x TG x TG = x TP + x PG Vel. = d/dt(x TG )  v TG = v TP + v PG Accel = d/dt (v TG )  a TG = a TP + a PG In any inertial frame the laws of physics are the same V PG (const) Gallilean transformations 0

36. N E GROUND N’ E’ AIR r PG r AG r PA a PG = (v PG ) = a PA + a AG v AG r PG = r PA + r AG v PG = (r PG ) = v PA + v AG P Looking from above 0 In any inertial frame the laws of physics are the same

37. AG PA AG V PG = V PA + V AG V PA = 215 km/h to East V AG = 65 km/h to North Tan  = 65/215  = 16.8 o Ground Speed of Plane

38. AG PA AG Ground Speed of Plane In order to travel due East, in what direction relative to the air must the plane travel? (i.e. in what direction is the plane pointing?), and what is the plane’s speed rel. to the ground? Do this at home and then do sample problem 4-11 (p. 74) V PA = 215 km/h to East V AG = 65 km/h to North

39. A motorboat with its engine running at a constant rate travels across a river from Dock A, constantly pointing East. Flow N A Compare the times taken to reach points X, Y, or Z when the river is flowing and when it is not. Your answer should include an explanation of what might affect the time taken to cross the river, and a convincing justification of your conclusion. I will quiz you on this next lecture X Y Z

40. Here endeth the lesson lecture No. III

41. Isaac Newton 1642-1727

42. Newton’s 1st Law If a body is at rest, and no force acts on it, IT WILL REMAIN AT REST. If it is moving with a constant velocity, IT WILL CONTINUE TO DO SO If a body is moving at constant velocity, we can always find a reference frame where it is AT REST. At rest  moving at constant velocity

43. An applied force changes the velocity of the body a  Fa  F Force If things do not need pushing to move at constant velocity, what is the role of FORCE??? Inertial mass The more massive a body is, the less it is accelerated by a given force. a =F 1 m F = a m The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed. a = F/m

44. Newton’s 2nd Law a =  F/m a a = = vector SUM of ALL EXTERNAL forces acting on the body m m _______  F = ma vector SUM of ALL EXTERNAL forces acting on the body = ma= ma = ma= ma

45. Newton’s 2nd Law  F=ma F = iF x + jF y + kF z a = ia x + ja y + ka z  F x =ma x  F y =ma y  F z =ma z Applies to each component of the vectors

46. Newton’s 2nd Law F = iF x + jF y + kF z a = ia x + ja y + ka z a x =  F x /m a y =  F y /m a z =  F z /m Applies to each component of the vectors a=  F/m

47. F A + F B + F C = 0 since a=  F/m and a = 0  F = 0  F x = 0  F y = 0 F C cos  - F A cos47= 0 = 220 sin 47 +170 sin28 = 160 +80  240 N F A sin 47 + F C sin  – F B = 0 F B = F A sin 47 +F C sin  220N ? N 170N  F y = 0  cos  = cos47   =28 0 220 170 170 cos  - 220 cos47= 0 (28 0 )

49. m2gm2g N T M2M2 m1gm1g T M1M1 Apply  F = ma to each body i.e  F x = ma x and  F y = ma y For m 2 Vertically  F y = -N + m 2 g = m 2 a y = 0  N = m 2 g Horizontally  F x = T = m 2 a For mass m 1 Vertical (only) m 1 g - T = m 1 a Analyse the equation! m 1 g - m 2 a = m 1 a m 1 g = m 2 a + m 1 a = a(m 2 + m 1 )

50. m2gm2g N T M2M2 m1gm1g T M1M1 Apply a=  F/m to each body i.e a x =  F x /m and a y =  F y /m For M 2 Vertically  F y = -N + m 2 g, = 0 since a y = 0  N = m 2 g Horizontally  F x = T so a = T/m 2  T = m 2 a Analyse the equation! For mass M 1 Vertical (only)  F y = m 1 g - T  F y = m 1 g – m 2 a Since a=  F /m (Newton 2) a = (m 1 g – m 2 a)/m 1 m 1 a = m 1 g – m 2 a a(m 1 + m 2 ) = m 1 g

51. 41% 26% 33% 2.5 x 10 3 N 3.9 x 10 4 N

52. 2.5 x 10 3 N T 3.9 x 10 4 N  F = ma m is mass of road train 5.0 x 10 3 N 3.9 x 10 4 N  F = ma a = (39 – 5) x 10 3 /4 x 10 4 a =  F/m a a = 0.85 m s -2

53. 2.5 x 10 3 N Don’t care! T  F = ma m is mass of trailer!  F is net force on TRAILER 2.5 x 10 3 N T a = 0 So  F= 0 T - 2.5 x 10 3 = 0 T = 2.5 x 10 3 N a = 0

54. 2.5 x 10 3 N 3.9 x 10 4 N  F = ma m is mass of roadtrain No driving force so  F = -5.0 x 10 3 N a = - 5.0 x 10 3 /4.0 x 10 4 = - 0.125 m s -2 Use v 2 = v o 2 + 2a(x – x 0 )  x – x 0 = 400 m v = 0 u = 20 m s -1,