2. Problems Chap 4 45, 53, Chap 5: 13, 19, 27, 45, 50, 53 Summary Lecture 4 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 5.7Newton 3 5.8Some examples 6.1-2Friction 6.4Drag force Summary Lecture 4 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 5.7Newton 3 5.8Some examples 6.1-2Friction 6.4Drag force Labs and Tutes start this week. Check times and places on notice board: level 2 physics podium. Thursday 12 – 2 pm “Extension” lecture. Room 211 podium level Turn up any time

3. 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

4. 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! ???????

5. 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

6. 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

7. 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.

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. Isaac Newton 1642-1727

15. 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

16. 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 ?

17. How to measure the Inertial Mass Balances the gravitational weight Which would work in a space ship away from gravity? m measures inertial mass

18. 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

19. 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

20. 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

21. 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 sin 28 = 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 )

22. A (M 2 – M 1 )g B (M 1 – M 2 )g C M 1 g D Unknown

24. 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 )

25. 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

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

27. 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

28. 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

29. 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,

30. Newton’s 3rd Law Forces come in pairs To every action there is an equal and opposite reaction. The action-reaction pairs DIFFERENT ALWAYS act on DIFFERENT bodies

31. N AG mg What is reaction pair to the weight force mg? Is it N (normal reaction)? N The reaction pair is the force of the apple on the Earth N is the force of the ground on the apple N GA The reaction pair to N GA (or N) is the force of the apple on the ground, N AG N GA NO!. N and mg act on the SAME body

32. According to stationary observer Rmg  F = ma Taking “up” as +ve R - mg = ma R = m(g + a) If a = 0 ==> R = mg normal weight If a is +ve ==> R = m(g + a) weight increase If a is -ve ==> R = m(g - a) weight decrease R is reaction force = reading on scales Measured weight in an accelerating Reference Frame accel a Spring scales Man in lift

33. According to traveller  F = ma R - mg = ma BUT in his ref. frame a = 0! so R = mg!! How come he still sees R changing as lift accelerates? Rmg R is reaction force = reading on scales Only if it is an inertial frame of reference! The accelerating lift is NOT! Didn’t we say the laws of physics do not depend on the frame of reference?

34. Here endeth the lesson lecture No. IV

35. Summary Lecture 5 6.1-2Friction 6.4Drag force Terminal velocity 6.5Taking a curve in the road Summary Lecture 5 6.1-2Friction 6.4Drag force Terminal velocity 6.5Taking a curve in the road Problems Chap 6: 5, 14, 29, 32, 33,

36. Staff-Student Liaison Committee This committee meets 3 times each semester, to consider reports from lectures, labs. and tutorials. The meetings are after 5 pm, and last ~1 hour. We need a rep. from this class to report comments from this lecture group.

37. Physics Phrequent Phlyers From 1 – 2 pm tomorrow Physics Lab. Level 3 Physics Podium building Measuring acceleration Checking projectile motion Quantifying circular motion Measuring mass with springs Realistic F1 Grand Prix calculations

39. mg Why doesn’t Mick Doohan fall over? Friction provides the central force In the rest reference frame

41. What is Friction Surfaces between two materials are not even Microscopically the force is atomic  Smooth surfaces have high friction Causes wear between surfaces  Bits break off Lubrication separates the surfaces

42. The Source of Friction between two surfaces

43. f F mg Static Friction As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero  does not move F is now greater than f and slipping begins If no force F No friction force f Surface with friction  As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly.

44.  f F f depends on surface properties. Combine these properties into a coefficient of friction  f   Nf   N  is usually < 1 Staticf < or =  s N Surface with friction Kineticf =  k N

45. f F f < f max (=  k N ) Static friction Kinetic friction Coefficient of Kinetic friction < Coefficient of Static friction Slipping begins (f max =  s N ) f max

46.  mg F f  At  crit F = f mg sin  crit = f =  S N Independent of m, or g. Property of surfaces only  S = tan  crit mg sin  mg cos   =  S mg cos  crit thus  S =

47. F1F1 F2F2 Making the most of Friction AF 1 > F 2 BF 1 = F 2 CF 1 < F 2 mg N N f1f1 f2f2 f =  S N Friction force does not depend on area! f =  S mg

48. So why do Petrol Heads use fat tyres? To reduce wear Tyres get hot and sticky which effectively increases . The wider the tyre the greater the effect. To reduce wear Tyres get hot and sticky which effectively increases . The wider the tyre the greater the effect. The truth! Friction is not as simple as Physics 141 says! T h e t r u t h ! F r i c t i o n i s n o t a s s i m p l e a s P h y s i c s 1 4 1 s a y s ! tribophysics

49. Force of Tyre on road Force of road on Tyre acceleration What force drives the car? Driving Torque

50. Braking force Friction road/tyres v d f v 2 =v o 2 + 2a(x-x o ) 0 = v o 2 + 2ad  F = ma =  s mg Max value of a is when f is max. Stopping Distance depends on friction  a max = -  s g  -f max = ma max  -  s mg = ma max vovo N mg f max =  s N

51. Braking force Friction road/tyres v d f v 2 =v o 2 + 2a(x-x o ) 0 = v o 2 + 2ad  F = ma =  s mg Max value of a is when f is max. Stopping Distance depends on friction  a max = -  s g  -f max = ma max  -  s mg = ma max vovo N mg f max =  s N

52. Thus since d min depends on v 2 !!Take care!! If v 0 = 90 kph (24 m s -1 ) and  = 0.6 ==> d = 50 m!!

53. r v F cent mg N F cent is provided by friction. If no slipping the limit is when F cent = f s(limit) =  s N =  s mg So that Does not depend on m So for a given  s (tyre quality) and given r there is a maximum vel. for safety. If  s halves, safe v drops to 70%….take care! Taking a curve on Flat surface

54. r N mg Nsin  Ncos  Banked Curve Vertically  F x = ma x =0 Ncos  = mg Horizontally For a given  there is a safe speed v. Recall that tan  =  s !!

56. Lateral Acceleration of 4.5 g The lateral acceleration experienced by a Formula-1 driver on a GP circuit can be as high as 4.5 g This is equivalent to that experienced by a jet-fighter pilot in fast-turn manoeuvres.

57. Albert Park GP circuit Central force provided by friction. mg N  = v 2 /Rg   = 4.3  = v 2 /Rg   = 4.3 mv 2 /R =  N =  mg R = 70 m mv 2 /R V=55 m s -1  for racing tyres is ~ 1 (not 4!). How can the car stay on the road?  for racing tyres is ~ 1 (not 4!). How can the car stay on the road?

58. Soft rubber Grooved tread Are these just for show, or advertising?

59. 200 km/h

60. Another version of Newton #2 p= mv =momentum F is a measure of how much momentum is transferred in time  t Momentum  p transferred over a time  t gives a force:-

61. Distance travelled in 1 sec @ velocity v Volume of air hitting each spoiler (area A) in 1 sec Area A m 2 mass of air (density  ) hitting each spoiler in 1 sec Momentum of air hitting each spoiler in 1 sec If deflected by 90 0, mom change in 1 sec Newton says this is the resulting force @ 200 kph v = 55 m s -1 A ~ 0.5 m 2  ~ 1 kg m -3 F ~ 3 x 10 4 N ~ 3 Tonne! = v m = v x A m 3 =  x v x A kg =  x v 2 x A kg m s -1 mv 2 /R =  N =  mg mv 2 /R =  N =  (m + 3000) g

62. VISCOUS DRAG FORCE DRAG

63. VISCOUS DRAG FORCE Assumptions low viscosity (like air) turbulent flow What is it? like fluid friction a force opposing motion as fluid flows past object

64. What does the drag force depend on? D  velocity (v 2 ) D  effective area (A) D  fluid density (  D   A v 2 D= ½ C  A v 2 D  velocity (v 2 ) D  effective area (A) D  fluid density (  D   A v 2 D= ½ C  A v 2 C is the Drag coefficient. It incorporates specifics like shape, surface texture etc. v

65. Fluid of density  V mV m Volume hitting object in 1 sec. =AV Mass hitting object in 1 sec. =  AV momentum (p) transferred to object in 1 sec. = (  AV)V Force on object = c onst  AV 2 Area A In 1 sec a length of V metres hits the object

67. V mg D D V V=0  F = mg - D  F = mg -1/2C  Av 2 D increases as v 2 until  F=0 i.e. mg= 1/2C  Av 2

68. 0mgAv1/2C dt dv m 2   F = mg –D D mg ma = mg -D D- mg dt dv m 

71. When entertainment defies reality

72. D= ½ C  Av 2 Assume C = 1 v = 700 km h -1 Calculate: Drag force on presidents wife Compare with weight force Could they slide down the wire?

73. D= ½ C  Av 2 Assume C = 1 v = 700 km h -1 Calculate: The angle of the cable relative to horizontal. Compare this with the angle in the film (~30 o ) 

74. In working out this problem you will prove the expression for the viscous drag force