2. Summary Lecture 5 Summary Lecture 5 5.4Inertial/non-inertial reference frames 6.1-2Friction 6.5Taking a curve in the road 6.4Drag force Terminal velocity Summary Lecture 5 Summary Lecture 5 5.4Inertial/non-inertial reference frames 6.1-2Friction 6.5Taking a curve in the road 6.4Drag force Terminal velocity Problems Chap 6: 5, 14, 29, 32, 33, Thursday 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time Thursday 12 – 2 pm PPP “Extension” lecture. Room 211 podium level Turn up any time

3. 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

4. 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 when lift accelerates? Didn’t we say the laws of physics do not depend on the frame of reference? Rmg R is reaction force = reading on scales Only if it is an inertial frame of reference! The accelerating lift is NOT!

5. 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

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

9. 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

10. The Source of Friction between two surfaces

11. 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.

12.  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

13. 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

14.  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 =

15. F1F1 F2F2 Making the most of Friction AF 1 > F 2 BF 1 = F 2 CF 1 < F 2 mg N N f crit1 f crit2 f crit =  S N Friction force does not depend on area! f crit =  S mg

16. 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

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

18. 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

19. 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

20. 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!!

21. 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

22. 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 !!

24. 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.

25. 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?

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

27. 200 km/h

28. 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:-

29. 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

30. VISCOUS DRAG FORCE DRAG

32. 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

33. 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

34. 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

36. 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

37. 0mgAv1/2C dt dv m 2   F = mg –D D mg ma = mg -D D- mg dt dv m 

40. When entertainment defies reality

41. 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?

42. 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 ) 

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