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Summary Lecture 5 5.4 Inertial/non-inertial reference frames 6.1-2 Friction 6.5 Taking a curve in the road Thursday 12 – 2 pm 6.4 Drag force Terminal velocity Problems Chap 6: 5, 14, 29 , 32, 33, PPP “ExtEnsion” lecture. Room 211 podium level Turn up any time Measured weight in an accelerating Reference Frame According to stationary observer R is reaction force R mg accel a = reading on scales F = ma Taking “up” as +ve R - mg = ma R = m(g + a) If a = 0 R = mg If a is +ve R = m(g + a) weight increase If a is -ve R = m(g - a) weight decrease normal weight Spring scales According to traveller R is reaction force R mg = reading on scales 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? Only if it is an inertial frame of reference! The accelerating lift is NOT! Why doesn’t Mick Doohan fall over? Friction provides the central force mg In the rest reference frame 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 The Source of Friction between two surfaces Static Friction If no force F Surface with friction N F No friction force f f mg As F increases friction f increases in the opposite direction. Therefore Total Force on Block = zero does not move As F continues to increase, at a critical point most of the (“velcro”) bonds break and f decreases rapidly. F is now greater than f and slipping begins Surface with friction N F f f depends on surface properties. Combine these properties into a coefficient of friction m fmN m is usually < 1 Static f < or = ms N Kinetic f = mk N f Slipping begins (fmax = msN ) fmax f < fmax (= mkN ) Coefficient of Kinetic friction < Coefficient of Static friction F Static friction Kinetic friction N f F q q At qcrit F=f mg sin qcrit = f = mS N mg = mS mg cosqcrit mg sin qcrit thus mS = mg cos qcrit mS = tan qcrit Independent of m, or g. Property of surfaces only Making the most of Friction A F1 > F2 fcrit = mS N B F1 = F2 fcrit = mS mg C F1 < F2 Friction force does not depend on area! N N F2 F1 fcrit1 mg mg fcrit2 So why do Petrol Heads use fat tyres? tribophysics To reduce wear? Tyres get hot and sticky which effectively increases m. The wider the tyre the greater the effect? What force drives the car? acceleration Driving Torque Force of Tyre on road Force of road on Tyre Stopping Distance depends on friction vo Braking force f N Friction road/tyres d v2 =vo2 + 2a(x-xo) 2 0 = vo + 2ad mg 2 v0 d 2a Max value of a is when f is max. fmax = msN = msmg F = ma -fmax = mamax -msmg = mamax amax = - msg 2 Thus since v0 d 2a 2 v0 dmin 2amax 2 v0 d min 2m s g dmin depends on v2!! Take care!! If v0 = 90 kph (24 m s-1) and m = 0.6 ==> d = 50 m!! Taking a curve on Flat surface v Fcent r N mg mv Fce n t r 2 Fcent is provided by friction. If no slipping the limit is when Fcent = fs(limit) = msN = msmg So that mv 2 μ s mg r v μ s gr Does not depend on m So for a given ms (tyre quality) and given r there is a maximum vel. for safety. If ms halves, safe v drops to 70%….take care! 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. Albert Park GP circuit Central force provided by friction. mv2/R = mN = mmg m = v2/Rg N m = 4.3 mv2/R mg m for racing tyres is ~ 1 (not 4!). How can the car stay on the road? Soft rubber Grooved tread Are these just for show, or advertising? 20 Another version of Newton #2 F ma dv m dt d(m v ) dt dp dt p= mv =momentum Momentum Dp transferred over a time Dt gives a force:- Dp F Dt F is a measure of how much momentum is transferred in time Dt Distance travelled in 1 sec @ velocity v =vm Volume of air hitting each spoiler (area A) in 1 sec Area A m2 mass of air (density ) hitting each spoiler in 1 sec @ 200 kph v = 55 m s-1 = v x A m3 A ~ 0.5 m2 = x v x A kg ~ 1 kg m-3 Momentum of air hitting each spoiler in 1 sec = x v2 x A kg m s-1 If deflected by 900, mom change in 1 sec Newton says this is the resulting force mv2/R = mN = mmg mv2/R = mN = m (m + 3000) g F ~ 3 x 104 N ~ 3 Tonne! VISCOUS DRAG FORCE VISCOUS DRAG FORCE What is it? like fluid friction a force opposing motion as fluid flows past object Assumptions low viscosity (like air) turbulent flow v What does the drag force depend on? D velocity (v2) D effective area (A) D fluid density () D A v2 D= ½ C 2 A v C is the Drag coefficient. It incorporates specifics like shape, surface texture etc. Vm Area A Fluid of density In 1 sec a length of V metres hits the object 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 = const AV2 Dp F Dt F = mg - D F = mg -1/2CAv2 V=0 mg D V D increases as v2 until F=0 mg i.e. mg= 1/2CAv2 D v 2 term V mg v term 2mg CA 2mg CA D F = mg –D ma = mg -D mg dv mg - D m dt dv m + 1/2CAv 2 mg 0 dt v [ 2gm (1 e cA 2m t cA )]1 / 2 2gm 1 / 2 v [ ] cA v [ 2gm (1 e cA 2m t cA )]1 / 2 When entertainment defies reality Calculate: D= ½ CAv2 Drag force on presidents wife Assume C = 1 Compare with weight force v = 700 km h-1 Could they slide down the wire? q Calculate: D= ½ CAv2 Assume C = 1 v = 700 km h-1 The angle of the cable relative to horizontal. Compare this with the angle in the film (~30o) In working out this problem you will prove the expression for the viscous drag force 1 F CAv 2 2