On the road

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On the road

  1. 1. On The Road
  2. 2. What happens when you drive too fast over a hill! • http://www.youtube.com/watch?v= QhjNmXpFnms&feature=related
  3. 3. What are we covering today? • Flying off hills • Sliding around corners • Staying on a racetrack
  4. 4. At what speed will the car leave the ground? What are the forces on the car when it is on the ground? Up S (reaction force) (mv2 )/r (Centripetal force) Down mg (weight) mg (mv2 )/rS + When the car is on the ground the forces are balanced mg = s + (mv2 )/r If the car is travelling at a speed greater than (V0) it will leave the ground At this point s = 0 as there is no reaction force from the ground
  5. 5. mg (mv2 )/r m v0 2 / r = mg m = mass V0 = is the first speed it takes to the air r = radius of the hill from the centre of curvature Re- arrange to make V0 the centre of attention to find the speed for take off V0 = g r Therefore a car would jump a hill of radius 6 metres if it is going a speed greater than 7.67 m/s http://www.animations.physics.unsw.edu.au/jw/circular.htm
  6. 6. On a roundabout http://h2physics.org/?cat=63 A vehicle of mass m and speed v in a circle of radius r around a roundabout r The centripetal force is created by the tyres contact friction on the road Friction Force F = mv2 r If F is greater than the grip available from the tyre and surface the car will slide – this will happen at a speed v0 and a corresponding force F0 The grip from the tyre is proportional to the weight of the vehicle and the road surface and the tyre combine to give the co-efficient of friction µ Sliding Force F0 = mv0 2 r http://phys23p.sl.psu.edu/phys_anim/mech/ embederQ2.20150.html
  7. 7. The grip from the tyre is proportional to the weight of the vehicle and the road surface and the tyre combine to give the co-efficient of friction µ Sliding Force F0 = mv0 2 r F0 = µmg The grip from the tyre can be broken down into its component parts µmg = mv0 2 r Rearrange to make v0 the centre of attention v0 = sqrt (µ g r)
  8. 8. Banked Track θ N1 N2 θ θ Car going around a banked track You can travel faster around a banked corner as a component of the cars mass keeps the car into the centre of the trackmg Resolving the vertical and horizontal components Horizontal component = (N1 + N2) sin θ Vertical component = (N1 + N2) cos θ The horizontal component act as the centripetal force mv2 /r = (N1 + N2) sin θ The vertical component balances the weight (N1 + N2) cos θ = mg using tan θ = sin θ / cos θ
  9. 9. Banked track (N1 + N2) sin θ (N1 + N2) cos θ = using tan θ = sin θ / cos θ (N1 + N2) cos θ = mgmv2 /r = (N1 + N2) sin θ mv2 mgr Reduces down to ...Reduces down to ... Tan θ = v2 /g rTan θ = v2 /g r Making v the focus V2 = g r tan θ http://phys23p.sl.psu.edu/phys_anim/mech/ embederQ2.20170.html
  10. 10. • http://phet.colorado.edu/en/simulation/rotation http://phet.colorado.edu/en/simulation/rotation

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