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  • 1. On The Road
  • 2. What happens when you drive too fast over a hill!
    • http:// www.youtube.com/watch?v = QhjNmXpFnms&feature =related
  • 3. What are we covering today?
    • Flying off hills
    • Sliding around corners
    • Staying on a racetrack
  • 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) (mv 2 )/r (Centripetal force) Down mg (weight) mg (mv 2 )/r S + When the car is on the ground the forces are balanced mg = s + (mv 2 )/r If the car is travelling at a speed greater than (V 0 ) it will leave the ground At this point s = 0 as there is no reaction force from the ground
  • 5. mg (mv 2 )/r m v 0 2 / r = mg m = mass V 0 = is the first speed it takes to the air r = radius of the hill from the centre of curvature Re- arrange to make V 0 the centre of attention to find the speed for take off V 0 = 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. 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 = mv 2 r If F is greater than the grip available from the tyre and surface the car will slide – this will happen at a speed v 0 and a corresponding force F 0 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 F 0 = mv 0 2 r http://phys23p.sl.psu.edu/phys_anim/mech/embederQ2.20150.html
  • 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 F 0 = mv 0 2 r F 0 = µmg The grip from the tyre can be broken down into its component parts µmg = mv 0 2 r Rearrange to make v 0 the centre of attention v 0 = sqrt (µ g r)
  • 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 track mg Resolving the vertical and horizontal components Horizontal component = (N 1 + N 2 ) sin θ Vertical component = (N 1 + N 2 ) cos θ The horizontal component act as the centripetal force mv 2 /r = (N 1 + N 2 ) sin θ The vertical component balances the weight (N 1 + N 2 ) cos θ = mg using tan θ = sin θ / cos θ
  • 9. Banked track (N 1 + N 2 ) sin θ (N 1 + N 2 ) cos θ = using tan θ = sin θ / cos θ (N 1 + N 2 ) cos θ = mg mv 2 /r = (N 1 + N 2 ) sin θ mv 2 mgr Reduces down to ... Tan θ = v 2 /g r Making v the focus V 2 = g r tan θ http://phys23p.sl.psu.edu/phys_anim/mech/embederQ2.20170.html
  • 10.
    • http://phet.colorado.edu/en/simulation/rotation
    http:// phet.colorado.edu /en/simulation/rotation