A car drives over a hill with a circular top that has a radius of curvature of 10m.How fast can it go at the top of the hill if it is not to leave the road?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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  • We know when the car is traveling on top of a curve, a centripital force (`mv^2/r` )will act on the car which will result in leaving the car from the road.
  • This force in increasing with the increase of velocity(`v` ) of the car.
  • Due to the friction normal force will act on the car at same direction of the centripital force.
  • Both normal and centripital force was balanced by the weight(`mg` ) of the car.
  • When the addition of centripital force and normal force exceed the weight of the car, the car will leave out from the road surface and lift up.
  • The maximum speed which can achive without leaving the car is the speed that centripital force = weight of car
  • In this maximum speed the car is about to leave and normal force can be considered as 0.


Centripital force = m*v^2/r

Weight of car    = m*g


mg = m*v^2/r

    v= sqrt(gr)

      = sqrt(9.81*10)

      = 9.9 m/s


So the velocity which can travel the car without leaving the road is 9.9 m/s



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