Centripetal Force the Road Centripetal Force The Road

Centripetal Force & the Road

Centripetal Force & The Road Learning Objectives Book Reference : Pages 26 -27 1. To show that the centripetal force is provided by real world forces such as tension, gravity & friction 2. To consider three particular cases of motion: • Over the top of a hill or humped back bridge • Around flat curves (roundabouts) • Around banked curves

Centripetal Acceleration : Recap During the last lesson we saw that an object moving in a circle has a constantly changing velocity, it is therefore experiencing acceleration and hence a force towards the centre of rotation. We called this the centripetal force: The force required to keep the object moving in a circle. In reality this force is provided by another force, e. g. The tension in a string, friction or the force of gravity.

Over the top 1 Consider a car with mass m and speed v moving over the top of a hill. . . S mg r

Over the top 2 At the top of the hill, the support force S, is in the opposite direction to the weight (mg). It is the resultant between these two forces which keep the car moving in a circle mg – S = mv 2 / r If the speed of the car increases, there will eventually be a speed v 0 where the car will leave the ground (the support force S is 0) mg = mv 02 / r v 0 = (gr)½ Any faster and the car will leave the ground

Around a Roundabout 1 On a level road, when a car travels around a roundabout the centripetal force required to keep the car moving in a circle is provided by the friction between the road surface and tyres Force of Friction F F = mv 2 / r friction velocity

Around a Roundabout 2 To avoid skidding or slipping, the force of friction F 0 must be less than the point where friction is overcome which occurs at speed v 0 Friction is proportional to weight and can be given by the coefficient of friction ( ): F mg F = mg At the point of slipping: F 0 = mv 02 / r v 0 = ( gr)½ mg = mv 02 / r

Banked Tracks 1 For high speed travel, race tracks etc have banked corners. In this way a component of the car’s weight is helping friction keep the car moving in a circle N 1 N 2 Towards centre of rotation mg

Banked Tracks 2 Without any banking the centripetal force is provided by friction alone. Banked corners allows greater speeds before friction is overcome The centripetal force is provided by the horizontal components of the support forces (N 1 + N 2) sin = mv 2 / r and the vertical components balance the weight (N 1 + N 2) cos = mg

Banked Tracks 3 Rearranging sin = mv 2 / (N 1 + N 2) r cos = mg / (N 1 + N 2) and since tan = sin / cos tan = mv 2 / (N 1 + N 2) r x (N 1 + N 2) / mg tan = mv 2 / mgr v 2 = gr tan Thus there is no sideways frictional force if the speed v is such that v 2 = gr tan

Problems 1 A car with mass 1200 kg passes over a bridge with a radius of curvature of 15 m at a speed of 10 ms-1. Calculate: a. The centripetal acceleration of the car on the bridge b. The support force on the car when it is at the top The maximum speed without skidding for a car with mass 750 kg on a roundabout of radius 20 m is 9 ms-1. Calculate: a. The centripetal acceleration of the car on the roundabout b. The centripetal force at this speed

Problems 2 A car is racing on a track banked at 25° to the horizontal on a bend with radius of curvature of 350 m a. Show that the maximum speed at which the car can take the bend without sideways friction is 40 ms-1 b. Explain what will happen if the car takes the bend at ever increasing speeds