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# Centripetal Force & the Road - PowerPoint PPT Presentation

Centripetal Force & the Road. Learning Objectives. Book Reference : Pages 26-27. Centripetal Force & The Road. To show that the centripetal force is provided by real world forces such as tension, gravity & friction To consider three particular cases of motion:

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### Centripetal Force & the Road

Book Reference : Pages 26-27

### Centripetal Force & The Road

• To show that the centripetal force is provided by real world forces such as tension, gravity & friction

• To consider three particular cases of motion:

• Over the top of a hill or humped back bridge

• Around banked curves

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.

### Centripetal Acceleration : Recap

Consider a car with mass circle has a constantly changing velocity, it is therefore experiencing acceleration and hence a force towards the centre of rotation.m and speed v moving over the top of a hill...

### Over the top 1

S

mg

r

• If the speed of the car increases, there will eventually be a speed v0 where the car will leave the ground (the support force S is 0)

• mg = mv02 / rv0 = (gr)½

• Any faster and the car will leave the ground

• ### Over the top 2

On a opposite direction to the weight (mg). It is the resultant between these two forces which keep the car moving in a circlelevel 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 = mv2 / r

friction

velocity

• To avoid skidding or slipping, the force of friction opposite direction to the weight (mg). It is the resultant between these two forces which keep the car moving in a circleF0 must be less than the point where friction is overcome which occurs at speed v0

• Friction is proportional to weight and can be given by the coefficient of friction ():

• F  mg F = mg

• At the point of slipping:

• F0 = mv02 / r mg = mv02 / r

•  v0 = (gr)½

• ### Around a Roundabout 2

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

### Banked Tracks 1

N1

N2

Towards centre of rotation

mg

• and the vertical components balance the weight

• (N1 + N2) cos = mg

• ### Banked Tracks 2

• Rearranging friction alone. Banked corners allows greater speeds before friction is overcome

• sin  = mv2 / (N1 + N2) r

• cos = mg / (N1 + N2)

• and since tan  = sin  /cos

• tan  = mv2 / (N1 + N2) r x (N1 + N2) / mg

• tan  = mv2/ mgr v2 = gr tan 

• Thus there is no sideways frictional force if the speed v is such that v2 = gr tan 

• ### Banked Tracks 3

• A car with mass 1200kg passes over a bridge with a radius of curvature of 15m at a speed of 10 ms-1. Calculate:

• The centripetal acceleration of the car on the bridge

• The support force on the car when it is at the top

• The maximum speed without skidding for a car with mass 750kg on a roundabout of radius 20m is 9ms-1. Calculate:

• The centripetal acceleration of the car on the roundabout

• The centripetal force at this speed

### Problems 1

• A car is racing on a track banked at 25 curvature of 15m at a speed of 10 ° to the horizontal on a bend with radius of curvature of 350m

• Show that the maximum speed at which the car can take the bend without sideways friction is 40ms-1

• Explain what will happen if the car takes the bend at ever increasing speeds