A car going round a roundabout experiences a force towards the centre of the roundabout. This force is provided by friction in the tyres.Limiting friction is proportional to the normal contact force, which on a flat road equals the car's weight: F = μmg where mg is the weight. So μmg must be greater than or equal to mv2/r. So v cannot be greater than √(μgr). Over the top of a hill the centripetal force F = mg - S = mv2/r where s is the normal contact force. S is zero when g = v2/r and the car loses contact with the road.On a banked road, the horizontal component of the normal reaction provides the centripetal force. v2 = gr tanθ so if the speed exceeds a certain value the radius has to increase and the car will move outwards.
On the Road
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Specification: AQA Physics (7408), A-level, 3.6.1.1 Circular motion. Topic: Circular Motion.
Key terms in this lesson
- Centripetal force
- The resultant force, directed towards the centre of a circle, needed to keep an object moving in that circular path.
More in Circular Motion
- Uniform Circular Motion
- Angular Displacement
- Angular Speed
- Centripetal Acceleration
- Proof - centripetal acceleration
- Centripetal Force
- The London Eye
- A Spin Dryer
All 9 lessons in Circular Motion · All AQA A-level Physics topics