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If a car is travelling over a curved hill, what is the maximum speed it can travel before losing contact with the road surface?

This is a question about circular motion that also includes some knowledge of Newton's laws. The car is travelling over a curved surface, which you can imagine as part of the circumference of a circle. In order for the car to be travelling in a circular path, the resultant of all the forces acting on it must provide the centripetal force (which acts towards the centre of the circle). The forces acting on the car are its weight, equal to its mass times the gravitational field strength (mg) acting vertically downwards, and the normal reaction force from the road surface, acting perpendicularly to the road surface. We shall call this R. The resultant of the two forces is equal to mv2 /r, where r is the radius of curvature of the circular path, and this force is the centripetal force. When the car is at the top of the hill, the reaction force is vertically upwards and the weight is vertically downwards, thus the resultant of the two that acts towards the centre of the circular path = mg - R. This resultant force provides the centripetal force and so mg -R = mv2 /r. If the car is to lose contact with the road, then there will no longer be a normal reaction force from the road acting on the car, therefore R=0. When this happens when the car is at the top of the hill, the centripetal force, mv2/r = mg. There is mass (m) present on both sides of the equation here which cancel out, leaving us with v2/r = g. We can rearrange this to make v, the speed at which the car loses contact with the road surface, the subject of the equation. We end up with v= (gr)1/2. The reason this is the maximum speed of the car when it loses contact is because this is when the entirety of the car's weight contributes to the centripetal force (since it acts in the same line as the centripetal force when the car is at the top of the hill) rather than some component of it.)

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2 years ago

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