Mr Toogood's Physics

The Bucket Trick

Water can be made to stay in an inverted bucket if it is spun faster than a certain minimum speed. This is because below this speed the water will fall because its weight will accelerate it at 9.81ms^-2. If it is rotated at an angular speed great so that the centripetal acceleration is equal to or greater than its free-fall acceleration, then at the top of the path the water will accelerate downwards at at least the same rate as the bucket.

When the bucket revolves faster than this rate the floor of the bucket pushes down on the water and provides it with enough centripetal acceleration to move in the same circular path as the bucket.

This is the same principle that is used to create artificial gravity in fairground rides and in (potential) space stations. For a space station to create artificial gravity it would need to rotate with a centripetal acceleration of 9.81ms^-2, so:

So for a space station with a radius of 100m would give an angular velocity of 0.31rad s^-1 and a period of revolution of 20s. The larger the space station, the slower it would have to rotate. There are, obviously lot of other practical considerations about building a space station like this, such as the huge mass.

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Cars Going Over Hills

When a car is travelling on a flat road, the support force from the road is in the opposite direction to and equal in magnitude to the weight of the car. As the weight ($mg$) and the support force ($S$) are in opposite directions, we should assign one of the forces as being negative, in this case, it is easier for us to make the support force, $S$, as being negative which we can state as:

When the car goes over a hill there is a resultant force acting on the car which causes it to move in a circular path. As there is now a resultant force, the sum of these two forces is no longer 0, and the difference is equal to the resultant force:

This resultant force is the centripetal force so:

Which tells us that the centripetal force which causes the car to travel in a circular path is the difference between the weight and the support force. The size of the support force can be calculated by rearranging the equation, the minus sign on the left in this case tells us that the support force is in the opposite direction to the weight and the centripetal force.

The faster the car goes, the greater required centripetal force. As the car’s weight is providing the centripetal force, then as the car gets faster then S must get smaller and smaller until it reaches 0. When S reaches 0 the car will lose contact with the road when the speed of the car reaches v₀.

So for a bridge, with a radius of r the maximum speed the car would be able to go before it became airborne would be:

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Going Round Bends

Cycle tracks have banked corners to allow the bikes to remain at high speeds as they go around the bends. Of a flat surface the centripetal force is provided by the frictional forces between the tyres and the road. When the track is banked, the support force from the surface provides some of the centripetal force, so less has to be provided by the friction.

Resolving horizontally $N \sin \theta+F\cos\theta=\frac{mv^{2}}{r}$ is the centripetal force acting on the car.

Resolving vertically $N \cos \theta=mg$ as it is opposite the weight.

At a certain speed the bike could travel around the track with no friction, in this case:

So the speed the bike would need to be travelling at for there to be no sideways friction is $v^{2}=gr\tan\theta$

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