Loop of death

JW Player goes here

A radius of the deadly circle curve has to be suitably short, so that the little sphere (or roller coaster carriages) will not fall down. In order to calculate its length let us analyse forces applied at the highest point of the circle.

At the highest point of the circle two forces affect the little sphere: gravitation force mgand reaction force of the track N.

The sum of these two forces should be smaller than the centripetal force acting on te spheres (which has the mass m). The centripetal acceleration equals to

Thus

In order for the ball not to break off the track, the reaction force N between the sphere and the track must be bigger than zero, since when it is N = 0, the sphere does not touch the track. If we assume that N > 0 than one reads

The kinetic energy of the ball is:

H and at the height of 2R, respectively, i.e.:

Having calculated v² from the above equation we can apply it in the following inequality. In conclusion, in order for the ball to follow the deadly circle curve it must be released from the height of:

As Professor Krzysztof Ernst analysed it in his book 'Einstein on a see-saw', the centrifugal force at the lowest point of a circle with a radius Ris 5mg. Combined with the gravitational force the overall acceleration would reach the level of 6g (a pilot of a jet plane loses temporary consciousness at the acceleration of 7g). That is why a radius of the bottom part of the deadly circle is a bit bigger and that is why the truck of roller-coaster train resembles Greek letter 'alfa'.

(In our calculations we disregarded the energy of spinning of the ball. As a matter of fact, if we use a ball, the height of the truck at the initial point should be 2.7R not 2.5R).