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• A 1 kg mass is attached to a 0.5 m long, stiff, light plastic thread that has a strength of 50 N. This is attached to frictionless, horizontal turntable. The mass is then spun in a circle with the rate of rotation gradually increasing. How fast is the turntable rotating when the thread breaks? How fast and in what direction does the mass move when thread breaks. How much kinetic energy does the mass possess? Can you say anything about what point in the rotation the thread is likely to break?

As the plane of rotation is horizontal and frictionless we can ignore gravity entirely. Using the standard circular motion equations:

Since all points in the circle are equivalent then we can say nothing about the point at which the rod breaks but when it does the mass will continue to move in the direction it was moving in immediately beforehand and so will move in a tangent to the circle until it falls off the turntable.

The experiment is repeated but this time the turntable is gradually tilted in to a vertical plane. If the thread cannot support compression what is the minimum rate of rotation that can be allowed for the vertical turntable? At what rotation rate do you expect the thread to break at now? How fast and in what direction does the mass move when thread breaks. How much kinetic energy does the mass possess? Can you now say anything about what point in the rotation the thread is likely to break?

Is there a minimum strength for the thread in order to achieve a vertical circle?

For the vertical circle we have to consider the effects of gravity. Assuming that the mass continues to rotate at a constant rate then the magnitude of the centripetal force must be constant (although the direction will change, always pointing towards the centre of rotation). At the top of the circle the weight of the mass combines with the tension in the thread to provide the required centripetal force, this means that the tension will be relatively small. At the bottom of the circle though the weight is pointing out of the circle and so the tension is much larger.

So at the top of the circle:

And at the bottom of the circle:

At the top of the circle the thread will lose all tension when the centripetal force is provided entirely by the weight of the mass:

If the rate of rotation falls below this then the mass will not be able to complete the circular path. Note that the rate of rotation is independent of the mass.

The tension is greatest at the bottom or the circle and this is where we can expect the thread to break

When the thread breaks we would expect the mass to move off at a tangent to the circle, i.e. horizontally. This would be the top of a parabola that the mass follows as it falls to the ground.

However as the rate of rotation is gradually increasing there is no assurance that the critical rotation rate will be reached at the very bottom of the circle. Also if the string is a ‘real world’ object then the point at which it breaks may well be affected by its history (how it has be stressed before) with small sections of a cotton like thread breaking before others. We can however expect the thread to break as the tension is increasing and so generally we can expect this to happen on the downward section of the circle and close to the bottom.

If there is a minimum strength for the thread then this will be found when the minimum rotation rate (at the top of the circle) matches the maximum rate (at the bottom).

So, in order to achieve a vertical circle, the minimum strength for the thread is twice the weight of the mass. Notice that this is independent of the length of the thread even though the minimum and maximum frequencies do depend on it.