27

The Tautochrone Problem

27

Abstract

The research statement of this Mathematics Extended Essay is the following: “On a frictionless wire, there are two beads in a uniform gravitational field; show that the shape the wire must take on, such that the two beads will reach the end of the wire at the same time when they are simultaneously released from rest, from any point on the wire, is half of an inverted cycloid cut at the non-inverted apex.” This is known as the Tautochrone Problem. In investigating this problem I used an energy approach, which allowed me to express the speed of the bead as a function of the height it dropped from during its motion. I then used the definition of instantaneous speed to express a fraction of the time of decent as a function of the speed of the bead at a certain height, and the infinitesimally small distance it covers during that time. The distance the bead covers in the infinitesimally small time interval was approximated to be a straight line. A method for determining the time of decent down any strictly decreasing and differentiable curve was then devised, which was ultimately used to prove that the curve that solves the Tautochrone Problem is a cycloid. A by-product of the proof states that the time of decent for a bead sliding down a cycloid is equal to where is the radius of the circle that defines the cycloid and is the strength of the gravitational field. Finally it was also argued that rolling bodies on a cycloid also exhibit tautochronous behaviour, but this is only true for bodies of the same moment of inertia.

Word Count: 274

Table of Contents

Heading / Page
Introduction / 3
Pre-examination of the Problem / 5
Defining the Curve / 6
Expressing Time / 8
An Example of Determining the Time of Descent / 14
Proving The Tautochronous Property of a Cycloid / 17
The Case with Rolling Bodies / 23
Further Investigations / 25
Bibliography / 26

Introduction:

From the Greek tauto, for “the same,” and chronos, for “time,” the Tautochrone Problem describes a frictionless ramp that causes a body to travel to the bottom in a set amount of time, regardless of the starting position of the body on the ramp.[1] This means that two bodies released at the same time, but starting at different positions on the Tautochrone, will arrive at the bottom at the same time.

This problem was first solved by the Dutch mathematician and physicist, Christiaan Huygens, and was published in his book Horologium oscillatoriumi in 1673.[2] It should be noted that this was before Leibniz and Newton published their discoveries of calculus.[3] Huygens attempted to use his discovery of the Tautochrone to make a more accurate pendulum clock that forced the pendulum to follow a tautochronous path, causing the period of the oscillating pendulum to be independent of the amplitude. This later proved to be impractical, due to the friction between the wire and the Tautochrone.[4]

This Extended Essay will focus on proving that the shape of the Tautochrone curve is a cycloid. The research statement of the essay will be as follows:

The reason why beads on a wire are chosen, as opposed to balls on ramps, is because for this extended essay, the rotational kinetic energies, of the bodies will be neglected. After the analysis, this point will be discussed in further detail.

My primary reason for deciding to investigate the Tautochrone is because of the way the curve seems unreal at first inspection, and its special property in the field of mechanics. I am personally fascinated by mechanics problems in general and I knew about this problem for quite some time, which is why I decided to investigate it further. The special property of the curve is so appealing that it is even mentioned in the novel Moby Dick by Herman Melville, where a soapstone slides inside a “try-pot”[5] shaped like a Tautochrone.[6]

It should be noted that for the sake of simplicity, this paper will focus on a calculus based proof and not the proof of Huygens, which relied on Greek geometry and Galileo’s laws of motion. Huygens’s proof makes use of limited mathematical tools, but despite that fact, it was remarkably still able to prove that the curve that solves the Tautochrone Problem is a cycloid.[7] By examining Huygens’s proof I personally believe he took a similar approach to me, in that he did initially speculate that the Tautochrone is a cycloid. He most likely came to that result through physical experimentation, and then proved his claim mathematically.

Pre-examination of The Problem:

In order to begin, one has to consider what the curve that solves the Tautochrone Problem would roughly have to look like. Consider a scenario where one bead starts further away from the end of the wire, than the other. The bead that starts further away from the end of the wire will have to be moving quicker at the end of its motion, than the other. Therefore, the bead further away has to experience a greater acceleration. This would imply the segment of the wire further away from the end of the wire would have to be steeper. Thus, the shape of the curve the wire will mimic will have to strictly decrease at a decreasing rate. By this logic, the Tautochrone curve would have to look something like figure 1 on the next page, where two beads, marked 1 and 2, are placed at arbitrarily points on the wire.

Figure 1

The arrow in figure 1 represents the direction of the uniform gravitational field, . Notice that bead 1 initially accelerates more than bead 2, due to the wire being steeper at that point.

Defining the Curve:

Now that an idea for what a Tautochrone should look like has been established, a cycloid can be formally defined. This coming definition will be used in a later proof to illustrate the cycloid’s tautochronous quality.

A cycloid is a curve defined by a point on the edge of a circle, which is rolling along a defined axis.[8] When working with Cartesian coordinates, it is difficult to explicitly state the equation of a cycloid, and therefore describe the vertical position of a bead, , in terms of its horizontal displacement, . It is much simpler to work with parametric equations, which describe the position of a bead in terms of a parameter , the angle through witch the cycloid-defining circle has rolled.

Figure 2

If the cycloid-defining circle is rolling along the x-axis and the cycloid defining point on the edge of the circle started from the point , then the centre of the circle can be described as , were is the radius of the circle and is measured in radians. The coordinate of the centre of the circle (the length of the line segment from 0 to A in the figure) comes from the fact that the arc length PA in figure 2 is equal to the coordinate of the centre of the circle, because the circle is rolling.

Now, determining the coordinate of point P is relatively simple. The x-coordinate is given by and the y-coordinate is given by , which can be determined by analysing figure 2 carefully.[9]

The equations above will be used later in the paper, but for now a method for determining the time of decent of a bead needs to be devised.

Expressing Time:

In order to determine the time of descent for a bead that starts at an arbitrary point on the wire, the way time passes has to be mathematically expressed. There are multiple ways of mathematically expressing how time passes; the method used in this paper will make use of the definition of instantaneous speed, which states that the instantiations speed of an object, is equal to the distance, , it covers over a time interval,, which is allowed to become infinitesimally small. Symbolically this would imply the following.

The speed of the bead at any given instant can be described using an energy approach. As the bead moves from a higher position to a lower position, the bead looses gravitational potential energy (PE) and gains kinetic energy (KE) and thus speed. (The red vector with the next to it stands for velocity, which describes a particle’s speed and direction.)

Figure 3

There are only two forces acting on the bead at any time: the force due to gravity, for “weight,” and the normal force, , the wire exerts on the bead. Because the normal force is always perpendicular to the instantaneous displacement of the bead, only gravity does work on the bead. This means, gravity is the only force responsible for the bead gaining energy as it descends.[10]

Figure 4

It should be noted that is equal to , where is the mass of the bead and is the gravitational field strength.

By the conservation of energy, it can be stated that after the bead looses gravitational potential energy, after falling through a height , the bead gains kinetic energy equal to the amount of gravitational potential energy lost.[11] From the definitions of potential energy, and kinetic energy, this mathematically implies,

(Potential Energy) (Kinetic Energy)

and therefore,

where is the speed of the instantaneous speed of the bead and where is the strength of the gravitational field. Notice that the speed of a bead is independent of its mass, and that is a scalar.

With the knowledge of instantaneous speeds, one is able to derive an expression that will give a means of determining the time of descent for a bead. Recall the definition of instantaneous speed is the following.

Now the following can be written.

Which implies,

where is a very small distance that the bead covers and is the very short time interval, in which the bead covers the distance .

In order to simplify the analysis, will be expressed in another way. If a very small portion of the curve is examined, one could argue that the small distance the bead has to travel for that small portion, is close to a straight line. Therefore the following can be written:

Figure 5

Where and are infinitesimally small distances covered by the bead in the horizontal and vertical direction respectively. Using the just derived equation, the amount of time required to slide down a fraction of the curve can be expressed as:

.

The equation above can now be integrated with respect to either or . It is also important to note in the equation above, only magnitudes of variables will be considered. The only problem with this approach is that in the picture that is usually used to solve a time of decent problem, is pointing downwards and is thus negative. In order to avoid problems involving negative values in square roots, the picture that is used to solve problems of time of descents will be inverted, such that it looks like the bead is sliding toward the top of the page.

Figure 6

It should be noted that time of decent problems can also be solved using a non-inverted picture, but one would also have to make negative, which can complicate the analysis, due to the square root.

By inspecting the right hand portion of the figure above, it can be seen that can be expressed as .

Now, in order to determine the total time required for a bead released from rest to travel between, the initial height, and , the final height, the equation found earlier will be integrated with respect to , from , to . This leads to the equation,

which is only valid for strictly decreasing and differentiable curves.

It should be noted that the equation above is not truly valid as the right most function being integrated does not exist when . This is an example of an improper integral. There are also a few situations where the end of the curve is horizontal, which leads to not existing, or becoming infinitely large. For this reason, the equation above should actually be altered to the following.

But it is cumbersome to repeat the limits in proofs, so they will be omitted but implied in this paper.

Below is a summary on how to solve time of decent problems, down strictly decreasing and differentiable curves of interest[12]:

v  Invert the curve. The curve can also be shifted parallel to to make the analysis simpler.

v  Apply

v  In some situations, recall that in the equation above, the integrated function on the right accepts values in the open interval .

An Example of Determining the Time of Descent:

In order to demonstrate the equation derived in the previous section, a simple situation where the answer is already known will be considered. Assume a bead slides down a straight wire, with a 45° incline from to as illustrated in the figure on the next page.