Brachistochrone and Tautochrone

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HW 3

This exercise is about a physics problem concerning falling bodies under constraint; what curve which restrains a frictionless body under only the force of gravity will give the shortest time to the ground http://old.nationalcurvebank.org//brach77/brach77.htm. In the demonstration we see that a falling body on a cycloid curve will reach the ground quicker than on a straight line. We will not be able to show that the cycloid is the fastest of all paths (requires mathematics beyond the scope of this course – but see last link listed belows), but we can show that it is quicker than a straight line. Additionally, we can show that the cycloid has a most unintuitive property at the end of the exercise. We will be using some physical laws; if you are not familiar with them, please accept their validity.

In this assignment I will state certain facts that you are to work for homework. They will be designated by #{number}.

The Brachistochrone To work this problem, we need a coordinate system to frame the problem. , we fix the starting point for the body at the origin (point O(0,0)) of a two-dimensional coordinate x-y system except we point the positive y axis “downward”. If we place the cycloid with parametric equations, x = at – a sin(t) and y = a – a cos(t), so that it “hangs” downward as in the figure, then the “bottom” of the curve which we will assume is the ground has coordinates B(aπ,2a) where “a” is the parameter in the parametric equations. Note that x(0) = 0, y(0) = 0, so the body on the cycloid is at the origin at t = 0. Also, x(π) = aπ, and y(π) = 2a, so the body is “on the ground” (point B), when t = π. So, the body is falling to the ground for 0 t π.

Let t also represent time and v(t) the velocity of the body. Then at time = 0, v(0) = 0, since we assume the body starts at rest. Looking at the above web site, you will see how the body should fall along the cycloid vs. the straight line. We now compute the work done by gravity in this situation in two different ways (see your physics book). The work done by gravity due to the change in potential energy is mgy, where m is the mass and g is the acceleration due to gravity. The work is also the difference in the kinetic energy which is ½ m v(t)2 – ½ m v(0)2 = ½ m v(t)2 since v(0) = 0. Thus, we have mgy = ½ mv2 for all t. Solving this equation for v, mgy = ½ mv2 → gy = ½ v2 → v2 = 2gy → v = , since v 0 because the y-axis is downward. Using s(t) to be the distance travelled by the body along the path of the curve, we have v(t) = . Solving this equation for dt, we dt = = . ds2 = dx2 + dy2. So, the total time to fall to the ground, designated Tf = = = (Equation A). For the cycloid, 0 aπ, so Tf = . We now use our parametric equations to simplify this integral.

Problem #1 Using the parametric equations for the cycloid given above show that: dx = a(1-cos(t)) dt and dy = a sin(t) dt

Problem #2 Using #1 show that ds2 = dx2 + dy2 = 2a2(1 – cos(t))dt2

Using #2, we obtain Tf = dt by substitution and using t for time and the parametric variable.

Problem #3 Using the integral above and the parametric equation for x(t), show the integral reduces to dt which is calculated by the Fundamental Theorem of Calculus to get Tcycloid = π.

Thus, we have the total time to get to the ground along the cycloid as π. We will use this result in the following two parts.

Falling along a straight line.

To find the time to fall along a straight line, we only need to find the parametric equations for a straight line from O to B. See Figure 11.12. From HW2 and the last set of notes, we know that a parametric system for a straight line from (0, 0) to (aπ, 2a) is given by x = 0 + t = t and y = 0 + ()t = and 0 t aπ. Using equation (4) p. 677-8 again, ds2 = dx2 + dy2, we have ds2 = dt2 + ( dt)2 = (1 + )dt2 = ()dt2.

Problem #4 Plugging the above results into Equation A, show we get Tf = dt. [This takes several steps.]

Problem #5 Computer the integral in #4 to get Tline = which is the time to fall along the line segment.

Since > π, then Tline > Tcycloid. Note that Tline / Tcycloid 1.185 so the cycloid is about 19% faster. Run http://old.nationalcurvebank.org//brach77/brach77.htm again to see this effect.

Tautochrone

We will now see that the cycloid has a very unintuitive property that is hard to believe; the time to fall to the ground along a cycloid is independent of where it starts. This makes the cycloid a tautochrone. Look at http://old.nationalcurvebank.org//tautochrone/tautochrone.htm to see a demonstration.

To calculate the time from a different starting point, we need to go back to the beginning and modify the physical results. Assume we start at A(x0, y0) a point on the cycloid lower than (0, 0) and we start at t0 > 0. The potential energy work is now mg(y – y0) instead of mgy, so we get ½ mv2 = amg (cos(t0) – cost(t)). Solving for v we get v = . The integral Tf becomes = .

#6 Show the above integral is using both the trig half-angle formulas, see page 4 in the “Calculus reference materials” in the “Course Information” Module.

#7 Using the integral from #6 with substitution u = cos(t/2) and c2 = cos2(t0/2) to get and then compute to get π .

The result says the time to fall from any starting point (x0, y0) to (a π, 2a) is always the same. This makes the cycloid a tautochrone.

You may find the video at https://www.youtube.com/watch?v=Cld0p3a43fU (rather long) of interest.