What is the position as a function of time for a mass falling down a cycloid curve?

In the brachistochrone problem and in the tautochrone problem it is easy to see that a cycloid is the curve that satisfies both problems.

If we consider $x$ the horizontal axis and $y$ the vertical axis, then the parametric equations for a cycloid with its cusp down is:

$$\begin{cases} x=R(\theta-\sin\theta)\\ y=R(\cos\theta-1) \end{cases}$$

A cycloid is the curve of fastest descent for an idealized point-like body, starting at rest from point A and moving along the curve, without friction, under constant gravity, to a given end point B in the shortest time. A cycloid is also the curve for which the time taken by an object sliding without friction in uniform gravity to its lowest point is independent of its starting point.

We already know the parametric equations for the the geometry of the path of the mass, but I would like to know the position of the object over time. I would like to know what is the solution to the equations of motion. It would be like specifying the arc length as a function of time $s=s(t)$ in the figure:

To sum up, what I mean is:

Given certain initial conditions, what are the expressions $x(t)$ and $y(t)$ that give the position as a function of time $\mathbf{\bar{r}}(t)=\left ( x(t),y(t)\right )$ for a particle falling down a cycloid curve?

1. Brachistochrone $$\boxed{\: \begin{matrix} x\left(\theta\right) = R\left(\theta-\sin \theta\right)\\ y\left(\theta\right) = R\left( 1-\cos \theta\right) \end{matrix}\:} \tag{b-01}$$

$$\omega= \dfrac{\,\theta \,}{t}=\dfrac{\mathrm{d}\theta }{\mathrm{d} t}=\sqrt{\dfrac{\,g\,}{R}} =\text{constant} \tag{b-02}$$

$$\boxed{\: \begin{matrix} & x\left(t\right) = R\Biggl[ \sqrt{\dfrac{\,g\,}{R}}\,t-\sin \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[\omega\,t-\sin \left(\omega\,t\right)\Bigr]\\ & \\ & y \left(t\right)= R\Biggl[1-\cos \left(\sqrt{\dfrac{\,g\,}{R}}\,t\right)\Biggr]=R\Bigl[1-\cos \left(\omega\,t\right)\Bigr] \end{matrix}\:} \tag{b-03}$$

$$s\left(t\right)=4R\Biggl[1-\cos\left(\dfrac{\theta}{2}\right)\Biggr]=4R\Biggl[1-\cos\left(\sqrt{\dfrac{g}{4R}}\,t\right)\Biggr]=4R\Biggl[1-\cos\left(\frac{\omega}{2}\,t\right)\Biggr] \tag{b-04}$$

Time of descent from point $\mathrm{A}(0,0)$ to lowest point $\mathrm{F}(\pi\,R,2R)$: from (b-02) with $\:\theta=\pi\:$ $$t\left[\mathrm{A}\rightarrow\mathrm{F} \right] = \pi\sqrt{\dfrac{\,R\,}{g}} \tag{b-05}$$

2. Tautochrone

$$t\left[\theta_{0}\rightarrow\theta\right]=\sqrt{\dfrac{\,R\,}{g}}\Biggl(\pi-2\arcsin\Biggl[\dfrac{\cos\left(\theta/2\right)}{\cos\left(\theta_{0}/2\right)} \Biggr] \Biggr) \tag{t-01}$$ Time of descent from $\:\theta_{0}\:$ to the lowest point $\:\theta=\pi\:$ : from (t-01) with $\:\theta=\pi\:$ $$t\left[\theta_{0}\rightarrow\pi\right]=\pi\sqrt{\dfrac{\,R\,}{g}}=\text{constant independent of \theta_{0}.} \tag{t-02}$$

Also from (t-01)

$$\cos\theta=\left( \dfrac{1+\cos\theta_{0}}{2}\right) \left[ 1+\cos\left(\sqrt{\dfrac{\,g\,}{R}}\,t \right) \right]-1 \tag{t-03}$$

$$s\left(t\right)=4R\Biggl[\cos\left(\dfrac{\theta_{0}}{2}\right)-\cos\left(\dfrac{\theta}{2}\right)\Biggr]=4R\cos\left(\dfrac{\theta_{0}}{2}\right)\Biggl[1-\cos\left(\sqrt{\dfrac{g}{4R}}\,t\right)\Biggr] \tag{t-04}$$

$$\theta\left(t\right)=\arccos\Biggl[\biggl(\dfrac{1+\cos\theta_{0}}{2}\biggr) \left[ 1+\cos\left(\sqrt{\dfrac{\,g\,}{R}}\,t \right) \right]-1\Biggr] \tag{t-05}$$

$$x\left(t\right)=R\big[\theta\left(t\right)-\sin\theta\left(t\right)\bigr] \tag{t-06}$$

3. Cycloid(properties)

• Thanks for your answer Frobenius. After I posted the question, I was able to find the solution to this problem myself. Then I also found the related problem of the Cycloidal Pendulum discussed in "Mechanics" by Arnold Sommerfeld and in "Solved Problems in Lagrangian and Hamiltonian Mechanics" by Claude Gignoux and Bernard Silvestre-Brac. – roy Jun 9 '17 at 19:53

Have you tried

$$\theta =\theta_{0}\cdot \cos \left(t\cdot \sqrt{\frac{g}{R}}\right)$$

where $\theta_{0}$ is the initial angle? The time dependence of the angle $\theta$ should work the same way like with the idealized pendulum and even better, since with the cycloid the simultaneous arrival independend of the starting position is not only an approximation but the exact solution.

Start with: $${\bf F}_{\rm net} = m\,\ddot{{\bf r}}$$ with: $${\bf r} = x(\theta)\hat{\bf x} + y(\theta)\hat{\bf y}$$ and a dot denoting a derivative with respect to time.

Then using the tangent vector: $$\hat{\bf T} = \frac{\frac{{\rm d}{\bf r}}{{\rm d}{\theta}}}{\left|\frac{{\rm d}{\bf r}}{{\rm d}{\theta}}\right|}$$ to define the normal vector: $$\hat{\bf N} = \frac{\frac{{\rm d}\hat{\bf T}}{{\rm d}{\theta}}}{\left|\frac{{\rm d}\hat{\bf T}}{{\rm d}{\theta}}\right|}$$

The forces acting are gravity: $${\bf F}_{\rm grav} = -g\,\hat{\bf y}$$ and a normal force: $${\bf F}_{\rm normal} = -\left({\bf F}_{\rm grav}\cdot\hat{\bf N}\right)\hat{\bf N}$$ Combine those to get ${\bf F}_{\rm net}$ and solve the resulting differential equation (using the initial conditions of course).

• Thanks for your answer. But that is just only one way to obtain a differential equation for $\theta(t)$. The most important part of my question is whether there is an analytical solution $\theta(t)$ to that differential equation. And if there is, what is that function. One can also write the Lagrangian and the Euler-Lagrange equation. I think that is an easier way. Nevertheless, what I am asking is what is the function solution of the differential equation. – roy Apr 24 '15 at 14:30
• @roy Well, you can't exactly just expect people to go around solving ODEs for you (unless you're lucky). I scribbled down a quick derivation of the DEs for $\theta(t)$ on a piece of paper and they look pretty messy... my hunch is that there is no analytic solution, but I could be wrong. It's certainly not one of the standard ODEs that I can solve in my head. – Kyle Oman Apr 24 '15 at 16:57
• Maybe you can post the DE in Mathematics and see if you will find a solution. – ja72 Nov 1 '15 at 0:40