# Motion on a frictionless vertical sinusoidal track

Today during lesson, my mechanics teacher provided a diagram of a "bowl" of the following shape:

The top left and the top right have the same height, and the top right part is horizontal.

He claimed that if you release a point particle from the top of the track on the left, it will take an infinite amount of time for it to reach the top right. His reason was that if the time is finite, and when you reverse it, the particle will start from equilibrium but spontaneously performs motion, which is impossible.

This reason seems too vague for me to understand the claim, so I tried to prove (or disprove) it mathematically, but I failed to do so. Can anyone explain this to me?

The following are my attempts to tackle the problem.

Since he did not specify the details of its shape, as an example I constructed a bowl which is a parabola $x^2$ on the left until the bottom, then a $1-\cos(x)$ on the right.

My first attempt was to ignore the left part and to consider it to have an initial speed of $\sqrt{2gh}$ at the bottom of the bowl. Denoting the horizontal displacement as $x$, I expressed $\theta$, the angle that the tangent at $x$ makes with the horizontal, as a function of $x$. ($\tan\theta=\frac{dy}{dx}=\sin x$.) Resolving forces shows that the force down the slope was $mg\sin\theta$, so the acceleration at any point must be $g\sin(\arctan(\sin x))$, which isn't very helpful when I try to integrate...

After reading Point particle moving on a frictionless semicircular hill, I also tried to express the energy at any point as a function of the angle, but I failed to find an expression for speed as a function of the angle.

(In fact in the very beginning I modelled the shape of the bowl as a part of the function $-e^{-x}\cos x$, but I gave up.)

The potential can be written $V(x)=mg(1-\cos x)$, so the total energy of the system can be written

$$E=\frac{1}{2}m\dot{x}^2+mg(1-\cos x)$$

Differentiating wrt x

$$m\dot{x}\frac{d\dot{x}}{dx}+mg\sin x=0$$ $$\dot{x}d\dot{x}=-g\sin x dx$$

$$\dot{x}^2=2g\cos x+C$$

We require $\dot{x}=0$ at $x=\pi$, so $C=2g$

$$\frac{dx}{dt}=\sqrt{2g(\cos x+1)}$$

$$\frac{dx}{\sqrt{\cos x+1}}=\sqrt{2g}dt$$

so the time to reach $x=\pi$ is

$$T(\pi)=\lim_{x\to\pi}\frac{1}{\sqrt{2g}}\int_{x'=0}^{x'=x}\frac{dx'}{\sqrt{\cos(x')+1}}$$

which diverges i.e. infinite time. In a sense, what this is really saying is that the particle never reaches $x=\pi$. This allows us to maintain that time reversal will produce the reverse trajectory. If the particle were to reach $x=\pi$ in finite time, time reversal would just produce the particle sat quite happily on top of the hill so there is a contradiction and the particle never reaches the top.

Your mechanics teacher's indirect proof relies on i) time-reversal symmetry and ii) determinism in Newtonian mechanics.

The first idea is that we can study the time-reversed problem. In the indirect proof we assume that the travelling time is finite. and hence we can assign the initial time (for the time-reversed problem) to be $t=0$.

Let $z$-axis points upward, let $s$ be the arclength from the smooth (time-reversed) horizontal starting point at $t=0$, and let

$$\tag{1} z~=~-h(s)$$

denote the profile of the hill as a function $h$ of arclength $s$ such that

$$\tag{2} h(s\!=\!0)~=~0.$$

Then the mechanical energy $E$ of the point particle is given as

$$\tag{3} 0~=~\frac{E}{m}~=~\frac{\dot{s}^2}{2}+gz.$$

In the first equality of (3), we used the initial conditions

$$\tag{4} \qquad s(t\!=\!0)~=~0, \qquad \dot{s}(t\!=\!0)~=~0.$$

Combining eqs. (1) and (3) yield the following first-order ODE

$$\tag{5} \dot{s}(t)~=~\sqrt{2gh(s(t))} ,\qquad s(t\!=\!0)~=~0.$$

Note that the initial value problem (5) has the trivial solution

$$\tag{6} s(t)~\equiv~ 0.$$

The second idea is to use the local uniqueness of the initial value problem (5) to argue that the trivial solution (6) is the only solution. This would contradict the fact that the point particle is moving between different points. A sufficient condition for local uniqueness is if the map $s\mapsto \sqrt{h(s)}$ is Lipschitz continuous.

• Alternatively, via explicit calculation: When $h(s)= {\cal O}(s^2)$ for $s\to 0$, the integrand $\frac{1}{\sqrt{2gh(s)}}$ in the time integral $\Delta t = \int_0^{s_0} \frac{ds}{\sqrt{2gh(s)}}=\infty$ has a pole at $=0$. – Qmechanic Sep 26 '14 at 21:36