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# 19.6: Motion on a Cycloid, Cusps Down

We imagine a particle sliding down the outside of an inverted smooth cycloidal bowl, or a bead sliding down a smooth cycloidal wire. We shall suppose that, at time $$t = 0$$, the particle was at the top of the cycloid and was projected forward with a horizontal velocity $$v_0$$. See Figure XIX.7.

This time, the equations of motion are

$\ddot{s} =g sin \psi \label{19.6.1}$

and

$\dfrac{mv^2}{ \rho} = mg \cos \phi - R. \label{19.6.2}$

By arguments similar to those made in Section 19.5, we find that

$\ddot{s} = \dfrac{gs}{4a} \label{19.6.3}$

The general solution to this is

$s = Ae^{pt} + Be^{-pt}, \label{19.6.4}$

where

$p = \sqrt{g/(2a)}. \label{19.6.5}$

With the initial condition given (at $$t = 0, s = 0, \dot{s} = v_0$$ ), we can find A and B and hence:

$s = v_{0} \sqrt{\dfrac{a}{g}} (e^{pt} - e^{-pt})\label{19.6.6}$

Again proceeding as in Section 19.5, we find for $$R$$:

$R = \dfrac{m}{4 \cos \psi} (4ga \cos 2 \psi - v^2_0) . \label{19.6.7}$

So – what happens?

If the constraint is two-sided (bead sliding on a wire) R becomes zero when $$\cos 2 \pi = v^2_0 / (2 /ga),$$ and thereafter R is in the opposite direction.

If the constraint is one-sided (particle sliding down the outside of a smooth cycloidal bowl):

1. If $$v^2_0\ > 4ga$$, the particle loses contact at the moment of projection.
2. If If $$v^2_0\ < 4ga$$ the particle loses contact as soon as $$\cos 2 \pi = v^2_0 /(2ga)$$, is very small (i.e. very much smaller than $$\sqrt(2ga)$$ ), this will happen when $$\psi = 45 \circ$$ ; for faster initial speeds, contact is lost sooner.

Example $$\PageIndex{1}$$

A particle is projected horizontally with speed v0 = 1 m s−1 from the vertex of the smooth cycloidal hill

$$x = a(2\theta + \sin 2 \theta$$

$$y = 2a \cos ^2 \theta ,$$

where $$a = 2$$ m. Assuming that g = 9.8 m s−2, how long does it take to get halfway down the hill (i.e. to $$y = a$$)?

Solution

We have to use Equation \ref{19.6.6}. With the numerical data given, this is

$$s = 0.451754(e^{1.565248t} - e^{-1.565248t}).$$

We can find $$s$$ from Equation 19.4.12, which gives us $$s$$ = 2.828427 m. If we let we now have to solve 6.26099 = $$\xi - 1 / \xi$$, or $$\xi^2 - 6.26099 \xi -1 = 0$$. From this, $$\xi$$ = 6.41683 and hence $$t$$ = 1.19 s.

I leave it to the reader to calculate R at this time – and indeed to see whether the particle loses contact with the hill before then. Perhaps the fact that I got a positive real root for $$\xi$$ means that we are all right and the particle is still in contact – but I wouldn't be sure of that. I leave it to the reader to investigate further.