A cycloid is the curve defined by the path of a point on the edge of circular wheel as the wheel rolls along a straight line. It is an example of a roulette, a curve generated by a curve rolling on another curve.

The cycloid is the solution to the brachistochrone problem (i.e. it is the curve of fastest descent under gravity) and the related tautochrone problem (i.e. [u]the period of a ball rolling back and forth inside this curve does not depend on the ball’s starting position[/u]).

The cycloid through the origin, generated by a circle of radius r, consists of the points (x, y), with $x = r(\omega t - \sin \omega t)$ $y = r(1 - \cos \omega t)$
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If its length is equal to that of half the cycloid, the bob of a pendulum suspended from the cusp of an inverted cycloid, such that the "string" is constrained between the adjacent arcs of the cycloid, also traces a cycloid path. Such a cycloidal pendulum is isochronous, regardless of amplitude. This is because the path of the pendulum bob traces out a cycloidal path (presuming the bob is suspended from a supple rope or chain); a cycloid is its own involute curve, and the cusp of an inverted cycloid forces the pendulum bob to move in a cycloidal path. $v_x=\frac{dx}{dt}=r\omega*(1+\cos\omega t)$ $v_y=\frac{dy}{dt}=r\omega \sin \omega t$
Combined the above two equations: $(v_x-r\omega)^2+v_y^2)=(r\omega)^2$
So $v_x=\frac{v_x^2+v_y^2}{2r\omega}=\frac{2g\Delta y}{2r\omega}=\frac{g\Delta y}{r\omega}$
and $\frac{dy}{dx}=\frac{v_y}{v_x}=\frac{\sin\omega t}{1+\cos\omega t}$

The time required to travel from the top of the cycloid to the bottom is $T_{1/4}=\sqrt\frac{r}{g}\pi$ $a_x=\frac{d^2x}{dt^2}=-r\omega^2\sin\omega t$ $a_y=\frac{d^2y}{dt^2}= r\omega^2\cos\omega t$

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