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 Author Topic: Cycloidal Pendulum  (Read 18953 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: September 06, 2009, 05:54:01 pm » posted from:Taipei,T\'ai-pei,Taiwan

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. the period of a ball rolling back and forth inside this curve does not depend on the ball’s starting position).

The cycloid through the origin, generated by a circle of radius r, consists of the points (x, y), with

$x = r(\omega t - \sin \omega t)$
$y = r(1 - \cos \omega t)$

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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