Equation of Cycloid

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Theorem

Consider a circle of radius $a$ rolling without slipping along the x-axis of a cartesian plane.

Consider the point $P$ on the circumference of this circle which is at the origin when its center is on the y-axis.

Consider the cycloid traced out by the point $P$.

Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.


The point $P = \tuple {x, y}$ is described by the equations:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$


Proof

Let the circle have rolled so that the radius to the point $P = \tuple {x, y}$ is at angle $\theta$ to the vertical.


Cycloid.png


The center of the circle is at $\tuple {a \theta, a}$.

From the diagram above, we see:

The $x$-coordinate is to the left of the center of the circle by $a \sin \theta$
The $y$-coordinate is below the center of the circle by $a \cos \theta$

It follows that:

$x = a \theta - a \sin \theta$
$y = a - a \cos \theta$

whence the result.

$\blacksquare$


Also see


Sources

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