The cycloid's trajectory is a combination of simple motions - rolling and translation. Because the circle rolls without slipping on the straight line, its translational velocity is equal to its rotational velocity.
In fact, at the point at which the circle touches the base, the velocities are exactly equal in magnitude and opposite in direction making this point the instantaneous center of rotation.
The green vectors in the image below shows the translational and rotational velocities, and the velocity of the tracing point is represented by the red vector. Further, the rotational velocity vector is further resolved along the perpendicular components $PG$ and $PO'$.
From the geometry of the configuration, we can easily see that $|PF|=2\cdot |HE|$.
Because $PE$ represents the circle's infinitesimal arclength (in time $dt$) and $HE$ is the component of that arclength along $PG$, we also know that $|HE|=d(|PG|)$. Therefore, the infinitesimal arclength $ds$ of the cycloid traversed by the tracing point $P$ in time $dt$ is,
$ds=|PF|=2\cdot |HE|=2\cdot d(|PG|)$
As the change in arclength is equal to twice the change in $PG$ in any given time interval, the total arclength of the semiarch of the cycloid is equal to twice the diameter of the circle.
We can use the almost same reasoning for hypocycloids which are curves traced by a point on a rolling circle of radius (say) $b$ inside within a larger circle of radius (say) $a$. For integer $q=a/b$, we get nice closed hypocycloids. The case for $q=5$ is shown below.
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| A pentoid ($q=5$) |
Note that because the circle rolls without slipping, the angular velocity of rotation of the rolling circle is $q$ times its angular velocity of its revolution. Consequently, the movement of the tracing point is a composition of (i) rotation of vector of length $a-b$ by angle $t$ in the counter-clockwise direction about the origin and (ii) rotation of a vector of length $b$ by an angle $(q-1)t$ in the clockwise direction about the origin.
The above in turn means that the velocity vectors of both the rotations noted above are the same and are proportional to $(a-b)$ times $\,dt$ (or equivalently $b$ times $(q-1)\,dt$) denoted by green vectors below.
Here again, from the geometry of the configuration, it is immediately clear that the infinitesimal arclength traversed by the tracing point in time $dt$ is $ds=|PF|=2|HE|$.
But, because the smaller circle rotates $q$ times faster, the magnitude of the velocity vector of any point on its circumference will be proportional to $b\cdot q\,dt=a\,dt$ which is represented by the cyan vector in the image above.
Therefore, $\displaystyle \frac{|HE|}{|H'E'|}=\frac{a-b}{a} \implies |HE|=\frac{a-b}{a}|H'E'|=\frac{a-b}{a}\,d(|PG|)$
We then finally have $\displaystyle ds=2\frac{a-b}{a}d(|PG|)$
Then, the arclength to be traversed by the tracing point to reach the top of the arch is $2(a-b)/a$ times its distance to the topmost point of the rolling circle.
While the demonstration given for the cycloid in Wikipedia is even nicer method to determine the arclength, it is always fun to arrive at the same results with different methods. As shown below, its is not very difficult to extend the demonstration quoted above to hypocycloids as well.
Hope you enjoyed this post.
Until then
Yours Aye
Me
