I'm interested in cycloid and make another problem about it.
If a unit circle rolls one lap on the straight line, the point on the circle draws a cycloid trajectory. Then what if a unit circle rolls one lap on the special curve $l$ so that the corresponding trajectory is a straight (horizontal) line?
I don't even know if the curve really exists, but try to find the curve on these settings:
$l: x=f(s), y=g(s)$ where $s \in [0,2\pi]$ means 'rolling distance on $l$'
$C$: unit circle rolling on $l$ with its center point $c(s)$
$p(s)$: the point on $C$ which makes the horizontal line trajectory we want.
$p(s)=(p_1(s),p_2(s))$ can be expressed by defining three vectors:
$p(s)=q(s)+u(s)+v(s)$ …. (2)
where
$q(s)=(f(s), g(s))$ is the tangent point at which $l$ meets $C$,
$u(s)$ is the unit vector from $q(s)$ to $c(s)$, and
$v(s)$ is the unit vector from $c(s)$ to $p(s)$
and make parametric equations below:
$\int_{0}^{s}{\sqrt{f'(u)^2+g'(u)^2}}=s$ …. (1)
$p_2(s)=0$ …. (2)
Eqn(1) is from the curve length formula. If $C$ rolls distance $s$, the curve length behind becomes $s$.
Eqn(2) is from the condition where $p(s)$ is on the horizontal line.
Finding exact form of $u(s)$, $v(s)$ and simplifying them result in the system of 1st order nonlinear ODEs below:
$(f'(s))^2 + (g'(s))^2 = 1$
$-g(s)+\sin(s)g'(s)+(\cos(s)-1)f'(s)=0$
so terrible equations… even WolframAlpha doesn't solve it. I cannot go further.
Are there some well-known keywords about the initial problem?
Best Answer
The "special curve" is a circle with twice the radius. This produces the Tusi couple.