Variables:
e -> point on a circle
r -> radius of said circle (constant)
x(t), y(t) -> parametric equations describing the position of point e
v -> velocity of the center of the circle
c -> constant speed of point e (also max speed of the circle; de/dt = c, 0 <= |v| <= c)
I'm looking to describe the motion of a point e that moves along the circumference of a circle, where the circle can be translated along the x-axis at any speed between 0 and c.
Point e has a constant speed of c, which means the following:
@ v = 0
x(t) = r*cos(t), y(t) = r*sin(t)
This is because when the circle is not moving, the point e just moves around the circumference.
@ v = c
x(t) = c*t, y(t) = y(0)
This is because when the circle is moving at the same speed of e (c), then e must have the same velocity vector as the center of the circle for it to remain on the circumference and maintain speed c.
I thought of something like this:
x(t) = r*cos(t)*(1 - v/c) + v*t
y(t) = r*sin(t)*(1 - v/c)
Since this reduces to the correct equations at v=0 and v=c, plus the angular speed is proportional to the velocity of the circle.
But this doesn't satisfy the constant velocity requirement where:
(dx/dt)^2 + (dy/dt)^2 = c^2
At least not as far as I can see.
You write
But if $x$ and $y$ are the coordinates of your point $e$, then $e$ in this case is moving with speed $r$, not speed $c$. I think you need to adjust this case to be $$ x(t) = r \cos(t\frac{c}{r})\\ y(t) = r \sin(t\frac{c}{r}) $$ so that $$ x'(t) = -r(\frac{c}{r}) \sin(t\frac{c}{r}) = -c \sin(t\frac{c}{r})\\ y'(t) = r (\frac{c}{r}) \cos(t\frac{c}{r}) = c \cos(t\frac{c}{r}) $$ from which it's evident that $x'(t)^2 + y'(t)^2 = c^2$ as desired.
In the case where the point is translating as well as rotating, we don't necessarily know what to put into the argument of the sine and cosine instead of $t\frac{c}{r}$, so let's just say 'it's some function $s(t)$' for now. Then we get, for a translational motion at speed $q$, with $0 \le q \le c$ the equations
\begin{align} x(t) &= qt + r \cos(s(t)) y(t) &= r\sin(s(t)) \end{align} and taking derivatives, we get \begin{align} x'(t) &= q - r \cos(s(t)) s'(t) y'(t) &= r\cos(s(t)) \end{align} Setting $x'(t)^2 + y'(t)^2 = c^2$ yields a constraint on the function $s$: $$ \left[q^2 -2rq \cos(s(t)) s'(t) + r^2 \cos^2(s(t)) (s'(t))^2\right] + \left[ r^2 \sin^2(s(t)) (s'(t))^2\right] = c^2. $$ Gathering the squared sine and cosine terms, and moving constants to the right hand side,we have $$ -2rq \cos(s(t)) s'(t) + r^2s'(t)^2 = c^2 - q^2 $$ Divide through by $r^2$ to get $$ s'(t)^2 -2\frac{q}{r} \cos(s(t)) s'(t) = \frac{c^2 - q^2}{r^2} $$ Now completing the square (i.e., rewriting $x^2 + 2bx$ as $x^2 + 2bx + b^2 - b^2 = (x+b)^2 - b^2$) we get \begin{align} s'(t)^2 &-2\frac{q}{r} \cos(s(t)) s'(t) + \left[\frac{q}{r} \cos(s(t))\right]^2 - \left[\frac{q}{r} \cos(s(t))\right]^2= \frac{c^2 - q^2}{r^2} \\ (s'(t)- &\left[\frac{q}{r} \cos(s(t))\right])^2 = \frac{c^2 - q^2}{r^2} + \left[\frac{q}{r} \cos(s(t))\right]^2\\ (s'(t)- &\left[\frac{q}{r} \cos(s(t))\right]) = \sqrt{\frac{c^2 - q^2}{r^2} + \left[\frac{q}{r} \cos(s(t))\right]^2}\\ s'(t)& = \sqrt{\frac{c^2 - q^2}{r^2} + \left[\frac{q}{r} \cos(s(t))\right]^2} + \left[\frac{q}{r} \cos(s(t))\right]\\ \end{align} which is a differential equation for $s$, but not one that admits a simple solution by any means. I mean, the solution can be expressed as an integral: \begin{align} s(t)& = K + \int_0^t \sqrt{\frac{c^2 - q^2}{r^2} + \left[\frac{q}{r} \cos(s(u))\right]^2} + \left[\frac{q}{r} \cos(s(u))\right] ~du \end{align} but that integral isn't likely to be easy to write out in elementary terms.