Suppose we have two objects $A$ and $B$, where $B$ is tied by a rope to $A$ with the slack pulled out, the initial tangential speed $c'(0)=(0,1)$ is perpendicular to $c(0)$ and $c(0) = (1,0)$, and assume that $A$ moves minimally as it "pulls" $B$, so for example it has a far larger mass. Suppose also that we can ignore any other forces acting on these two objects besides the force of the rope pulling on $B$. I think we can then parametrize the movement of $B$ by a curve at time $t$ as

a $c(t) = r(t) (\cos(\varphi(t)), \sin(\varphi(t)))$,

and the force of the acting on the rope is always in the direction of the center, so we can describe $c''$ as

b $c''(t) = k(t) (\cos(\varphi(t)), \sin(\varphi(t)))$.

If we just "let" things be, $B$ will orbit around $A$ at a constant speed in a circle. Suppose however that we pull the rope with some additional force so that $B$ starts getting closer to $A$, or that we start pulling it at a fixed speed (I'm not quite sure what additional force would correspond to this, but I suppose it might need a strong/infinite impulse at the start, so that the change of speed is immediate? I'm not even really sure this makes sense/is possible, hence me asking this question). I imagine all of this as a friction-less pulley system at $B$, with us pulling the rope perpendicularly to the orbiting plane.

Two observations I've made, which I would appreciate if someone checked the validity of:

If $c'\cdot c'' = 0$ were to be true and also $c' \neq 0$, then $c \cdot c' =0$ by b, and so the orbit must be a sphere. This means that if $r'$ is not zero, then $c' \cdot c'' \neq 0$, and so the magnitude of $c'$ will be changing.

Expanding out $c''$, if b is to be true, then we can acquire the condition: $2 r'(t) \varphi'(t) + r(t) \varphi''(t) = 0$. If $\varphi$ were to be linear, this reduces to $C r'(t) = 0$, again showing that the orbit is a sphere - this means that the angular speed will also be changing.

I think my methods of showing both of the above might be unnecessarily complicated and I would appreciate a simpler way to show it. Especially the second seems intuitively true - based on the fact that the speed will be increasing and the radius getting smaller, but I don't know how to show it.


Lastly, how exactly can $r(t)$ look - assuming we can pull with any amount of force at any time, but only in the direction towards $A$? What if the rope can only ever get shorter? Is it possible for $r'$ to be constant? Is it possible to satisfy $r'(0) = 0$, and then $r'(x) = -1$ on $\langle \epsilon, 1 \rangle$ for some $\epsilon > 0$ ? Assuming $r'$ is constant, and using b and $2 r'(t) \varphi'(t) + r(t) \varphi''(t) = 0$, then for say $r(t) = 1-t$, we get $\varphi' = C(1-t)^{-2}$, and so overall $\varphi = C(1-t)^{-2} + D$. Maybe we could now show that from this it follows that the force is always "pulling" $B$ towards $A$, at least while $t \in \langle 0,1\rangle$ that is. That should be equivalent to showing that $k(t)$ is always negative.

Also, I would like to know whether these sorts of problems couldn't be solved without using forces etc, and simply using only the knowledge of the starting tangential velocity of $B$, and the constant velocity of $r'$. Maybe we could think of these things in terms of more along "the object is moving at a tangential velocity, the rope is pulling at a constant velocity, and so we simply sum the velocities in some way" - maybe we could approximate the movement of $B$ at certain intervals $0+n \epsilon$. Working with the position of $B$ as a sequence $p(n)$ we would approximate the "previous velocity" by $p(n+1) - p(n)$, project this "velocity" onto the plane that is normal to the rope, and then finally add the velocity by which the rope is shortening. Something like that.


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