# Deriving the vertical displacement $y$ as a function of $t$ if we disturb a (stable) equilibrium of two opposing forces [closed]

I was thinking about a situation as I sketched above, somewhere in outer space. In this sketch, we see a mass $m$ attached to a red rope (considered massless). The mass makes a circular motion with radius $r$ and speed $v$ around the middle point of a vertical tube, in a plane perpendicular to the tube, through which the rope is led. On the other side of the tube, the rope comes out again and is connected to a charged metal mass (which for simplicity we'll also give the value $m$) placed in a uniform electric field (how this field is produced is of no importance), parallel to the vertical part of the rope, in such a way that it can't let the metal mass rotate and pulls at the rope downwards with a constant force $F_E$. The rope can move through the tube without friction (how can it be else as the rope has zero mass?). The circulating mass pulls on the rope with a centrifugal force $F_{cf}=\frac {mv^2} r$. $L=mvr$ is the angular momentum or the rotating mass m. The two forces are in equilibrium, so $F_E=\frac{mv^2}r$. At this point, the vertical distance $y$ is zero.

The whole structure is placed in an inertial frame in outer space and is via the thin blue lines connected to a mass (the striped area below) which is big enough not to affect the momenta of the two masses, and small enough not to exert noticeable gravitational forces on the masses. The reason I placed the construction in space instead of on Earth is that in this way the figure traced out by the rotating rope is not a cone (caused by the downwards vertical gravitational force exerted on the rotating mass) but a circular plane perpendicular to the tube and no changing angle of the rope with a horizontal or vertical line is involved if the circular motion changes. Instead of gravity pulling the two masses downwards, the electric field only pulls the non-rotating mass "down", with a value independent of $y$, just as would be the case in an (approximate) uniform gravity field. Again for simplicity.

Now my question is: How depends $y$ for the mass in the electric field on the time $t$ if we disturb the equilibrium of forces?

We see that (if we take $r=1$, which isn't really necessary but I like the number 1) we can't pull the mass lower than $y=-1$. In fact, we can't pull it to that point because of special relativity, and we'll stay far enough from $y=-1$ so special relativity won't kick in. The force due to the electric field stays the same. It's $F{cf}$ that varies. If we choose the top of the tube as the origin for a horizontal x-axis going through the top (x lies between zero and infinity), then according to the conservation of angular momentum if we change $r=1$ to $xr=x$, $v$ will have to change to $\frac v x$ (we could also change $v$ into $xv$ and $r=1$ into $\frac 1 x$ but in this case infinity would lie at the top of the tube and infinity can't lie somewhere), and the expression for $F_{cf}$ becomes:

$$F_{cf}=\frac{mv^2} r {\frac 1 x}^3=\frac{mv^2}{x^3}$$

If we now consider the y-axis the force on the non-rotating m becomes:

$$F(y)=my''=-mv^2+\frac{mv^2}{y+1}^3$$

m Drops out and you might just as well put v equal to 1 now like we did with $r$, so the equation for $F(y)$ becomes:

$$y''=-1+{\frac 1 {(y+1)^3}}=-1+(y+1)^{-3}$$

This is the equation in its clearest form (r and v can always be put back in). My problem is solving this equation. It's not a second order ordinary differential equation, because of the factor $(y+1)^{-3}$. But I think it's one in closed form.

You can, of course, proceed ass follows:

$$y'=-t+(y+1)^{-3}t+v(0)=-t+(y+1)^{-3}\rightarrow y=-{\frac 1 2}t^2+{\frac 1 2}(y+1)^{-3}t^2+y(0)$$

but this gives an answer for y(t) which contains y(t) itself. And $(y+1)^{-3}$ is off course not a constant, so this approach won't work.
I draw a qualitative picture how the solutions for y(0) smaller than zero (but relatively far away from -1), and y bigger than one would look like:

They look like vertically displaced sinus or cosine functions, which they are of course not. If you pull the mass a little downward and let it go with zero velocity, the mass will move past y=1 after which it stops and returns to the initial point (no friction whatever). Here stands the dark curve for.
The light curve stands for the case where you let the mass go from some point above y=1. In a sense they are each other's inverses. So my question isn't really a physical one but a mathematical one (but it nonetheless belongs here). The physics is clear, but I'm curious to know what the exact mathematical formula for the function y(t) is. Anyone?

## closed as off-topic by sammy gerbil, Jon Custer, honeste_vivere, ACuriousMind♦Sep 10 '17 at 1:13

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• It's a complicated central force problem. 1) A simpler setup that is equivalent to yours is having the rotating mass rest on a frictionless horizontal surface. This way you don't need to be in space, nor to have electrical charge and field. 2) "charged metal mass": but what might make your problem simpler is to assume this charge to be massless. 3) It might be easier to find a solution with a Lagrangian approach and/or using an additive perturbation, $r = r_0 + x$, instead of a multiplicative one $r = xr_0$. – stafusa Jul 25 '17 at 8:35

It's enough to find the trajectory of the rotating mass, as $\Delta y = \Delta r$.

At least when ignoring the mass of the hanging charge (which adds to the inertial of system, but not to the angular momentum), this problem should have a solution in terms of known functions.

We know, though (Bertrand's theorem), that the linear ($F=kr$) and inverse square law ($F=k/r^2$) are the only central forces that guarantee closed orbits, so the solution here won't be that simple. (Though perhaps $r(t)$ alone, and thus $y(t)$ might be periodic.) And according to the answers and comments found here, the solution (unfortunately not given) is indeed rather complicated.

The solution when a particle is attracted to the origin with constant acceleration $\mu$ is given in Example 2, page 85, of Whittaker's Treatise:

But it looks rather cryptic to me ($\zeta(s)$ denotes Weierstrass zeta-function, but I'm not sure, for example, whether $\wp$ denotes the Weierstrass elliptic function).

• @stafusa-Thanks for the answer! But don't you think that in this case, the mass makes a closed orbit too? You pull the mass down [not too close to y=-1, the minimum value of y ; if you'd pull the mass to the point where the velocity of the rotating mass gets close to c (which requires infinite energy in fact), the mass goes to infinity, so y(t) shoots up to infinity never to return to the initial point again) ), which subsequently experiences a force upwards, moves beyond y=1, returns to its starting position, and the cycle repeats itself. It's a periodic motion, though a complicated one. – descheleschilder Jul 25 '17 at 10:56
• @descheleschilder, we cannot use $v\to c$ leading to some divergence, because that's a relativistic effect, and the equations of motion we're using are Newtonian. But you're right about $y(t)$: even when the orbit of the rotating mass isn't periodic, it might be that the motion in the $r$ direction alone (which matters to $y$) can be. I'll update my answer. – stafusa Jul 25 '17 at 11:41
• @stafusa-Thanks. I've already mentioned that we can't choose y too close to -1 (r=0, or x, as I use for the fractional distance of the vertically moving mass) because of relativistic effects. Do you think it is possible to construct this system (the construction with the rotating mass on a vertical plane is indeed much easier, and you don't have to go in space!) in such a way that if y has returned to its begin point then also the rotating mass has? – descheleschilder Jul 27 '17 at 5:40
• $\Delta y$ is indeed equal to $\Delta y$, but the force acting on the rotating mass is only the force due to the mass moving vertically, while the force on the vertically moving mass consists of two components: the constant downward force (in the construction on Earth) $F_gravity=mg$ and the varying upward force $\frac {mv^2}[r}$. – descheleschilder Jul 27 '17 at 5:52
• In my comment in which I wrote about a vertical plane, I was indeed mistaken. I confused "vertical" with "horizontal". I understand what you mean with the much easier construction. And I understand your remark about v approaching c. All clear now! – descheleschilder Jul 29 '17 at 13:05