Intractable physics/trigonometry problem

tl;dr The real problem is in bold at the bottom.

This is derived from a couple of much simpler problems I gave my students, and which sparked some further discussion.

Two balls with equal charge are suspended from very lightweight rods. They are suspended from a single point and can only move in one vertical plane, so they repel each other until the rods form a certain angle. Then a horizontal electrical field is applied. This causes the whole assembly to rotate in one direction, and also pushes the balls slightly towards each other. Increase the field strength and they rotate further, but are also pushed closer.
What is the maximum angle between the vertical and one of the rods?**

In my example I had balls of 5.9 mg, rods of 45.5 mm and a charge for each of the balls of 5.9 nC. I started out with the balls suspended from strings, but, a little late, realized that with the field applied they’d just rotate out of the plane of the paper sketch, so for this extended problem it’s necessary to limit movement to one plane.

I’ve found a solution for that situation of about 100 degrees, but my approach required graphing the field strength as a function of angle between the two balls and as a function of the angle the whole assembly rotates, and using computer tools to add the inverse functions together. Actually transforming the equations involved into their inverses is beyond me, but I’m curious if it’s at all possible.

**It boils down to transforming

tan a * (sin a)^2 = c/sqrt(k + b^2)

so a is a function of b, instead of the easier b is a function of a, which is what I did.**

A closed-form solution does exist, but it’s ugly as sin and involves the roots of a cubic polynomial. Let x = tan a. It is then the case that

x/√(1 + x^2) = tan a / √(1 + tan[sup]2[/sup] a) = tan a / sec a = sin a.

Thus, your equation can be rewritten as

x[sup]3[/sup]/(1 + x[sup]2[/sup]) = Q,

where Q is the right-hand side of your original equation; or

x[sup]3[/sup] - Q x[sup]2[/sup] - Q = 0.

It’s possible to analytically solve a general cubic polynomial in terms of cube roots and square roots, but the results are not nice. If you calculate the discriminant of this polynomial, you’ll find that it only has one root rather than three, so that’s a small mercy at least. If you are so hell-bent on getting a closed-form solution that you decide to write out x as a function of c & b, then a will be the inverse tangent of x.

But less ugly than tan?

That’s just sec. :o

I’m gonna barf cos all y’all’s bad puns are makin’ me sick. IOW: y’all need to cot it out now. :barfing smilie:

All these puns are so bad, I’m about to grab my cosh.

(Eh, not really. I’m just being hyperbolic.)

I think I should apologize for derailing this thread without first complimenting MikeS for his high-quality response.

Kinda went off on a tangent?

While a closed-form solution apparently does exist for this one, it’s worth noting that it’s really easy, using trig functions, to end up with what’s called a transcendental equation, which can’t be solved in closed form.

@Xema Agreed. I looked a the OP early in the day, scratched my head, & clicked [next]. Bravo MikeS!!

I’m gonna need to meditate on that a bit.

I think by definition the equation in the OP is a transcendental equation as would any equation that involved trig functions as the definition of a transcendental equation is one that contains a transcendental function AFAIK. I guess what you mean though by “transcendental equation” is one whose solution does not have a closed-form.

FWIW these days it is as easy as typing the OP’s equation into an appropriate bit of software to find the closed form solutions and they are total messes.

One simple example of such an equation is Kepler’s equation, relevant for orbital mechanics: M = E - e*sin(E). It’s often the case that M and e are known, and you want to find E, but there’s no direct inversion of the equation possible (fortunately, it’s not difficult to solve numerically using techniques like Newton’s method).

Yeah. I’ve occasionally considered learning the general formula for cubic polynomials, just so I can mess with the students’ minds and claim they too have to learn it, but it’s too much of a mess. And I’m not going even going to create this horrible monster using the computer … or maybe I am …

Thanks a million!

That’s nothing. I once fully expanded out the general solution for the quartic polynomial (which is to say, I made Maple do it for me), and it ran, literally, to twenty pages.

EDIT: Thankfully, there’s no general solution for quintics, so this madness ends here.

For what it’s worth, I think this expression should be:
tan a * sin a = c/sqrt(k + b^2)
which can be massaged into a quadratic function of cos a:
cos[sup]2[/sup]a + {c/sqrt(k + b^2)}cos a - 1 = 0
which is easily solved and has only one physically relevant solution. To make comparison easier: my a here is the half-angle between the rods at equilibrium and b is the Coulomb force. I am adding a to another angle corresponding to the overall rotation of the symmetry axis (i.e., the angle between down and the “new down”), and them maximizing this sum w.r.t. b.

It’s entirely possible I messed things up rearranging the involved formula in a rushed back of the envelope manner.

No, I think I had it correct the first time. What we want is for the sum of forces on a ball to be zero. That means the Coloumb force from the other ball (F), pluss the force from the rod (S), plus the new down force (D) has to equal zero.

Trig then gives us F = D tan a (a still being the half angle)

But F is the Coloumb force, so it is also k*q^2/r^2

Now r is 2lsin a, which gives us the (sin a)^2 term.

Wait, you’re saying I’m not able? Impossible!

I don’t even know if this is a joke, but…

What Chronos is alluding to is the fact there’s no equivalent to the quadratic formula for polynomials with terms raised to the fifth power. There’s no, and there can never be, a formula expressed in terms of addition, subtraction, multiplication, division, and taking roots (that is, an algebraic solution or a solution in radicals) which works to find the zeroes of all polynomials with terms raised to the fifth power.

You can find the solution of some such polynomials that way, and you can always find a solution to every polynomial using numerical methods (that is, making guesses and refining them), but there are no, and can never be, equivalents to the quadratic formula beyond quartics, or polynomials where the highest term is raised to the fourth power. Figuring this out was a significant achievement of Nineteenth Century mathematics.