Help with an Irodov problem (Problem 3.3 electrodynamics )

[Jahnavi]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

 

[Charles Link]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

That does simplify the problem, and that scenario does look like it would eliminate the inconsistencies. :)

 

[Jahnavi]

That does simplify the problem, and that scenario does look like it would eliminate the inconsistencies. :)

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

Feel free to say NO

 

[kimbyd]

Does everyone agree that if approach velocity is a constant instead of being dependent on x (as given in the question ) , the two spheres are in equilibrium at all times and the problem has a solution ?

Yes. That would work.

 

[Charles Link]

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

Feel free to say NO

I actually think the author tried to be a little fancy, and didn't realize what he was proposing was unphysical. In the author's case, $\frac{dq}{dt}$ is a constant, while for your case $\frac{dq}{dt}$ is dependent on position.

 

[Jahnavi]

I actually think the author tried to be a little fancy, and didn't realize what he was proposing was unphysical. In the author's case, $\frac{dq}{dt}$ is a constant, while for your case $\frac{dq}{dt}$ is dependent on position.

You didn't answer my question

 

[Charles Link]

You didn't answer my question

The only thing missing is the word "No". :)

 

[Jahnavi]

The only thing missing is the word "No". :)

Very bad .

OK , one more chance . I might have something up my sleeve .

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

 

[Charles Link]

Very bad .

OK , one more chance . I might have something up my sleeve .

Would you believe me if I say my interpretation is actually what the author intended but somehow the problem got messed up somewhere ?

It certainly is a possibility. One other thing the author could have said is to avoid $x \approx 0$, and to only consider the region where $\frac{dv}{dt}$ is small, so that the approximation of equilibrium between the electrostatic force and the gravitational force is reasonably good. (So that any additional unbalanced horizontal force can be small, and the acceleration remains small). $\\$ In general, I think most of us are in agreement that the author's "empirical" formula $v=\frac{dx}{dt}=-\frac{a}{\sqrt{x}}$ simply can not work all the way to $x=0$.

 

[Jahnavi]

@haruspex , @Charles Link , @ehild , @kimbyd this is the actual question along with the original problem in Russian .

 

[Charles Link]

@Jahnavi So who is the translator who introduced a $v=-\frac{a}{\sqrt{x}}$? Was it a student who assisted with the publication? Is the problem submitted by the OP an English and later version of the same book? This is getting interesting. :)

 

[Jahnavi]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

In recent editions of the book , this has been corrected .

I also have a copy of this book . It has same question as given in the OP

 

[Charles Link]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

In recent editions of the book , this has been corrected .

I also have a copy of this book . It has same question as given in the OP

I give you +2 for this. A very good finding. :)

 

[Jahnavi]

I give you +2 for this

 

[ehild]

I don't know , but the question in the OP is actually what is given in the English editions of the book and for years this has confused both teachers and students as to why they could assume that the spheres were in equilibrium .

Congratulation Jahnavi!
Somebody has changed the problem, and then followed the original solution that the acceleration was zero. In the original question it was because of the constant velocity. The English version explained it with equilibrium.
Moral: do not change Russian problems. Usually they are rather good, you can only mess up them.
I wonder how the problem was corrected in the recent version? To v=const?

 

[Jahnavi]

In the original question it was because of the constant velocity. The English version explained it with equilibrium.
Moral: do not change Russian problems.

No

In fact in the initial editions of the Russian version , it was given that v = a/√x . But , in the recent editions it has been corrected .

 

[ehild]

No

In fact in the initial editions of the Russian version , it was given that v = a/√x . But , in the recent editions it has been corrected .

My bad. I am totally confused. Was there an original Russian version with v=a/√x, and it was changed to v=const, and then an English version with v=a/√x, and then a new English version with v=?
When I was young, we used quite a few Russian textbooks and problem books ( usually translated to Hungarian, or in original, as we studied Russian from age 10) and I found them rather good.

 

[Jahnavi]

Firstly I would say that this is perhaps a rare problem you would find in this book with some issue . It is an excellent , superb book .Very accurate ! And mind boggling too .

I guess that the initial Russia version actually had v = a/√x which was later corrected to v = constant .

But the problem is that the English editions keep printing the initial Russian version . Even now the new English editions at my place has the same wording as given in the OP .

The copy of the book I am having also has the same incorrect version

 

[ehild]

It is amazing to see the English text of the problem with the solution which originally belonged to the corrected Russian version
http://exir.ru/3/resh/3_3.htm

 

[rude man]

But as per the solution we need to take net acceleration every instance 0. Why is it so?

Who says it is so? With x = x(t), neither a(x) nor a(t) is necessarily constant. But I'm referring to the original problem as posted in the OP's post 1 and I gather that has eo ipso been questioned. But the math looks menacing even so.

 

[rude man]

EDIT: possible approach iff v = (const)/√x. I should have looked at all the prior posts so the following is wheel-spinning ... I don't read Russian but post 40 sure looks like v = 0.55 something/s i.e. constant.

1. Let length of pendula = L instead of 1. I don't lke losing the ability to check dimensions as I stumble along.
2. Change v = a/√x to c/√x. "a" should be reserved for acceleration.
So one mass starts at x = -x₀/2 and the other at x = x₀/2.
With these mods, taking the mass at x= x₀/2,
Σ F_x/m = a_x = kq²(x)/mx² - gx/2L,
v_x = ∫ a_x dt = ∫ a_x/v_x dx = cx^-1/2
winding up after some grief with
q²(x) = (m/2k)(gx³/L - c²)
So we have q(x).
To get q(t) solve dx/dt = -cx^-1/2 giving x(t) = {x₀^3/2 - (3/2)ct}^2/3.
Then substitute for x in q(x) and that's it.
Still very laborious. Lots of opportunities for mistakes!
I wonder if we could cheat and ignore gravity?