# Interesting problem involving pendulum, collisions, and electric charges

This problem was invented by the physics professor at the community college near me which he used on a past final exam:

A pendulum consists of a massless rod of length L and a conducting ball of mass m at the bottom of the rod. A charge of +q is placed on this ball and the pendulum is held at a position displaced theta degrees to the left of the vertical. A second pendulum identical to the first (except that a charge of -q is placed on the ball) is mounted from the same pivot point and held at a position displaced theta degrees to the right of vertical. Then the two balls are released from rest at the same time. The two balls collide without deformation (i.e. elastic collision) and the two charges are neutralized. The question is, what angle are the pendulums relative to the vertical when they achieve their maximum height after the collision?

The professor posted this solution (which I don't agree with):

Let $y_i$ and $y_f$ be the heights of the pendulum balls above the ground at their initial and final positions (final being the max height achieved) and let r be the initial distance between the two pendulums. The work required to assemble the two charges from infinity is $k(+q)(-q)/r = -k q^2/r$. So the total initial energy is $2mgy_i – kq^2/r$ and the total final energy is $2mgy_f$ (since there no longer is any charge). Since the initial and final energies must be equal we see that $y_f < y_i$ a result that the professor admits is not intuitively obvious. He then goes on to compute the final angle (a computation I won't bore you with).

I have no training in physics other than the freshman physics course I took in college 45 years ago, but still this result did not sit well with me. As the balls are falling the balls are accelerated faster than they would be with gravity alone and this energy has to go somewhere. So I conclude that the final position will be higher than the initial position.

I attempted to explain in an email to the professor why his computation was not correct. I suggested that it was improper to treat the E field as electrostatic as he did because the rapid charge neutralization violates the electrostatic assumption of stationary or slowly moving charges. In electrostatics we have path independence meaning that the work required to move a collection of particles from position A to position B is the negative of the work required to move them back to position A. If we consider A to be the initial pendulum position and B to be any lower position, there is an electric force from A to B, but in the reverse direction there is no electric force because the charges have been neutralized. So we don't have path independence and we can't pick an arbitrary zero reference (such as infinity in this case).

I also suggested that we can't even compute the final height (or angle) with the information given because the size of the balls were not specified. Smaller balls would mean that the two charges get closer together before neutralizing and thus more energy would be gained on the way down.

Apparently the professor does not agree with my objections and is sticking with his belief that his original answer is correct. I'm hoping that I get some replies that will either solidify my position or point out the error of my ways.

Also I had some other concerns about the question that I didn't even mention to the professor. I seem to recall that an accelerating charge radiates energy, so some energy will be lost as the two charges are accelerated downwards towards the collision. I'm not so worried about that, because the balls will still be gaining energy overall due to the electric force. Perhaps the acceleration will be small enough that this energy loss could be neglected in a practical situation. However what about the charge neutralization? An elastic collision is an idealization which implies that the balls only touch for an infinitesimal time which means that the charges have to neutralize in essentially zero time. Would charges moving that fast produce an infinite B field? Is an elastic collision a practical idealization in this situation? In a real world example of such a construction would the energy loss due to the neutralization be small enough to ignore?

## 4 Answers

Your professor's final result is incorrect, and your reasoning about energy is largely correct. Your professor is taking point charges from infinity and putting them together, but is neglecting to consider the energy it takes to assemble a point charge. This is extremely problematic. First, it takes infinite energy to assemble a point charge. Proof:

Consider an isolated metal sphere. This will have some capacitance. Referencing the expressions on this page capacitance of metal spheres

The energy of the sphere is

$U=\frac{1}{2}CV^2=\frac{1}{2}qV$

and the voltage on a sphere is related to both the charge and the radius of the sphere

$q=RV/k$

To assemble a point charge, the charge $q$ should be constant while the radius of the sphere $R$ goes to zero. The only way for this to be possible is if the voltage goes to infinity. Looking at the expression for the energy, it's clear that assembling a point charge requires infinite energy. In your professors treatment of the problem, two point charges eliminate themselves. This would release infinite energy... evidently this is problematic.

To fix your professor's problem, you have to consider both the energy between the spheres, as well as the energy necessary to assemble the charges on a sphere. The exact energy is a complicated problem, but for small spheres with a large separation between them, we can use an approximate expression:

$U=kq^2/R-kq^2/r$

This is the negative energy your professor identified correctly, together with the energy of the individual spheres. This will always be positive.

As a result of this additional energy, the final height of the two pendulums will be greater than the initial height.

You're correct that you can't predict the final height of the balls without knowing the size of the balls. You're also correct that the smaller balls have more energy. Even though the situation isn't static, it falls well within a static or quasistatic approximation, such that you can use static equations to predict the energy of the spheres, and the dynamics of the situation. Any kind of radiation is negligible, along with magnetic forces between the charges. It's a fairly standard result that charges have to be relatively close to the speed of light to have non-negligible magnetic forces between them, compared to electric forces.

I also suggested that we can't even compute the final height (or angle) with the information given because the size of the balls were not specified. Smaller balls would mean that the two charges get closer together before neutralizing and thus more energy would be gained on the way down.

This seems the biggest issue. I don't have much problem ignoring the non-static issues here. But the distance over which the charges interact cannot be ignored. To just remove them and their energy seems incorrect.

To calculate the height of rebound I would model the drop and the rebound as two separate phases. As the balls approach, energy would be released from both GPE and EPE, but as the balls rebound, that energy would go only into GPE.

As you mention, the rebound point does not matter for the GPE calculation, but is essential to gather the EPE released.

A simpler argument against your professor's position

1. At the moment of collision, the charged spheres are moving faster than similar uncharged spheres due to the electrical attraction.
2. After the collision and neutralization, the spheres will rebound with the same faster velocity due to the elasticity of the collision.
3. With no electrical attraction to impede their upswing, the spheres will reach a higher height due to their larger starting velocity.

Now to fully work through this problem so you get to see my beautiful artwork!

First, the setup:

In the diagram, $r$ is the initial distance between the centers of the spheres, $r_s$ is the radius of each sphere, $h_i$ is the initial height of the spheres, $m$ is the mass of each sphere, and $q$ is the charge on each sphere. It's a bit hard to see, but the spheres have opposite charge.

The total energy in the system in its various states will be given by $$E = GPE + EPE + KE$$ with gravitational potential energy $GPE$, electrical potential energy $EPE$, and kinetic energy $KE$. Initially, the total energy is $$E_0 = 2mgh_i - \frac{kq^2}{r}.$$

Next, the spheres swing down and collide. Just before the collision, the situation is as so:

The spheres are at their lowest point but have not neutralized yet. The energy of the system is given by $$E_1 = -\frac{kq^2}{2r_s} + 2\left(\frac{1}{2}mv^2\right) = E_0.$$ Since gravitational and electrical forces are conservative, energy is conserved ($E_1 = E_0$). This equation is somewhat inaccurate because the charges on each sphere will move towards the other one since the spheres are conducting. What this means is that the distance $2r_s$ is an overestimate and the spheres will have a larger velocity $v$ than given in this equation. However, that doesn't change the final result.

Next, the charged spheres neutralize each other.

The electrical potential energy disappears into heating the spheres due to the electric current generated. The total energy is now purely kinetic: $$E_2 = mv^2.$$

Finally, the spheres rebound and reach their peak height $h_f$.

The total energy is now purely gravitational potential energy and is conserved from the previous step: $$E_3 = 2mgh_f = E_2.$$

Now to find the final height: $$E_0 = E_1$$ $$2mgh_i - \frac{kq^2}{r} = -\frac{kq^2}{2r_s} + mv^2$$ $$mv^2 = 2mgh_i - kq^2\left(\frac{1}{r} - \frac{1}{2r_s}\right)$$ I'll leave the equation like this for later convenience.

Next, the neutralized equations: $$E_2 = E_3$$ $$mv^2 = 2mgh_f$$ $$h_f = \frac{mv^2}{2mg}$$

Substituting $mv^2$ from above: $$h_f = \frac{2mgh_i - kq^2\left(\frac{1}{r} - \frac{1}{2r_s}\right)}{2mg}$$ $$h_f = h_i - \frac{kq^2}{2mg}\left(\frac{1}{r} - \frac{1}{2r_s}\right)$$

Now $r > 2r_s$ since the spheres cannot overlap. This means that the expression in the parentheses is negative: $$r > 2r_s \implies \frac{1}{r} < \frac{1}{r_s} \implies \frac{1}{r} - \frac{1}{2r_s} < 0.$$ Therefore, the final height will be higher than the starting height.

When I said above that the equation after the collide picture was inaccurate, the effective distance between the spheres would be less than $2r_s$, which would result in a lower electric potential energy, thus a larger kinetic energy, and thus a higher final height.

In the limit of the spheres being much smaller than the distance between them, $$r_s \ll r \implies h_f = h_i + \frac{kq^2}{4mgr_s}$$

About your other concerns:

• The energy loss from accelerating charges is tiny when the charges are moving much slower than the speed of light. Look up synchrotron radiation.
• The magnetic field generated by the instantaneous current can be ignored since the spheres aren't moving for that instant, so there are no extra forces generated (hooray for spherical cows on frictionless planes colliding elastically in a vacuum!).

I'm switching variable names to avoid subscripts, so I chose r for the radius of the balls and d as the distance between the two balls in their initial position. Taking a simple minded approach of computing the initial potential energy I first considered the work required to assemble a single isolated charge of charge q:

$$dw = V dQ = \frac {kQ}{r} dQ$$ So the total work to assemble the charge is $$w = \int_0^q \frac {kQ}{r} dQ = \frac {kq^2}{2r}$$ We have to assemble both charges, and then bring them from far apart to distance d of each other giving the total potential energy: $$U = kq^2(\frac{1}{r} - \frac{1}{d})$$ If I understand David's notation, this is the same as the approximation that David suggested for what is actually a complicated problem. From this we get the answer to the pendulum problem: $$\Delta h = \frac {kq^2}{mg} (\frac{1}{r} - \frac{1}{d})$$ Since d can't really be smaller than r, $\Delta h$ is positive and indeed the final position is higher, as everybody seems to agree.

It seems reasonable assume $d >> r$ which simplifies the result nicely: $$\Delta h = \frac {kq^2}{mgr}$$ However Mark H using a different method (which seems reasonable to me) arrives at: $$\Delta h = \frac {kq^2}{4mgr}$$ Since these two results differ by a factor of 4, they can't both be correct. Can anyone point out which method (or perhaps both) is incorrect?

• One factor of two comes from the two masses. Substitute 2m for m to fix this one. The rest of the height difference is the result of electrical energy dissipation. You can figure out this energy loss by writing the electric potential energy before and after the collision. After the collision the electric potential energy is 0. Before the collision it's kq^2/2R. This much energy would have to be dissipated into heat. It can't do work, because the spheres aren't moving at the instant the charges dissipate. – David Nov 27 '15 at 0:08
• Oops, the first factor of two (2m vs. m) was a silly mistake I might have fixed if I just looked over my work more carefully, but I don't think I would have figured out where the other factor of two was going. Thanks David for the clear explanation, as now I believe I understand the problem pretty well :) – pmennen Nov 27 '15 at 5:42