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Suppose that we have two rods of length $l_1, l_2$ connected at one end but free to rotate. These rods have charge density $\lambda$ uniformly distributed, so the total charge of rod $i$ is $\lambda l_i$. Given an interior angle between them $\theta$, what is the force between the two rods? What is the torque?

Idea for a solution: find the force on a particle with charge $d\lambda$ from a finite-length rod then integrate over the whole rod. However, it's been a while since I last took an E&M class, so I'm not sure how to do that. I'm also not sure what the torque would be; $k\sin \theta$ makes the most sense until you realize that that would also mean small torque for small angles, which is wrong. This indicates the torque must be discontinuous at $\theta = 0$. Something proportional to $\cot \frac{\theta}{2}$ is my best bet.

Edit: this is not homework, though it appears to be. I just thought of it and couldn't easily solve it.

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Note that the homework tag also apply to non-homework that is homework-like. –  Qmechanic Mar 14 '13 at 5:34
    
@jclancy, I like the question :-) –  Siva Mar 22 '13 at 19:47
    
Are you saying that the $l_2$ string is connected to the end of $l_1$? –  Ataraxia Mar 23 '13 at 17:08
    
@ZettaSuro Indeed. –  jclancy Mar 23 '13 at 22:06
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3 Answers

up vote 6 down vote accepted
+50

Since this is a homework-type problem, here are some

Hints for the force

  1. The electrostatic force $d\vec F$ on a small segment $dl$ of the rod given the field $\vec E$ of the other rod is $$ d\vec F = \lambda\, dl \,\vec E $$

  2. Determine the field of one rod, and use the above expression to integrate the force it exerts on the other rod.

  3. This is a 2D problem since, by symmetry, the force is in the plane containing both rods.

Hints for the torque

  1. Take the point at which both rods are joined as the origin, then the torque on a small segment of one rod due to the field of the other is $$ d\vec \tau = \vec r\times d\vec F $$ where $\vec r$ is the vector pointing from the origin to the segment.

  2. Integrate to find the total torque.

Addendum 1.

I started solving this problem, then I got to integrals I needed to compute to determine the electric field of one rod at a general location on the other, and I realized that there is no solution to this problem as stated. The force between the two charge elements at the end of each rod where they are joined is infinite (because the distance between them is zero). You can make this problem well-posed by considering a charge densities $\lambda_1, \lambda_2$ that goes to zero sufficiently quickly at the point where the rods are joined so that the integrals you have to perform don't diverge, but that would be a different problem.

Addendum 2.

I pointed out in the comments that the integrals you would need to compute to obtain the force between uniformly charged rods with joined endpoints diverges because of the charges that are at zero distance from one another at the vertex (place where the rods are joined).

Here are the integrals you need to compute as requested. Take rod 1 to be on the positive $x$-axis with left-hand endpoint at the origin, and take rod 2 to be angled at some $\theta$ measured counterclockwise from the positive $x$-axis with it's left endpoint at the origin as well. Let $r_1$ denote the radial coordinate along rod 1 and let $r_2$ denote the coordinate along rod 2 with $r_1=r_2=0$ at the origin. Given any two points on the rods located at radial distances $r_1$ and $r_2$, the squared distance between these points is $$ d^2 = r_1^2+r_2^2-2r_1r_2\cos\theta $$ which comes from some basic vector algebra, or equivalently, the law of cosines. Now, let's say we're computing the force on rod $2$ by rod $1$, then we also need the expression for the vector pointing from the point at position $r_1$ on rod 1 to the point at position $r_2$ on rod 2. This is given by $$ \vec d = (r_2\cos\theta -r_1)\,\hat x+r_2\sin\theta\,\hat y $$ The force of charge element $dq_1 = \lambda_(r_1)dr_1$ one charge element $dq_2 = \lambda_2(r_2) dr_2$ at the positions $r_1$ and $r_2$ along the rods is given by Coulomb's Law; $$ d\vec F = \frac{1}{4\pi\epsilon_0}\frac{dq_1dq_2}{d^3}\vec d = \frac{1}{4\pi\epsilon_0}\frac{\lambda_1(r_1)dr_1\lambda_2(r_2)dr_2 ((r_2\cos\theta -r_1)\,\hat x+r_2\sin\theta\,\hat y)}{(r_1^2+r_2^2-2r_1r_2\cos\theta)^{3/2}} $$ and the desired force is given by and integral over the lengths of both rods from $0$ to the length of each rod; $$ \vec F = \frac{1}{4\pi\epsilon_0}\int_{0}^{\ell_1} dr_1\int_0^{\ell_2} dr_2 \frac{\lambda_1(r_1)\lambda_2(r_2) ((r_2\cos\theta -r_1)\,\hat x+r_2\sin\theta\,\hat y)}{(r_1^2+r_2^2-2r_1r_2\cos\theta)^{3/2}} $$ Now you can simply play around with the function forms of $\lambda_1$ and $\lambda_2$ to determine what sorts of charge distributions give convergent integrals!

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See my addendum. –  joshphysics Mar 22 '13 at 20:20
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Can you post the integrals that you got? I want to see what would happen if you assumed that the charge densities were some nonzero constant when greater than some distance $\epsilon$ away from the join point, and zero inside, then let $\epsilon \to 0$. I guess if that does nothing I also want to experiment ith what happens with exponential and quadratic distributions. –  jclancy Mar 22 '13 at 21:55
    
The distance between the two rods goes linearly as you approach the hinge. (More precisely, the distance between a point on rod1 and the point on rod2 which is closest to the point on rod1). So I would think any distribution of charge which is quadratic of higher variation close to the hinge will avoid the divergence. "Regulating" the integral by $\epsilon$ is also a good idea as it should let you extract the qualitative features from the finite part without worrying about the infinity. –  Siva Mar 23 '13 at 1:11
    
See addendum 2. –  joshphysics Mar 23 '13 at 10:10
    
Great, thanks so much! This is precisely what I wanted. –  jclancy Mar 23 '13 at 19:14
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As joshphysics' answer showed, the force would indeed be infinite in the case of uniform lineic distributions, but the torque does not need to be. Using the same conventions joshphysics did, let's compute the elementary torque $d\vec{T}$ experienced by a piece $dr_2$ of rod 2 at position $\vec{r_2}$ from the joining point and integrate it over rod 2. It can be written $$ \begin{array}{rcl} \vec{T} &=& \int_0^{\ell_2}d\vec{T} \\ &=& \int_0^{\ell_2}\vec{r_2}\wedge\lambda_2dr_2\int^{\ell_1}_0 \frac{1}{4\pi\varepsilon_0} \frac{\lambda_1dr_1}{({r_1}^2+{r_2}^2-2r_1r_2\cos\theta)^{3/2}}\,(\vec{r_2}-\vec{r_1})\\ \vec{T} &=&\frac{1}{4\pi\varepsilon_0} \int_0^{\ell_2}\int_0^{\ell_1} \frac{\lambda_1\lambda_2\,r_1r_2\sin\theta\,dr_1dr_2}{({r_1}^2+{r_2}^2-2r_1r_2\cos\theta)^{3/2}}\hat{z} \end{array} $$

To make the integration easy, let's suppose that $\theta=\pi/2$, so that we have first to compute

$$ \int_0^{\ell_1}\frac{r_1\,dr_1}{({r_1}^2+{r_2}^2)^{3/2}} = \left[-\frac{1}{\sqrt{{r_1}^2+{r_2}^2}}\right]_0^{\ell_1} = \frac1{r_2}-\frac1{\sqrt{{\ell_1}^2+{r_2}^2}} $$

so that finally

$$ \vec{T} = \frac{\lambda_1\lambda_2}{4\pi\varepsilon_0} \int_0^{\ell_2}\left(1 - \frac{r_2}{\sqrt{{\ell_1}^2+{r_2}^2}}\right)dr_2\,\hat{z} $$

that can be integrated similarly as the previous one to give, if I'm not mistaken,

$$ \vec{T} = \frac{\lambda_1\lambda_2}{4\pi\varepsilon_0}\left(\ell_2+\ell_1 -\sqrt{{\ell_1}^2+{\ell_2}^2}\right)\hat{z} $$

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If only there was more than one bounty to give. I am a little surprised that the force can be infinite but the torque finite, but on the other hand it makes sense - there should be no torque at $\theta = \pi$. –  jclancy Mar 23 '13 at 22:15
    
+1: Cool. Makes quite a bit of sense given how torque scales with distance. I didn't check the integration, but I'll take your word for it ;) –  joshphysics Mar 23 '13 at 22:34
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Physical intuitiion

I imagine that $\theta=\pi$ is a stable equilibrium and $\theta=0$ will be an unstable equilibrium. So the torque should be zero at $\theta=0$ and slowly (linearly) increase for small $\theta$, but in a direction which drives the system away from equilibrium. So $\tau \sim k \, \sin \theta$ seems like a reasonable conjecture for perturbations from the equilibria.

Find the potential energy of the system

Consider a small element in rod1 and find it's potential energy due to rod 2. Thsi will involve an integration over rod2. Now, integrate over rod1 to find the total potential energy of the system. Both integrations are over scalars... no mess with multiple components and projections. You will get an expression for the system's potential energy as a function of the angle between the rods.

Finding the torque

Fix your coordinate sustem with the origin at the hinge and (say) X-axis along rod2. Then, effectively, you only have 1 degree of freedom: the angle $\theta$ between the rods. To find the torque $\tau$, differentiate the expression for potential energy w.r.t $\theta$.

If you want to find the force, you could invert the expression $\int_0^L F(r) \, r\, dr = \tau$ to find $F(r)$ and then integrate that along the rod to find the tangential component of the force.

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It seems to me that the radial component of the forceshould be zero in some nice frame... maybe when you sit in the frame where the hinge is stationary, but I'm not quite convinced. –  Siva Mar 22 '13 at 19:45
    
I don't understand why the torque should be small for small $\theta$. Shouldn't the opposite be true, i.e. the torque should drive the rods from the $\theta = 0$ equilibrium extremely quickly? However I agree the radial component of the force should be zero. –  jclancy Mar 22 '13 at 22:00
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When the two rods are parallel, the system has maximum potential energy. That means the torwue must be very small. But like I said, the equilibrium is unstable. So the torque will start small and as it slowly pushes the rods out of equilibrium, it will gather in strength. Imagine a ball rollong off a curved mountaintop (say: inverted parabola to leading order in the taylor expansion). It will start off very slowly, with a small force because the mountain is almost horizontal at the top (near an extrema). As the slope (of the mountain ~ potential energy) increases, so will the force. –  Siva Mar 23 '13 at 0:56
    
OK, thanks, I understand now. –  jclancy Mar 23 '13 at 0:58
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