A Physics Problem Recently Sent to Me

Here is a problem someone asked me about recently.

I have come to find out that my solution is not the simplest possible one, but here it is anyway.

Assume that the middle of the string moves with constant speed. The position and velocity of the right ball is:

x=Lcosθy=vtLsinθx˙=Lsinθθ˙y˙=vLcosθθ˙x = L \cos \theta \\ y = v t - L \sin \theta \\ \dot{x} = -L \sin \theta \dot{\theta} \\ \dot{y} = v - L cos \theta \dot{\theta}

The kinetic energy TT , potential energy V V , and Lagrangian are:

T=mL2θ˙2+mv22mvLcosθθ˙V=keq22Lcosθ=keq22LsecθL=TV=mL2θ˙2+mv22mvLcosθθ˙keq22LsecθT = m L^2 \dot{\theta}^2 + m v^2 - 2 m v L \cos \theta \dot{\theta} \\ V = \frac{k_e q^2}{2 L \cos \theta} = \frac{k_e q^2}{2 L } \sec \theta \\ \mathcal{L} = T - V = m L^2 \dot{\theta}^2 + m v^2 - 2 m v L \cos \theta \dot{\theta} - \frac{k_e q^2}{2 L } \sec \theta

Equation of motion:

ddtLθ˙=Lθ \frac{d}{dt} \frac{\partial{\mathcal{L}}}{\partial{\dot{\theta}}} = \frac{\partial{\mathcal{L}}}{\partial{\theta}}

Evaluating these equations results in:

θ¨=keq24mL3secθtanθ \ddot{\theta} = - \frac{k_e q^2}{4 m L^3} \sec \theta \tan \theta

Suppose the physical parameters are such that the oscillations are small. In this case:

θ¨keq24mL3θγ=keq24mL3 \ddot{\theta} \approx - \frac{k_e q^2}{4 m L^3} \theta \\ \gamma = \sqrt{\frac{k_e q^2}{4 m L^3} }

This corresponds to simple harmonic motion with angular frequency γ \gamma . Writing out the general form for the solution:

θ=Acos(γt)+Bsin(γt)θ˙=Aγsin(γt)+Bγcos(γt)\theta = A \cos(\gamma t) + B \sin(\gamma t) \\ \dot{\theta} = -A \gamma \sin(\gamma t) + B \gamma \cos(\gamma t)

Assume that the balls start at rest. Applying initial conditions θ(0)=0 \theta(0) = 0 and θ˙(0)=vL \dot{\theta}(0) = \frac{v}{L} results in:

θ=vγLsin(γt) \theta = \frac{v}{\gamma L} \sin(\gamma t)

The maximum value of θ \theta is:

θmax=vγL=vL4mL3keq2 \theta_{max } = \frac{v}{\gamma L} = \frac{v}{L} \sqrt{\frac{4 m L^3}{k_e q^2}}

The minimum distance between the charges is:

Dmin=2Lcosθmax=2Lcos(16πϵ0mLv2q2) D_{min} = 2 L \cos \theta_{max} = 2L \cos \Big(\sqrt{\frac{16 \pi \epsilon_0 m L v^2}{q^2}} \Big)

We can work this into an alternate form which matches the form expected in an answer key. Since the argument of the cosine function is small (by assumption), we can represent the cosine as a two-term Taylor expansion:

cosα1α22Dmin2L(18πϵ0mLv2q2)=2L(q28πϵ0mLv2q2) \cos{\alpha} \approx 1 - \frac{\alpha^2}{2} \\ D_{min} \approx 2 L \Big( 1 - \frac{8 \pi \epsilon_0 m L v^2}{q^2}\Big) = 2 L \Big( \frac{q^2 - 8 \pi \epsilon_0 m L v^2}{q^2}\Big)

A bit more manipulation:

Dmin=2L(q28πϵ0mLv2q2)=2L(q2βq2)=2L(q2βq2)(q2+βq2+β)=2L(q4β2q4+βq2) D_{min} = 2 L \Big( \frac{q^2 - 8 \pi \epsilon_0 m L v^2}{q^2}\Big) = 2 L \Big( \frac{q^2 - \beta}{q^2}\Big) = 2 L \Big( \frac{q^2 - \beta}{q^2}\Big) \Big(\frac{q^2 + \beta}{q^2 + \beta}\Big) = 2 L \Big( \frac{q^4 - \beta^2}{q^4 + \beta q^2}\Big)

Since β \beta is presumed to be small:

Dmin=2L(q4β2q4+βq2)2L(q4q4+βq2)=2L(q2q2+β)=2L(q2q2+8πϵ0mLv2) D_{min} = 2 L \Big( \frac{q^4 - \beta^2}{q^4 + \beta q^2}\Big) \approx 2 L \Big( \frac{q^4}{q^4 + \beta q^2}\Big) = 2 L \Big( \frac{q^2}{q^2 + \beta}\Big) = 2 L \Big( \frac{q^2}{q^2 + 8 \pi \epsilon_0 m L v^2}\Big)

Note by Steven Chase
3 months, 3 weeks ago

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@Steven Chase the solution of book which I shared with you was correct or incorrect???
Because according to @Karan Chatrath he said , the solution is incorrect.

A Former Brilliant Member - 3 months, 3 weeks ago

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I'm pretty satisfied with the solution. If he would like to comment, I would be happy to hear it.

Steven Chase - 3 months, 3 weeks ago

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Well, I did not say that the solution is incorrect. Please do not misquote me. I said that the solution provided in the book made no sense to me. It was just a few lines with minimal explanation.

Coming to the solution provided by @Steven Chase, I must say that it is pretty elegant. I solved the problem using a Newtonian formulation and by using Cartesian coordinates instead of the θ\theta coordinate. The resulting equations of motion I obtained were pretty big and I saw no other way to proceed apart from a numerical route.

Thank you for sharing this note.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath Ok sir. Sorry.
Now I have retracted what I said.
I want to ask you a single correct question.

Choose the correct option:

The book solution is
a)Correct
b)Incorrect

It is compulsory to answer the question.

A Former Brilliant Member - 3 months, 3 weeks ago

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@A Former Brilliant Member If I had to choose only among these two options, I choose the second one. This is because the initial speed of each ball is not vv. The balls are initially separated by 2L2L and are at rest. That is what my solution predicts. Moreover, if you notice the expressions for velocity components provided in this note, they also indicate that the initial speed of each ball is zero unlike the solution provided in the book. Beyond this point, nothing else made sense to me while reading the solution provided in your book.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath One thing to keep in mind is that in the book solution, they take a frame of reference coinciding with the center of the string. In that case, the initial velocity of each ball is v v .

Steven Chase - 3 months, 3 weeks ago

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@Steven Chase Okay, I missed that key phrase, now that you draw attention to it. And this also means that I stand corrected about my stance about the solution being incorrect. But I cannot help but think that even if we were to take an inertial frame of reference at rest, such as I have, the energy conservation principle should yield the answer. I wonder why isn't that so.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath Right, it is not clear whether the mid point is supposed to have constant speed or not. In constant-energy systems with no external forcing functions, all parts generally oscillate or change speed somehow. In my conception of the problem, there is external forcing. See my other comment

Steven Chase - 3 months, 3 weeks ago

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What book is this? I wonder.

Krishna Karthik - 3 months, 3 weeks ago

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Not sure. All I have is that image

Steven Chase - 3 months, 3 weeks ago

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@Neeraj Anand Badgujar To elaborate further:

Taking ϵo=m=q=1\epsilon_o=m=q=1, L=2L=2, v=0.1v=0.1 and solving the equations of motion (derived in cartesian coordinates) numerically, I get for each ball:

Notice the initial speed of each ball is zero. Moreover, at the instant of minimum separation, the y component of the velocity of each ball is 0.1 and the x-component is zero. So using this information, if we were to apply the energy conservation principle, then:

PEinitial+KEinitial=PEfinal+KEfinalPE_{initial} + KE_{initial} =PE_{final} + KE_{final}

KQ22L=KQ2d+mv2\frac{KQ^2}{2L} = \frac{KQ^2}{d} + mv^2

This is different from the solution provided in the book. It seems weird to me why the energy conservation principle does not work for the masses here, but I think that the equation of energy conservation must also account for the energy provided to the system due to the prescribed motion.

@Steven Chase what do you think? For reference, I am attaching the snapshot shared with me earlier.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath I think conservation of energy is not working because while applying it, we can't say about the velocity of y direction of particles, but it 100% guaranteed that x direction velocity will be zero. Therefore I think the question is not as much easy as the author of the book thinks.

A Former Brilliant Member - 3 months, 3 weeks ago

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You are right that it is not as easy as it looks. The answer provided in the solution is correct but the method used makes no sense. Hence my initial reservation about calling the book solution incorrect. But these extra checks have brought out some interesting features of this problem.

As to why the energy conservation principle does not work, it is not because of a lack of knowledge of the Y velocity components. I have used the Y velocity components while computing the final kinetic energy. I think that it does not work because the energy used to drive the system into motion is not accounted for. How that is done, I do not know yet.

The fact that the initial kinetic energy is zero and the final kinetic energy is not, completely contradicts the solution of the book.

Karan Chatrath - 3 months, 3 weeks ago

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I believe that the book solution is correct. The author just decided to look at things from different reference frame and, since the problem assumes constant velocity of the middle of the thread, it's an inertial reference frame and all the laws of physics are completely valid. What more, this approach is more natural taking into account the way the problem is posed. He also correctly reasons that the velocity is 0 at the moment of minimum distance because the charged particles are constrained to move along the circle.

Regarding energy conservation, the equation obtained by @Karan Chatrath with respect to stationary observer is not correct because it's obvious that the energy is periodically provided to and taken away from the system in order to maintain the constant velocity of the thread. Energy in more general sense, seen as one of the integrals of equations of motion, equals: Lθ˙θ˙L=const.\frac{\partial L}{\partial \dot{\theta}}\dot{\theta} - L = \text{const.} If one evaluates this expression using Lagrangian derived by @Steven Chase, one gets an equation identical to the one in the book.

Uros Stojkovic - 3 months, 3 weeks ago

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@Uros Stojkovic @Uros Stojkovic Sir but when author is applying energy conservation the velocity of particles in xx direction is zero and it should be. But the velocity in yy direction is not necessary to be zero, and the author has assumed it to be zero. So is it correct??

A Former Brilliant Member - 3 months, 3 weeks ago

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@A Former Brilliant Member It's correct. Velocity components (in the reference frame fixed at the middle of the thread) are dependent on each other because the motion of the charged particles is constrained to follow the circle of radius ll.

Uros Stojkovic - 3 months, 3 weeks ago

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Another question is whether we are to assume that the velocity of the anchor point is constant. That is what I have assumed in my solution, but it is not clear whether that assumption is allowed. One thing we could do is just create a new problem with the following features:

1) The middle anchor point has constant upward velocity v v
2) The balls are initially at rest
3) Each ball is connected to the anchor point by a rigid rod
4) The rods are hinged, and can rotate freely about the anchor point

If you set up the problem this way, I believe the dynamics are described by that equation with the secant/tangent. And we need not limit ourselves to small angles.

Steven Chase - 3 months, 3 weeks ago

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Yes, I made the same assumptions when I worked out the equations of motion in Cartesian coordinates.

The rigid rod assumption is an interesting one and it aligns with the notion that the string remains taut throughout the motion.

Keeping in mind your related comment, I also agree that there is some form of external forcing. The question is then how can the work done by this external force be quantified. This is what eludes me.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath How about we make the following additional changes?

1) Give the anchor point mass
2) Remove the external forcing
3) Initialize the anchor point with some speed while the balls are at rest

I think that would be a nice problem with no ambiguity

Steven Chase - 3 months, 3 weeks ago

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@Steven Chase Sounds like a fun problem proposition. Would the anchor point only have an initial velocity vv along the Y direction, or would it have a constant velocity vv along the Y direction throughout time?

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath It would only have initial velocity v v . As the system evolved, that would change. Energy conservation would apply

Steven Chase - 3 months, 3 weeks ago

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@Steven Chase Sounds great. I would have offered to post it myself but I can only do so after several hours as I will be away from my laptop for the rest of my evening right until tomorrow. If you do post it in the meantime, I will get to it when I can.

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath That's fine, I'll wait for you to post it. Should be fun

Steven Chase - 3 months, 3 weeks ago

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@Karan Chatrath @Karan Chatrath But is there any rule of staying away from laptop????

A Former Brilliant Member - 3 months, 3 weeks ago

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