Double pendulum roller coaster FIXED

My last post was wrong. I’m to blame. But in thinking about it and talking about it with with lots of helpful friends I ended up learning a ton. Here’s the upshot: There were kinks in the roller coaster loop that led to integration mistakes on the part of Mathematica. Thanks to a great suggestion from my friend Craig I smoothed those out:

The blue track is the one with kinks in it. The orange one is the used for the simulation in this post

The blue track is the one with kinks in it. The orange one is the used for the simulation in this post

And now the simulation animation looks like this (there’s some extra annotation that I’ll talk about below):

The green dot is the center of mass of the system. The orange arrow is the normal force. The purple arrow is the direction that the center of mass is traveling.

The green dot is the center of mass of the system. The orange arrow is the normal force. The purple arrow is the direction that the center of mass is traveling.

Note first that the system never gets above the dotted green line, which was my (mistaken) idea from the last post. This post will try to talk about what I learned about whether the normal force does any work (which was my mistaken explanation in the last post).

The track exerts a force on the red ball to keep it on the track. Gravity and the connection to the first black ball are yanking on that ball and the track does whatever it has to in order to ensure that the red ball stays on the track. My argument from the last post boils down to this: The normal force is an external force to the system of three balls. That system has a center of mass that I can pretend the external force acts on. If the center of mass is moving perpendicularly to the normal force (as would happen with just the red bead), there would be no work. But if the center of mass is moving at times slightly parallel to the normal force, then there would be some work. It turns out there’s really nothing wrong with that description. However, assuming that changes the kinetic energy of the system is wrong. What it does (as again my friend Craig suggested) is it changes the translational kinetic energy of the system (basically the kinetic energy of the system if you replaced it with all the mass being at the center of mass). However, the total kinetic energy of the system is both the translational and rotational kinetic energy. What I intend to discuss here is that the effect on the rotational kinetic energy due to the normal force is exactly the opposite of the effect on the translational kinetic energy.

First a quick plot. This shows in blue the time derivative of the translational kinetic energy of the system (subtracting out the effects of gravity) and in orange the work per unit time that the normal force does on the system:

The comparison of the rates of change of the translational kinetic energy (without gravity effects) and the work that the normal force does on the system.

The comparison of the rates of change of the translational kinetic energy (without gravity effects) and the work that the normal force does on the system.

Actually, you don’t see the orange because, to the accuracy of the thickness of the lines, the orange is completely underneath the blue.

Let’s try to understand what’s going on. Consider first the time rate of change of the kinetic energy of the system:

\frac{d}{dt}\left(\frac{1}{2}\sum_i m_i v_i^2\right)=\frac{d}{dt}\left(\frac{1}{2}\sum_i m_i \vec{v}_i\cdot \vec{v}_i\right)

The derivative can come right into the sum, and the vector product rule gives us:

\frac{d\text{KE}}{dt}=\sum_i m_i \vec{a}_i \cdot \vec{v}_i=\sum_i \vec{F}_i\cdot \vec{v}_i

where I’ve used Newton’s second law in the last step. Now the normal force is only “attached” to the red bead, but that’s the one bead that’s guaranteed to be moving perpendicular to the normal force. So the contribution to the time change of kinetic energy due to the normal force is indeed zero! Hence my last post is wrong.

But what about this business with the translational kinetic energy? We tell our students all the time that they can think of all forces as acting on the center of mass. In other words, the change of momentum of the center of mass is due to the collection of all external forces. Those forces will do work if they act, at least partially, in the direction that the center of mass is traveling. In the animation above the purple arrow shows the direction that the center of mass is traveling. You can see that it doesn’t always point along the track. That means that it’s not always perpendicular to the normal force. Hence work is done on the center of mass. But that just affects the translation of that center of mass, not any rotation about it. To see the whole story, let’s redo the last calculation using a coordinate system centered on the center of mass. For those variables, I’ll use primes. First I’ll start with an expression for the kinetic energy:

\text{KE}=\frac{1}{2}\sum_i m_i (\vec{v}_\text{cm}+\vec{v}_i')\cdot(\vec{v}_\text{cm}+\vec{v}_i')

Now when you do the FOIL of that dot product, two of the terms go to zero (that’s the beauty of using the center of mass, by the way) and you’re left with:

\text{KE}=\frac{1}{2} M V^2+\frac{1}{2}\sum_i m_i \vec{v}_i'\cdot \vec{v}_i'

where M is the total mass of the system and V is the velocity of the center of mass. Now, let’s consider doing a time derivative of that. For the first term you’ll get exactly what I was talking about above. In other words you’ll get the dot product of the total external forces and the velocity of the center of mass.

\frac{d}{dt}\text{KE}=\sum_i \vec{F}_i\cdot \vec{v}_\text{cm}+\sum_i m_i \vec{a}_i\cdot \vec{v}_i'

Now here’s a trick. Let’s re-express the velocity back into the normal frame (and use Newton’s second law again) for the second term above:

stupidequation

Here’s where the magic happens. The normal force is only applied to the particle on the track. But it’s velocity is perpendicular to the normal force by definition. So the first term in the parenthesis yields a zero. What we’re left with is:

-\sum_i \vec{F}_i\cdot \vec{v}_\text{cm}

which is exactly the opposite of the change to the translational energy. In other words, you can either say that, yes, the normal force does some work, but it changes the translational kinetic energy by exactly an amount that is the opposite of how it changes the rotational kinetic energy, or you can just say that the normal force does no work. You decide.

Thoughts? Here are some starters for you:

  • Thanks for this, I was totally at a loss for figuring out the mistakes in the last post.
  • I’m glad you figured this out for yourself, just know that the rest of us knew this all along and have been laughing at you for your last post for a few days now.
  • Wait, it doesn’t work!? I’ve already starting building it in my backyard!
  • How did you figure out the normal force? Did you determine the accelerations of all the particles and subtract all known forces, starting with the last black dot and moving up to the red dot. Or did you use Lagrange multipliers to figure out the normal force more directly, and, if so, how did you figure out the constraint equations for the track? (yes, yes, and it’s a long but interesting story involving me jumping out of bed this morning and trying something that worked!)
  • So how would you say it? Does the track do work on the system?
  • How is it that you were willing to believe that the track could help you violate energy conservation? What, are you some sort of “momentum is king” kind of guy or something?
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About Andy "SuperFly" Rundquist

Professor of physics at Hamline University in St. Paul, MN
This entry was posted in mathematica, physics. Bookmark the permalink.

7 Responses to Double pendulum roller coaster FIXED

  1. It is a general problem with almost all intro textbooks that they make very serious errors in the treatment of the energetics of extended systems (as opposed to point systems). The spatial integral of the net force along the path of the center of mass is equal to the change in the translational kinetic energy. This is the “work-energy theorem” and, being an integral of dp/dt = Fnet, is really a momentum equation in disguise. In fact, the integral of Fnet,x*dx_cm is equal to the change in 0.5Mv_cm,x^2, and similarly for y and z (and the sum of these is the work-energy theorem), and the validity of all three of these integrals shows vividly that this is all about momentum, not energy. Most textbooks do not clearly distinguish between the work-energy theorem and the true Energy Principle (aka First Law of Thermodynamics).

    Some years ago John Jewett published a very nice sequence of five articles in The Physics Teacher about these issues, and Serway and Jewett is one of the intro textbooks that gets it right. On my home page www4.ncsu.edu/~basherwo are a couple of AJP articles from the early 1980s that offer a number of striking examples of these matters, including an explanation of a seeming paradox in work and energy in sliding friction that arises from mistakenly thinking of a sliding block as a single particle whereas it actually deforms at the interface with the table and has a large number of internal degrees of freedom, associated with the motion of the many atoms in the block.

    • Andy "SuperFly" Rundquist says:

      Thanks for the reference. I got caught up in rederiving everything for myself and didn’t do a good search for other sources.

  2. I’m not absolutely positive that it’s the same issue you uncovered, but here’s something similar. In earlier editions of our Matter & Interactions textbook we had an incorrect derivation of the motional emf associated with dragging a bar at constant speed v along conducting rails a distance L apart, in the presence of a magnetic field B. I think it’s wrong in many intro textbooks. In the new 4th edition we do it right.

    Electrons move through the moving bar, following a path that is a slanting line, composed of components of the motion parallel to and perpendicular to the rails. The magnetic force on the moving electrons does no work since in F = qvxB the force is always perpendicular to the displacement. However, if you mathematically express this zero work as the work done by the component of magnetic force parallel to the rails plus the work done by the component of magnetic force perpendicular to the rails, you find that the first piece, which is negative, has the same magnitude as the work you do (and the bar doesn’t gain kinetic energy), and the second piece, which is positive, is associated with the motional emf vBL that drives a resistor connected across the rails. The total work done by the magnetic force is indeed zero.

    • Andy "SuperFly" Rundquist says:

      I’ve struggled when I’ve taught that, even when I use a text that gets it right. The students often say “but it’s just easier to think of it as the b-field doing work as I can get the answer faster” and I’m trying to get them to appreciate how subtle the whole process is. Another feature of that same material is how often motional emf is the cause of electron motion and not and induced electric field, even though mathematically you get the same result.

      For this material, I think the students would want to say that the gravitational potential energy gets converted to kinetic energy, and then separately say that some of that would be rotational and some translational. I like this because you can figure out how much of each (though admittedly it would be hard to predict without doing the full blown integration first.)

      • What we do in our textbook is treat motional emf in the chapter on magnetic force, a couple of chapters before getting to Faraday’s law, where we emphasize that where there is a time-varying magnetic field there is a curly electric field (after which we introduce how to calculate the resulting emf from the rate of change of the flux). This separation in time and space of the two phenomena then lets us talk about the differences in the two phenomena, that even if there is a changing flux, Faraday’s law does not apply if there is no time-varying magnetic field; instead it’s a situation of motional emf.

  3. bretbenesh says:

    I blame Mathematica for the mistake in your last post. You are blameless.

  4. Pingback: Double pendulum roller coaster | SuperFly Physics

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