# Feynman exercises bead parabola accelerometer

1. Aug 24, 2011

### rude man

In the subject accelerometer (see link) the answer given is a = kgx. I get a = 2kgx. I'm wondering if anyone cares to discuss this problem. The answerer used an energy argument that I'm not sure I followed, whereas I used a simpler force-balance method.

F = force on bead by parabola and is normal to the tangent at the quiescent point
(F is in 1st quadrant on the exercise picture)

θ = angle F makes with the x axis. 0 < θ < π/2
Fcosθ = ma
Fsinθ = mg

So F = mg/sinθ
a = Fcosθ/m = (g/sinθ)cosθ = g/tanθ
But θ = π - π/2 - arc tan(dy/dx) = π/2 - arc tan(dy/dx)
where dy/dx = 2kx
Applying tan(c-d) = [tan(c) - tan(d)]/[1 + tan(c)tan(d)] I get
tanθ = 1/2kx
giving a = g/tanθ = 2gkx.

http://www.feynmanlectures.info/

Select "Exercises" and then "Bead parabola accelerometer".

Last edited: Aug 24, 2011
2. Aug 24, 2011

### K^2

a = -2kgx is the correct answer.

Think about it this way, at the bead's position, vector a+g is perpendicular to the curve (to be balanced by normal force). That gives you a/g = -dy/dx = -2kx.

Edit: Energy argument can be used, but energy is NOT CONSERVED when you change acceleration. There is an external force involved.

Instead, find the minimum of the energy. E = mgy - max = mgkx² + max. The energy is minimized when dE/dx = 2mgkx + ma = 0. That gives you the same solution, a = -2gkx.

Last edited: Aug 24, 2011
3. Aug 25, 2011

### rude man

Hooray! I feel pretty good getting your corroboration. That means the answer in the Feynman exercise book is wrong. Can't wait to tell them. Feynman would turn over in his grave!

I'm glad you agreed that energy conservation was not the way to go.

4. Aug 25, 2011

### codelieb

Unfortunately, the http://www.feynmanlectures.info/exercises/bead_parabola_accelerometer.html" [Broken] confuses a lot of people. Many, like you, write to me about it, convinced that the answer to the problem is "a = -2kgx" instead of "a = -kgxa," and so far, I have convinced every single one of them that they are wrong. I am glad to have the opportunity now to straighten this out in a public forum so that, hopefully, I will get fewer emails about it in the future!

Let's begin with a description of the behavior of the system, as seen in the accelerated reference from of the wire: The bead starts at rest at the origin. When the acceleration begins, the bead begins to climb up the wire. It reaches equilibrium at x = -a/2kg, where it also achieves maximal velocity. It continues to climb up the wire until it reaches x = -a/kg where it has zero velocity. It then reverses its direction, and descends down the wire until it reaches x = 0 with zero velocity. The bead oscillates harmonically between x=0 and x = -a/kg, as long as the acceleration continues. (As states in the problem, the bead slides on the wire without friction, so there is no damping of the oscillation.)

The problem asks you to "Find the relationship between the acceleration a of the wire and the bead’s maximum horizontal displacement x relative to the wire." The maximum horizontal displacement occurs when a = -kgx, not when the bead is in equilibrium which occurs at a = -2kgx).

If you follow the above link to the problem, you can find a correct solution by Sukumar Chandra posted with it. However, I would like to offer another solution here, which I find conceptually, and also mathematically, simpler: Using Einstein's principle of equivalence, we can see that the accelerated system is equivalent to one at rest in a gravitational field whose acceleration is the sum of the vectors -a and g. You can think of the parabola being upright (as shown in the figure) and gravity being tilted (clockwise), or you can think of gravity being vertical and the plane of the parabola being tilted, i.e. rotated about the origin (counterclockwise), by the angle arctan[a/g]. Then the answer to the problem becomes obvious! It's clear that the equilibrium position occurs where gravity is perpendicular to the wire, or in other words, where the derivative of the parabola equals -a/g. So for equilibrium you get y' = 2kx = -a/g, or a = -2kgx (which is the first calculation made by K^2). However, as described above, the bead does not stop at its equilibrium position - it keeps going. Obviously, it will reach its extreme position, with zero velocity, when it's gravitational potential is the same as when it started (think of a pendulum), and that must occur on a line that intersects the origin, where the bead started, and which is perpendicular to gravity, i.e. the line y = (-a/g)x. So for the extreme position you get y = kx^2 = (-a/g)x, or a = -kgx.

Mike Gottlieb
www.feynmanlectures.info
www.basicfeynman.com

Last edited by a moderator: May 5, 2017
5. Aug 25, 2011

### K^2

In that case, the maximum achieved height depends on the acceleration time profile, a(t). The problem does not state the way the acceleration is "turned on". Only that the beads rests at the bottom of parabola when a=0. Suppose, then, that instead of having a(t)=θ(t) (Heaviside step function.) like you assumed, we choose some other profile for a(t). Lets see what happens if instead of turning on acceleration instantly, we build up to it. After all, that's far more realistic.

In rest-frame of the parabola, we can write down the Lagrangian.

$$L = \frac{1}{2}m(\dot{x}^2+\dot{y}^2) - mgy - max$$

With constraint f(x,y)=y-kx²=0

$$\frac{\partial L}{\partial x} = -ma$$
$$\frac{\partial L}{\partial y} = -mg$$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = m\ddot{x}$$
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} = m\ddot{y}$$
$$\frac{\partial f}{\partial x} = -2kx$$
$$\frac{\partial f}{\partial y} = 1$$

That gives us 2 equations in addition to equation of constraint for the three unknowns.

$$m\ddot{x} + ma + 2kx\lambda = 0$$
$$m\ddot{y} + mg - \lambda = 0$$

Differentiating the constraint, we obtain.

$$\dot{y} = 2kx\dot{x}$$
$$\ddot{y} = 2k\dot{x}^2 + 2kx\ddot{x}$$

That results in the system of equations.

$$m\ddot{x} + ma + 2kx\lambda = 0$$
$$2kmx\ddot{x} + 2km\dot{x}^2 + mg = \lambda$$

And finally, just one equation.

$$4k^2x^2m\ddot{x} + m\ddot{x} + 4k^2xm\dot{x}^2 + 2kxmg + ma = 0$$

Not sure if there is a clean analytic solution, but lets just take a look at some numeric results. For sake of simplicity, I took m=1kg, k=1, g=10m/s².

Plots are attached.

Figure 1. a(t)=5*θ(t) m/s². This is the test case to compare to your solution. It does, in fact, show turning point at x=-a/(gk) = -0.5m. Fantastic. Moving on.

Figure 2. a(t)=(5*t) m/s² for t<1s and 5m/s² for t>=1s. The turning point is now around -0.35m. The bead does continue to oscillate, but it never gets to the -a/(gk) mark. So the solution is wrong for this case. Can we do better?

Figure 3. a(t)=(5-5 Exp(-t/5)) m/s². Lets see what happens if we turn up acceleration gradually. What do you know? The maximum value attained by x is just over -a/(2gk). That really should not surprise you. If you increase acceleration gradually, the bead is always near equilibrium, and so does not attain high speeds needed to overshoot the equilibrium.

Why does conservation of energy method fail for all but the first case? Because energy is not conserved in an accelerated frame of reference in general. Yes, under constant acceleration, we can pretend that there is a potential function. But that is not generally true otherwise. Use of energy conservation for this problem is completely misleading.

So what's the conclusion? The way the problem is stated, it has no definite solution. It simply makes no sense to ask how high the bead will climb without being very specific about how the acceleration is turned on. It makes sense to ask for an equilibrium position, but nothing else. Especially when you are talking about using the setup as an accelerometer, and in any realistic situation there will be friction, and equilibrium position will be used.

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6. Aug 25, 2011

### codelieb

K^2,

In my opinion, the only thing relevant in your reply is summarized in your Figure 1, and your notes to same, which validate my answer (and invalidate yours and rude man's).

In this problem, the acceleration starts instantaneously, which is just as much to say that a force is applied to the system starting at some time t and continues to be applied to it for a while. I see nothing particularly unphysical about that. It is the same kind of approximation that is used tacitly in solving all elementary physics problems of this sort. In fact, I would say it is a lot less unphysical than assuming that there is no friction between the wire and the bead, which you don't seem to mind - at least you did not include friction in your analysis.

As far as I can tell, from what you posted in this thread before I chimed in, your mistake had absolutely nothing to do with the problem not specifying how the acceleration starts. To force your answer out of the problem you had to make a(t)= 5-5 Exp(-t/5), which is not exactly a natural assumption, is it? (If you don't like instantaneous accelerations, then wouldn't it be more natural to assume that the bead is not released until after the acceleration is constant? Or do you rule out the possibility of constant acceleration as well?)

It's an easy problem, with a simple solution, and you and rude man got it wrong the exact same way lots of other people do, by confusing the equilibrium position and the extreme position of the bead. In fact that type of error is commonly made by elementary physics students, as documented in the works of Hestenes and others who study elementary physics education.

Your response gives the impression of someone trying to hide their embarrassment at making a rudimentary gaff with a smokescreen of irrelevant mathematical analysis, numerical calculations and graphs. To that I say: Sour grapes, sir! Now I agree with rude man that Feynman might be turning over in his grave... over Ph.D's in physics who can not admit when they make a simple mistake, and instead try to cover it up with a bunch of highfalutin hooey!

Mike
"The first principle is that you must not fool yourself--and you are
the easiest person to fool. So you have to be very careful about
that. After you've not fooled yourself, it's easy not to fool other
scientists. You just have to be honest in a conventional way after
that."

Richard Feynman

Last edited: Aug 25, 2011
7. Aug 25, 2011

### rude man

K^2's point is perfectly valid: the time nature of the input function was not defined in the problem, and given an arbitrarily low da/dt there would be no oscillations at all.

When you call a device an "accelerometer" it makes sense that the device could be used in some way as a real device, whereas looking at a time-varying sine wave would give one a headache! Also, I feel Chandra was remiss in not pointing out the oscillaltory nature of the solution, even though the question did not explicitly ask for it. Aren't we supposed to learn something from these exercises and their "correct" solutions?

I also would comment that a more civil response from a major physics board's (not this one) moderator would have been - er - nice.

8. Aug 25, 2011

### codelieb

Dear rude man,

At the moment I am involved in a number of book projects, including printed, ebook, and multimedia editions of The Feynman Lectures on Physics (FLP), the second edition of Feynman's Tips on Physics, and a new exercise book for FLP, with about 1000 exercises covering all material in FLP Vols. I-III. So, I am busy. Therefore I will preface my response to you by saying it is my last comment on this subject.

You may be right about that in some sense, but there are many real devices in which a measurement is derived from the amplitude of an oscillating signal, in a similar manner as this one might be used, so I do not feel that this problem, as an example for elementary physics students (who may one day have to build such instruments), is in any way inappropriate.

Sukumar Chandra is a teacher, but I do not think he is YOUR teacher! He provided a correct solution to the problem that is both simple and elegant, and, knowing him, as I do (because he has contributed quite a number of problems and solutions to The Feynman Lectures Website, and we have corresponded about physics) I would say only that, in the field of classical mechanics, at least, his physical insight and perception are so clear, that the motion of the bead was obvious to him, and his only fault in not mentioning it in his solution to this problem was in assuming that it was equally obvious to the rest of us!

Mike

Last edited: Aug 25, 2011
9. Aug 25, 2011

### rude man

"Default Input"

I must remember that concept. Next time I get a wrong answer I will simply restate the problem with my custom-made "default input". What inspiration!

10. Aug 25, 2011

### codelieb

That would be an inspiration derived from K^2, who restated the problem with a(t)= 5-5 Exp(-t/5) to fit (somewhat) his and your answer, based on your initial solutions, posted in this thread, which, in fact, assume that the acceleration is constant, and which apparently confuse the equilibrium position of the bead with its extremal position. Brilliant! I wish you the best of luck in your academic career.

Last edited: Aug 25, 2011
11. Aug 25, 2011

### KingNothing

Summary of thread: K^2 and rude man point out a completely valid flaw in the way a question is posed, and codelieb defends it anyway rather than admitting the flaw.

12. Aug 25, 2011

### rude man

Codelieb: (1) what happened to your promise of no further ripostes?
(2) My input was not K^2's exponential, athough it was a perfectly resonable example. If you read mine, it said "arbitrarily low da/dt." I should even have said, ANY FINITE da/dt would make the problem as stated incomplete and unsolvable.

Oh, that's right, I forgot the "default" input of infinite jerk! (Hmmm ... no, I won't say it).

Last edited: Aug 25, 2011
13. Aug 25, 2011

### KingNothing

What is or is not the correct "default" input is completely irrelevant. A well-formed problem statement doesn't have any implicit parameters; rather, they are all explicit.

14. Aug 25, 2011

### Dickfore

The point is that when the system is accelerated, there is extra potential energy due to the inertial force:
$$U = -m g y - m a x$$
The initial and final kinetic and potential energies are:
$$T_{i} = 0, U_{i} = 0$$
$$T_{f} = 0, U_{f} = -m g y_{0} - m a x_{0} = -m \left(a x_{0} + g k x^{0}\right)$$
(the final kinetic energy must be zero if this is the maximum displacement. If it is not zero, then the bead will keep on moving).
According to the Law of Conservation of Energy ($T_{i} + U_{i} = T_{f} + U_{f}$), we must have:
$$U_{f} = 0 \Rightarrow a x_{0} + g k x_{0} = 0 \Rightarrow \begin{array}{l} (x_{0})_{1} = 0 \\ (x_{0})_{2} = -\frac{a}{k g} \end{array}$$
The first solution corresponds to the intial poin. The second solution is the physical one and we have:
$$x_{\mathrm{max}} = -\frac{a}{g k} \Rightarrow a = -k g x_{\mathrm{max}}$$

Using the condition that the normal force, the inertial force and the force of gravity are in equlibrium is not physical here as we are required the maximum displacement, and not the equilibrium position. Once the bead reaches this position, it will slide down to the origin and keep oscillating.

Last edited: Aug 25, 2011
15. Aug 25, 2011

### K^2

You are missing the point. I admit that I missed the bit that asks for maximum position, found equilibrium instead, and so my response was, in fact, wrong. The point is that the problem is malformed, and it would not take a lot to fix it. You have two options.

1) Ask for equilibrium instead of maximum position. That's a more natural problem for something titled "accelerometer," but you seem to have some problem with it. Fine.

2) Explicitly state that acceleration goes from a=0 to a=a0 at t=0. Then the question of maximum bead displacement becomes completely valid. That's all you need to do to fix the problem.

Just as a note, I wasn't trying to "fit" the acceleration to my solution. If I was, you better believe it wouldn't "somewhat fit". The problem of optimal control for this Lagrangian is trivially solved numerically. If I was trying to prove validity of my earlier reply, I could easily give you a(t) that takes the bead from x=0 to x=xeq in some time t and keeps it there from then on. That was not what I was trying to do.

16. Aug 26, 2011

### codelieb

Okay, you whiners . I have reworded the posted problem for you. It now reads:

An accelerometer is made of a piece of wire in the shape of a parabola y = kx^2 with a bead on it that can slide without friction, as shown in the drawing. The bead is initially attached to the wire at the lowest point of the parabola. The wire is accelerated with a constant acceleration parallel to the x-axis, and then the bead is released. Find the relationship between the acceleration a of the wire and the bead’s maximum horizontal displacement x relative to the wire.

Had it been worded this way in the first place, rude man and K^2 would have produced the same wrong answer, for the same reasons, only then they could not pull the same red herring out of their hat, and instead they'd have to blame the problem for neglecting to mention whether or not the system was in a vacuum, or whether the wire deformed under acceleration, or whether the bead was to be considered a point mass or not, or whether we should take into account the fact that gravity is not constant across the entire wire, or whether there were electromagnetic fields that might affect the wire, ....

Apparently NO physics problem is "well-formed" because we can always invent "implicit parameters" that are unstated. This is especially handy for avoiding embarrassment by distracting people after you have given the wrong answer to an elementary physics problem.

17. Aug 26, 2011

### rude man

Codelieb - I don't like you putting words in my mouth. Had you made it clear to begin with that a step acceleration was the input, anyone would have been alerted to a time-variant solution.

Your wording is still ridiculous. Are we supposed to imagine the thing moving with acceleration and then some hardy little gremlin sitting on the accel releases the bead? Why don't you just swallow your Ueberego and indicate a step input of acceleration?

18. Aug 26, 2011

### K^2

Considering the fact that I explained in the very first post what the problem with using energy conservation is, and made it very clear that I'm looking for equilibrium solution, you are way out of line. Yes, I misread the statement of the problem. Wooo, I'm so embarrassed. You, on the other hand, apparently weren't aware of how to work with conservation laws in an accelerated frame. Which do you think is a real cause to be embarrassed?

19. Aug 26, 2011

### codelieb

This really HAS to be my final comment in this thread.

As for my own understanding, or lack thereof, I am not exactly sure what you are criticizing. The solution posted with the problem on the Feynman Lectures website isn't mine - it's Sukumar Chandra's. The only solution from me you have seen is the one I posted here. In that solution, I found a way of looking at the problem that is intuitively appealing, mathematically correct, and makes the behavior of the bead and the answer to the problem quite obvious. That's the kind of solution we like at Caltech: physically appealing, simple, direct, to the point, and, of course, correct.

Kudos again to Sukumar Chandra, with his excellent physical intuition: the man did not have to solve a Lagrangian or do a numerical simulation to see how the bead moves: he saw it right in his head! That is the kind of ability we like to teach at Caltech.

I am going to end my comments with another problem, for your amusement, one that is not posted on The Feynman Lectures Website. This one is from the original Feynman Lectures course, and, amongst the simpler problems, it is one of my favorites. Here it is:
The pivot point of a simple pendulum having a natural period of 1.00 second is moved laterally in a sinusoidal motion with an amplitude 1.00 cm and period 1.10 seconds. With what amplitude should the pendulum bob swing after a steady motion is attained?​

The thing that makes this problem interesting is this: you have to solve it without using any calculus or differential equations (or integral equations or difference equations, etc), and without iterative numerical methods. You can use only algebra, geometry, trigonometry, dimensional analysis, Newtonian mechanics, and, of course, your physical intuition! The answer does not have to be exact, but it should be at least a very close approximation.

(I didn't make this problem up, so don't blame me if you don't like it. The kids at Caltech have been solving it for decades.)

Good luck!

Last edited: Aug 26, 2011
20. Aug 26, 2011

### Dickfore

rude man's and K^2's answer is wrong. codelib's gave the correct answer. I am reporting this thread for intentional offensive behavior.

21. Aug 26, 2011