Test your knowledge of inertial forces

[Studiot]

Well this is the danger of being entirely theoretical.

If anyone did a calculation they would find that there is no normal force.

 

[A.T.]

If anyone did a calculation they would find that there is no normal force.

I guessed so, because such a very specific value for omega was given. But was too lazy to check.

 

[Dale]

Well this is the danger of being entirely theoretical.

If anyone did a calculation they would find that there is no normal force.

First, that is not true in general and would depend on the radius, which was not given. Second, since you didn't give the radius a calculation was not possible.

 

[Studiot]

Apologies, the string is 2m long and the mass is 2kg.

 

[Ken G]

Well this is the danger of being entirely theoretical.

If anyone did a calculation they would find that there is no normal force.

Actually, there is no possible way to set up that experiment where the normal force will be precisely zero (even if all the necessary information is included), because of one more force on the mass-- air resistance. That will require a small normal force to allow for the balance between air resistance and static friction. All this means is, if you try to spin the cone too fast, no steady-state solution will be possible, so any steady state requires a nonzero normal force. The "zero normal force steady-state solution" is an idealized situation that is impossible to achieve in practice if the motion of the mass will be due to the spinning cone.

But even that doesn't matter-- the prescription for calculating constraint forces allows them to be zero, it is not any kind of bother when they come out that way in idealized problems that do contain all the required information.

The real question is: what does this have to do with erroneous statements about what inertial forces are?

 

[Dale]

Apologies, the string is 2m long and the mass is 2kg.

With those values the normal force is actually negative 9.9 N. I.e. 9.9 N inwards. I assume this is unintended.

 

[Ken G]

Yes, I get that omega^2 * L can never exceed g / cos(theta) with that kind of setup, where L is the length of the string. Here 1/cos(theta) = 2, so omega^2 * L must not exceed 19.6 m/s². Here it is 22.8 m/s².

 

[Studiot]

Mentor note:
This post, and several that follow, were made in another thread. There were moved to this thread because this thread is where discussions of this problem belong.[/color]

Why don't you write again the complete question in one post, give us your two solutions (inertial frame, co-rotating frame with fictious forces) and show us how the later "will definitely get you the wrong answer"?

Why?

Well first point is that both D'Alembert's and Newton's methods will lead to an invalid conclusion if inappropriately applied.

The question was specifically constructed (not by me) to demonstrate the particular point that it is posible to step outside the obvious assumptions and thereby obtain the 'wrong' answer.

Unfortunately instead of having an adult discussion about Physics I felt I was being personally attacked in the other thread.

The relevant equation, by either method, is

R = 4.93 - \frac{{\sqrt 3 }}{4}{\omega ^2}

I make the ω at which R becomes zero the 3.374 I posted.

Various values for R, including negative ones were offered, but none if I recall correctly equal to zero.

Whirling beyond this speed simply causes enough tension in the string to lift the mass of the cone.
Of course R remains zero or is non existant. It surely is never negative.

 

[Dale]

Whirling beyond this speed simply causes enough tension in the string to lift the mass of the cone.
Of course R remains zero or is non existant. It surely is never negative.

The problem stated that the mass was attached to the cone. Therefore the mass could not lift off of the cone and the normal force could become negative. Of course, the problem did not specify the method of attachment so in all likelyhood the problem was either inconsistent or overdetermined.

Zero was not correct.

 

[A.T.]

In this recent thread I offered a problem (posts 53 and 54) where D'Alembert's method (fictious forces) will definitely get you the wrong answer.
https://www.physicsforums.com/showthread.php?t=531470

Well first point is that both D'Alembert's and Newton's methods will lead to an invalid conclusion if inappropriately applied.

This triviality is quite a backpedaling compared to "D'Alembert's method will definitely get you the wrong answer". How is that problem relevant to this thread, which about comparing the two approaches?

It surely is never negative.

A negative numerical result for a contact force which is defined as compressive is perfectly valid, if interpreted correctly. It means that the bodies will separate if not attached to each other, so the initial situation stated in the question will change. If attached to each other, a negative compressive force gives the tension in the attachment.

 

[Dale]

If attached to each other, a negative compressive force gives the tension in the attachment.

The question referred to here explicitly stated that it was attached.

 

[Studiot]

Originally Posted by Studiot
Well first point is that both D'Alembert's and Newton's methods will lead to an invalid conclusion if inappropriately applied.

This triviality is quite a backpedaling compared to "D'Alembert's method will definitely get you the wrong answer". How is that problem relevant to this thread, which about comparing the two approaches?

I am sorry you consider this a trivial point, it makes proper considered discussion difficult.

I am trying to show Harrylin that it is possible to find problems where misapplication of either method will lead to the wrong answer, but not in the obvious way I think he originally expected.

The example I presented has been used by the University of Newcastle to undergraduate and postgraduate students to bring home exactly this point. It's veracity has been well tried and tested by many experts.

The point is that although both methods can be applied to any mechanical system model, more than one model may be needed to cover the full range of the real system's operation.

That is the case in my example.

It is also the reason why both you and Dale have yet to obtain the correct solution.

Further the issue is not necessarily one of frames of reference, which posters here seem to be concentrating on.

Originally Posted by Studiot
It surely is never negative.

A negative numerical result for a contact force which is defined as compressive is perfectly valid, if interpreted correctly. It means that the bodies will separate if not attached to each other, so the initial situation stated in the question will change. If attached to each other, a negative compressive force gives the tension in the attachment.

The question referred to here explicitly stated that it was attached.

The mass is not attached to the cone by adhesive which would be needed to obtain a negative reaction force.
It is attached by the string which would be unecessary if adhesive were to be used.

 

[harrylin]

[..] both D'Alembert's and Newton's methods will lead to an invalid conclusion if inappropriately applied.

The question was specifically constructed (not by me) to demonstrate the particular point that it is posible to step outside the obvious assumptions and thereby obtain the 'wrong' answer.

[..] The relevant equation, by either method, is

R = 4.93 - \frac{{\sqrt 3 }}{4}{\omega ^2}

I make the ω at which R becomes zero the 3.374 I posted.

Various values for R, including negative ones were offered, but none if I recall correctly equal to zero.

Whirling beyond this speed simply causes enough tension in the string to lift the mass of the cone.
Of course R remains zero or is non existant. It surely is never negative.

That looks interesting, but I don't know that topic and the calculations are missing... Do you by any chance have a reference?

 

[Studiot]

At last someone with some good manners and an open mind.

 

[A.T.]

I am trying to show Harrylin that it is possible to find problems where misapplication of either method will lead to the wrong answer,

Misapplication of methods will give the wrong answer for any problem. Unless you make two errors that cancel each other.

 

[D H]

It appears from your diagram that the mass is suspended from a string, and lies against the cone. If so, there are three real forces on the mass-- the tension from the string (which points along the string), gravity (which is downward), and the normal force from the cone (which is perpendicular to the cone surface).

You are talking about a tetherball physics problem. That is not the correct setup for this problem.

The cone would be a meaningless complication were this the correct setup; you might as well just have a vertical pole. The angular displacement of the string from the vertical for tetherball rotating at 3.374 rad/sec and at a distance of 2 meters from the pivot point would be 64.58 degrees assuming g=9.80665 m/s². However, the surface of the cone is displaced only 30 degrees from the vertical. What is the correct setup? Studiot said, emphasis mine,

A (point) mass m is attached to a smooth cone of 60 angle.

Here "attached" essentially means "bolted to". Attached where? He didn't say. He later did say "2 m", but measured with respect to what, and along what line?

Studiot, when you are specifying a problem, you need to be very specific about the setup and about what quantities are to be found. Along what line is that "2 m" distance measured? Is gravity in the picture? Do you want the forces acting on the point mass by the attachment mechanism resolved into normal and tangential components? You didn't specify any of these. We can't read your mind.

 

[Dale]

The mass is not attached to the cone by adhesive which would be needed to obtain a negative reaction force.
It is attached by the string which would be unecessary if adhesive were to be used.

None of that was specified in the problem. This is Physics Forum, not Psychics Forum. You cannot post an inconsistent or incomplete problem and then claim failure of some analysis technique by the resulting "mistakes". The question itself was bad.

Also, you don't know if I did the analysis in a rotating frame or in an inertial frame. As it happens I did the analysis in an inertial frame. Shall we therefore reject inertial frames?

 

[harrylin]

[Dynamics 4.2]

I suppose that that is of the same book that you cited before?

About your calculation example: [moved to thread where that comment belongs!]

I agree with the terminology there, as I see no reasonable alternative to calling the reaction force due to inertia "inertial force" - as it surely also has been called in physics from the start. NewSpeak is evil. But since that expression has been hijacked to mean something subtly different, it is necessary to clarify when the implied meaning is the proper, original meaning. That's all that I'll say about that. The day will come that we'll have to clarify when we mean a small animal with "mouse".

Harald

 

[Dale]

At last someone with some good manners and an open mind.

I would very much like to see the previous page with the problem description. I suspect it is much more complete than your rendition.

 

[D H]

At last someone with some good manners and an open mind.

Please.

The attachment in that post does not contain the problem setup. However, there enough information is supplied so as to reverse engineer the problem. The values 4.93 and 4.93√3 in equation (iii) respectively represent mg·sin(θ) and mg·cos(θ), where θ is 30 degrees. Presumably these values have units of Newtons. From this, and using g=9.80665 m/s², one can deduce that the mass is 1.00544 kg. That is a goofy value for a textbook physics problem, but it is easily explainable by looking at the value 4.93. I'm going to assume that this is a typo. A value of 4.903 yields a nice value of 1 kg for m.

What about r, and what does it represent? The values ω²/4 and √3ω²/4 in equation (iii) respectively represent mrω²sin(θ) and mrω²cos(θ). From this, and using the previously reverse engineered value of m=1 kg, one arrives at r=1/2 meter. (The value is something different if the value of 4.93 was not a typo.) Finally, from the free body diagram, it is obvious that r is the distance between the axis of rotation and the point mass. For a point mass attached by a string to the apex of a cone with a half angle of 30 degrees, one can deduce that the length of the string is l = 2r = 1 meter.

Apologies, the string is 2m long and the mass is 2kg.

Try again. I don't want you to show the pages from that book that describe the setup; we're bordering on copyright infringement in this thread as it is. You can describe the setup for us. Take great care in doing so; you did not describe the problem well before.

 

[Ken G]

You are talking about a tetherball physics problem. That is not the correct setup for this problem.

The cone would be a meaningless complication were this the correct setup; you might as well just have a vertical pole.

That's my point, it is a tetherball problem only if the angular velocity is above the limit discussed earlier, and it appears to be in this case (though the constraints have emerged somewhat fitfully). That is inconsistent with the description that the cone is providing the angular velocity. So that's the claim I'm making-- the problem is internally inconsistent.

But more to the point, the numbers could easily be fixed up to reduce the normal force to zero and be only at the boundary of the tetherball problem. But in any real experiment where the cone is providing the velocity (I presumed by static friction, because otherwise you'd need no string), there would need to be some normal force to get that friction. So the angular velocity could never really reach the limit, only very close to it, especially once air resistance is thrown in. We're not disagreeing, the problem does not succeed as any kind of "trap" for anyone who understands inertia and forces.

 

[D H]

It is the tetherball problem. It just wasn't well specified.

 

[Studiot]

Oh dear J'adoube.

How very embarrassing



I took this problem at face value and simply posted the values provided. Normally I would have reworked the problem completely with different values.
It seems there is an arithmetic error in the book

More in a moment, but I do not want to detract from the very important point of mechanics the authors made, since their analysis appears to be correct.

Following the problem statement I posted the length of the string as 2m and the mass of the particle as 2kg.

I agree with the author's analysis and have worked through the equations to their solution.

I can only obtain their solution equations for R and T if the length of the string (L) is 1 metre and the mass is 1 kilogram.

 

[Ken G]

What "important point of mechanics" are you talking about? It looks like a perfectly routine first-year problem to me, I see no profound messages here. Certainly nothing revealing about inertia-- you'll note that the critical angular velocity where the normal force goes to zero doesn't even depend on the mass of the object.

 

[Studiot]

you'll note that the critical angular velocity where the normal force goes to zero doesn't even depend on the mass of the object.

pardon?

 

[D H]

In Studiot's favor, this underlying message is a very important one, one that when ignored has resulted in loss of life, mission failure, and all other sorts of mayhem. It is one part of the hot button topic in the modeling and simulation world for the last five to ten years. That hot button topic is "simulation verification, validation, and accreditation" (google that exact phrase). A big part of this is "verification, validation, and accreditation of simulation models" (you can google that exact phrase, also).

Suppose that you have been assigned to develop a simulation of a Rube Goldberg device. Being a good lazy software engineer, you do not want to develop from scratch a physical model of each one of the components in this Rube Goldberg contraption. Suppose one part of this contraption involves a tethered ball on a rotating cone, and suppose that your corporate suite of physical models contains a very nice model of such a device. So you just plug this existing model into your simulation without checking to see if that model truly is applicable to the problem at hand.

Now if you had just read the fine documentation on this model you would have come across the assumptions and limitations section that clearly stated that this model is suitable for small rotation rates only. The authors of that package did not test whether the normal force was directed inward because such a situation could never arise in the original application of the model. With no way to validate that test, they intentionally elected not put the test into their model.

Suppose the cone in your tethered ball on a cone in your Rube Goldberg device can spin up to a high rotation rate. Your simulation does not cover this case so your sim does not show the damage that ensues when the cone spins out of control. Who is at fault? Well, you are, or whoever accredited this model for use in this new simulation. That freebie model should never have been used as-is.

 

[Ken G]

pardon?

Which part of that did you not understand?

 

[Ken G]

Who is at fault? Well, you are, or whoever accredited this model for use in this new simulation. That freebie model should never have been used as-is.

Even so, the simple analysis yields a normal force of zero, or negative, in the inappropriate situations. That's something the user should notice if they are serious about what they are doing. In other words, it doesn't require some deep appreciation for the mysteries of inertia, it just requires that someone has a clue, an interest in actually mastering their own craft rather than just faking their way through. The lesson is true, we all must constantly ask ourselves "does this make sense" at every stage of a calculation-- but that goes almost without saying for anyone who has done calculations and wants them to mean something.

 

[Dale]

Even so, the simple analysis yields a normal force of zero, or negative, in the inappropriate situations. That's something the user should notice if they are serious about what they are doing. In other words, it doesn't require some deep appreciation for the mysteries of inertia, it just requires that someone has a clue, an interest in actually mastering their own craft rather than just faking their way through. The lesson is true, we all must constantly ask ourselves "does this make sense" at every stage of a calculation-- but that goes almost without saying for anyone who has done calculations and wants them to mean something.

Although, if the ball is attached to the cone, as was stated, then a negative force is appropriate and makes sense.

 

[Ken G]

Although, if the ball is attached to the cone, as was stated, then a negative force is appropriate and makes sense.

Exactly. So one must always know what one is doing, but the forces come out what they would need to. It's much like with computer programming-- don't blame the computer when it does what it is asked to do, the user has to make sure they are posing the problem they think they are posing, so the "does this make sense" test must be applied often. (By the way, the problem is more interesting if the mass is hung from a string. It's hard to tell, the language used is very vague, but it looks like there is supposed to be a string, despite the use of the term "attached"-- it is attached by a string? If it hangs from a string, but comes to equilibrium against the surface of the cone via friction and the rate of the cone's rotation, then it is only "attached" to the side of the cone for the slower rotation-- for rotation past the critical limit, the mass will begin a very chaotic stick-slip kind of motion that would be very difficult to analyze.) Anyway, the problem was supposed to show us how to separate "real" forces from "inertial" ones, and that I would say is a complete red herring here, because no one needs to invoke inertial forces at all, though they may certainly choose to use that language for the ma term if they are clear about it. All the same, massless objects never have inertial forces, so the whole idea that there would be inertial forces on the massless string (which is where this all started) is clearly wrong.

 

[D H]

Even so, the simple analysis yields a normal force of zero, or negative, in the inappropriate situations. That's something the user should notice if they are serious about what they are doing. In other words, it doesn't require some deep appreciation for the mysteries of inertia, it just requires that someone has a clue, an interest in actually mastering their own craft rather than just faking their way through. The lesson is true, we all must constantly ask ourselves "does this make sense" at every stage of a calculation-- but that goes almost without saying for anyone who has done calculations and wants them to mean something.

While it might be simple to see the problem in this simple case, seeing the problem in a complex system is, well, complex. Seeing the problem that will lead to loss of life, mission failure, or some other catastrophe ahead of time is getting harder and harder ss systems become ever more complex. Failing to see the problem led to the crash of the initial flight of the Ariane 5, the loss of the Mars Climate Orbiter, the crash of the Mars Polar Lander, the failure of the DART (Demonstration of Autonomous Rendezvous Technology) mission, the loss of Milstar-2 F1, two Shuttle flights, and others, and that is just in the aerospace field.