Test your knowledge of inertial forces

[D H]

Hi Studiot, my understanding agrees with Doc Al, A.T., and Ken G's understanding. Inertial forces are also known as fictitious forces. They are frame-variant and only arise in non-inertial reference frames, they are not part of any 3rd law interaction, and they are always proportional to the mass.

Add me to that list. What you are calling an "inertial force" in this thread is not what almost everyone else means by the term "inertial force".

There is a very good reason to use consistent terminology, Studiot. Inconsistent terminology leads to unnecessary arguments that hopefully terminate with an Emily Litella "Oh, that's very different ... never mind" moment. Consistent terminology avoids all that.

 

[Doc Al]

So where did it come from? and What do you call it?

Why give it a name? A pushes on B, thus B pushes back on A. Newton's 3rd law.

Prof DH explicitly called it the "inertia reaction"

I was trying to show how this might arise and I think he is correct although I also think he left out a step or two.

I would be really helpful if folks would stop trying to 'prove me wrong' and just work through the logic with me - I am perfectly happy to say you are correct if it turns out that way, but so far it has not.

This is not a question of 'logic', but of using the standard meaning of terms. Prof DH wants to use his own personal terminology to refer to the reaction force as an 'inertial force', because it's 'due to inertia'. Fine. But that's not standard usage. (At least not in any standard classical mechanics book that I own.)

For the standard meaning of 'inertial force', read the first couple of paragraphs here (even Wiki gets it right): http://en.wikipedia.org/wiki/Fictitious_force"

 

[Doc Al]

And do you agree that the two extracts you quoted were speaking of entirely different things? Yes or no?

 

[Studiot]

Doc Al

because it's 'due to inertia'.

Was that a response to my question

Where did it come from?

If so does that mean you agree it (ultimately) comes from the inertia of the the reacting object in the cases shown.

Obviously if the reacting object was restrained then there would still be a third law reaction, but its ultimate source would be different (non inertial).

Do you not think this to be significant?

 

[Studiot]

And do you agree that the two extracts you quoted were speaking of entirely different things? Yes or no?

I have already said that.

What is the point of identical examples?

They are of course different in the fact that in one case we are considering the effect on object A and in the other case the effect on object B.

That does not mean that we cannot bring them to a common comparison by considering the effect on object A in both cases.

 

[D H]

Prof DH explicitly called it the "inertia reaction"

Just to clarify, that "Prof DH" is not me. Studiot is talking about Jacob Pieter Den Hartog, a professor of mechanical engineering. The book Studiot is referencing is "Mechanical Vibrations", copyright 1934, 1940, 1947, 1956, 1985. The 1985 edition is a Dover Books republication of the 1956 edition. So old, outdated terminology is most certainly a possibility here, as is a discrepancy between the terminology used in mechanical engineering versus that used in physics.

Studiot, the multiple pages of argument that comprise the bulk of this thread result from very contradictory meanings of the term "inertial force" as used in that book and the term "inertial force" as used in physics. Since the name of this site is www.physicsforums.com[/url] rather than [PLAIN]www.outdatedconceptsinmechanicalengineering.com , I would venture that the physics definition wins. At least at this site.

 

[Dale]

Studiot, I do not agree that 3rd law pairs come from inertia. Third law pairs exist even in static situations where there is no acceleration or between idealized massless objects. In either case it is hard to see the relevance of inertia. I have a hard time justifying calling a 3rd law pair an inertial force on that grounds.

I would stick with the standard usage of the term "inertial forces". It avoids confusion, and there doesn't seem to be a good reason to go to the non-standard usage.

 

[Studiot]

Good evening Dale,

I think underlying this whole discussion and also the difficulty many students struggle with is the question of how can an object apply a force at all?

The whole point of matter possessing properties is for those properties to be deployed when suitable conditions prevail.

Remember that this is the classical physics thread so I do not expect to discuss exchange particles or spacetime curvature here.

Consider the block in my example 2.

It suffers a rightward force from a push rod.

Situation 1)
The block is restrained against a rigid foundation to its right so does not move.
I can measure the reaction force on the push rod by means of a spring scale. I can see why the block can do this - the other side of the block is pushing against the foundation.

Situation 2)
The block is restrained against a spring foundation to its right so does move.
I can measure the reaction force on the push rod by means of a spring scale. I can see why the block can do this - the other side of the block is pushing against the foundation.
However the force measured on the spring scale to right (the spring foundation) is not the same as the force impressed by the push rod.

Situation 3)
The block has no restraint to its right so does move (accelerate) indefinitely so long as the rod remains in contact.
I can measure the reaction force on the push rod by means of a spring scale.

In all three situations the reaction force measured by my spring scale is the same, yet the source of this reaction force is clearly different.

 

[D H]

In all three situations the reaction force measured by my spring scale is the same, yet the source of this reaction force is clearly different.

The source of the reaction force is clearly the same in all three scenarios. It is the normal force exerted by the block at the point where your push rod touches the block.

 

[Dale]

In all three situations the reaction force measured by my spring scale is the same, yet the source of this reaction force is clearly different.

I don't really know what it means for something to be a force's source. I know what it means for a force to be a certain kind, like gravitational, or frictional, or mechanical, is that what you mean by source? In that case, the source is the same in all three situations, the mechanical compression of the structures.

 

[Ken G]

I don't quite understand what you mean by the question however.

Change?

Change from what? Before and after the cut?,

Yes exactly, change before and after the engine is cut off.

You have already stated there was no inertial force before the cut, but there is after so there is a change there.

I asked for any changes in forces on the spring, we agree there are changes in the forces on the rocket. This is a good device for understanding the forces on the spring both before and after the cut, to look and see if there are any changes.

 

[Studiot]

Here is a new example for you experts to chew over.

A (point) mass m is attached to a smooth cone of 60 angle. The cone rotates at an angular velocity of 3.374 rads/sec.

What are the forces (real and inertial) acting on the mass?

 

[Andrew Mason]

Here is a new example for you experts to chew over.

A (point) mass m is attached to a smooth cone of 60 angle. The cone rotates at an angular velocity of 3.374 rads/sec.

What are the forces (real and inertial) acting on the mass?

The real forces are tension, normal force and gravity. They sum (vectorially) to the mass x acceleration of the object (ie. the centripetal force that keeps the object moving in a circle).

There is no inertial "force". The inertial "force" only appears if you analyze it in the frame of reference of the body, which is a non-inertial reference frame.

AM

 

[Dale]

What are the forces (real and inertial) acting on the mass?

I agree with Andrew Mason.

 

[Ken G]

Here is a new example for you experts to chew over.

A (point) mass m is attached to a smooth cone of 60 angle. The cone rotates at an angular velocity of 3.374 rads/sec.

What are the forces (real and inertial) acting on the mass?

The issue hardly requires "chewing", it is a simple "free body diagram" from first-year physics. 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). Both the string tension and the normal force are set entirely by the requirement for these three forces to add up to give to a net force, called the centripetal force, whose magnitude must be w²r, where w is the angular velocity and r is the distance from the central axis of the cone. The direction of this net centripetal force is toward the axis. In a rarely used language where the centrifugal force is regarded as an "inertial force", it will have the same magnitude but opposite direction as the net centripetal force, because in this use of the term inertial force (being applied in the inertial frame of the observer here), it is envisaged as a balancing force that must make all the forces on the mass add to zero (that's how d'Alembert used the concept, it is still seen today as a useful way to talk about the -ma term in F-ma=0). Please note the key point however: being -ma, the "inertial force" is always zero on all massless objects, like the band in your original problem.

 

[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.

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.