Like what IS energy? relative to force

[Dale]

So the result what you get when you take only a part of this system is... partial.
Adding/substracting energy to/from one of the elements, changes the energy of the whole system.
So taking only one element and ignoring the others is not correct.

For the third or fourth time, there is no requirement in physics that a system must be isolated. This argument is fallacious.

In physics you are allowed to define the system as you wish. You can, in fact, take only one element and ignore others. It is correct. The laws of physics work correctly for the one element system as well as the whole isolated system.

For a single body in a two body central force problem there is a net force with no energy transfer. Your original statement is false in that respect and your arguments supporting it are fallacious. Please stop.

 

[Andrew Mason]

OK
So how exactly you are calculating resultant force here and you get non zero result?

There is no outward force on the rotating body. If your diagram is showing such an outward force it is wrong. There is only the tension of the rope on the body. This tension accelerates the body toward the centre of rotation. It is called centripetal acceleration.

AM

 

[Malverin]

For the third or fourth time, there is no requirement in physics that a system must be isolated. This argument is fallacious.

In physics you are allowed to define the system as you wish. You can, in fact, take only one element and ignore others. It is correct. The laws of physics work correctly for the one element system as well as the whole isolated system.

For a single body in a two body central force problem there is a net force with no energy transfer. Your original statement is false in that respect and your arguments supporting it are fallacious. Please stop.

That is written in the books

http://en.wikipedia.org/wiki/Rotating_reference_frame

Newton's law in the rotating frame then becomes

Treat the fictitious forces like real forces, and pretend you are in an inertial frame.

— Louis N. Hand, Janet D. Finch Analytical Mechanics, p. 267

Obviously, a rotating frame of reference is a case of a non-inertial frame. Thus the particle in addition to the real force is acted upon by a fictitious force...The particle will move according to Newton's second law of motion if the total force acting on it is taken as the sum of the real and fictitious forces.

— HS Hans & SP Pui: Mechanics; p. 341

This equation has exactly the form of Newton's second law, except that in addition to F, the sum of all forces identified in the inertial frame, there is an extra term on the right...This means we can continue to use Newton's second law in the noninertial frame provided we agree that in the noninertial frame we must add an extra force-like term, often called the inertial force.

— John R. Taylor: Classical Mechanics; p. 328

I said what I have to say.

 

[Dale]

Completely irrelevant. None of the discussion above has involved non-inertial frames.

You made a mistake, plain and simple. But for some reason, instead of simply saying "oops" and moving on you want to double down with irrelevancies and fallacies.

The application of a force does not imply a transfer of energy, nor does the application of a net force.

 

[Malverin]

Completely irrelevant. None of the discussion above has involved non-inertial frames.

You made a mistake, plain and simple. But for some reason, instead of simply saying "oops" and moving on you want to double down with irrelevancies and fallacies.

The application of a force does not imply a transfer of energy, nor does the application of a net force.

http://en.wikipedia.org/wiki/Work_(physics)

Cartesian coordinates

dWx = Fx * dX

dWy = Fy * dY

dWz = Fz * dZ

Polar coordinates

dW_radial = F_radial * dR

dW_tangential = F_tangential *dθ

Indexes show that the change in energy is due to movement/deformation along the coordinate axes

I leave the comments to you.

 

[mic*]

Polar coordinates

dW_radial = F_radial * dR

dW_tangential = F_tangential *dθ [/B]

The centripetal force being discussed is the radial force or F_radial above, and in that instance dR = 0 so dW_radial = 0

Do you concur with that Malverin?

 

[Dale]

http://en.wikipedia.org/wiki/Work_(physics)

Note that if F is perpendicular to v then F.v=0 even if both F and v are non-zero. Therefore you can have a force and movement without a transfer of energy. Your own citation directly contradicts your claim.

 

[Malverin]

The centripetal force being discussed is the radial force or F_radial above, and in that instance dR = 0 so dW_radial = 0

Do you concur with that Malverin?

If there is not movement/deformation , there is no work done.

That is well known.
Discussion about orbiting, was pointed to forces, not displacements.

 

[Malverin]

Note that if F is perpendicular to v then F.v=0 even if both F and v are non-zero. Therefore you can have a force and movement without a transfer of energy. Your own citation directly contradicts your claim.

Speaking about acting force, I always meant the component of the force parallel to the movement/deformation.
I thought that is clear.

 

[Dale]

Speaking about acting force, I always meant the component of the force parallel to the movement/deformation.
I thought that is clear.

It was never clear to me.

So then, do you agree that the original statement was incorrect as written (i.e. without any discussion about components and movement)?

When you apply a Force, you transfer Energy

 

[mic*]

Speaking about acting force, I always meant the component of the force parallel to the movement/deformation.
I thought that is clear.

There is a force that is radial. Everyone is agreeing on that.

There is no radial movement, as per my last post. You agreed. This is movement parallel to the force.

There is only motion perpendicular to the force, as per Dale's last post. (EDIT: 2nd last post)

Are you just being argumentative for kicks?

 

[Malverin]

It was never clear to me.

So then, do you agree that the original statement was incorrect as written (i.e. without any discussion about components and movement)?

Yes it is inaccurate. I have agreed with that

Malverin #35 Dec9-13, 08:02 PM
P: 43
Quote Quote by DaleSpam View Post
Note that, as you say, there remain 2 forces despite the fact that energy is no longer being transfered. Therefore your original statement "When you apply a Force, you transfer Energy" is clearly wrong.
It is inaccurate yes.
When resultant force is not zero is more accurate.
Because there will be movement or/and deformation.

 

[Malverin]

There is a force that is radial. Everyone is agreeing on that.

There is no radial movement, as per my last post. You agreed. This is movement parallel to the force.

There is only motion perpendicular to the force, as per Dale's last post. (EDIT: 2nd last post)

Are you just being argumentative for kicks?

I said. Discussion about orbiting was not about displacements but forces.

If resultant force is zero, displacement doesn't matter, right?

 

[Dale]

Yes it is inaccurate. I have agreed with that

Even if you say "resultant force" instead of "force" it is still wrong. Do you agree? Your previous statement did not fix the problem.

Furthermore, changing it to "resultant force" introduces new problems since you can have a 0 resultant force with a transfer of energy. For example, if a vehicle is going at a constant velocity on a level straight road then the propulsive force from the road is opposed to the viscous drag force from the air. These two forces yield a 0 resultant force, but energy is being transferred from the vehicle to the air. Note that this scenario involves forces parallel to the direction of motion.

 

[Malverin]

Even if you say "resultant force" instead of "force" it is still wrong. Do you agree? Your previous statement did not fix the problem.

Furthermore, changing it to "resultant force" introduces new problems since you can have a 0 resultant force with a transfer of energy. For example, if a vehicle is going at a constant velocity on a level straight road then the propulsive force from the road is opposed to the viscous drag force from the air. These two forces yield a 0 resultant force, but energy is being transferred from the vehicle to the air. Note that this scenario involves forces parallel to the direction of motion.

Yes it is so. Friction and drag take away energy when forces are in equilibrium.

 

[Andrew Mason]

Yes it is so. Friction and drag take away energy when forces are in equilibrium.

Things have gotten a bit off topic. But you seem to still think that a body in gravitational orbit around another body does not experience a net force. I am not sure why you think this but it is wrong. There is no force other than the force of gravity and it is only in one direction (ie. toward the other body).

AM

 

[Malverin]

Things have gotten a bit off topic. But you seem to still think that a body in gravitational orbit around another body does not experience a net force. I am not sure why you think this but it is wrong. There is no force other than the force of gravity and it is only in one direction (ie. toward the other body).

AM

I have attached a drawing. You can see the forces there.

 

[Bandersnatch]

I have attached a drawing. You can see the forces there.

In your picture(which by the way lacks the force keeping the axle in place), the only force acting on the body is the force provided by the tension of the rope(F_b in your pic). F_r acts on the rope itself, not on the body, and is what keeps the rope taut.

The reaction forces are applied at the same point, but act in opposite directions, so on different bodies. Otherwise you could argue that it's impossible to kick a ball, since the leg-ball system has got two equal and opposite forces exactly cancelling each other at the point of contact.


Back to the picture. You could have zero net forces acting on the body, if you chose to use a rotating reference frame. In which case you'd have to add centrifugal force to the drawing, exactly matching F_b in magnitude.
In an inertial reference frame, the mere fact of there being a rotation makes it clear that there has to be an unbalanced force accelerating the body.

Note that if you'd do what others suggested and simplify things by having the force of gravity acting between two bodies rather than using a rope, it would remove the confusing bits while producing the same results.

 

[russ_watters]

I have attached a drawing. You can see the forces there.

Much of the problem here is that you are misusing words. The term "resultant force" does not apply in this thread. You mean "net force". But you are still misusing it: net force is a force that causes acceleration. But you are describing equal and opposite fore pairs in that context, when the reality is that force pairs ALWAYS sum to zero, regardless of motion or acceleration.

What is frustrating here is that it isn't clear if you are purposely being evasive or are just very confused. That's why the thread hasn't been locked yet. Either way, you need to try harder to listen and learn rather than trying to teach or demonstrate your knowledge.

 

[russ_watters]

Speaking about acting force, I always meant the component of the force parallel to the movement/deformation.
I thought that is clear.

Given that the discussion was about circular motion, where the force is perpendicular to displacement, that is highly dubious.

 

[Malverin]

In your picture(which by the way lacks the force keeping the axle in place), the only force acting on the body is the force provided by the tension of the rope(F_b in your pic). F_r acts on the rope itself, not on the body, and is what keeps the rope taut.

The reaction forces are applied at the same point, but act in opposite directions, so on different bodies. Otherwise you could argue that it's impossible to kick a ball, since the leg-ball system has got two equal and opposite forces exactly cancelling each other at the point of contact.


Back to the picture. You could have zero net forces acting on the body, if you chose to use a rotating reference frame. In which case you'd have to add centrifugal force to the drawing, exactly matching F_b in magnitude.
In an inertial reference frame, the mere fact of there being a rotation makes it clear that there has to be an unbalanced force accelerating the body.

Note that if you'd do what others suggested and simplify things by having the force of gravity acting between two bodies rather than using a rope, it would remove the confusing bits while producing the same results.

I am not arguing about the force acting on the body.
I said that if you change the energy of the body, the energy of the whole system (body, rope, axle) will change.
And any analysis about energy here have to consider them too.

 

[Malverin]

Given that the discussion was about circular motion, where the force is perpendicular to displacement, that is highly dubious.

When you have a rotational motion and energy is changed, it can be due to radial movement and/or change in rotation speed.

 

[russ_watters]

When you have a rotational motion and energy is changed, it can be due to radial movement and/or change in rotation speed.

The uniform circular motion scenario being discussed does not involve a change in any kind of energy. This is the root of the issue: you've made several statements saying or implying you think it does.

 

[Malverin]

The uniform circular motion scenario being discussed does not involve a change in any kind of energy. This is the root of the issue: you've made several statements saying or implying you think it does.

All started when I said that when there is a net(resultant) force it will cause movement/deformation and do work.
Then others gave me an example that there is a net force by circular motion, but no energy change.

My statement was (and still is) that rotating body is a part of the rotational system and when we consider its energy we have to consider all its elements, not only the rotating body alone.

If someone has understand something else, I apologize I have not explained it clear enough.
If you (and the others) think that my statement about energy of rotating system is not correct, please
show(explain) why the change of energy of the rotating body will not affect energy of the rope or axle(or another body) in the rotating system.

I think, disscusion about rotation became so long, because I have not asked my questions clear enough.

 

[russ_watters]

Expanding on the bit about force pairs:

By Newton's third law, all forces applied at a point come in equal and opposite pairs (sum to zero).
http://en.wikipedia.org/wiki/Newton's_laws_of_motion#Newton.27s_third_law

Since it is always true that forces come in pairs, it is redundant - and as we see here often confusing - to draw both halves of the force pair.

An unbalanced "net" force is not referring to such force pairs, but situations where multiple unequal force pairs exist (or just one force pair and nothing opposing it). So when analyzing forces (drawing free body diagrams), only half of each force pair is taken into account; the force that is applied to the object. The force the object applies back is left out.
http://en.wikipedia.org/wiki/Free_body_diagram
http://www.physicsclassroom.com/class/newtlaws/u2l2d.cfm

In the second link, the first diagram describes a box being lifted by a rope. Notice that the tension force is only one force, not two, and it is pointing up. We draw-in the force of the rope acting on the box, but not the equal and opposite force of the box acting on the rope. Same goes for gravity: we draw the force of gravity pulling down on the box from earth, but not the force of the gravity pulling up on the Earth from the box.

So for the revolving tether-ball or circular orbit scenario, there is one force pair, so there is only one vector drawn on a proper diagram; from the object, pointing to the center of the orbit/revolution.

 

[russ_watters]

All started when I said that when there is a net(resultant) force it will cause movement/deformation and do work.
Then others gave me an example that there is a net force by circular motion, but no energy change.

My statement was (and still is) that rotating body is a part of the rotational system and when we consider its energy we have to consider all its elements, not only the rotating body alone.

That is still wrong.

If someone has understand something else, I apologize I have not explained it clear enough.
If you (and the others) think that my statement about energy of rotating system is not correct, please
show(explain) why the change of energy of the rotating body will not affect energy of the rope or axle(or another body) in the rotating system.

I'm sorry, but that makes no sense. In these scenarios, the rope and axle neither have nor apply energy. And if the tethered object is sped-up (like a slapped tether-ball), the forces on the axle and in the rope tension go up, but they still do no work/apply no energy.

 

[Malverin]

That is still wrong.

It can be wrong yes.
Where I can read information that proves it wrong?
Thank you for your answers.

 

[russ_watters]

It can be wrong yes.
Where I can read information that proves it wrong?
Thank you for your answers.

Sorry/see my late edit.

 

[Malverin]

That is still wrong.

I'm sorry, but that makes no sense. In these scenarios, the rope and axle neither have nor apply energy. And if the tethered object is sped-up (like a slapped tether-ball), the forces on the axle and in the rope tension go up, but they still do no work/apply no energy.

If there is a speed-up, then rotational speed Ω is changed.

If the rope has mass, its kinetic energy will change.

If there is a sped-up , there is a torque applied to the axle. It will change its energy either.

Is that so?

 

[Dale]

My statement was (and still is) that rotating body is a part of the rotational system and when we consider its energy we have to consider all its elements, not only the rotating body alone.

This is simply not true. I think this is the 4th time that I have corrected you on this point. There is no requirement that you consider the entire isolated system. In mechanics you are free to define your system boundaries as you desire. You are not obligated to continue expanding your system until it becomes an isolated system.

Please stop repeating this false assertion which I have corrected multiple times already.

 

[Malverin]

This is simply not true. I think this is the 4th time that I have corrected you on this point. There is no requirement that you consider the entire isolated system. In mechanics you are free to define your system boundaries as you desire. You are not obligated to continue expanding your system until it becomes an isolated system.

Please stop repeating this false assertion which I have corrected multiple times already.

OK
I see we will go nowhere here.
So I will not post at this thread anymore

 

[voko]

Much of this discussion could have been avoided, had Malverin just said that "when a force does work, it transfers energy". But I guess that would have been too trivial for him to say :)

 

[russ_watters]

If there is a speed-up, then rotational speed Ω is changed.

If the rope has mass, its kinetic energy will change.

If there is a sped-up , there is a torque applied to the axle. It will change its energy either.

Is that so?

If we consider and we're discussing the mass of the rope and axle, sure. But we were discussing the ball - we don't know anything about the rope and axle, so now you are making up the scenario as you go along. Moreover, the mass of rope is generally much smaller than the ball so it is negligible and the axle is often not moving (such as with a tetherball).

This is goalpost shifting; you are far, far away from what was originally being discussed and your statements about it. From earlier:

there is a force applied to the body, this is the force of the rope, that keeps the body at constant distance from the axis. At the other end of the rope there is the same magnitude force with the opposite direction.

This is only true if you assume the rope is massles. If it is not, the force varies along the rope.

This is a real shame. You had a good learning opportunity here and chose to be argumentative instead.