Question about something i just read

[J.C.MacSwell]

Why?

 

Conservation of momentum

[Johnny5]

Conservation of momentum

 

Ok, explain this mathematically please. I want to see how you analyze the problem.

 

Thanks

[J.C.MacSwell]

Ok' date=' explain this mathematically please. I want to see how you analyze the problem.

 

Thanks[/quote']

 

The force on the leftmost particle (upwards) is equivalent to having the force on the middle particle (upwards) plus half of that force on the leftmost particle (upwards) plus half of that force on the rightmost particle (downwards) which resolves to a translational F=ma plus a rotational moment.

 

As the "bar" accelerates the point of contact of a constant upward force on the original particle will shift to the right (with the particle) reducing the rotational moment. It will eventually oscillate left and right (constant frequency) while continually accelerating upwards at greater velocity.

[DavidAngelMX]

Look, originally, Newton's Laws were intended to describe the motion (or change of it) of a POINT particle (an idealization of a rigid, non-dimensional body). Newtons second law states that "The change of movement state of a particle (ie it's acceleration) is proportional to the sum of external forces applied to it"

 

Obviously Newton's laws work even if the object is not a point particle, but only for the center of mass; for the rest of the body you have to take several considerations, like taking the body as a system of poit particles, or a continuous, and use some math to deduce the consecuences of applying an external force.

 

So if the question is "why don't Newton's laws consider that a (generalized) system could present rotational (and not only translational) change of motion depending on the direction, with respect of the system's center of mass, in which the force is being applied?", the answer is that Newton's laws are more basic and siplified than that, as they describe the effects of forces acting upon POINT PARTICLES, and the generalized effect on particle systems has to be derived from them.

[Johnny5]

The force on the leftmost particle (upwards) is equivalent to having the force on the middle particle (upwards) plus half of that force on the leftmost particle (upwards) plus half of that force on the rightmost particle (downwards) which resolves to a translational F=ma plus a rotational moment.

 

 

Ok' date=' just to make sure I followed you, let the net applied force to the system have a magnitude F. Suppose that X is the portion of F that goes into translating the center of mass. Let the remainder be Y. Therefore, X+Y=F. And you are also saying that X=Y.

 

In order to figure out how the entire line mass moves (not just the center of mass of the line mass) we can break the force F into three forces. One with magnitude X, which pushes the CM upwards, and then one which pushes the leftmost particle up (which will have magnitude Y/2), and a third force which pushes the rightmost particle DOWNwards.

 

So if I understand you correctly, you are saying that what happens by applying just one force F to the leftmost particle, will be the same thing which happens if three forces act simultaneously on the object at different places. On with magnitude Y/2 upwards on the leftmost particle, another with magnitude Y/2 downwards on the rightmost particle, and a third with magnitude X upwards on the center of mass particle (assuming there is a particle there (and there isn't because the number of protons is even but ignore this for now), and Y/2+Y/2+X=F. <--- Is this what you are saying?

 

Assuming that is what you are saying, what is the simplest way to mathematically check if what you are saying is true?

 

Would you switch to watching the motion in the CM frame (which translates with the body after the push), and prove that in this frame there is only pure rotation?

 

Can you do a mathematical analysis of the problem in various frames of reference? Intuitively you know you can break the force into three, but exactly how do you infer that from Newton's laws? I have so many questions about this one problem. Next you say:

 

 

 

As the "bar" accelerates the point of contact of a constant upward force on the original particle will shift to the right (with the particle) reducing the rotational moment. It will eventually oscillate left and right (constant frequency) while continually accelerating upwards at greater velocity.

 

Can you explain this part in greater detail? What did you assume was responsible for the applied force? A particle? Another line mass? Also, what oscillates? Are you treating the line mass as purely rigid? Or almost rigid?

 

It's not an easy problem, perhaps not even well posed, but still I want to work on this.

 

Thank you very much, regards.

[J.C.MacSwell]

Ok' date=' just to make sure I followed you, let the net applied force to the system have a magnitude F[b']. Suppose that X is the portion of F that goes into translating the center of mass[/b]. Let the remainder be Y. Therefore, X+Y=F. And you are also saying that X=Y.

 

.

I think this is where you are going wrong. ALL of the force acts to translate the centre of mass.

 

F=X, if X+Y=F then Y=0

 

Picture 100 particles. Any configuration. Some glued together rigidly. Some totally detached. Some attached by elastics.

 

Now apply a force upwards on any particle or particle group. Call that force F.

 

F will equal ma, where a will be the acceleration upwards and m will be the mass of the system.(elastics and all)

 

Redo the math for the same thing adding a mass the size of Jupiter to the system where Jupiter is 100 miles away and is totally unaffected.

 

F will equal ma, where a will be the acceleration upwards and m will be the mass of the system including Jupiter.

 

Redo the math for the same thing adding a mass the size of Jupiter to the system where Jupiter is 100 miles away and is the only body affected by the force (upwards again whatever that means in the reference frame you are using)

 

F will equal ma, where a will be the acceleration upwards and m will be the mass of the system including Jupiter.

 

If F accelerates the mass in translation and in rotation it doesn't matter.

 

F will equal ma.

(if you sum the "rotational accelerations in the upward, or any direction they will cancel out)

 

If this seems odd it is probably because you intuitively feel that since F may be doing work to stretch elastics or cause rotations etc that maybe some of F is "used up" in this regard.

 

 

It is not.

 

What is different and does need to be divided up is the energy or work done by the force F.

 

The work done by F is very different in each of the above cases.

[Johnny5]

I think this is where you are going wrong. ALL of the force acts to translate the centre of mass.

 

I don't follow. I am picturing a bar in space. I am floating next to it, and am at rest with it. Then I use my finger to push the leftside of the bar, in a direction perpendicular to the bar. That bar is going to start spinning isn't it? If not, then yes I am doing something wrong. Why won't the bar start spinning?

[J.C.MacSwell]

The force on the leftmost particle (upwards) is equivalent to having the force on the middle particle (upwards) plus half of that force on the leftmost particle (upwards) plus half of that force on the rightmost particle (downwards) which resolves to a translational F=ma plus a rotational moment.

 

As the "bar" accelerates the point of contact of a constant upward force on the original particle will shift to the right (with the particle) reducing the rotational moment. It will eventually oscillate left and right (constant frequency) while continually accelerating upwards at greater velocity.

What I was trying to show here is that any force acting on a rigid body that is not acting through the centre of mass is equivalent to an equal force in the same direction acting through the centre of mass plus a rotational moment.

[Johnny5]

What I was trying to show here is that any force acting on a rigid body that is not acting through the centre of mass is equivalent to an equal force in the same direction acting through the centre of mass plus[/b'] a rotational moment.

 

So then you are saying that it does spin?

 

Regards

[J.C.MacSwell]

I don't follow. I am picturing a bar in space. I am floating next to it, and am at rest with it. Then I use my finger to push the leftside of the bar, in a direction perpendicular to the bar. That bar is going to start spinning isn't it? If not, then yes I am doing something wrong. Why won't the bar start spinning?

 

It will spin and translate (wave good-bye to it)

 

It will move away in the direction of the force and you will move in the opposite direction. Depending on the alignment of the force with respect to your centre of mass you may spin also.

[Johnny5]

What I was trying to show here is that any force acting on a rigid body that is not acting through the centre of mass is equivalent to an equal force in the same direction acting through the centre of mass plus[/b'] a rotational moment.

 

See I don't think this is right. Won't some of the applied force go into translation, and some go into rotation?

[Johnny5]

It will spin and translate (wave good-bye to it)

 

It will move away in the direction of the force and you will move in the opposite direction. Depending on the alignment of the force with respect to your centre of mass you may spin also.

 

Yes, it will move away, I know that. My center of mass, and it's center of mass will separate, in the center of mass frame (me+it) I know this (assuming Newton's third law true). But, how can one applied force F, at the end of a rigid body, be equivalent to the same force acting in the same direction at the center of mass, plus a rotational moment? <--- thats what i don't get.

 

Doesnt that violate conservation of energy?

[J.C.MacSwell]

See I don't think this is right. Won't some of the applied force go into translation, and some go into rotation?

 

All of it will go into translation.

[Johnny5]

All [/b'] of it will go into translation.

 

But my intuition says the bar is gonna spin? What on earth???

 

Here is how I am viewing this.

 

Let there be a master frame, in which the center of mass of the universe is at rest.

 

Let my center of mass, and its center of mass be at rest in the master frame, and furthermore neither me, nor the bar is spinning in the master frame.

 

Then I push the end of the bar. Now, the center of mass of the bar has accelerated in the master frame, but my intuition also tells me that the bar will be spinning in the master frame. What, if anything, is my intuition doing wrong?

 

I seem to be convinced, that the bar will now be spinning in the master frame, as well as having its center of mass be translating.

[J.C.MacSwell]

Yes' date=' it will move away, I know that. My center of mass, and it's center of mass will separate, in the center of mass frame (me+it) I know this (assuming Newton's third law true). But, how can one applied force F, at the end of a rigid body, be equivalent to the same force acting in the same direction at the center of mass, plus a rotational moment? <--- thats what i don't get.

 

[b']Doesnt that violate conservation of energy[/b]?

 

It takes more energy to maintain the force.

 

If you and I were in outer space with bars and we were equally powerful and I pushed with all my might through the centre of mass and you pushed with all your might on the end as described on an equivalent bar then I would exert more force.

[Johnny5]

It takes more energy to maintain the force.

 

If you and I were in outer space with bars and we were equally powerful and I pushed with all my might through the centre of mass and you pushed with all your might on the end as described on an equivalent bar then I would exert more force.

 

Can you do this with numbers?

 

Suppose we push on the bar in opposite directions, both of us through the center of mass of the bar.

 

We push equally hard, because we are equally strong.

 

Now focus on things in the center of mass frame of the bar.

 

The bar remains at rest in this frame.

 

You move away from me, and I move away from you, and the bar remains at rest in the CM frame.

 

The force upon me is 2F, and same on you.

 

I exterted a force on you of F, and an equal force was reflected back upon me, and you exerted a force on me of F, and an equal force was exerted back upon you, so the total force upon me was 2F, and letting M denote my mass, my acceleration in the CM frame (bar at rest) is found through:

 

2F = M a

 

Now, move me to the edge of the bar, and we both push simultaneously, equally hard. You say you exert a greater force. Explain this to me, i expect you to say something about torque.

[J.C.MacSwell]

All [/b'] of it will go into translation.

 

Maybe this will help:

 

Assuming the force does not act through the centre of mass.

 

All of the force will also go into rotation. The moment will depend on the magnitude of the force and the "lever" due to the alignment of the force wrt the centre of mass.

[J.C.MacSwell]

Can you do this with numbers?

 

Suppose we push on the bar in opposite directions' date=' both of us through the center of mass of the bar.

 

We push equally hard, because we are equally strong.

 

Now focus on things in the center of mass frame of the bar.

 

The bar remains at rest in this frame.

 

You move away from me, and I move away from you, and the bar remains at rest in the CM frame.

 

The force upon me is 2F, and same on you.

 

I exterted a force on you of F, and an equal force was reflected back upon me, and you exerted a force on me of F, and an equal force was exerted back upon you, so the total force upon me was 2F, and letting M denote my mass, my acceleration in the CM frame (bar at rest) is found through:

 

2F = M a

 

Now, move me to the edge of the bar, and we both push simultaneously, equally hard. You say you exert a greater force. Explain this to me, i expect you to say something about torque.[/quote']

 

2F? It's only F.

[Johnny5]

Maybe this will help:

 

Assuming the force does not act through the centre of mass.

 

All of the force will also go into rotation. The moment will depend on the magnitude of the force and the "lever" due to the alignment of the force wrt the centre of mass.

 

Yes that helps' date=' but can you prove that it is true for me? Will energy be conserved?

 

Let the applied force at the end of the bar be F, perpendicular to the bar.

 

The torque is found from this:

 

[math'] \tau = torque = \vec r X \vec F [/math]

 

The center of mass is going to be accelerated. And you are telling me the acceleration of the CM of the bar in the original rest frame of (me +bar) is found from

 

F=ma

 

where m is the mass of the bar.

 

Does this violate conservation of energy?

[J.C.MacSwell]

 

Now' date=' move me to the edge of the bar, and we both push simultaneously, [b']equally hard[/b]. You say you exert a greater force. Explain this to me, i expect you to say something about torque.

 

If equally hard means equal force...

 

I said if we were equally powerful.

[Johnny5]

2F? It's only F.

 

I push on you plus the bar with force F.

 

If you didnt push back, then there is a force on me of F.

 

But you also pushed me with force F, so its 2F.

[J.C.MacSwell]

I push on you plus the bar with force F.

 

If you didnt push back' date=' then there is a force on me of F.

 

But you also pushed me with force F, so its 2F.[/quote']

 

Nope

[Johnny5]

Nope

 

Aren't you forgetting about Newton's third law?

[J.C.MacSwell]

I push on you plus the bar with force F.

 

If you didnt push back' date=' then there is a force on me of F.

 

But you also pushed me with force F, so its 2F.[/quote']

 

Lets say you're standing still and your weight is 200 lb.

 

The floor is pushing you upwards with a force of 200 lb.

 

Now lets say you push back with a force of 200 lb.

 

Now is the force 400 lb or 200 lb?

[Johnny5]

Lets say you're standing still and your weight is 200 lb.

 

The floor is pushing you upwards with a force of 200 lb.

 

Now lets say you push back with a force of 200 lb.

 

Now is the force 400 lb or 200 lb?

 

I am standing on my bathroom scale, it reads 200 lb. Ok.

 

My weight is W=Mg = 200 lb

 

where M is my inertial mass.

 

Free body diagram of the forces upon me:

 

There is my weight W down, and the Normal force of the earth upon me.

 

Let N denote the normal.

 

The sum of forces on me is N+W, and this must equal zero, since I am not accelerating.

 

I have to go. I will think about this later, and finish up tomorrow.

 

Thank you