What is conserved in collisions: linear momentum or angular momentum?

[bobie]

The angular velocity vector of an object, representing its total rotation.

What is total or net ω, isn't ω always the net velocity? if the bykewheel in the video is spinning at 5 rps, what is net velocity?

 

[A.T.]

What is total or net ω, isn't ω always the net velocity?

"ω" is what you define "ω" to be. I didn't even use that symbol, so I don't know why you ask me.

 

[bobie]

"ω" is what you define "ω" to be. I didn't even use that symbol, so I don't know why you ask me.

Sorry, but that is what I studied:

In physics, the angular velocity is defined as the rate of change of angular displacement and is a vector quantity (more precisely, a pseudovector) which specifies the angular speed (rotational speed) of an object and the axis about which the object is rotating... Angular velocity is usually represented by the symbol omega (ω).

Were you not referring to that, anyway when is 'that' net?

 

[Dale]

No work means no energy spent, no effort on her muscles

Absolutely not. No work means no energy transferred to another system. Efficiency is the ratio of work done over energy spent, and it is not generally equal to 1. Unless you are dealing with a machine with 100% efficiency you cannot equate work done with energy expended.

The human body is a remarkably inefficient machine. In many cases its efficiency is 0 meaning that energy is spent without any work done.

 

[A.T.]

The human body is a remarkably inefficient machine.

For example: Muscles are not optimized for applying static forces. Therefore some birds can lock their wing joints for soaring, and relax the muscles which would have been used otherwise to stabilize the joint.

 

[Dale]

some birds can lock their wing joints for soaring, and relax the muscles which would have been used otherwise to stabilize the joint.

That is cool! I didn't know that.

 

[Dale]

bobie, if you are interested in the KE of a rotating object, then you can use this formula: $KE = \frac{1}{2}I \omega^2$. Together with the other formulas I provided, you can calculate all of the details if you like. I would recommend starting with a highly simplified scenario.

 

[A.T.]

That is cool! I didn't know that.

Prime example is the Albatross:
http://faculty.weber.edu/rmeyers/meyers-albatross.pdf

It stays airborne for weeks without flapping its wings once, using dynamic soaring. Having the muscles tensioned all this time would be a tremendous waste of energy.

 

[bobie]

bobie, if you are interested in the KE of a rotating object, then you can use this formula: $KE = \frac{1}{2}I \omega^2$. I would recommend starting with a highly simplified scenario.

My first scenario was simple: a bikewheel (m= 2, r= .4, v = 10) 'paddle' hits a ball at rest (m= 2 or another 'paddle'), are you saying that is not momentum that is tranferred as in ordinary collisions, but KE and that this is tranformed into P=20?

No work means no energy transferred to another system.

I know that. If I hold up a ball I do effort, no work. If I push a ball on the floor, the ball has inertia, I have to win resistance, I give KE to move it , it moves, I have done work.
What you say on birds is interesting, but, please explain how it applies here:
The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel, the wheel has inertia when it is at rest, wiki says that a spinning wheel has extra inertia and resists changes, how can you conclude that she has done no work?or you mean that the work on the wheel is being transferred to the platform?

 

[bobie]

. The platform is more complicated. The angular velocity in x and y is 0, so the x and y torques do no work on the platform, but the angular velocity in z is non-zero, so the z torque does work on the platform.

The girl does no work on the wheel but does work on the platform. correct?

The formula for work is F*d(rad), could you show me how you calculate the work done on the platform? what we need here is only L_w = 1.6 Js, or do we need the details of m,r,ω ? How do we find out the value of F in this example?

 

[A.T.]

If I push a ball on the floor, the ball has inertia, I have to win resistance, I give KE to move it , it moves, I have done work.

You can apply a force to a moving object, without doing work.

The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel,

You can apply a torque to a rotating object, without doing work.

 

[bobie]

You can apply a force to a moving object, without doing work.
You can apply a torque to a rotating object, without doing work.

Sure, but here the situation is different, AT,
If an object is not moving and then moves is it impossible that nobody did work.
If an object is moving in one direction and then moves in two directions work must have been done. If you throw a ball with spin you do work both for translation and rotational energy.
The wheel is spinning on one plane and the girl rotates it in another direction in which it was not moving, does that motion come free?
But the main argument on my side is that that is the only action the girl is doing, and the motion of the platform can be originated only by the work she is doing on the wheel.
There is no other source of energy around. Am I wrong?


Thanks for your help, could you please answer my post #53, it probably escaped your attention

 

[A.T.]

If an object is moving in one direction and then moves in two directions work must have been done.

Wrong. You can change the direction of the velocity vector without doing work. That has been explained several times to you already.

Thanks for your help, could you please answer my post #53, it probably escaped your attention

See my post #46.

 

[bobie]

You can change the direction of the velocity vector without doing work. That has been explained several times to you already..

You said that many times, but that is an abstract principle, could you apply it here?

Can you, please, answer directly the practical,simple questions:
who does the work that sets the platform spinning? how? what is the value of the work done? how you derive it from the given parameters?
That is the only way to make me understand. In many threads, when I asked these simple questions I got only abstract, apodictical principles that are surely clear to you scientists.

As to post 46-53 I gave you a quote from wiki that says that angular velocity is given by ω, what is wrong?

Thanks a lot for your patience

 

[A.T.]

You said that many times,...

And you just keep contradicting it. So what is the point of telling you something.

 

[bobie]

And you just keep contradicting it. So what is the point of telling you something.

I am not contradicting it, I am only asking you to show me how it works in practice.
I am saying, all right, the girl does no work, who does it?
The outcome of the video seems to contradict that, I do not know what is happening, ( a scientist as great as Laithwaithe raised the issues and he was not an ignorant)
if you know what is happening , just tell me, if you wish.
Just repeating that the girl tilting the wheel does no work does not explain anything of what we see in the video.
Thanks for your posts.

 

[A.T.]

I am not contradicting it

Yes, you are. You said: "If an object is moving in one direction and then moves in two directions work must have been done." which simply doesn't follow from the definition of work.

Laithwaithe raised the issues and he was not an ignorant

On this topic he was. His misconceptions are explained here:

https://www.youtube.com/watch?v=tLMpdBjA2SU

 

[bobie]

On this topic he was. His misconceptions are explained here:

I had seen that video (and hundred of others) and it does not explain misconceptions. Of course he was wrong when he went as far as to question the validity of the laws of motion, but most of issues still have no explanations, I hoped that the issue in the OP video had found a full description in these last 40 years.

What happens here is that force to lift the wheel is applied to accelerate it horizontally and that acceleration is shifted by 90° , i.e. upwards, by the gyroscope propriety. You should note that, when he starts to roteate it, his own weight increases as he is pushing/leaning on his foot to push the wheel around.
You surely know that if you accelerate the horizontal motion the gyroscope goes up (seems lighter) if you slow it it gets 'heavier' and falls down. He feels less effort as (a) it is easier to do work horizontally than vertically, (b) he is diluting his work, that is all.
For other reasons even when you swing a stone on a string it goes up.

If you are interested in this intriguing issues I'd be glad to discuss them with you,
see here , for example ( at 1:45/49)
why does the horizontal rotation stops immediately when the weight is lifted?

 

[A.T.]

I had seen that video (and hundred of others) and it does not explain misconceptions.

Works fine for me.

 

[Dale]

My first scenario was simple: a bikewheel (m= 2, r= .4, v = 10) 'paddle' hits a ball at rest (m= 2 or another 'paddle'), are you saying that is not momentum that is tranferred as in ordinary collisions, but KE and that this is tranformed into P=20?

No, I never said anything of the sort. Momentum and energy are separate quantities and cannot be transformed into each other. They are each individually and separately conserved for an isolated system.

I know that. If I hold up a ball I do effort, no work.

When you make a mistake and are corrected, it accomplishes nothing to say "I know that". If you make mistakes then obviously you don't know it well enough to avoid the mistakes.

If I push a ball on the floor, the ball has inertia, I have to win resistance, I give KE to move it , it moves, I have done work.

"Win resistance" is not a physics term. It doesn't mean anything that I am aware of.

If you apply a force on an object and that object's mechanical energy increases then you have done work on the object. If you apply a force on an object and that object's energy does not change then you have done no work on the object, regardless of the fact that you exerted a force on the object and regardless of the fact that the object has inertia and regardless of whatever energy you (as an inefficient machine) may have expended.

The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel,

How much is "some" KE? Use the formula I provided above and calculate the KE in the beginning and ending and then tell me exactly how much KE was transferred to the wheel.

 

[Dale]

The girl does no work on the wheel but does work on the platform. correct?

Yes.

The formula for work is F*d(rad), could you show me how you calculate the work done on the platform? what we need here is only L_w = 1.6 Js, or do we need the details of m,r,ω ? How do we find out the value of F in this example?

Use the formula I posted above: $KE = \frac{1}{2} I \omega^2$ where I is the moment of inertia and ω is the angular velocity.

 

[Dale]

Some useful formulas for rotational mechanics:
$L=I\omega$
$\tau=dL/dt$
$W=\int \tau \cdot \omega \; dt$
$KE = \frac{1}{2}I \omega^2$

I think that all of the questions you have posed can be covered with those, but I could have missed one or two.

 

[bobie]

I think that all of the questions you have posed can be covered with those, but I could have missed one or two.

Thanks, DaleSpam, you are a great tutor, really a gold mine. But I warn you, I'll use up all your patience.
If some of the questions are not relevant here, please tell me and I start separate threads.
I'll address the problems one by one , taking my time to digest your inputs, and I'll try to avoid hasty replies that contains silly slips like the one I comment in next post.

 

[bobie]

When you make a mistake and are corrected, it accomplishes nothing to say "I know that". If you make mistakes then obviously you don't know it well enough to avoid the mistakes.

"Win resistance" is not a physics term. It doesn't mean anything that I am aware of. .

In the post you mention I had already stated the correct principle, that confirms I knew it

(I have learned that you can make effort/ spend energy and do no work)[/I], but I suppose that no-work implies no-energy/effort..

What happened next was a banal inversion of terms, I meant:
no effort implies no work, in logical terms: 'effort is a necessary condition but not sufficient'.
Of course at a hasty reading the gross mistake catches the attention. I regret that.

What I meant is the following:
I can push sometyhing (make effort) but if it doesn't move , no work, but nothing can move if I make no effort. In another thread I was told I can easily rotate a 100-ton anti-rolling gyro, which is a limit case of what we are discussing here. Now, if you allow, that is really unbelievable, and (despite a formula that can only describe a principle) requires a minimum of explanation.

They said contrasting things in other threads : small...infinitesimal... work up to zero effort

original bold... If you grasp the golden axles (in your previous .gif) and apply a torque, always at right angles to the rotation of the flywheel it will precess from rotation on the xy plane to rotation on the xz plane with no expenditure of energy

Now, no energy spent means no effort (and that obviously implies no work). Can you give a clear, rational explanation why you or (even) I should be able to make a 100-ton material object move easily anyway, in any direction, in any context ?

There other cases where apparently an exceptional feat is performed, like by a lever, a screw/hydraulic jack, (or even like the one in the video posted by AT), but there some effort is done, diluted and apparently magnified; here you are stating that no effort is done no energy is spent , so zero cannot be diluted or multiplied, zero effort = zero motion.

Moreover, ff the wheel (offers no resistance, better:) has no inertia in that direction, if I spend no energy tilting it by a radiant, I should spend no energy (0*10) tilting it by 10 radiants, and it should spin forever on a frictionless gimbal, which does not happen, as you can see in the link I gave in my previous posts.

This is the core problem of this (and other threads), I hope you can clarify all that.
Thanks again

 

[A.T.]

I should be able to make a 100-ton material object move easily anyway, in any direction, in any context ?

Define "move easily". Or better: Stop using such vague terms. It just makes your questions ambiguous, and any answer useless.

here you are stating that no effort is done no energy is spent

If the sum of potential and kinetic energy of the gyro increases, then some energy went into the gyro. But even if no energy goes into the gyro, muscles or some other force providers might still dissipate energy.

 

[jbriggs444]

I can push sometyhing (make effort) but if it doesn't move , no work, but nothing can move if I make no effort. In another thread I was told I can easily rotate a 100-ton anti-rolling gyro, which is a limit case of what we are discussing here.

You were not told that. You were told that causing rotating a 100-ton gyro to precess through 90 degrees could be done with negligible work done, not that it would be "easy" or that it would not involve application of significant torque over significant time. Nothing was said about how much "effort" would be required.

The relevant principle was that a non-zero torque applied at right angles to a non-zero rotation does zero work.

The 100 ton gyro may be a limiting case of this same situation. But I doubt that you have the same limit in mind that I do. Please spell out the sense in which you think that the two scenarios are analogous and what scale factor you are considering to be approaching a limit.

 

[bobie]

degrees could be done with negligible work done, not that it would be "easy" or that .

We might might have a play-off between negligible and easy

 

[jbriggs444]

We might might have a play-off between negligible and easy

"negligible" in this context means "can be made arbitrarily small"
"easy" -- you've been asked to quantify already.

To me the distinction is between "effort" (whatever that may mean) and "work" (which has a well-defined meaning).

 

[Dale]

'effort is a necessary condition but not sufficient'.

Yes, using energy is necessary for doing work, not sufficient. The amount of external work done is a lower bound on the amount of energy used, and a 100% efficient machine will achieve that lower bound.

In another thread I was told I can easily rotate a 100-ton anti-rolling gyro, which is a limit case of what we are discussing here. Now, if you allow, that is really unbelievable, and (despite a formula that can only describe a principle) requires a minimum of explanation.

You may not believe it, but it is true. To remove the effort vs work confusion, let's consider designing a machine for doing this and see if we can come up with an ideal 100% efficient machine and see how much work it does.

In this case, such a machine is almost trivial to design, simply attach the gyro to the floor or some pedestal using a frictionless bearing, and make sure that the axis of the gyro is not vertical. Since the axis is not vertical, gravity will provide a torque in the horizontal plane which will cause the gyro to precess.

The floor provides a force without using any energy, so the work done by this machine is 0. This can further be seen in the fact that the KE of the gyro is unchanged as is the gravitational PE. Thus a 100-ton gyro can have its axis rotated (called precession) without expending any energy.

 

[bobie]

You may not believe it, but it is true...
a 100-ton gyro can have its axis rotated (called precession) without expending any energy.

Thanks, Dalespam for clarifying the confusion and wiping off the vague, ambiguous, non scientific "negligible". No /zero energy is surely more acceptable than "arbitrarily small", which poses questions about" how negligible" is the energy required to rotate a toy vs. an anti-roll gyro.

I believe that it is true, if I find a rational, plausible explanation for the aspects that seem to contradict that, which I have already pointed out:
- if it takes 0 energy to rotate the gyro by 90° it should take 0*4 energy to rotate it by 360° or 3600°, why does it stop in its track when the weight is lifted ( at 1:45/49) ?
Do you know of any other case in nature where you can make something move with zero energy?

 

[bobie]

To remove the effort vs work confusion,...

I have a unique chance here to understand the concept of "work": its utility/necessity escapes me.
Effort is of course a lay substitute for energy, so let's talk about "energy spent" vs. "work".

What is the purpose, utility of introducing this distinction?
if I push a 1-Kg ball and spend 1/2J it moves at 1m/s, I have done 1/2J of work , right?
if the ball is at 1 cm from a wall it stops after a fraction of a second, but I still did work
if the ball is at 1 mm or a micron from a wall, have I done work? where do you draw the line? can you draw a line at all? It becomes a sort of philosophical issue:
in a cradle the first ball does no work on the second but does work on the last? If it is so, work can travel , like energy, through matter? [edit: what if there is 1mm distance (or less/more) between each ball? can all energy be tranferred in a tiny distance?]
if the ball hits the wall, 1/2J of energy is tranferred to the wall but it has done no work;
if it makes a tiny dimple it has done work, what if the dimple is invisible?

Thanks again!
Edit: added in red the omitted figure, sorry for the mistake

 

[Nugatory]

What is the purpose, utility of introducing this distinction?
if I push a 1-Kg ball and spend 1J it moves at 1m/s, I have done 1J of work , right?

If you push a 1kg ball and spend 1J, it won't be moving at 1m/sec, it will be moving at $sqrt{2}$ m/sec. This is true whether you use a very gentle push over a very long distance, or a very hard push over a very short distance, or anything in between.

if the ball is at 1 cm from a wall it stops after a fraction of a second, but I still did work
if the ball is at 1 mm or a micron from a wall, have I done work?

Yes. But I do not want to hear another question from you until after you have calculated the force required to do 1J of work (accelerate the 1kg ball from rest to a speed of $\sqrt{2}$ m/sec over a distance of 1cm, 1mm, and one micron. You need to do that before you can take on your next questions:

where do you draw the line? can you draw a line at all? It becomes a sort of philosophical issue:

in a cradle the first ball does no work on the second but does work on the last? If it is so, work can travel , like energy, through matter?

The next to last ball does work on the last ball. What's traveling through the balls is a force: each ball pushes on the one next to it. This force transfers energy and momentum to the last ball in the cradle.[/QUOTE]

 

[A.T.]

so let's talk about "energy spent" vs. "work".

Energy is more general. Mechanical work is one form of energy transfer.

if the ball hits the wall, 1J of energy is tranferred

That is not very likely. In an inelastic collision energy will be dissipated as heat. In an elastic collision the ball will bounce back, and still have energy.

 

[jbriggs444]

Thanks, Dalespam for clarifying the confusion and wiping off the vague, ambiguous, non scientific "negligible". No /zero energy is surely more acceptable than "arbitrarily small", which poses questions about" how negligible" is the energy required to rotate a toy vs. an anti-roll gyro.

So you do not wish to use "negligible" or "arbitrarily small". But apparently you are comfortable with the idealizations of perfect efficiency, frictionless bearings and the limiting case of high rotation rates and low precession rates so that "zero" is accurate.

- if it takes 0 energy to rotate the gyro by 90° it should take 0*4 energy to rotate it by 360° or 3600°, why does it stop in its track when the weight is lifted

Precession requires the application of torque. Remove the torque and you stop the precession.
The application of torque does not imply the supply of non-zero work.

Do you know of any other case in nature where you can make something move with zero energy?

If I understand this remark correctly, you are complaining that the precession of the gyroscope preserves the original (rapid, on-axis) rotation while superimposing an additional (slow, off-axis) rotation of its own. You have referred to this kind of situation several times with phrases like "rotating on two axes".

You note that an object that is moving in such a fashion clearly has more kinetic energy than one that is simply spinning in place. You complain that this ought not occur if no work has been done to supply that extra energy.

That is a fair complaint. But it flies in the face of your earlier unwillingness to use terms such as "negligible" and "arbitrarily small".

The amount of external work that need be supplied varies with the precession rate that is imposed. If you are willing to allow one hour to precess the gyroscope through 90 degrees, the kinetic energy associated with the precession rate is small. If you are willing to allow one day it will be smaller. If you are prepared to allow an arbitrarily long time the required work becomes arbitrarily small.

It was this fact that led me to use terms such as "negligible" or "arbitrarily small" in the first place.

 

[Dale]

Thanks, Dalespam for clarifying the confusion and wiping off the vague, ambiguous, non scientific "negligible". No /zero energy is surely more acceptable than "arbitrarily small", which poses questions about" how negligible" is the energy required to rotate a toy vs. an anti-roll gyro.

There was nothing vague, ambiguous, or non-scientific about jbriggs444's use of the term "negligible". He was saying the same thing that I was and, in a scientific paper, would have been clearly understood by the scientific community. I assumed an ideal frictionless bearing. He recognized that such a bearing is not possible, but through careful engineering you can get so close that the difference can be neglected, hence "negligible".

- if it takes 0 energy to rotate the gyro by 90° it should take 0*4 energy to rotate it by 360° or 3600°,

That is correct.

why does it stop in its track when the weight is lifted ( at 1:45/49) ?

Because the torque causing the precession is lifted.

Do you know of any other case in nature where you can make something move with zero energy?

Yes, many.

 

[Dale]

What is the purpose, utility of introducing this distinction?

The purpose is to focus the discussion on quantitative mechanical concepts that can be rigorously analyzed (work) rather than on subjective or biological things like muscle fatigue or calories burned (effort spent).

if I push a 1-Kg ball and spend 1/2J it moves at 1m/s, I have done 1/2J of work , right?

No. You are a very inefficient machine. If you spend 1/2 J of energy then you will do much less than 1/2 J of work.

if the ball is at 1 cm from a wall it stops after a fraction of a second, but I still did work
if the ball is at 1 mm or a micron from a wall, have I done work? where do you draw the line? can you draw a line at all? It becomes a sort of philosophical issue:

I despise philosophical issues and I am not about to enter into such a debate. If you are interested in learning physics, then I am glad to help. If you are done learning physics, then I am glad to close the thread.

in a cradle the first ball does no work on the second but does work on the last? If it is so, work can travel , like energy, through matter? [edit: what if there is 1mm distance (or less/more) between each ball? can all energy be tranferred in a tiny distance?]
if the ball hits the wall, 1/2J of energy is tranferred to the wall but it has done no work;
if it makes a tiny dimple it has done work, what if the dimple is invisible?

An impulsive force certainly can do work. However, this seems very off-topic for the current thread.

 

[bobie]

Because the torque causing the precession is lifted..

Thanks, Dalespam, if I apply a torque to a merry-go-round, when I lift the torque it goes on spinning for a while until it stops because of friction. What I meant is : here we have an almost frictionless gimbal, why does it behave differently?

As to work, no philosophy, practical questions: if a ball hits a wall on an inelastic collision the KE is dissipated as heat, no work done, but the joule is also the unit of heat, so: why if it is passed on to another ball work is done and if it is transformed into heat no work is done?. But if this is off-topic, I'll ask these questions in another thread.

Thanks again.

 

[Dale]

Thanks, Dalespam, if I apply a torque to a merry-go-round, when I lift the torque it goes on spinning for a while until it stops because of friction. What I meant is : here we have an almost frictionless gimbal, why does it behave differently?

In a merry-go-round the torque is applied parallel to the angular momentum. In the gimbal video the torque is applied perpendicular to the angular momentum.

If you go back to the first two equations I posted in post 72 you can see that a torque parallel to the angular momentum will cause the angular velocity to increase, while a torque perpendicular to the angular momentum will cause precession. Both the observed behavior of the merry-go-round and the observed behavior of the gimbal are explained by the same equations. They do not, in fact, behave differently.

But if this is off-topic, I'll ask these questions in another thread.

I think that is best. Mixing the angular momentum topics with the linear collision topics is likely to be confusing.

 

[A.T.]

...merry-go-round, when I lift the torque it goes on spinning for a while ...here we have an almost frictionless gimbal, why does it behave differently?

The same rule applies to both cases:
If the angular momentum vector changes, then there must be a non-zero net torque.

A vector has magnitude and direction:
-To change angular momentum magnitude you need a torque. (changing merry-go-round RPMs)
-To change angular momentum direction you need a torque. (changing gyro axis orientation)

 

[Dale]

I assumed an ideal frictionless bearing. He recognized that such a bearing is not possible, but through careful engineering you can get so close that the difference can be neglected, hence "negligible".

I should also mention that I also assumed that the angular velocity about the axis of the gyro is so great that the angular momentum can, at all times, be considered exactly parallel to the axis of the gyro. That assumption allows the moment of inertia to be treated as a scalar rather than a tensor.

Again, you cannot do that in reality, but through careful engineering you can get pretty close. I would make the ideal assumptions, perform the analysis, and say "0" while jbriggs444's would not make the ideal assumptions, would perform the exact same analysis, and say "negligible".

The scientific community would understand either approach. Mine would tend to be used for textbooks, and jbriggs444's would tend to be used for manuscripts.

 

[mattt]

- what happens when she turns the plane of rotation by 90°

Imagine you call girl's (at the beginning) left as X-axis, girl's backwards as Y-axis and upwards is Z-axis.

Then at the beginning (I mean, when the wheel is spinning vertically) the total angular momentum of the system (system = platform + girl + wheel) is equal to the angular momentum of the wheel, which is non-zero, something like (L, 0, 0) with respect to that XYZ-frame.

If she were isolated in outer space (no need of the platform in this case), as she tries to change the plane of rotation (just as she does in the video) she herself would start to spin, both with a Z-component (negative or downwards) and with a X-component (positive). Why?

Because as she is trying to change the plane of rotation of the wheel, what is happening with the angular momentum of the wheel is that it is changing from something like (L,0,0) to something like (L-a, 0, b) with "a" and "b" being positive given numbers. So the angular momentum of the girl (in outer space we don't need platform at all) will change in exactly the opposite sense, from (0,0,0) at the beginnig to something like (a, 0, -b).

In the real case of the video, she is not in outer space, so the Earth (the friction of her feet with the Earth surface) does not allow her that X-component of her spin ( the "a" in (a, 0, -b) ).

 

[bobie]

Yes, many.

Could you give me a few examples, so that I can put this into a general frame?

. In the gimbal video the torque is applied perpendicular to the angular momentum.
... a torque parallel to the angular momentum will cause the angular velocity to increase, while a torque perpendicular to the angular momentum will cause precession.

In video #18 (here:http://www.gyroscopes.org/1974lecture.asp) we can see an instance when a perpendicular torque moves the gyro in that direction but it keeps moving when the torque is lifted, what is the difference? still no energy spent?

 

[A.T.]

Do you know of any other case in nature where you can make something move with zero energy?

"Make something move" is vague. You don't "make the gyro move", you just change it's state of movement. And that is possible without energy, for example: Applying a force perpendicular to velocity changes the velocity vector, without doing work or requiring energy.

 

[bobie]

": Applying a force perpendicular to velocity changes the velocity vector, without doing work or requiring energy.

This is basically the same case: L pseudovector, p, v vector, any other examples? DaleSpam said there are many

 

[A.T.]

any other examples?

Examples for what exactly? Not in vague laymen terms please.

 

[Dale]

Could you give me a few examples, so that I can put this into a general frame?

Any time the initial energy is equal to the final energy you can devise an ideal machine which will accomplish the motion with no expenditure of energy. Some of those will involve energy which is input and then extracted, or others can be entirely "iso-energetic".

For example, sliding a block to a different location on a frictionless level surface. Changing direction at a constant speed on a level surface. Circular orbits. Etc.

In video #18 (here:http://www.gyroscopes.org/1974lecture.asp) we can see an instance when a perpendicular torque moves the gyro in that direction but it keeps moving when the torque is lifted, what is the difference? still no energy spent?

This is called torque free precession. It occurs when the moment of inertia changes over time, e.g. when it is spinning about an axis which is not an axis of symmetry as is the case here.

In such cases you cannot treat the moment of inertia as a scalar and you have to use the whole tensor. I recommend that you learn the simple cases before the complicated ones. Please concentrate on the equations I posted in post 72. Do you understand those? If so, then try to apply them to some simple scenarios involving simple rigid bodies spinning about their axis of symmetry.

Please stop simply posting random videos to analyze. The number of videos available on the internet far outstrips my desire to analyze them. Furthermore, you need to study simple cases and not seek complicated ones. Look to textbooks or coherent presentations, not random videos.

 

[bobie]

, you need to study simple cases and not seek complicated ones. ... not random videos.

I know, you're telling me to wind this up!
Please, just tell me when to stop.
For future posters' sake I'd hate do cause the closure.
Thanks

 

[bobie]

Do you know of any other case in nature where you can make something move with zero energy?

"Make something move" is vague.
Applying a force perpendicular to velocity changes the velocity vector, without... requiring energy.

This example is not appropriate, because the body is already moving, and you are not increasing its speed or KE (same applies to orbits, etc.).

I have repeatedly and clearly explained what I mean :
if an object is rotating in one plane it has k KE, if you make it rotate in 2 different planes it has undeniably KE > k, somebody must have given it some KE and therefore must have spent some energy. Is that vague to you? Is this wrong or arguable in any case?

The link I gave in post # 92 is not a random video, I posted to show visually what seemed so hard to get across: the gyro (after the jerk/twist by Laithwaite,) spins in two different planes at the same time. Can we say that it required no energy?

In this case it goes on spinning longer than in the previous (post #80: at 1:45/49) because friction/inertia/ or other force does not slow it down, (or as Dalespam says : you can use a scalar and not the whole tensor), but can we say that in the previous case it required less or no energy? In both cases the force has been applied perpendicularly to the plane of rotation.
That is what I was talking about.

Thanks , anyway, for your efforts.

 

[jbriggs444]

I have repeatedly and clearly explained what I mean :
if an object is rotating in one plane it has k KE, if you make it rotate in 2 different planes it has undeniably KE > k, somebody must have given it some KE and therefore must have spent some energy. Is that vague to you? Is this wrong or arguable in any case?

A proper treatment of rotation in three dimensions does not use the term "rotation in 2 different planes". Rigid rotation in three dimensions is always a vector. It has roll rate, pitch rate and yaw rate. There is a single plane that is perpendicular to the rotation vector.

But rotation in three dimensions is complicated. The rotation vector of an object can change even when it is under no external torques. An easy experiment is a pencil that is rapidly spinning about its long axis and given a slow rotation from end to end as it is tossed into the air. From an untutored perspective, one could describe it as having "rotation in two planes" (which both change over time). The perspective that Dale would use would describe it as rotating about a single plane not aligned with any axis of symmetry. That plane can change over time.

Despite the change in the rotation vector over time, angular momentum is conserved. Because in this regime, the objects moment of inertia is not just a simple scalar. It is a tensor (as Dale pointed out in post 90).

The link I gave in post # 92 is not a random video, I posted to show visually what seemed so hard to get across: the gyro (after the jerk/twist by Laithwaite,) spins in two different planes at the same time. Can we say that it required no energy?

In this case it goes on spinning longer than in the previous (post #80: at 1:45/49) because friction/inertia/ or other force does not slow it down, (or as Dalespam says : you can use a scalar and not the whole tensor), but can we say that in the previous case it required less or no energy? In both cases the force has been applied perpendicularly to the plane of rotation.

In the case of the jerk/twist, the force will NOT have been applied perpendicularly to the plane of rotation. It may have been applied perpendicularly to the plane in which the object was rotating initially. It will not, in general, be perpendicular to the plane in which the object is rotating after having been jerked/twisted. Accordingly, the jerk/twist can do work and can achieve a change in rotational kinetic energy.

I have been careful to talk about what can happen as we approach an ideal case where the applied force is gentle and and at right angles to the momentary axis of rotation. Dale has been careful to talk about what does happen in the ideal limiting case where the applied force is gentle and at right angles to the momentary axis of rotation.

You've been asked before to become comfortable with ordinary linear mechanics rather than going on about gyroscopes. In linear mechanics the analogue to a jerk/twist is that of an impulsive force. That is usually discussed in the context of collisions.

Suppose that you have a ball moving from west to east on a pool table as it is hit by the cue ball moving from south to north and striking the target ball exactly at right angles. The target ball will continue to move eastward at the same rate as before. But it also acquires a velocity component in the northward direction.

The impulsive force was applied at right angles to the ball's motion. No work should have been done (you suggest). So where did the extra kinetic energy come from?

Try to answer this riddle before we come back to jerks and gyroscopes.

 

[bobie]

Suppose that .. But it also acquires a velocity component in the northward direction.
The impulsive force was applied at right angles to the ball's motion. No work should have been done (you suggest). So where did the extra kinetic energy come from?
Try to answer this riddle before we come back to jerks and gyroscopes.

I am not the one, who ever suggested that no work is done, jbriggs, where did you get that impression? I have been explicitly accused of denying that:

And you just keep contradicting it. So what is the point of telling you something.

It is no riddle, the ball will move in a north-east direction and the angle will be determined by a parallelogram and the M/m ratio , and the its speed and KE will increase accordingly. What is the problem? Changing the direction and increasing KE of the vector required KE/work/energy.
Now suppose your ball hits perpendicularly the edge of a spinning wheel on gimbals like in video #18, it starts to rotate or not, does it gain speed and KE in that direction or not, does the total KE of the gyro increase or not, has work been done or not?