What is the energy/work required to turn something?

[frankencrank]

All correct.

This is often true, but not always. There are many "passive" forces, and you have already been given several examples. Tension in a rope, gravity, magnets, electrostatic forces, etc. can all apply a force without doing work or requiring energy input.

Actually, all materials stretch some when force is applied. This means work is being done even though the distance may be small in a kevlar rope compared to a rubber band. Physics problems usually assume no stretching to make solutions easier and help people to see concepts but in many applications it is necessary to take into account these concerns.

 

[Dale]

You are just grasping for straws now. Even if you have a very elastic rope that stretches a great deal in order to achieve a specific tension, once it has stretched it can continue to apply that tension indefinitely without further energy input. In that condition it can be used to turn an object without energy for an arbitrary amount of time and through an arbitrary arc. The energy is required for stretching the rope, not for using the rope to turn something.

Energy is simply not required to turn. Any turning mechanism which uses energy does something else besides turning (e.g. heat, high velocity exhaust, material stress and strain, etc.)

 

[frankencrank]

You are just grasping for straws now. Even if you have a very elastic rope that stretches a great deal in order to achieve a specific tension, once it has stretched it can continue to apply that tension indefinitely without further energy input. In that condition it can be used to turn an object without energy for an arbitrary amount of time and through an arbitrary arc. The energy is required for stretching the rope, not for using the rope to turn something.

Energy is simply not required to turn. Any turning mechanism which uses energy does something else besides turning (e.g. heat, high velocity exhaust, material stress and strain, etc.)

Grasping at straws, not at all. It depends upon what that rope is attached to. It it is attached to another equal particle then you are right. Then we have a spinning disk scenario. But, that is not the scenario we were examining. We were looking at turning a single mass. So, in this instance, if the other end of the rope (or rubber band, or whatever), if it is attached to a central pole, will cause the central pole to move and deviate as the mass moves around it and work will be performed and energy will be lost. The amount lost will depend on the amount of deviation which will depend upon the stiffness of the pole.

 

[Dale]

Energy is simply not required to turn. Any turning mechanism which uses energy does something else besides turning (e.g. heat, high velocity exhaust, material stress and strain, etc.)

if the other end of the rope (or rubber band, or whatever), if it is attached to a central pole, will cause the central pole to move and deviate as the mass moves around it and work will be performed and energy will be lost. The amount lost will depend on the amount of deviation which will depend upon the stiffness of the pole.

This is no different than a stretchy rope. The energy is lost to material stress and strain here, not turning. I already mentioned this above.

I challenge you to find any example where the energy lost goes into the turning itself and not into something else.

 

[frankencrank]

This is no different than a stretchy rope. The energy is lost to material stress and strain here, not turning. I already mentioned this above.

I challenge you to find any example where the energy lost goes into the turning itself and not into something else.

The example cited works for your challenge. Yes, the energy is lost in material stress and strain of the pole. But, it is the pole that is providing the force to do the turning. Therefore, energy is being lost from the system and so there is an energy cost to the turning. If the turning had not occurred, the energy would not have been lost.

If you can find me a real world example (other than a spinning disk) where it is possible to exert a force without also causing an concomitant material stress and strain loss then you have a point. This is the real world, it is impossible. Those losses can be small, but they will always be there.

 

[Dale]

If you can find me a real world example (other than a spinning disk) where it is possible to exert a force without also causing an concomitant material stress and strain loss then you have a point. This is the real world, it is impossible.

Circular orbits for one. (I also don't think it is reasonable for you to exclude a spinning disk)

I think you are completely missing my point. Let's say that you have a real-world turning mechanism that requires some amount of energy, Etotal. Let's say further that you determine how much heat energy, Eheat, is generated and how much kinetic energy (e.g. rocket exhaust), Ekinetic, and how much mechanical strain energy, Estrain, etc. Then you could determine how much energy was required for the turn itself by:

Eturn = Etotal - Eheat - Ekinetic - Estrain - Eetc

Do this for ANY proposed turning mechanism and you will find that Eturn = 0 and so there is no energy required for the turn itself. And you have already been given examples where Etotal = 0, further indicating that there is no energy required for turning. And in addition the conservation of angular momentum inherently implies that turning does not require energy.

Further, none of the Etotal, Estrain, etc are constant for all turns. Only Eturn = 0 is constant for all turns, and therefore only Eturn = 0 represents any kind of general rule for the amount of energy required to turn.

 

[frankencrank]

Circular orbits for one. (I also don't think it is reasonable for you to exclude a spinning disk)

I think you are completely missing my point. Let's say that you have a real-world turning mechanism that requires some amount of energy, Etotal. Let's say further that you determine how much heat energy, Eheat, is generated and how much kinetic energy (e.g. rocket exhaust), Ekinetic, and how much mechanical strain energy, Estrain, etc. Then you could determine how much energy was required for the turn itself by:

Eturn = Etotal - Eheat - Ekinetic - Estrain - Eetc

Do this for ANY proposed turning mechanism and you will find that Eturn = 0 and so there is no energy required for the turn itself. And you have already been given examples where Etotal = 0, further indicating that there is no energy required for turning. And in addition the conservation of angular momentum inherently implies that turning does not require energy.

Further, none of the Etotal, Estrain, etc are constant for all turns. Only Eturn = 0 is constant for all turns, and therefore only Eturn = 0 represents any kind of general rule for the amount of energy required to turn.

You can look at it however you want. I look at it as kinetic energy is being lost from the system in the form of heat due to the turning. I take that as the turn requiring energy as it would not be there except for the turn.

Actually, except under special circumstances, even orbiting satellites require energy to maintain their orbits, although the losses would be extremely small. The losses come about because of tidal drag from the gravitational tether. These losses are probably zero, however, in the case of geosynchronous orbits over the equator, because there would be no tidal drag. The moon is gaining energy from the Earth because of tidal drag due to the Earth's rotation being faster than the moons rotation but if the Earth did not rotate, the moon would be losing energy and coming closer to the Earth due to these tidal forces and no one would doubt these forces and losses exist.

 

[Dale]

You can look at it however you want. I look at it as kinetic energy is being lost from the system in the form of heat due to the turning. I take that as the turn requiring energy as it would not be there except for the turn.

So why don't you answer your own question then. What is "the energy required to turn something"? At a minimum it should be expressable as a function of the arc of the turn or the angular velocity of the turn. Otherwise it certainly isn't energy required for the turning.

Actually, except under special circumstances, even orbiting satellites require energy to maintain their orbits, although the losses would be extremely small. The losses come about because of tidal drag from the gravitational tether.

Tidal lock is hardly uncommon, and even if it were it is still an example of turning without energy loss.

 

[frankencrank]

So why don't you answer your own question then. What is "the energy required to turn something"? At a minimum it should be expressable as a function of the arc of the turn or the angular velocity of the turn. Otherwise it certainly isn't energy required for the turning.

Tidal lock is hardly uncommon, and even if it were it is still an example of turning without energy loss.

I have answered it. I did it earlier in this thread. It is not possible to know the energy losses assoicated with a turn without knowing the physical properties of the object that is providing the force to cause the turn. If one assumes a rigid object, then the losses are zero. Under almost all other circumstances it seems there must be some, albeit usually very small, losses.

 

[Dale]

It is not possible to know the energy losses assoicated with a turn without knowing the physical properties of the object that is providing the force to cause the turn.

That's fine, use whatever variables you need to describe the relevant physical properties.

 

[YellowTaxi]

Interesting discussion. (or at least I think so ;-)

from the original question:

...
But, take the case of a single particle, say a spaceship in deep space. It will tend to move in a straight line unless a force is applied to change its direction. Fire a rocket normal to the direction of travel and the spaceship will travel in a circle. Clearly, the energy required to do this is not zero. Wouldn't the same analysis apply to turning a car or bicycle or ourselves? How do we calculate the energy cost of turning a single moving particle in a circle knowing the mass, speed, and turning radius through an arc of x radians?

I liked this topic and thought it might be interesting to try and make some sense of what's going on by transfering to the point of view of a second spaceship initially traveling along-side this first at exactly the same speed.

So in effect both spaceships are then standing still out in space..

.) At t=0 The rocket motors on the side of spaceship #1 are turned on, and it starts to move away from us guys watching from inside our 'straight line' spaceship #2

i) After a short time we see it also starts to move backward as it moves away from us.
Because it's accelerated to move in a circle (relative to the fixed stars).

ii) After the same time again we notice it's a little further out, and also dropped back behind us some distance,
iii) And after the same interval again we see it starts to come back into us, behind the rear of our spaceship #2, But it's not catching us up at all. And it's really quite a long way back.
iv) And after the same time interval again we see it's ended up directly behind the rear of our spaceship #2, but a long way back.

If we trace out the path of spaceship#1 (as seen from inside our spaceship#2), it actually looks like a perfect cycloid. It's definitely not a circle.
http://mathworld.wolfram.com/Cycloid.html

Spaceship#1 moved away from us initially, dropped behind some, then finished directly behind us by quite a distance. (And of course it kept on tracing this cycloid for as long as the rocket motors were turned on).

It's only a circular path when viewed from one very privileged frame. But from the frame of reference of the original spaceship trajectory it's going to be a perfect cycloid.

So the question can now be re-phrased: how much energy is actually required to make an object move in a perfect cycloid ?
http://mathworld.wolfram.com/Cycloid.html

My maths and physics is really rusty but maybe if you take the co-ords given by wolfram as the equation of motion you can compute the energy required.
And I'm pretty shure it won't be zero.

Hope that was clear. Sorry if you think that's just junk that I've posted, but I thought it was quite a good way to maybe get a meaningful solution to frankencrank's interesting question.

 

[Dale]

It's only a circular path when viewed from one very privileged frame. But from the frame of reference of the original spaceship trajectory it's going to be a perfect cycloid.

Your comment is quite true, but energy and momentum are frame-variant quantities so you cannot really switch frames and answer the question. The frame of reference where it goes in a circular path is the only frame where the spaceship is simply turning and not changing speed. In any other frame it is obvious that energy is involved since the speed is changing and therefore the kinetic energy is also changing.

.

 

[YellowTaxi]

Your comment is quite true, but energy and momentum are frame-variant quantities so you cannot really switch frames and answer the question. The frame of reference where it goes in a circular path is the only frame where the spaceship is simply turning and not changing speed. In any other frame it is obvious that energy is involved since the speed is changing and therefore the kinetic energy is also changing.

Thanks for the reply.
But I didn't switch frames half way through did I. I said let's start from the position where initial kinetic energy = 0 and see what happens.
Your last sentence agrees that kinetic energy is required for any frame which doesn't see the motion as a circle even though its the same motion. So we can say the "No energy input required" is purely frame dependent. Although I understand what you said, I think that's a pretty strange law in physics, and it must cause occasional problems or confusion. Like here for instance ;-)

Another point,
I think your comment earlier to frank~~ was that circular orbits require no energy input and continue indefinitely. How is a circular orbit any different from an elliptical orbit around the same planet? They both return to the same starting point (indefinitely) but velocity for the ellipse (kinetic energy) is obviously changing all the time...
Unless of course you move to the vantage point where the ellipse looks like a perfect circle. But then the circular orbit will look like an ellipse... :-(

 

[Dale]

Thanks for the reply.
But I didn't switch frames half way through did I. I said let's start from the position where initial kinetic energy = 0 and see what happens.

Yes, I know you didn't switch frames halfway through your analysis, and in the frame you chose your analysis seems correct to me. My point wasn't that you did anything wrong, but simply that because you did your analysis in a different frame it does not answer the original question (since the answer itself is frame variant).

Your last sentence agrees that kinetic energy is required for any frame which doesn't see the motion as a circle even though its the same motion. So we can say the "No energy input required" is purely frame dependent. Although I understand what you said, I think that's a pretty strange law in physics, and it must cause occasional problems or confusion. Like here for instance ;-)

Why is that strange? All sorts of things are frame variant: time, space, velocity, momentum, wavelength, etc. Why should energy be frame invariant? Note that invariance and conservation are different concepts.

I think your comment earlier to frank~~ was that circular orbits require no energy input and continue indefinitely. How is a circular orbit any different from an elliptical orbit around the same planet? They both return to the same starting point (indefinitely) but velocity for the ellipse (kinetic energy) is obviously changing all the time...
Unless of course you move to the vantage point where the ellipse looks like a perfect circle. But then the circular orbit will look like an ellipse... :-(

As frank noted earlier there can be some small amount of energy lost to tidal strains. The kind of orbit without such strains is when the bodies are tidally locked to each other, which implies a circular orbit.

 

[YellowTaxi]

All sorts of things are frame variant: time, space, velocity, momentum, wavelength, etc. Why should energy be frame invariant? Note that invariance and conservation are different concepts.

True, but this questionable (IMHO) law has nothing whatsoever to do with relativity theory. And as far as I know gen rel has trouble dealing with the energy conservation law anyway. not sure why, maybe black holes or whatever..

I think the basis of the idea is that no energy's required to travel in the circle because it's always recovered when the object returns to its starting point, and with the same speed and orientation. And partly because it doesn't appear to speed up or slow down (erm, if viewed from that frame where it DOES look like a circle, not a cycloid or anything else..)

Anyway, whatever, I realized after re-reading frankenkrank's idea of using a single rocket motor to give the ship a circular orbit that it wouldn't work [Or at least it would be very difficult to make it work]. The ship would likely just move in a parabola. You would need at least one more rocket to make the spaceship rotate at just the right angular speed - ie a speed identical to the required orbit angular speed. I hadn't thought of that before, so at least I've learned something from this topic...:-)
It made me realize why a ball hanging on a string wobbles so much at each end of the swing. It's trying to get rid of it's spin before it travels back the other way. And why when a racing motorcyclist falls off his motorbike in mid-corner, he'll spin into the gravel trap rather than simply fly there face first...

 

[Dale]

True, but this questionable (IMHO) law has nothing whatsoever to do with relativity theory.

You are right, relativity (special nor general) is not needed here. If you remove "time" from my previous list, all of the other things are frame-variant in Newtonian physics. I shouldn't have mentioned time because it is irrelevant as you mentioned and just adds needless confusion to this thread.

 

[YellowTaxi]

You are right, relativity (special nor general) is not needed here. If you remove "time" from my previous list, all of the other things are frame-variant in Newtonian physics. I shouldn't have mentioned time because it is irrelevant as you mentioned and just adds needless confusion to this thread.

Mentioning gravity at all was the first mistake, as frank said some time ago..

 

[IwanWatt]

I have been following this discussion with some interest. I considered a spacecraft moving in a straight line at a given velocity. I tried to work out how much energy would be needed to make it turn, say through an angle of 90 degrees on a circular arc, keeping the speed, or magnitude of the velocity constant. Of course, I can see that theoretically, no work is done. On the other hand, the turn will require a force that can be geneerated a number of ways. By using a rocket burn, the manouver will require energy to be used up. On the other hand, if the rocket could shoot out a non-extensible line to a massive asteroid somewhere nearby and use the asteroid as a fixed point around which to rotate, the turn might be completed with zero or at least much smaller energy expenditure.
Following on from this, what about another example more related to biology? What if we considered a bird flying along at constant speed and making a turn, maybe to go after some food? Another example might be a fish swimming in the sea? It appears reasonable to suggest that the bird/fish/animal has to expend some energy to make a turn. (Although I can accept that, according to physics, this energy expenditure is not mandatory)
Is there some way we could make an estimate of the energy expenditure required by a bird or a fish? I have been considering a bird of mass 0.5kg and flying at 11m/s. Is there some way I could make some realistic estimates of the energy cost of turning and compare that with the normal cost of flying/swimming in a straight line? Even an order-of-magnitude calculation would be useful to me.