What is the energy/work required to turn something?

[frankencrank]

Normally one would think that the work required to move something in a circle is zero since the force is directed normal to the direction of movement and kinetic energy doesn't change. This is easy to understand using a spinning disk, where every particle has a cohort what is changing momentum in the opposite direction such that momentum is conserved.

But, take the case of so 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?

 

[Doc Al]

But, take the case of so 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.

The force of the rocket pushing the spaceship in a circle does no work on the spaceship. (Does the energy of the spaceship change?) But it certainly takes energy to produce the rocket thrust.

Wouldn't the same analysis apply to turning a car or bicycle or ourselves?

Depends on what's doing the pushing. If it's a passive force like friction, no energy is required to maintain the force.

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?

The force making something go in a circle does no work. But creating and maintaining that force might require energy.

 

[frankencrank]

The force of the rocket pushing the spaceship in a circle does no work on the spaceship. (Does the energy of the spaceship change?) But it certainly takes energy to produce the rocket thrust.

Depends on what's doing the pushing. If it's a passive force like friction, no energy is required to maintain the force.

The force making something go in a circle does no work. But creating and maintaining that force might require energy.

But, that is exactly the point. The speed of the object does not change so the energy does not change but the velocity does, because the direction changes. We are changing the momentum without changing the kinetic energy. Changing momentum take energy. The energy must come from somewhere. If there is no source of the energy (like a rocket on the space ship) then a "passive" energy like friction should slow the object down. It seems to me the amount of energy that must be added or would otherwise be lost should be calculable but I can't figure out how to do it.

 

[Doc Al]

But, that is exactly the point. The speed of the object does not change so the energy does not change but the velocity does, because the direction changes. We are changing the momentum without changing the kinetic energy. Changing momentum take energy.

No it doesn't. It takes force, not energy.

A satellite orbits the earth, its momentum continually changing. Does it require energy to maintain its orbit?

 

[frankencrank]

No it doesn't. It takes force, not energy.

A satellite orbits the earth, its momentum continually changing. Does it require energy to maintain its orbit?

Where does the force come from? We are not talking satellites here, we are talking what it takes to change something from moving in a straight line to a circular path without invoking gravity. Even invoking gravity requires energy as gravity comes from mass and mass is energy. How can one develop a force without there being some energy involved. How can the passenger in a spaceship know whether they are sitting on the ground in a gravitational field, an elevator accelerating up, or a spaceship moving in a circle. The passenger feels exactly the same thing in each scenario yet some involve substantial "work" in the classic sense and some do not, depending on the frame of reference but we should be able to get the same answer if we can account for the differences. How do you explain the squeal of tires when a car is cornering if there is no energy transfer/loss involved?

Further, it is not clear to me that satellites are not requiring energy to make their circles. The moon speeds up because its orbit is slower than the Earth is spinning so energy is transferred between the two from the tidal influences. Other satellites that orbit faster than the Earth slow down I suspect for the same reason. It is not obvious because the energy requirements are small compared to the energy they contain. The classical physics approach usually requires objects to be rigid, to simplify the solution. Rigid bodies do not exist in the real world.

Anyhow, I don't think this is as straight forward as it is sometimes presented. I would like to understand what is going on whether it is straight forward or not.

 

[Doc Al]

Where does the force come from? We are not talking satellites here, we are talking what it takes to change something from moving in a straight line to a circular path without invoking gravity. Even invoking gravity requires energy as gravity comes from mass and mass is energy.

You are really stretching here.

How can one develop a force without there being some energy involved. How can the passenger in a spaceship know whether they are sitting on the ground in a gravitational field, an elevator accelerating up, or a spaceship moving in a circle. The passenger feels exactly the same thing in each scenario yet some involve substantial "work" in the classic sense and some do not, depending on the frame of reference but we should be able to get the same answer if we can account for the differences.

I don't know what you're talking about here or how it relates to your question.

How do you explain the squeal of tires when a car is cornering if there is no energy transfer/loss involved?

The tires slip a bit and work is done. I don't see the relevance to your original question.

Further, it is not clear to me that satellites are not requiring energy to make their circles. The moon speeds up because its orbit is slower than the Earth is spinning so energy is transferred between the two from the tidal influences. Other satellites that orbit faster than the Earth slow down I suspect for the same reason. It is not obvious because the energy requirements are small compared to the energy they contain. The classical physics approach usually requires objects to be rigid, to simplify the solution. Rigid bodies do not exist in the real world.

Again, I don't see the relevance of tidal forces to your question. Note that work is being done and energy is being transferred. If something moves in a circle at constant speed, no work is done on it.

Anyhow, I don't think this is as straight forward as it is sometimes presented. I would like to understand what is going on whether it is straight forward or not.

It's not clear what your question is. (But I agree--rarely are things easy and straightforward!)

 

[frankencrank]

You are really stretching here.

I don't know what you're talking about here or how it relates to your question.

The tires slip a bit and work is done. I don't see the relevance to your original question.


Again, I don't see the relevance of tidal forces to your question. Note that work is being done and energy is being transferred. If something moves in a circle at constant speed, no work is done on it.


It's not clear what your question is. (But I agree--rarely are things easy and straightforward!)

Let's just go to the spaceship question. Let's assume we have a spaceship of 100 kg mass moving at 20 m/sec in deep space so there are no extraneous influences on the ship that might cause it to deviate from a straight line course. Then we decide we want to go home so we fire a rocket that has a force of 10 Newtons and we ensure it is always perpendicular to the current direction of motion to turn us around 180º so we can go home.

1. Even though we are not increasing the total energy of the spaceship in this maneuver, can we calculate from this how much energy the rocket is delivering to space (how much of its total available energy has it lost)?

2. Can we calculate from this what the turning radius of the spaceship would be?

 

[americanforest]

But, that is exactly the point. The speed of the object does not change so the energy does not change but the velocity does, because the direction changes. We are changing the momentum without changing the kinetic energy. Changing momentum take energy. The energy must come from somewhere. If there is no source of the energy (like a rocket on the space ship) then a "passive" energy like friction should slow the object down. It seems to me the amount of energy that must be added or would otherwise be lost should be calculable but I can't figure out how to do it.

Changing the momentum doesn't necessarily require any work to be done. In uniform circular motion, the momentum is constantly changing but there is no work done by the centripetal force because W=\vec{F} \bullet {\vec{D}}. The energy depends on the magnitude of the velocity, not its direction.

PS. Is there a better way to do dot product in latex?

 

[frankencrank]

Changing the momentum doesn't necessarily require any work to be done. In uniform circular motion, the momentum is constantly changing but there is no work done by the centripetal force because W=\vector{F} \bullet \vector{D}. The energy depends on the magnitude of the velocity, not its direction.

True, if one is making the assumption they are dealing with a rigid body, which involves zero deflection and zero internal frictional losses. That is not true in the real world and it clearly is not true in the special situation of the spaceship question where the circular motion would not occur unless the rocket expends energy. In that instance it is clear that energy is required for circular motion despite the fact that the total energy of the system has no changed.

There has to be a way of calculating this, otherwise I don't see how NASA could have ever hit the moon.

 

[Doc Al]

Let's just go to the spaceship question. Let's assume we have a spaceship of 100 kg mass moving at 20 m/sec in deep space so there are no extraneous influences on the ship that might cause it to deviate from a straight line course. Then we decide we want to go home so we fire a rocket that has a force of 10 Newtons and we ensure it is always perpendicular to the current direction of motion to turn us around 180º so we can go home.

1. Even though we are not increasing the total energy of the spaceship in this maneuver, can we calculate from this how much energy the rocket is delivering to space (how much of its total available energy has it lost)?

Interesting question. I'd think it would depend on the nature of the rocket.

2. Can we calculate from this what the turning radius of the spaceship would be?

Assuming we can pretend that the mass doesn't change (which can't be true since the rocket is firing), since we have the centripetal force we can calculate the radius. (Set F = mv^2/r.)

 

[frankencrank]

Interesting question. I'd think it would depend on the nature of the rocket.


Assuming we can pretend that the mass doesn't change (which can't be true since the rocket is firing), since we have the centripetal force we can calculate the radius. (Set F = mv^2/r.)

I agree, the mass of the rocket would have to change but we can assume it doesn't for the purposes of this problem just like people assume they are dealing with rigid bodies (which don't exist in the real world) when dealing with these kinds of problems.

Since we can calculate the radius of the circle and we know the speed of the spaceship we can know the time that the rocket must fire to turn the spaceship around. From this can we calculate how much energy left the rocket during a 180º turn and, at the same time, the power of the rocket?

 

[americanforest]

I agree, the mass of the rocket would have to change but we can assume it doesn't for the purposes of this problem just like people assume they are dealing with rigid bodies (which don't exist in the real world) when dealing with these kinds of problems.

Since we can calculate the radius of the circle and we know the speed of the spaceship we can know the time that the rocket must fire to turn the spaceship around. From this can we calculate how much energy left the rocket during a 180º turn and, at the same time, the power of the rocket?

The spaceship can't turn 180 degrees by launching one rocket orthogonally to its velocity because of momentum conservation. Assuming it is much heavier than the rocket which it fires the deflection angle from the spaceship's original path would be significantly less than 90 degrees. If the masses are equal the deflection is 90 degrees.

What do mean by "how much energy left the rocket"? The rocket is what is being fired off of the spaceship right? The spaceship loses the kinetic energy it gave to the rocket, which depends on the rocket velocity.

 

[frankencrank]

The spaceship can't turn 180 degrees by launching one rocket orthogonally to its velocity because of momentum conservation. Assuming it is much heavier than the rocket which it fires the deflection angle from the spaceship's original path would be significantly less than 90 degrees. If the masses are equal the deflection is 90 degrees.

What do mean by "how much energy left the rocket"? The rocket is what is being fired off of the spaceship right? The spaceship loses the kinetic energy it gave to the rocket, which depends on the rocket velocity.

No, you misunderstand. The thrust of the rocket by definition would always be delivered at 90º to the direction of movement at the time so there would be no slowing of the spaceship. Such a force, if continued for a period of time, would eventually result in a 180º turn as it replicates a centripetal force. It matters not what the mass of the rocket is.

 

[americanforest]

No, you misunderstand. The thrust of the rocket by definition would always be delivered at 90º to the direction of movement at the time so there would be no slowing of the spaceship. Such a force, if continued for a period of time, would eventually result in a 180º turn as it replicates a centripetal force. It matters not what the mass of the rocket is.

Oh, I see. I though you meant actually firing a physical missile at a 90 degree angle.

 

[frankencrank]

Oh, I see. I though you meant actually firing a physical missile at a 90 degree angle.

Another way of looking at the problem would be to fire an infinite number of tiny bullets directed at 90º from the direction of motion. We would know the energy of each bullet and the reactive thrust of each bullet and could solve the problem, I guess, that way. I will let the experts here though comment on that approach.

 

[rcgldr]

In the case of the rocket, we seem to be forgetting that the rocket turning in a circle via thrust from it's own engine is going to be accelerating spent fuel outwards, resulting in a large increase in kinetic energy of the spent fuel, and therefore work is being done.

If the rocket were captured by a frictionless circular track that was extremely massive (or otherwise virtually imobile), then the rocket would follow the circular track with no consumption of energy.

 

[frankencrank]

In the case of the rocket, we seem to be forgetting that the rocket turning in a circle via thrust from it's own engine is going to be accelerating spent fuel outwards, resulting in a large increase in kinetic energy of the spent fuel, and therefore work is being done.

If the rocket were captured by a frictionless circular track that was extremely massive (or otherwise virtually imobile), then the rocket would follow the circular track with no consumption of energy.

Is it right to say that that "work" done on each individual particle spewed out by the rocket when summed is the "equivalent" of work done to change the direction of the spaceship? We certainly can know the energy of each particle and the sum of the energy spewed into space. The energy per unit time can also be expressed as power. Would these two ways of looking at this problem give us the same answer?

It seems the only thing we know here is the thrust of the rocket. Does it matter how the thrust is generated as to what the energy spewed into space is (I assume different types of rockets have different efficiencies in this regards)? Is there some way thrust can be converted into energy?

If there is no one to one correlation of thrust to energy is there a possibility of knowing what the "best case" scenario would be, a rocket with 100% efficiency perhaps?

 

[rcgldr]

Efficiency for rocket engines, is rated at impulse per unit of fuel.

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

The power produced by a rocket engine would require knowledge of the actual rate of kinetic energy of the spent fuel which would equal the spent fuels terminal velocity (relative to the rocket), times the rate of mass of spent fuel ejected from the engine (mass of spent fuel ejected per unit time).

This link provides some "common" efffective exhaust velocities.

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

Another link, with an equation for calculating exhasust velocity:

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

 

[Dale]

The rocket is a poor example. In the case of a rocket firing thrusters to make a circular turn the rocket expends energy by burning fuel, this energy does not go into increasing the KE of the rocket, but into increasing the KE of the exhaust. All of the energy in the rocket fuel goes into the exhaust and none into the rocket in this instance. That is essentially why this type of maneuver is never used.

The bottom line is that circular motion does not require energy, but you certainly can have examples of circular motion where energy is expended.

 

[rcgldr]

A better example would be two objects of equal mass connected by a string, rotating in space. Each object moves in a circle and absent of friction, there is no consumption of energy and the objects will continue to circle indefinitely.

 

[frankencrank]

A better example would be two objects of equal mass connected by a string, rotating in space. Each object moves in a circle and absent of friction, there is no consumption of energy and the objects will continue to circle indefinitely.

It is not "fair" to change the problem to make the solution easy. You have now described a two particle system where momentum is conserved. I described that system in the original post. The original question/problem is one particle system. Is it possible to change the momentum of that particle without expending energy? If one does change the momentum of this one particle system, is it possible to work backwards and determine how much energy was expended in making the change?

 

[Dale]

It is not possible to change the momentum of an isolated one particle system at all, regardless of energy. Your rocket example does not constitute a one particle system either.

 

[kamerling]

The amount of energy expended depends on the speed of the exhaust. If you use a rocket fuel that uses U kg/sec of fuel with an exhaust speed of v_e, you will produce a thrust of
F = U * v_e. This will use up a power of (1/2) U * v_e^2. Of course all this power wil be added to the energy of the exhaust.

 

[frankencrank]

It is not possible to change the momentum of an isolated one particle system at all, regardless of energy. Your rocket example does not constitute a one particle system either.

This is how I see it. The same outside force applied to a single moving particle or object can result in only 3 outcomes. 1. It increases or decreases the speed without changing the direction of motion. This results in a change in both kinetic energy and momentum. 2. It increases or decreases the speed and also changes the direction. This results, again, in a change in both kinetic energy and momentum. 3. It changes the direction without changing the speed. This results in no change in kinetic energy but does result in a change in momentum, since momentum is a vector quantity.

The outside force is exactly the same in each instance, only the direction of the applied force is different yet in most instances it is seen as doing work and in the one single instance it is not. Yet, it seems that energy must have been expended in generating the force. Just because no work was done does not mean that energy was not expended. I am trying to figure out how to quantify the energy expended in this one instance where no work is done but energy must have been expended.

 

[frankencrank]

The amount of energy expended depends on the speed of the exhaust. If you use a rocket fuel that uses U kg/sec of fuel with an exhaust speed of v_e, you will produce a thrust of
F = U * v_e. This will use up a power of (1/2) U * v_e^2. Of course all this power wil be added to the energy of the exhaust.

If we don't know the speed of the exhaust can we calculate the expected energy requirement by looking at the deviation.

What if we looked at a series of collisions between moving balls. In such a collision, momentum is conserved. In this series though in each collision one ball, when it hits the other comes to a stop and transfers all of its momentum to the other. The other, is simply deviated in course, picking up the y component momentum and losing a small amount of x component momentum to keep total momentum and energy constant. The energy cost of each collision would be the energy lost in the ball that comes to rest. With a series of collisions the moving ball would travel in a circle.

Is that a valid way of looking at the problem?

 

[frankencrank]

If we don't know the speed of the exhaust can we calculate the expected energy requirement by looking at the deviation.

What if we looked at a series of collisions between moving balls. In such a collision, momentum is conserved. In this series though in each collision one ball, when it hits the other comes to a stop and transfers all of its momentum to the other. The other, is simply deviated in course, picking up the y component momentum and losing a small amount of x component momentum to keep total momentum and energy constant. The energy cost of each collision would be the energy lost in the ball that comes to rest. With a series of collisions the moving ball would travel in a circle.

Is that a valid way of looking at the problem?

How about this way of looking at the problem. the energy required depends upon how the force is generated.

Lets take the case where a massive object is swinging around a center pole unbalanced by an object on the other side. Depending upon the tensile strength of the pole there will be a certain amount of deviation. That deviation causes certain frictional losses in the pole and will generate heat losses. The amount of those losses will depend upon the mechanical properties of the pole. It would be the same for a bicycle turning a corner, the losses would depend upon the mechanical properties of the tires and track/road it is on. So, if the force is coming from a relatively rigid object where there would be minimum deviation, the losses would be very small. Whereas, if the force is coming from an insubstantial object (a rocket spewing particles into empty space) the losses could be very large.

Since perfectly ridgid objects do not exist in the real world, circular motion can never be lossless. But, depending upon the environment, it could be very small.

Does this make sense?

 

[frankencrank]

As I have been thinking about this overnight I believe I understand the issues now. I would like to get some confirmation from the experts here.

The amount of loss in turning an object depends upon the amount of work involved in providing the necessary forces. Work equal to force x distance. In a "perfect" world, using rigid bodies, force does not cause distortion so the work involved is zero. But, in the real word all objects react to applied force with some distortion, some more than others. So, the amount of work required to turn an object depends entirely on the environment and cannot be calculated without knowing the specifics of the environment.

In general though we could rank the energy required to turn an object of any given mass from lowest energy to highest energy based upon the "rigidity" of the environment thus:

0. rigid body in completely rigid system (zero energy required, does not exist in real world)
1. Steel wheels on steel tracks (typical train)
2. rubber wheels on asphault (typical car)
3. a water environment (typical boat)
4. an air environment (typical airplane)
5. empty environment (spaceship, particle in a cyclotron)

Does anyone have any thoughts on this? If I have it wrong I would like to know it so I can (eventually) get it right.

 

[Dale]

This is how I see it. The same outside force applied to a single moving particle or object can result in only 3 outcomes. 1. It increases or decreases the speed without changing the direction of motion. This results in a change in both kinetic energy and momentum. 2. It increases or decreases the speed and also changes the direction. This results, again, in a change in both kinetic energy and momentum. 3. It changes the direction without changing the speed. This results in no change in kinetic energy but does result in a change in momentum, since momentum is a vector quantity.

All correct.

Yet, it seems that energy must have been expended in generating the force.

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.

 

[Dale]

As I have been thinking about this overnight I believe I understand the issues now. I would like to get some confirmation from the experts here.

The amount of loss in turning an object depends upon the amount of work involved in providing the necessary forces. Work equal to force x distance. In a "perfect" world, using rigid bodies, force does not cause distortion so the work involved is zero. But, in the real word all objects react to applied force with some distortion, some more than others. So, the amount of work required to turn an object depends entirely on the environment and cannot be calculated without knowing the specifics of the environment.

In general though we could rank the energy required to turn an object of any given mass from lowest energy to highest energy based upon the "rigidity" of the environment thus:

0. rigid body in completely rigid system (zero energy required, does not exist in real world)
1. Steel wheels on steel tracks (typical train)
2. rubber wheels on asphault (typical car)
3. a water environment (typical boat)
4. an air environment (typical airplane)
5. empty environment (spaceship, particle in a cyclotron)

Does anyone have any thoughts on this? If I have it wrong I would like to know it so I can (eventually) get it right.

I don't think this is correct. A circular orbit is about the most lossless circular motion you can imagine, certainly much less losses than steel wheels on steel tracks, even though the environment is empty.

 

[frankencrank]

I don't think this is correct. A circular orbit is about the most lossless circular motion you can imagine, certainly much less losses than steel wheels on steel tracks, even though the environment is empty.

Actually, I think I did get it wrong as I was thinking about this out on my bike ride. However, I think the general principle is correct. What I got wrong, I believe, is the particle in the cyclotron should have the fewest losses of those that I listed, not the most. The reason is the force is coming from magnetic fields that are coming from a very massive and rigid mechanism. And, as you point out, a satellite in orbit also would have minimal losses for much the same reason, the Earth is relatively rigid compared to the forces it is seeing.

So, my revised rank order of energy cost of turning is:

1. particle in cyclotron, satellite in orbit (a satellite in geosynchronous orbit might have zero losses as it would be similar to particle of a spinning disk and doing no work regarding tidal forces, etc.)
2. steel wheels on steel tracks
3. rubber wheels on asphault
4. boat on water
5. airplane in air
6. spaceship in space.

 

[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 dependant. 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 dependant. 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.