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What is the energy/work required to turn something?

by frankencrank
Tags: energy or work, required, turn
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frankencrank
#1
Mar14-08, 11:04 AM
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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?
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Doc Al
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Mar14-08, 11:36 AM
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Quote Quote by frankencrank View Post
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
#3
Mar14-08, 01:10 PM
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Quote Quote by Doc Al View Post
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
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Mar14-08, 01:21 PM
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What is the energy/work required to turn something?

Quote Quote by frankencrank View Post
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
#5
Mar14-08, 02:40 PM
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Quote Quote by Doc Al View Post
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
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Mar14-08, 03:07 PM
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Quote Quote by frankencrank View Post
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
#7
Mar14-08, 03:25 PM
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Quote Quote by Doc Al View Post
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 180 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
#8
Mar14-08, 03:33 PM
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Quote Quote by frankencrank View Post
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 [tex]W=\vec{F} \bullet {\vec{D}}[/tex]. 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
#9
Mar14-08, 03:41 PM
P: 21
Quote Quote by americanforest View Post
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 [tex]W=\vector{F} \bullet \vector{D}[/tex]. 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
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Mar14-08, 03:49 PM
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Quote Quote by frankencrank View Post
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 180 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 [itex]F = mv^2/r[/itex].)
frankencrank
#11
Mar14-08, 04:04 PM
P: 21
Quote Quote by Doc Al View Post
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 [itex]F = mv^2/r[/itex].)
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 180 turn and, at the same time, the power of the rocket?
americanforest
#12
Mar14-08, 04:33 PM
P: 218
Quote Quote by frankencrank View Post
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 180 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
#13
Mar14-08, 04:55 PM
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Quote Quote by americanforest View Post
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 90 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 180 turn as it replicates a centripetal force. It matters not what the mass of the rocket is.
americanforest
#14
Mar14-08, 05:06 PM
P: 218
Quote Quote by frankencrank View Post
No, you misunderstand. The thrust of the rocket by definition would always be delivered at 90 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 180 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
#15
Mar14-08, 05:13 PM
P: 21
Quote Quote by americanforest View Post
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 90 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
#16
Mar14-08, 07:28 PM
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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
#17
Mar14-08, 07:57 PM
P: 21
Quote Quote by Jeff Reid View Post
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
#18
Mar14-08, 09:23 PM
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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


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