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What is the energy/work required to turn something?

  1. Mar 14, 2008 #1
    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?
     
  2. jcsd
  3. Mar 14, 2008 #2

    Doc Al

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    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.
     
  4. Mar 14, 2008 #3
    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.
     
  5. Mar 14, 2008 #4

    Doc Al

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    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?
     
  6. Mar 14, 2008 #5
    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.
     
  7. Mar 14, 2008 #6

    Doc Al

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    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!)
     
  8. Mar 14, 2008 #7
    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?
     
  9. Mar 14, 2008 #8
    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?
     
    Last edited: Mar 14, 2008
  10. Mar 14, 2008 #9
    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.
     
  11. Mar 14, 2008 #10

    Doc Al

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    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].)
     
  12. Mar 14, 2008 #11
    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?
     
  13. Mar 14, 2008 #12
    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.
     
    Last edited: Mar 14, 2008
  14. Mar 14, 2008 #13
    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.
     
  15. Mar 14, 2008 #14
    Oh, I see. I though you meant actually firing a physical missile at a 90 degree angle.
     
  16. Mar 14, 2008 #15
    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.
     
  17. Mar 14, 2008 #16

    rcgldr

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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.
     
  18. Mar 14, 2008 #17
    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?
     
  19. Mar 14, 2008 #18

    rcgldr

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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
     
    Last edited: Mar 14, 2008
  20. Mar 14, 2008 #19

    Dale

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    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.
     
  21. Mar 14, 2008 #20

    rcgldr

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    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.
     
  22. Mar 15, 2008 #21
    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?
     
  23. Mar 15, 2008 #22

    Dale

    Staff: Mentor

    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.
     
  24. Mar 15, 2008 #23
    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.
     
  25. Mar 15, 2008 #24
    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.
     
  26. Mar 15, 2008 #25
    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?
     
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