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Kinetic Energy

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Ok, KE=1/2*m*v^2. What about when v is a vector? It seems it would take energy to change the direction of v. Assuming you are the only mass in the universe, i.e. me, all me, nothing but me then your potential energy didn''t change. So you must have added or removed kinetic energy from the system. Assuming your speed and mass didn''t change, just your direction then 1/2*m*|v|^2 is the same. So I''m confused.

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Guest Anonymous Poster
quote:
Original post by LilBudyWizer
Ok, KE=1/2*m*v^2. What about when v is a vector? It seems it would take energy to change the direction of v.

It takes a force, but no energy. Think of the moon going around earth...

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When you speak of kinetic energy, you are often thinking of it in terms of work, via the work-energy theorem, which states that:

W_net = (DELTA)K

We know that work is independent of path, provided we are talking about conservative forces (most of which are... friction is basically the only nonconservative force you ever need worry about). So because of the fact that work is independent of path, this leads to the v in 1/2mv^2 being speed rather than velocity.

In other words, the kinetic energy of an object does not depend on the direction it is heading, but instead, only on the magnitude of its velocity (well, and mass, but that''s a given ).

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Ok, I can accept that. I guess I have trouble with the idea that you could produce a force without energy. Since velocity is a contineous function in the real world then to alter the velocity it seems the force has to be applied over time. Since you have a velocity you moved during that time and thus have a force acting over a distance.

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Guest Anonymous Poster
dE = F.dr

Changing only the direction of motion means F is always directed orthogonal to dr, thus F.dr = 0.

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The solution is rather simple.....It sounds as though you are mixing scalar(speed) quantities with vector(velocity) quanties. Scalar quantities have magnitude only, whereas vector quantities have magnitude and direction. In the equation:
K=1/2mv^2
even if v was a vector quantity, once it is squared it still gives us a positive value. Thus K for the given v will always be positive. The only way K can be negative is if you are finding (delta)K when final v is less than initial v.:
(delat)K=(1/2mvf^2)-(1/2mvi^2)
This would only imply a loss on kinetic energy.

[edited by - Moorphene on March 4, 2003 11:46:48 AM]

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lol.. no idea what this post is supposed to prove

if your the only thing in the universe.. and your moving in a straight line with some velocity(v) and kinetic energy(.5mv^2).. there is no reason for your direction to change..

so why is anymore energy needed in the system?

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Just because you are the only mass doesn''t mean energy doesn''t exist. Einstein showed that the two are equivalent, so you could generate a force using some of the energy. Refering to your question about velocities, you are taking the dot product for the velocity, so v.v = mag(v)^2. Furthermore, think about this- A charged particle will move around in the path of a circle in a uniform magnetic field. No work is being done, however, the direction of the velocity is always changing. It''s kinda wierd, but it makes sense. The direction you are travelling really doesn''t change the amount of energy that you have.

Brendan

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If you are the only thing in the universe, motion is undefinable.

Motion is always measured relative to another object (i.e. in a reference frame). There is no such thing as ''absolute'' movement versus a universal co-ordinate system except in games.

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I guess I''m just confusing energy with fuel. If you are in orbit, land then return to the same orbit then you have the same potential and kinetic energy at the end as when you started. Even ignoring air friction it seems that would take a considerable amount of fuel. Half of that trip takes little fuel assuming you can survive the landing. Perhaps I should think of force over time as more closely related to fuel consumption than changes in potential/kinetic energy or work.

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Forces can act (Remember all forces act in pairs, NEVER alone! Newton''s third law says this.)
wihout causing movement, work, or changes in potential or kinetic energy.

Think of a rock sitting on the ground. It is experiencing a constant gravitational force pulling it down and a constant normal force from the ground holding it up. No energy is expended, no work done, and no movement occurs. Now, remove the gound, and the lack of a normal force allows the gravitational force to accelerate the object. Work is done, and potential energy is converted into kinetic energy.


Note: There are actually four forces in the above example...The rock also exterts a gravitational force and a normal force against the ground. That''s why after the removal of the ground, there is still a balanced pair of forces.

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LilBudyWizer: One clear way of thinking about this problem is to think in vectors. If you have changed the direction of the object, then the components of its velocity vector have changed (relative to some arbitrary coordinate basis) and as such, the speed in one coordinate direction has increased while the speed in the other has decreased. The kinetic energy associated with these component vectors will have changed, however the total kinetic energy of the body will remain the same (provided its speed is constant throughout the manouvre). So, if the kinetic energy before the event is the same as the kinetic energy after the event, then there was not net work done on the object during the event. This doesn't mean work wasn't done... just no net work... so, in a frictionless environment, a ball can collide with a wall, impart momentum (energy into the wall, stored as potential energy) and then the wall imparts momentum to the ball in the form of kinetic energy. Because this only occurs for the component of velocity perpendicular to the wall, the ball appears to bounce off... (or glance off if the component parallel to the wall was non-zero) in a new direction. There were changes in kinetic energy but they were all balanced out so that there is no net change.


Cheers,

Timkin

[edited by - Timkin on March 4, 2003 11:46:21 PM]

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Terms like energy and work have quite specific meanings in Physics. Unfortunately these words are used quite loosely in our everyday language which causes some confusion. Energy is not some special quality about an object, it is simply a number one can calculate based on properties and conditions of an object, etc. It turns out that this number happens to not change when conditions of the object changes. This is the conservation of energy of course. As long as we keep track of all the types of energy (kinetic, potential, rest mass, thermal, etc) we will find that this number does not change. Work is done on an object when a force ( or a component of that force) acts in the direction of motion. This is the definition of work and it happens to be equivalent to a change in kinetic energy. This is why a force that only changes the direction of an ojbect (acting at right angles to the motion) does no work and why there is no change in kinetic energy. The confusing part is when you imagine yourself pushing on an object to change its direction. Because of the way our muscles operate, we actually expend energy or work to do this, but the kinetic energy of the object that we push will not change. It''s really a matter of separating semantics from the physical description.

On another note, vectors are quantities that are independent of coordinate systems. That''s why we use them in equations of physics because it will not depend on the coordinate system used to describe the phenomena. The components of the vector will of course change depending on the choice of the coordinate system as well as the orientation of the coordinate axes. However, no matter what coordinate system is used, the magnitude of the vector is always the same.

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You feed fuel into a combustion chamber and combust it. That hopefully substantially decreases the density, i.e. mass per unit volume. There is only one way for that gas to go and that is out the nozzle. The rate at which you feed in fuel determines the volume of gas exiting the nozzle per unit time. The size of the nozzle lets you calculate the speed of the gas exiting the nozzle. You know the mass you are feeding in which equals the mass exiting once you are operating at a constant state. So assuming the initial velocity is effectively zero and the you know the average distance from combustion to the nozzle you have an acceleration. The mass and acceleration gives you a force. So the fuel consumption is related to the force and not the work or change in energy.

That is simply counter-intuitive to me. It just seems that if you are buring fuel you should be gaining energy. You may not be burning enough fuel to get off the ground so obviously it isn''t true.

Some of you may be wondering how I go from the original question to this. Originally it came out of the Gravity Area question. Being my usual self I wandered far and wide from the question asked Specifically you have the gravitational field and gravitational potential with the gravitational field being the negative gradiant of the gravitational potential. So I can figure out how to plot a path though a vector field given an initial position and velocity. At least numerically. I also understand that a line integral through a conservative vector field depends only on the starting and ending points and not the path through it. Your fuel consumption is certainly going to depend upon your path though the field though. That is where I have a problem.

Initially I thought maybe you could use conservation of energy. I know the potential energy isn''t dependant upon the path so I thought maybe the kinetic energy does. I know enough to know it doesn''t, but thinking about a problem you are bound to chase a few wild geese. It has been 20 years since I had a calculus based statics/dynamics physics course. My engineering class pretty much only uses high school math.

So at some level I figure I gotta integrate a force over time. What is a Newton second though? Mass times velocity, but if that is useful why isn''t there a unit for it? So that makes me think that is a screwed up idea. 1/2mv^2 looks a whole lot like the integral of mv with respect to v. v is a vector though so that integral would be a vector. A dot product of the velocity with the force would reduce it to a scalar. The dot product would seem to be a line integral through a vector field. The fuel consumed is certainly a scalar. That is kinetic energy though and that doesn''t determine my fuel consumption. That is really my point of confusion.

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Guest Anonymous Poster
quote:
Original post by LilBudyWizer
So at some level I figure I gotta integrate a force over time. What is a Newton second though? Mass times velocity, but if that is useful why isn''t there a unit for it?


Mass times velocity is called linear momentum, and yes, it is a very useful quantity indeed.

quote:
So that makes me think that is a screwed up idea. 1/2mv^2 looks a whole lot like the integral of mv with respect to v. v is a vector though so that integral would be a vector.


The integral would be ∫ mv⋅dv , where both v and dv are vectors. The dot product makes the integral scalar. And as said before, when changing only the direction, v and dv will be perpendicular, so v⋅dv = 0.

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Guest Anonymous Poster
Didn''t work...

The integral would be S mv.dv , where both v and dv are vectors. The dot product makes the integral scalar. And as said before, when changing only the direction, v and dv will be perpendicular, so v.dv = 0.

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Well, thanks everyone. I think I have a handle on what I need to do. Now the problem is to figure out a least cost versus least time path through a gravitational field given limited fuel and maximum thrust. I assume in the real world speed would be much more of a problem and gravity much less, but then just how entertaining would interstellar travel be in the real world. Whee, start the trip with your son asking "Are we there yet" and end it with your great-grandson asking. So I figure interstellar distances a whole lot less, speed a whole lot more and gravity a whole lot stronger.

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quote:
Original post by dexterace

On another note, vectors are quantities that are independent of coordinate systems. That''s why we use them in equations of physics because it will not depend on the coordinate system used to describe the phenomena. The components of the vector will of course change depending on the choice of the coordinate system as well as the orientation of the coordinate axes. However, no matter what coordinate system is used, the magnitude of the vector is always the same.


Which is exactly what I said in the previous post... in fact, most of what you said was just repeating what I said in different words... or did you think I was suggesting something else?

Timkin

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LilBudyWizer: When you''re burning fuel to produce thrust, you''re converting chemical potential energy into kinetic energy. For the rocket flying around in the gravitational field, you''ve got three forms of energy: chemical potential, gravitational potential and kinetic. The first is always decreasing while the engines are running. The other two are varying depending on the orientation of the thrust vector relative to the velocity of the craft and the gradient of the gravitational potential field.



quote:
Original post by LilBudyWizer
So at some level I figure I gotta integrate a force over time. What is a Newton second though?



Units of ''Impulse''

quote:
Original post by LilBudyWizer
1/2mv^2 looks a whole lot like the integral of mv with respect to v.


It is, which is what the AP was trying to write.

quote:
Original post by LilBudyWizer
v is a vector though so that integral would be a vector.



If you don''t like thinking about taking the integral over the magnitude... then consider the result... 1/2 m v 2. If v is a vector then v 2 = v .v = |v |2 = v2.

Cheers,

Timkin

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