View Full Version : conservation of momentum

darryl

12-30-2008, 01:18 AM

Was reading about the flyball governor and it reminded me about this. Let's say you have weights on arms and you spin the whole thing about an axis. Get it to any stable speed, then make the arms shorter by whatever means. The speed will increase, same as an ice skater doing a spin. As they pull their arms in, their speed of rotation increases. So, now the trick is to build a mechanism which makes use of this.

I don't see it useful as a flywheel system in itself because it would be limited in what amount of energy it could shuffle back and forth, but as a moderator of sorts, possibly in conjunction with a flywheel, it could be useful. Anyone know of anything which makes use of this 'phenomenon'?

An adjustable tensioner on the spring does that.

RancherBill

12-30-2008, 02:01 AM

What about a green car?

When they brake they have some charging system that converts the braking energy. It would seem reasonable to have the arms out during initial braking and retract them at the end of the braking cycle to charge caps or other short term battery to provide a boost for the next acceleration cycle. It would be separate from the regular propulsion / storage battery.

chief

12-30-2008, 02:45 AM

Bill,

I'm not sure what an electric car has but I believe you are talking and dynamic or regenerative braking, there is no mechanical type linkage or governing system in these setups.

darryl

12-30-2008, 03:51 AM

Actually, what Bill suggested is somewhat along the lines I had been thinking. Suppose that this 'mechanism' was linked through a clutch to the driveshaft at some point. While cruising, the 'arms' are in, and the device rotates at whatever speed is given to the driveshaft by the engine for the speed the car is going. To slow the car and thus absorb some of the energy, the arms are swung outwards in a controllable manner. Pressure on the brake pedal brings this on, and also operates the brakes in the same manner as is done now. The driver would know by a dashboard light when the actual brakes are being activated so he can control the direction the braking energy goes when the brake pedal is being pushed.

The device itself will want to slow its own rotation as this happens, It will slow the driveshaft as well, slowing the car. The clutch is operated automatically when the rpm of the driveshaft must slow further than the rpm the device has reached with the arms fully out. The car can come to a complete stop in the normal fashion. When you want to accelerate, the initial movement of the gas pedal brings the clutch in, and the momentum of the device transfers energy to the driveshaft. Once the clutch is fully engaged, the arms are brought inwards, effectively transferring more rotational energy to the driveshaft as it tries to speed up. At some point it is fully 'retracted', at which point it just coasts along at whatever speed the driveshaft is driven to by the cars engine.

This clutch could be an electrical generator, which is 'let out' by shorting it's output to a variable extent. This energy can go back into the battery pack, or it can go towards running the electric motor, if the car is an electric or a hybrid.

What you are proposing is just a much more complicated version of a regular flywheel. A standard flywheel which is the armature of a generator that produces electric energy by converting angular momentum in the flywheel to a synthesized frequency controlled AC current is a direct analog of your idea but with only one moving part, the flywheel.

J Tiers

12-30-2008, 08:37 AM

It has the advantage that one can in theory accept input energy at a lower RPM, and be able use it at a higher RPM, as for instance a generator system. There would be no speed-up gearing, etc.

How it would work out in reality is more questionable.

I don't know what ratios are attainable. And it is not a zero input action to pull in the weights, since you exert a force over a distance.

A.K. Boomer

12-30-2008, 09:42 AM

Why not incorporate it in reverse, instead of soaking up power at different speeds use it to deliver --- Variable speed transmission (aid)

I have to add there really is no free lunch, just a way of transferring the "stored" power at different speeds.

Lew Hartswick

12-30-2008, 09:54 AM

It may well be that the Formula 1 boys are working on this at the

very moment. That is one of the rules for next year. Some form of

regenerative braking, as far as I know it can be either mechanical or

electrical. It should be intresting. Too bad about Honda leaving the

scene.

...lew...

Circlip

12-30-2008, 10:06 AM

The common mechanisms that used this form of technology were steam engines and gramaphones, two machines that could be set and left to their own devices so why unecessarily complicate the equasion by introducing humans into the chain?

The great god CAR has had soooo many "Saving aids" added that driving is /has become a dying art. We have seat belts, cruise control, Sips, Dips not to mention Crips and Gits, the last two coming with varying degrees of stupidity already built in.

Regards Ian.

RancherBill

12-30-2008, 10:57 AM

Darryl has a good idea.

What I was getting at is that an alternator/generator's output is not linear. At higher RPM you get more output. When you are braking there is a bunch of bonus energy that is available for a short period of time. The rate of storage by batteries is a constant.

Thus, you have a bonus of all this braking / electrical energy and the batteries can't take the charge fast enough and as a result some energy is dissipated to the etherworld via heat. Using the arm mechanism you can create a huge amount of "temporary" energy to charge some special batteries or caps. The energy in the caps is quickly used in the next acceleration cycle.

I am thinking of the example of urban driving where you stop at a sign and accelerate within say 20-30 seconds. The benefit I see is that you might not need "universal" main batteries that function under 100% of circumstances. The main batteries function for 95% of circumstances and the caps work 5% of the time. In this arrangement the main batteries might cost less or alternately you get better acceleration when using this temp energy in the caps.

A hybrid hybrid.

fasto

12-30-2008, 04:36 PM

The main batteries function for 95% of circumstances and the caps work 5% of the time. In this arrangement the main batteries might cost less or alternately you get better acceleration when using this temp energy in the caps.

A hybrid hybrid.

See this (http://www.maxwell.com/ultracapacitors/products/modules/bmod0063-125v.asp).

If you have to ask the price, you can't afford it.

I use a smaller Maxwell capacitor in one of my products as an energy storage device for short term power cuts, and it works very well.

I wouldn't invest too much money in ultra capacitors or flywheels. There is new battery technology being introduces as we speak that will wipe out the advantages of those systems. Toshiba has a new battery that can be recharged to 80 percent of capacity in 2 minutes and 90 percent in 5 minutes.

Tokyo-based Toshiba (OTC: TOSBF) announced that it planned to release a fast charging lithium ion battery early next year for use in industrial vehicles and other applications.

The company said its Super Charge ion Battery can recharge to 90 percent of full capacity in less than five minutes.

Toshiba said it plans to make the battery a mainstay of its industrial systems and automotive products businesses, with the first system to be shipped in March 2008

http://www.google.ca/search?hl=en&q=toshiba+fast+charge+batteries&meta=

That should make regen braking a very much more effective strategy because the battery will be able to soak up the energy as fast as it is generated (even faster).

BillH

12-30-2008, 05:58 PM

I wouldn't invest too much money in ultra capacitors or flywheels. There is new battery technology being introduces as we speak that will wipe out the advantages of those systems. Toshiba has a new battery that can be recharged to 80 percent of capacity in 2 minutes and 90 percent in 5 minutes.

That should make regen braking a very much more effective strategy because the battery will be able to soak up the energy as fast as it is generated (even faster).

I think Ultracaps would be much more useful for electric powered airplanes being 1/10th the weight of a Lithium Ion, and supposedly a much greater lifespan.

darryl

12-30-2008, 06:13 PM

Evan has a point, and if I may expand on what I believe it it- if there's going to be any electrical conversion going on as part of any drive train scheme, then why add more mechanicals to the system if it can be done more simply with maybe two motor/alternator sets and electronics- especially if up-to-date battery technology is brought into the equation. Part of the problem has always been supplying enough torque for start-up and hill climbing, etc, without being hard on the batteries and motors, or requiring too much weight in the motors, etc. My idea would give a large torque at a variable rpm using mostly mechanical means, and getting it from the momentum stored in spinning masses. As such, the torque available would be very high in theory, to the limit of strength of materials, bearings, housings, etc. Consider how much instantaneous torque you could get if you tried to stop a flywheel very quickly- certainly lots.

It would not be a simple mechanical system, of course, and that is usually a no-no when it comes to long life systems, especially those that are subject to all kinds of possible abuse, etc. Somehow I think that drivable vehicles in the hand of the public fits into this category :)

My idea has drawbacks- one is complexity, one is size- the ability to swing weights outwards from a spinning shaft would take up room- another is possible design flaw. As J Tiers said, there's the issue of pulling in the spinning weights- exerting force over distance- I'll have to think about whether that can even be done without using up all the energy that would otherwise be available to output.

Many years ago I had the idea to use one of these 'variable flywheels' on each side of differential gearing. Each would have it's own electronically controlled electric motor to spin it up. If both were spinning at the same rpm but in opposite directions, the output side of the gearing would be at 0 rpm. This part connects to the drive shaft. As one is made to speed up and the other to slow down, there becomes a net rpm delivered to the shaft. High torque can be delivered because it's waiting in the spinning masses. The motors would be sized only to supply the long-term average power demand.

I have seen systems like this where only the motors and not flywheels of any sort, are connected through differential gearing to the drive shaft. These motor/generator sets would shuffle energy from one to the other as an output rpm is called for- in operation one is the drive motor, and the other is slowed as it feeds energy to the drive motor, aided by the battery. Only when these motors are at different speeds can the output shaft turn. Still seems like a good transmission idea to me, even if virtually no flywheel effect is available in the system.

ckelloug

12-30-2008, 06:27 PM

As Jtiers pointed out, there is work done in moving the weights inward and as a corolary there is work provided by the flywheel to move them outwards.

The principle here is analagous to several electronic systems:

A flying capacitor voltage doubler fills up two capacitors in parallel to a given voltage and then connects them in series to briefly release the energy at twice the voltage.

A boost converter stores energy in a capacitor and releases it through an inductor to produce a voltage higher than that input.

Thinking about this, it seems like a wonky mechanical transistor in series with a leaky capacitor. From this analogy, it appears to me that will have to be coupled to the output via a torsional spring and flywheel so that you get an oscillatory action which allows the device to periodically transfer power to the output.

I think that some similar device must be used in a clock escapement somewhere to convert the stored energy in the mainspring to fast enough motion to turn the hands. This is a leap of the imagination however.

What I am fairly certain of is that it is easier to perform that action electrically in modern times.

As Jtiers pointed out, there is work done in moving the weights inward and as a corolary there is work provided by the flywheel to move them outwards.

Um, not quite. What is happening is a change in entropy rather than energy. When the weights move outward total angular momentum remains the same(neglecting any losses) and total stored energy is unchanged. Entropy however increases as the sum total energy is now occupying a larger volume which is exactly what is happening as the universe expands. In a closed system it is not possible to reverse entropy using only the energy present in the system. So, if the weights are in the out position it requires an energy input to pull them back in. That energy input is what decreases entropy within the system.

To further illustrate it is possible to completely destroy all angular momentum in the system if we allow entropy to increase at the maximum possible rate. Just cut the weights loose.

Rich Carlstedt

12-30-2008, 09:57 PM

Thanks for detailed explanation Evan !

thats good !

Putting it in simple terms, you can see this exact action in a tether ball.

If you propell it forward at say 10 feet per second, you have the balls' weight, (say one pound) and speed to give you 10 footpounds of energy.

If the cord suddenly locks to the pole, the ball starts to wind up and seems to go faster, BUT IT ISN"T. It is still going a 10 FPS, but because the circumference is now shorter, the RPM increases, but the speed stays the same.

No energy is lost or transferred--- Until the ball reaches the end (of it's rope !) and the 10 FPS is applied as torque to the pole

You can see this with a torque wrench also. If you had a one foot long torque wrench, you would apply 10 pounds of pressure to reach 10 FT/LB

With a 6 inch long Torque wrench, You need to apply 20 pounds of force to get a 10 FT/LB level.

Rich

This idea was reviewed in Popular Science back in the 60's

when a flywheel powered bus was supposed to have been built.

When the bus wanted to stop, a transmision "increased" the ratio for braking, so the net effect was a speed UP of the flywheel.

I think the mecahnics did it in.

A better way, but beyond my ability would be to use a generator to brake the wheels and supply the energy back to the flywheel in the form of increased frequency pulsed electricity.

there you go Evan !

wierdscience

12-30-2008, 10:19 PM

Enter the Perm-o-drive system-

http://www.permo-drive.com/tech/index.htm

This system is being applied to fleet vehicles already as well as US military vehicles.It uses a hydraulic pump in the drive line to boost up pressure in an accumaltor during deceleration and releases that energy to accelerate.It's basically a large hydraulic pump/motor.

Best part is it is adaptable to existing vehicles and in testing has demonstrated a possible 35-40% fuel savings mainly in vehicles like garbage trucks and buses that do alot of stop-start driving.

A Perm-o-drive unit about to be installed in a big rig-

http://img.photobucket.com/albums/0903/wierdscience/PERMODRIVE.jpg

I love hydraulics, but that little jewel must cost upwards of 20grrr doesn't it? Guess the payback would be quick enough at a couple hundred gallons a week though.

Re:Flywheels:I seem to recall a Popular Mechanics article from my childhood in the 70's or early 80's on flywheel assisted cars. The two points that have stuck were gyroscopic effects on handling, and safety.

Some institute was working on a flywheel that would disintegrate into fluff if it was oversped.

There is an image that always makes me giggle.

ckelloug

12-31-2008, 12:17 AM

Evan,

Are you arguing no work was done in the movement of the weights in either direction? In that moving them either direction involves a force applied over a distance, this cannot be correct.

It is true that the angular momentum stored by the flywheel is constant regardless of the position to which the affixed weights are moved after the system begins rotating. It is false however that no work is done moving the weights (which is neglected in the average freshman physics textbook).

In order for the weight to have accelerated and moved from its last position to it's current one, the motion has to have consumed energy because work was done. In the case of moving the weight inwards, you are providing an external force.

The force on a weight required for it to stay in place in the system is F_r=mv^2/r where v is the velocity of the weight and r is the radius of its current orbit. If the force on the weight varies from F_r then the weight has to move because it has an applied force. Conversely, if the weight moved then it has an applied force. Either way of stating it, If the weight moved then work has been done somewhere because Energy=Work=Integral of F over distance.

When the weight moves outward, the same argument applies. If the weight was being held in place and is released, then work is done to retain the weight in it's circular orbit as it moves.

The sign of the work done on the system is positive because an external force moved the weights inwards. The sign of the work is negative if the weights move outwards indicating that the system did work to prevent the weights from leaving their orbit due to a lack of centripetal force.

From Section 19.4 of the Feynman lectures on Physics, the kinetic energy stored in the flywheel under these circumstances is not constant (which I had forgotten). The angular momentum is constant but the energy is not.

Since the flywheel changes velocity when the weights move in either direction, a torque is exerted on the flywheel either positive or negative depending on the direction of motion of the weights. Inward movements of the weights exerts a positive torque on the flywheel accelerating it and outward movements exert a negative torque decelerating it.

Since purely centripetal forces do not apply torques, a force must have been applied somewhere to provide the torque to change the angular velocity of the flywheel. In short, positive or negative work are done depending on the direction of the motion to overcome the effects of Coriolis Force while retaining the weights in circular motion. Thus entropy is not the correct explanation: work was done either on or by the system depending on the direction of motion of the weights.

Since between you, me, and Richard Feynman, only one so far has a Nobel Prize in Physics, I'm going to have to defer to Feynman on this one.

--Cameron

Some institute was working on a flywheel that would disintegrate into fluff if it was oversped.

It may turn into "fluff" alright but all the energy doesn't go away. If that flywheel was storing 100 kilowatt hours that 100 kilowatts is very suddenly on the move. The housing is going to become extremely hot.

Are you arguing no work was done in the movement of the weights in either direction

Not exactly. Energy is transferred between components within the system. The system as a whole still has the same amount of energy.

I have a problem with your interpretation of Feynman's lecture. Allowing the weights to move outward does not change the angular momnentum and we are agreed on that. But, if it changes the total system energy where does that energy go? There is no reason why we cannot posit a closed rotating system that operates in this manner and in fact the Earth-Moon system is an example. Eventually the Earth will be tidally locked to the Moon and will transfer some of it's angular momentum to the moon in doing so. The Moon will in turn move away from the Earth as this takes place. Right now the radius of it's orbit increases about 4cm per year. Aside from losses in internal friction from gravitational tides the angular momentum will remain the same. Excepting the losses the energy of the system must also remain the same as there is nowhere for it to go.

Thus entropy is not the correct explanation: work was done either on or by the system depending on the direction of motion of the weights.

Any time that the energy in a system becomes distributed over a larger volume the entropy of the system has increased. That is an axiom.

NickH

12-31-2008, 04:19 AM

How do you controllably "Let" the weights move outwards?

If you use any kind of damping or braking you use energy.

If you use electrical braking & store the energy you might as well expand that system & do away with the moving weights.

When you do this by spinning on a revolving chair & pull in your legs to speed up & let them out to slow down you will see that as well as having fun you are using energy whilst speeding up and slowing down,

Regards,

Nick

How do you controllably "Let" the weights move outwards?

In a thought experiment we only need be concerned with the actual energy exchanges that are required by physics to take place. Technical details such as designing a working implementation are merely engineering problems.

We can allow the weights to move outward by simply releasing them from the edge of the flywheel using a mechanism that requires arbitraily low force to activate. They then move out until stopped by a restraint such as a rod with an end stop. Since the energy inputs to such devices can be made arbitrarily low we can set them at zero and ignore them.

ckelloug

01-01-2009, 10:23 PM

[Quote=Evan]

The system as a whole still has the same amount of energy.

[Quote]

This is the flaw in your argument. The weights are moved by an external force which is opposed by the coriolis force and thus requires work be done. Thus the energy done by the work changes the kinetic energy of the system.

Feynman says specifically on pg 19-8 that the kinetic energy of the system is not constant. It looks constant from casual inference but it is not.

This is the hangup. Only the angular momentum is constant in the system.

The work overcoming the Coriolis force which is always opposite the movement of the weights is provided externally by the force moving the weights. Thus the external force that moves the weights does work on the system via overcoming the Coriolis force as it moves the weights.

This problem is usually considered simplistically where the fact that angular momentum is conserved is to be demonstrated. As a result, the fact that work is done by an external force moving the weights is neglected.

As an example, lets work out the kinetic energy of a weight on a string at two different radii such that the angular momenta are equal and note that the energies differ.

For Example for the same mass m at radii r and r/2 and angular velocity w, the moments of inertia are:

I1=m*(r)^2

I2=m*(r/2)^2=m*(r)^2/4

To make angular momenta equal: p1 must equal p2, thus omega has to change which we all realize works from the flyball governor and previous discussions.

Angular momentum for the first particle on a string is

p1=I1*w1

Angular Momentum for the second particle on a string is

p2=I2*w2

But if I1*w1=I2*w2 then

I1*w=4*I2*w=I2*(4*w)

Since the I's are a function of radius and mass only, this 4 term has to go to the angular velocity which is 4 times the original one when the radius is halved in order to keep angular momentum constant.

Thus the angular momentum balance:

I1*w = w*m*(r)^2 = I2*4w = 4*w *m*(r/2)^2=(4w)*m*(r)^2/4

The 4 here cancels the 4 in the denominator of I2 making the angular momenta equal.

Rotational Kinetic Energy is I*omega^2

In notation:

E=I*w^2

E1=I1*w^2 therefore:

E1=m*(r)^2*w^2

E2=I2*(4*w)^2 therefore

E2=m*(r/2)^2*(4*w)^2

E1=m*(r)^2*w^2 != E2=m*(r/2)^2*(4*w)^2

E1!=E2

Thus I1*w=I2*(4*w) but the energies are not equal and thus energy to move the weights has to account for the discrepancy.

I failed to point out last time that the Coriolis force is always in the tangential direction. When you factor that in, you have applied a force perpendicular to the direction of motion in order to move the weights. Thus work is done in moving the weights and a force applied perpendicular to the radius of a rotating flywheel is a torque. The torque from moving the weights acts upon the system to change the angular velocity and thus the kinetic energy stored in the flywheel.

Thus as usual, energy is conserved, you just have to account for all the energy. And the energy here doesn't go into entropy, it applies a torque to the system to change its angular velocity.

--Cameron

J Tiers

01-01-2009, 10:47 PM

In a thought experiment we only need be concerned with the actual energy exchanges that are required by physics to take place. Technical details such as designing a working implementation are merely engineering problems.

Not quite....

It would, for instance be quite in the realm of a thought experiment to allow the weights to move outwards while restrained by a dashpot..... then there would be energy dissipated in that dashpot, which would have to be accounted for.

Cameron,

I still don't see how the total energy can change in a closed system. We need only consider the case of the weights moving out, not in. There are several options here but they all have one thing in common. No external energy is required to make the weights move out. We can imagine a mechanism that is holding the weights latched in the "in" position. The mechanism is so finely tuned that the amount of energy to trip it (perhaps like a mousetrap) is very ,very small.

This is not zero of course, but wait: If we apply some engineering to this problem we can reduce the amount of energy required. In fact, with really good engineering we can reduce it to the point that all it takes for the weights to trip is the possibility than an electron will change quantum state. Or perhaps not. But sometimes it will and the weights will be released to fall to a new, further removed position where they are then held by a limit device such as a rod or rope. In fact, because of the way quantum mechanics works sometimes the weights will be tripped even when the electron doesn't change state as long as we aren't looking.

This means that we can ignore the energy required to activate any such mechanism because, as I said, the amount of energy required can be arbitrairily reduced infintely close to zero.

It means that the only energy in the system before the weights move and after they move is is the energy the system contained to begin with.

There are some caveats if we depart from a Newtonian view point. There will be some gravitational radiation that carries away some energy and there may be some frame dragging also. There will be a continuous loss of temperature no matter how cold the system becomes because it can only approach absolute zero, not reach it.

But, the kinetic energy of the system must remain the same unless it is converted to some other form. Even then the total energy of the system remains the same since energy cannot be created or destroyed, only made to change form.

Paul Alciatore

01-02-2009, 01:05 AM

Evan,

What I think you are missing in your thought process is that if the weights are released from an inner position, even with a totally negligable amount of energy, and allowed to fly outward to some kind of mechanical stop, then they will slam into that stop with some force. That force will distort the metal or other material of the weights or the stops or whatever and will produce heat or some other mechanical form of energy which will not add to the rotational energy of the system. Hence, it does not take any energy input to the system to bring the weights further out, energy is created in doing so.

On the other hand, if they are brought closer in, it will require an external energy input to accomplish this. Or, as in the case of the ball on a cord winding about a pole, that energy can be subtracted from the original rotational energy of the system. If you have ever sat in a rotating chair with weights in your outstretched arms, you will quickly see how much energy or work it takes to bring them closer to the center. We used to have such fun in the breaks between physics classis. When our arms were relaxed to allow the weights to return to the outer positions, that energy was again released and had to be overcome with muscular effort which undoubtedly created heat inside my body.

Angular momentum was preserved. Energy was not if you consider only the rotation of the weights.

I don't have a Nobel prize either, but I also think the Feynman text is correct.

ckelloug

01-02-2009, 03:00 AM

It means that the only energy in the system before the weights move and after they move is is the energy the system contained to begin with.

Lets draw the system boundary around the flywheel with a single movable weight. Let's assume perfect frictionless weight mechanism and 0 energy trip mechanism.

All of the the energy stored in the system is in the form of rotational kinetic energy. With a closed system boundary, the energy has to come from that stored energy as there is no other place for it to come from.

From Feynman's explanation, you actually have to exert a torque on the moving mass for it to stay in circular motion while being moved. The torque comes from the fact that a mass which is moving in a straight line in flywheel coordinates is moving in a curved line in an inertial reference frame.

This can be shown by differentiating the angular momentum of the moving mass as it moves along the radius of the flywheel in a rotating coordinate system.

The torque is given on pg 19-9 as 2m*w*r*dr/dt.

While intuition would tell us that there is no tangential Force component, calculus tells us that there is only a tangential component when moving radially and Feynman gives it as 2*m*w*v_r where v_r is the radial velocity of the moving mass.

A radial force component on a rotating object is a torque and thus it is necessary for the mass to apply a torque to the flywheel and the flywheel to apply a torque to the mass in order for the mass to move in a straight line along the radius while remaining in circular motion on the flywheel. It is this torque that changes the angular velocity of the flywheel.

In that a torque was applied to the flywheel, it has to have undergone shear. If it underwent shear then the shear did work on the flywheel. Likewise, the mass underwent compression and perhaps shear too. It also has to have had work done on it.

If we give our flywheel a shear modulus, we can calculate the shear forces it underwent and compute the shear work done in it. The net result won't be large but it will be the heat that the flywheel gained from the torque caused due to exerting the Coriolis force on the mass during the movement.

Also, since the mass in the Feynman derivation says that it moves at speed v_m, it neglects acceleration and deceleration of the mass. These will also cause compressive and tensile forces in the mass resulting in heat.

In short, I agree with Paul that the "lost" energy shows up as heat.

In conclusion, if there is no external force on the weight and it is released to move radially in a slot in the flywheel, it exerts a torque on the flywheel that slows the flywheel down via the Coriolis force. This torque creates shear forces in the flywheel during movement of the mass and the work done on the flywheel by the shear forces is manifested as heat. Likewise, compressive forces are induced in the moving mass and the work done by these compressive forces also shows up as heat.

In general, moving a mass either direction along the radius of a flywheel requires energy due to the effects of the Coriolis force which uniformly opposes motion in any direction in a rotating system.

What I think you are missing in your thought process is that if the weights are released from an inner position, even with a totally negligable amount of energy, and allowed to fly outward to some kind of mechanical stop, then they will slam into that stop with some force. That force will distort the metal or other material of the weights or the stops or whatever and will produce heat or some other mechanical form of energy which will not add to the rotational energy of the system. Hence, it does not take any energy input to the system to bring the weights further out, energy is created in doing so.

There is no reason to assume that the process of allowing the weights to assume a new position will result in a conversion of rotational kinetic energy to another form. We can posit that the materials are perfectly elastic so that no conversion to heat takes place. Glass is a perfectly elastic material as long as any strain produced is below the yield point.

Even so, energy cannot be created or destroyed and even if energy of rotation does convert to heat that is still kinetic energy and it is still system energy.

For the purpose of this thought experiment we can even make the position of the weights temperature dependent and account for any such conversion that way.

If you reread my arguments and questions I am not referring to the energy balance of components of the system, only total system energy. I am not trying to complicate the matter, but rather simplify it. If the rotational energy becomes less with a change in position of the weights then where does that energy go? It cannot simply disappear.

Cross post Cameron. But, what then if we assume a perfectly elastic material?

Cameron,

Any answer that requires some sort of loss mechanism at work isn't a very satisfactory answer unless it can show that :

A. Such loss is unavoidable.

B. Such loss is independent of the materials in the system.

C. Such a loss is dependent on the system geometry.

Now we are talking about loss mechanisms suddenly instead of pure mechanisms that operate because of system geometry.

[further edit]

Note also if the system rotational energy is converted to heat because of the change in geometry of the weights to an outward position then my assertion that this results in increased entropy is correct. If it doesn't but the total energy content of the system remains as momentum that is also contained within a larger volume then entropy has still increased.

Bottom line

You are trying to tell me that energy of motion is converted to heat in some invariable manner but at the same time this does not represent an increase in entropy.

Please explain how that is so.

Paul Alciatore

01-02-2009, 10:31 AM

There is no reason to assume that the process of allowing the weights to assume a new position will result in a conversion of rotational kinetic energy to another form. We can posit that the materials are perfectly elastic so that no conversion to heat takes place. Glass is a perfectly elastic material as long as any strain produced is below the yield point.

Even so, energy cannot be created or destroyed and even if energy of rotation does convert to heat that is still kinetic energy and it is still system energy.

For the purpose of this thought experiment we can even make the position of the weights temperature dependent and account for any such conversion that way.

If you reread my arguments and questions I am not referring to the energy balance of components of the system, only total system energy. I am not trying to complicate the matter, but rather simplify it. If the rotational energy becomes less with a change in position of the weights then where does that energy go? It cannot simply disappear.

Cross post Cameron. But, what then if we assume a perfectly elastic material?

Well, here we go again.

If the materials are perfectly elastic, then the weights will reach the stops and bounce back. If it is perfectly elastic as you suggest, they will continue to bounce over and over with no end (you did say "perfectly" elastic). You will need some in-elastic substance to absorb the energy of motion and convert it to some other form of energy.

Yes, heat is a form of energy, it's even motion of the atoms. If you are going to call it system energy, you have expanded the system as we were only considering the rotational energy that could be converted back and forth. Heat energy is lost energy in this context as the original system has no way of converting it back to rotational. So, yes energy is always conserved in a closed system. But that is not a factor here.

And when you move the weights inward, you still must provide some energy source from outside the rotational system or steal it from the rotational energy as with the winding up ball. But the rotational energy does change while the angular momentum does not.

Oh, and entropy does always increase in any real system. The universe is running down. This is not in contradiction to anything I have said.

J Tiers

01-02-2009, 10:40 AM

From another point of view, Evan seems to be proposing an infinite energy sink.

If, as is true, force has to be exerted to pull in the weights, then some work is done to do that, and energy is required.

if the weights are released in some way, and move to their original positions THE QUESTION THEN BECOMES WHAT HAPPENED TO THAT ENERGY?

if there is no energy consequence at all, then we can repeat the operation of pulling in and releasing indefinitely, and energy is sucked in and destroyed.

That is impossible.

So the energy consequences of letting the weights fly out, then pulling in and repeating must be accounted for. It can't be dismissed with a "hand wave".

So the energy consequences of letting the weights fly out, then pulling in and repeating must be accounted for. It can't be dismissed with a "hand wave".

What hand wave?

I am only talking about the weights moving out. I already stipulated it takes external energy to bring them in.

This is about accounting for all the energy in the system. Cameron has stated that entropy does not account for the reduction in energy when the weights move out. He has also agreed with Paul that the difference is accounted for by conversion to heat. That is a classic example of an increase in entropy in a closed system.

It cannot be both ways at the same time.

Paul,

Perfectly elastic does not mean no loss. Glass IS a perfectly elastic material but you cannot build a perpetual motion machine because of that property.

Yes, heat is a form of energy, it's even motion of the atoms. If you are going to call it system energy, you have expanded the system as we were only considering the rotational energy that could be converted back and forth.

No, I haven't. I initially maintained that entropy increases when the size of the system expands in the same manner it does when the universe expands. The energy is distributed over a larger area. I have since repeatedly referred to "total system energy" since the actual question at hand is whether an increase in entropy accounts for anything regarding the system. I also stipulated "neglecting any losses" so attributing the change in energy to losses is not answering my question.

My words:

Um, not quite. What is happening is a change in entropy rather than energy. When the weights move outward total angular momentum remains the same(neglecting any losses) and total stored energy is unchanged. Entropy however increases as the sum total energy is now occupying a larger volume which is exactly what is happening as the universe expands. In a closed system it is not possible to reverse entropy using only the energy present in the system. So, if the weights are in the out position it requires an energy input to pull them back in. That energy input is what decreases entropy within the system.

NickH

01-02-2009, 01:29 PM

But, what then if we assume a perfectly elastic material?

Then the balls just bounce in & out forever,

you can't have them perfectly elastic and yet magically for your thought experiment also be infinitely damped upon impact.

You have to have a consistant set of rules even for a thought experiment if it is to have any meaning.

Are you suggesting that you can have translation of energy of any kind without losses? You might as well take the next step and go for greater than unity efficiency:D

Nick

BillH

01-02-2009, 01:36 PM

All you can do is alter the rotational rpm, exactly like an ice figure skater does when they bring their arms inward and speed up the rotation. You can not generate more energy from the currently stored energy in the system.

What you CAN do is tap some of the waste energry and bring it back into the system, much like some radial engines that had turbines on the exhaust that drove a hydraulic motor connected to the crank shaft to recover some lost energy.

Then the balls just bounce in & out forever,

you can't have them perfectly elastic and yet magically for your thought experiment also be infinitely damped upon impact.

You have to have a consistant set of rules even for a thought experiment if it is to have any meaning.

Are you suggesting that you can have translation of energy of any kind without losses? You might as well take the next step and go for greater than unity efficiency

Glass is perfectly elastic. It helps to know what that means.

From the Corning Museum of Glass

Push, pull or twist a piece of glass hard enough, and it will bend or stretch. Not very much, admittedly, but some bending or stretching is possible. Watch the reflections in a large window when a strong wind is blowing on it and you can observe the way the window bends from the force of the wind. Glass is an unusual material in this respect, not because it bends or stretches—most materials do—but because it returns exactly to its original shape when the bending or stretching force is removed. This characteristic of glass classifies it as a perfectly elastic material. If you apply an increasing force, the glass breaks when the force reaches the ultimate strength of the glass. But at any point short of breakage, the glass will not deform permanently.

http://www.cmog.org/dynamic.aspx?id=5752

Everyone is so far skating around my point about entropy which is my original point and has not yet been addressed correctly.

philbur

01-02-2009, 01:51 PM

I'm confused and I know I'm going to regret this but. If contraction is not possible with out external input how does the ice skater manage to do so. Also I was of the impression that a contracting universe was not a mathematical impossibility, where would the external input coming from. I thought all that is required is a conservation of energy. What have I missed?

Phil

My words:

Um, not quite. What is happening is a change in entropy rather than energy. When the weights move outward total angular momentum remains the same(neglecting any losses) and total stored energy is unchanged. Entropy however increases as the sum total energy is now occupying a larger volume which is exactly what is happening as the universe expands. In a closed system it is not possible to reverse entropy using only the energy present in the system. So, if the weights are in the out position it requires an energy input to pull them back in. That energy input is what decreases entropy within the system.

The skater is supplying stored external energy via muscular action. It isn't a closed system.

philbur

01-02-2009, 02:00 PM

But the skater is part of the spinning system. he is the universe. Also what about the contracting universe.

Phil

The skater is supplying stored external energy via muscular action. It isn't a closed system.

darryl

01-02-2009, 02:33 PM

I just thought of this- whether it serves as a model to aid in understanding much of this or not, I don't know- but here goes.

Hang the weights on springs connected to the central axis. Speed it up to some rpm where the springs are about half stretched. We'll assume the springs are perfectly elastic and bearing friction, air friction etc are not factors. The rpm is stable, and the weights have stabilized at a certain distance from the axis of rotation. We'll assume that the springs can only shorten or lengthen along a radial line and won't have any ability to 'wrap' around the axis. We have some string attached so we can pull the weights towards the axis. The mechanicals of how this is done is not important at this time.

The system is spinning- now pull the strings. You will be exerting force over distance to brings the weights inwards, so you have done some work. What happens to that effort? Now release the strings- the weights move outwards and eventually reach an equillibrium again. Pull the strings in again, doing more work.

Explain the behaviour of this model.

philbur

01-02-2009, 02:52 PM

If the work done as you pull the balls in comes from the spining ball then you have taken rotational energy from the system and converted it into work. Wether you get it back as you let the strings out again depends on what you used that work for during the contraction. If it ended up as heat then repeated operations will slow the system down to zero.

I think.

Phil

I just thought of this- whether it serves as a model to aid in understanding much of this or not, I don't know- but here goes.

Hang the weights on springs connected to the central axis. Speed it up to some rpm where the springs are about half stretched. We'll assume the springs are perfectly elastic and bearing friction, air friction etc are not factors. The rpm is stable, and the weights have stabilized at a certain distance from the axis of rotation. We'll assume that the springs can only shorten or lengthen along a radial line and won't have any ability to 'wrap' around the axis. We have some string attached so we can pull the weights towards the axis. The mechanicals of how this is done is not important at this time.

The system is spinning- now pull the strings. You will be exerting force over distance to brings the weights inwards, so you have done some work. What happens to that effort? Now release the strings- the weights move outwards and eventually reach an equillibrium again. Pull the strings in again, doing more work.

Explain the behaviour of this model.

Pulling the strings is an external input of energy. Letting the weights out is energy cost free. The system will speed up in exactly the same manner that you can pump a swing sitting in the seat.

ckelloug

01-02-2009, 07:00 PM

Evan,

Sadly, your latest argument is dead on arrival. I've already shown that there is a Coriolis force acting at right angles to the motion of the weight when moving it either direction. So to say that moving the weight outward is free demonstrates that you didn't follow the previous discussion. I believe your hangup is that you appear to believe that a spontaneous process can consume no energy. In short, your original entropy argument had more merit than this completely incorrect argument but in it's presented form, your entropy argument also failed to explain exactly where the energy goes in the closed system.

I've also shown that a definite amount of energy is consumed moving the mass either direction with an external force and I've shown that the kinetic energy between the two states is not equal.

What I haven't shown is how the energy is conserved in a closed system. It's a long derivation and I haven't completed it yet due to other business.

Following the last comments you made about a assuming a perfectly elastic system and my comments about the shear induced in the flywheel: without doing the math, I would have to conclude that the energy used in moving the weight outwards has to be manifested as a torsional oscillation in the flywheel since your added conditions on the problem remove the dissipative elements that would turn those oscillations to heat and return the system to pure rotational motion.

Jedi mind tricks will not work on the coriolis force. It's still there. As Feynman says, walk a radial line on a carnival carousel.

Regards,

Cameron

philbur

01-02-2009, 07:11 PM

Yes that's the same answer as given in post 41, but where's the flaw in post 42.

Phil

Pulling the strings is an external input of energy. Letting the weights out is energy cost free. The system will speed up in exactly the same manner that you can pump a swing sitting in the seat.

J Tiers

01-03-2009, 12:08 AM

Letting the weights out is energy cost free.

Can't be..... if the action of pulling them in uses energy, then there has to be energy consequence to letting them move out..... You'd have to do work to pull them in a second time.

You COULD use the energy available as they move out to raise a weight, for instance, increasing the potential energy of the weight.

So if you FAIL to use the energy, but the weights move out, there must definitely be an energy consequence. otherwise you would not need to do any work to pull them in the second time......

I think you actually explained the matter in a prior post, did you forget about that?

Can't be..... if the action of pulling them in uses energy, then there has to be energy consequence to letting them move out..... You'd have to do work to pull them in a second time.

You COULD use the energy available as they move out to raise a weight, for instance, increasing the potential energy of the weight.

So if you FAIL to use the energy, but the weights move out, there must definitely be an energy consequence. otherwise you would not need to do any work to pull them in the second time......

I think you actually explained the matter in a prior post, did you forget about that?

The work done the second time is added to the work done the first time. Heven't you ever swung on a swing? There doesn't have to be any relation at all between the work done on each occasion. The system has no memory.

I have already from the outset allowed there will be losses and specifically mentioned that. I am interested in what happens in the system not including the losses. There are always losses and I also mentioned a few of the possible relativistic losses. The reason for bringing up such things as a perfectly elastic system is to remove the focus on the losses which is what I originally wrote.

Losses are not the point here. What is is the energy balance of the system as the geometry changes and where the energy winds up, so to speak. If all we have is a closed system similar to Einstein's accelerating elevator then there is no place for energy to go unless it is radiated out of the system by some obscure mechanism. If all the energy that is removed from the energy of rotation is accounted for by the losses then that is entropy at work.

BillH

01-03-2009, 01:15 AM

There is no free lunch unless you harness the energy of something else that you spent nothing to create.

J Tiers

01-03-2009, 01:28 AM

The work done the second time is added to the work done the first time. Heven't you ever swung on a swing? There doesn't have to be any relation at all between the work done on each occasion. The system has no memory.

Well it would if it were a spring, as has been somewhat proposed, unless I misunderstood...... and the system has memory in that if work has been done on it, that energy is present "inside the boundaries" now, and unless it evaporates, the system 'remembers" that the energy is present.

Point being that

1) it takes work done to pull in the weights.

2) the weights CAN do work in the process of 'flying outwards".

3) If the weights are allowed to fly out to some stop, afterwards you cannot get them to do the work in (2) again except by pulling them in by doing more work.

4) Therefore there MUST be some change of energy state involved in letting them fly outwards without doing any work, as you apparently suggested earlier.

Losses are not the point here. What is is the energy balance of the system as the geometry changes and where the energy winds up, so to speak.

Well losses may be the point if the energy ends up in them (it will, eventually)...... that is all part of doing the accounting on "where the energy winds up".

The question about the weights is not so much about the fact that the work done in pulling them in comes from an external source, as it is on where that energy goes if you allow the previously pulled-in weights to fly out "without doing any work"/ "energy cost free".

I think that is exactly the sort of question you just asked.

It has been asserted that it costs energy for the weight to move outward. I want to know what that cost is and more precisely where it ends up. Regardless of whether moving the weights out does or doesn't cost energy that places no bounds on andding energy to the systen by pulling in the weights. The events are not connected. Either state can be considered alone using either state as the initial state. We also need not consider repetitions of the states. The same thing must happen each time we cycle the system through the same configuration.

J Tiers

01-03-2009, 01:57 AM

It has been asserted that it costs energy for the weight to move outward. I want to know what that cost is and more precisely where it ends up. Regardless of whether moving the weights out does or doesn't cost energy that places no bounds on andding energy to the systen by pulling in the weights. The events are not connected. Either state can be considered alone using either state as the initial state. We also need not consider repetitions of the states. The same thing must happen each time we cycle the system through the same configuration.

case 1: Allow the weights to go from state 1 (pulled in) to state 2 (out) while extracting work from them (as is plainly possible from examining the flyball governor)

Case 2: Allow them to go from state1 to state2 without apparently extracting any work from them. You are asserting that there is no energy change associated with that.

case 3: Pull weights in , from state 2 to state 1. Plainly there is an energy chage, since external input is required.....

Assume frictionless operation etc..... why is there energy required to go from state 2 to 1, but absolutely no change in energy from state 1 to 2?

Paul Alciatore

01-03-2009, 02:51 AM

The weights moving in and out are opposite processes and will involve the same amounts of energy but in opposite directions. One requires that some energy be supplied (moving in) and the other produces excessive energy which has to go somewhere, probably heat (moving out). If you first move them in by supplying energy and then move them back out by whatever process, you will return the ROTATING system to the exact state as it was before these two processes were began. Same rotational speed, same angular momentum, and same overall energy if you are talking about the energy of rotation only. The rotational energy was greater while the weights were in the inner position and less again in the outer position, but the angular momentum was not changed. Angular momentum is conserved throughout the entire process. All this assumes no friction in the rotational motion.

The energy that must be supplied to pull the weights in is exactly equal to the energy that must be dissipated when they move out. It MUST be dissipated somewhere if the weights are to remain in the outer position and not bounce back.

The quoted definition of glass as a "perfectly elastic" substance is talking about it's ability to return to the original shape. It is not intended to exclude heat losses. When glass bends, as it does, heat will be generated as the layers, the atoms rub against each other or pull apart or change relative positon. The fact that there is no permanent deformation does not mean that there is no conversion of energy to heat. And also there is air friction. The fact is, that in a real system, the weight will eventually stop bouncing back and forth and come to rest. The energy of motion (radial motion) will have been converted to heat in the weight itself, in the stop that it hits, and in the air it is passing through. This heat energy can not be converted back to mechanical energy with 100 percent efficiency and this is precisely where the entropy of the system has increased, if only temporairly. This heat energy will eventually dissipate into the surroundings and will be permanently lost to the rotating system. So in the end, the entropy of the universe has increased but the entropy of the rotating system has not increased in any appreciable manner.

Evan is hung up on entropy. Entropy can not decrease in a closed system and in all real, CLOSED systems it will increase due to losses from processes like friction. But this is NOT a closed system. We are supplying external energy to bring the weights in and energy is lost when they move out. Entropy is actually irrevelant in this experiment. If the system is cooled via external means, the entropy of the rotating system would/could actually decrease.

Energy and angular momentum are the relevant considerations. And the fact that angular momentum is conserved while energy is not created from nothing tell us that we will not get any "free lunch" from a mechanism that employs rotating weights that move in and out. In fact, to quote one of my favorite authors, "TINSTAAFL." (There Is No Such Thing As A Free Lunch PERIOD).

This does not mean that the mechanism is not useful. Perhaps we should return to the original question.

The rotational energy was greater while the weights were in the inner position and less again in the outer position, but the angular momentum was not changed. Angular momentum is conserved throughout the entire process. All this assumes no friction in the rotational motion.

The energy that must be supplied to pull the weights in is exactly equal to the energy that must be dissipated when they move out. It MUST be dissipated somewhere if the weights are to remain in the outer position and not bounce back.

That description doesn't agree with reality. The Earth is transferring angular momentum to the Moon and the Moon is moving further out as a consequence. The energy of motion is not dissipated, it is transferred. The connection between the two is gravity. Gravity is analogous to a stretched spring. It requires more energy to place something in a higher orbit than to place it in a lower orbit. This is directly analogous to transferring angular momentum from the flywheel to the weights.

Another aspect that contradicts your description is the conservation of angular momentum AND energy that takes place in an elliptical orbit. The mathematical rule for orbital velocity is that the orbiting body sweeps out equal area in equal time along the radius of the orbit. That is Kepler's second Law. Newton proved it using his theory of gravitation which is both correct for non-relativistic velocities and is still used today for navigating spacecraft.

Neither the angular momentum or the energy of an orbiting body and it's primary changes in an elliptical orbit. The total energy of the system remains the same regardless of the distance from the primary to the satellite in the course of an orbit. Furthermore, no transfer of energy or angular momentum takes place during the course of an elliptical orbit excepting the very small gravitational effects.

If we insert energy to the system in the form of acceleration it can either increase or decrease total system energy. There is no requirement that one balance the other. The energy input may be unidirectional and either pump the system or damp it. The effects of energy input on such a rotating system are not required to be complimentary. They are also not intuitive. Transfers of angular momentum between orbiting bodies results in changes in orbital period. The amount of change is proportional to the ratio of the masses. It does not represent a change in total system energy according to Kepler's Laws of planetary motion and Newton's laws of Gravitation.

Lower, closer and faster orbits represent lower energy states, not higher. Higher, further and slower orbits represent higher energy states, not lower. This is in accordance with conservation of angular momentum AND conservation of energy. If we wish to change the energy state of the system we must apply acceleration regardless of the final energy state of the system. An input of energy can be employed to either raise or reduce the energy state. It is not required to be complimentary.

darryl

01-03-2009, 06:37 AM

Looking at it strictly from the springs point of view- each spring has been stretched because a weight tied to one end is pulling on it. The process of spinning up the system has then stored some energy in each spring because a force had to be exerted over a distance. Not only did we have to add momentum to the weights, we had to add some energy to the springs as well in order for the weights to have moved outwards.

Pull the weights in and the amount of energy needed to do this is lessened by the exact amount that the springs contribute as they return closer to their length in the unstretched state. Let the weights out and the amount of energy taken out of the springs is returned to them again.

Two things to consider- the rpm of the system will rise and fall as the weights are moved in and out, and the springs will alternately absorb and release energy at the same time. A weight hanging on a spring can be made to bounce- will this not be the same in this rotating system? In other words, pull the weights in while the mechanism is spinning, then let go of the strings- the weights would oscillate outwards and inwards as the springs flex (yes or no?) and the rpm of the system would rise and fall in response.

Getting back closer to the original question, to use this mechanism we would be taking torque off the shaft to turn our vehicles drive shaft, so momentum is being converted into work to accelerate the vehicle. Here's where another aspect of this is fuzzy for me- how much of this work must be supplied by pulling the weights in as opposed to being supplied by the momentum of the system? Am I better off jabbing a stick into the ground and pushing off with it to speed up my vehicle (somewhat equivalent to pulling in the weights), or am I going to get some 'amplified' assist from the momentum of the variable flywheel system?

Here's another model to consider which is similar to pulling the weights in, then letting them out. Have a rope with a weight attached to it, then lift it. Force over distance, work is done, arm gets tired. Now lower the weight- does your arm get refreshed, or do you do still more work-( lowering the weight in this case is equivalent to letting the weights outwards in the flywheel system, not simply by releasing the strings but by controlling them until the equilibrium is reached again.)

It occurs to me that a model of this variable flywheel system might make a good physics project for a student. Hook up some instrumentation and test it out.

Here's another model to consider which is similar to pulling the weights in, then letting them out. Have a rope with a weight attached to it, then lift it. Force over distance, work is done, arm gets tired. Now lower the weight- does your arm get refreshed, or do you do still more work-( lowering the weight in this case is equivalent to letting the weights outwards in the flywheel system, not simply by releasing the strings but by controlling them until the equilibrium is reached again.)

You aren't doing work on the object being lowered. You are absorbing the potential energy of the object. That is the exact opposite of doing work on it. It is doing work on you. If you were able to absorb that energy instead of wasting it as friction then there would be no loss of energy. There would be a conversion of potential energy to some other form.

Here is another way to look at lowering the weight. Say you have an equal weight suspended from a cable that runs over a pulley. It requires no work to maintain an object at a particular position any more than a table does work on a book sitting on it. To lower the weight in question you connect your handy cable and with your Acme frictionless cable system you apply an infinitesimal amount of force to makes the weights change position. You then lock your cable with your weight in the new position and unhook. The subject weight has been placed in a new lower energy position and your weight now has the potential energy the subject weight had. Was any work done?

J Tiers

01-03-2009, 11:14 AM

You are still hand waving a bit....

You need to account strictly for the energy.

if the system is your rotating shaft and weights...... go from state2 to state 1. in what I called state 1 energy has therefore just been input to the system by pulling in the weights.

now allow it to go to state 2 in some way. your "released latch", for instance..... the weights simply fly out with no friction or resistance.*

In state 2 now, the system is mechanically at the same condition it was before it went to state 1.

But a process put in energy to the rotating system, and so far that energy has not been accounted for.

Account for it.

* if the weights do work as they fly out, the energy can be accounted for

"Account for it."

The system gains energy and rotates faster. Just like pumping a swing.

J Tiers

01-03-2009, 12:26 PM

"Account for it."

The system gains energy and rotates faster. Just like pumping a swing.

Faster than it would have otherwise, you mean...

I won't ask for the math.

Paul Alciatore

01-03-2009, 12:45 PM

You aren't doing work on the object being lowered. You are absorbing the potential energy of the object. That is the exact opposite of doing work on it. It is doing work on you. If you were able to absorb that energy instead of wasting it as friction then there would be no loss of energy. There would be a conversion of potential energy to some other form.

Evan,

Before you make such a statement, sit in a freely rotating chair with some weights in your hands, spin yourself or have someone do it, and move the weights inward toward yourself. You will observe that it takes work (Force X Distance) to do so. This is not the same thing as a satellite orbit.

Evan,

Before you make such a statement, sit in a freely rotating chair with some weights in your hands, spin yourself or have someone do it, and move the weights inward toward yourself. You will observe that it takes work (Force X Distance) to do so. This is not the same thing as a satellite orbit.

But, if I have my handy acme frictionless pulley with weight attached brimming with potential energy and I attach the line to the presently extended weight I can, using an infinitesimal amount of energy, reel in the extended weight by trading off the potential energy as I lower my weight. Potential energy is traded from one weight to the other. It remains as potential energy and is not converted to kinetic energy or heat etc. It also recaptures the kinetic energy of the extended weight while exact balance of the weights is maintained by using my Acme cam shaped pulley.

It is the same as a satellite orbit. The rules are the same everywhere.

ckelloug

01-04-2009, 01:50 AM

Well,

I got out the dynamics books and cranked out some math. Fortunately, one of the books had an example of a marble in a track on a rotating disk. Unfortunately, they assume constant angular velocity which means that in our problem we'd have to assume a motor inputting energy to keep the angular velocity constant. Strangely, reading Feynman carefully, he stipulates without proof that moving the mass along the radius of the flywheel shouldn't cause a change in angular velocity which doesn't make sense unless he's assuming that the flywheel has a much higher moment than the marble or something.

The following derivation really isn't helpful but. . .

for constant omega:

ma= m* r_double_dot(t)-m*omega^2*r(t) in the radial direction + 2*m*omega*r_dot in the tangential direction.

The solution is thus a system of differential equations which spells messy. The above derived equation matches the book but I don't have an easy cross check for the following work. Assuming I haven't made any mistakes or forgotten anything, I noticed the following:

looking in the tangential direction, we can solve this differential equation for the acceleration of the mass:

m*r_double_dot(t)=m*omega^2*r(t)

Solving by inspection where r is the position of the marble,

r(t)=e^(omega*t)

Let's call the force on the marble F(t)

F(t)=m*omega^2*e^(omega*t)

So, if we integrate F(t) * r(t) then we should get work.

W=Integral m*omega^2*e^(omega*t)* e^(omega*t) dt

W=Integral m*omega^2*e^(2*omega*t) dt

This is just painful enough to integrate that I'll leave it at that.

The tangential coriolis force is then the following since the speed of motion is set by solving the differential equation for the speed of the particle:

2*m*omega^2*e^(omega*t).

The tangential displacement S= 2*Pi*r(t)*omega*t

differentiating with respect to t, dS/dt=2*Pi*r(t)*omega

The work done by the coriolis force therefore must be

W=Integral 4*Pi*m*omega^3*e^(omega*t) dt

Integrating this out is also painful so I'll stop.

Conclusion for a marble running in a track with constant angular momentum in the system:

There is definitely a force on the marble pointing to the outside if it is free to roll in the track. Since the marble is free to roll in the track, it appears no force provided by the system radially can either push or oppose the marble moving directly down the track. Since there is no constraint on the marble in the axial direction, no work can be done by the marble on the system via an axial force freely moving the weights out. Work is definitely done axially to pull the marble inward.

The marble moves outwards because it is being forced into circular motion by the Coriolis force at ever increasing radius while there is no portion of the flywheel providing a centripetal force to keep the radius constant.

The marble would like to fly off the flywheel tangentially but is prevented from doing so by the coriolis force provided by the side of the track while it moves. If the track provides a tangential force to the marble, the marble has to provide a tangential force to the track. That is the torque provided by the coriolis force which accelerates and decelerates the flywheel to keep angular momentum constant.

The coriolis force acts when the weight is moved either direction which definitively proves that work is done to move the weight either direction and because omega is non-zero, it is guaranteed to do work moving weights in either direction.

With no dissipative elements and an ideal elastic flywheel, the torque provided by the coriolis force moving the weight outward has to be converted at least in part to torsional vibrations in the flywheel. This mechanism shows that moving the weight out cannot be done without an energy consequence whether or not this mechanism accounts for all of the energy.

There is one other thing that hasn't been postulated yet and that is that there is some way to view a portion of moving the weights as a way of converting potential energy to kinetic energy. I cannot see any way of making this work however because letting the weights out reduces the kinetic energy and pulling them in against the centrifugal force(which is real in a rotating coordinate system) increase kinetic energy.

With Gravity which is a conservative force, raising the weight creates potential energy and dropping it creates kinetic energy. Pushing against the Centrifugal Force in our system (analyzed in rotating coordinates) increases kinetic energy so it's definitely not the same circumstance as gravitational kinetic and potential energy.

I'm going to have to give up on this problem as I don't have any more to add to it. The hint that this is actually a hard problem comes from the fact that my dynamics book and Feynman both dance around this problem and use it as examples being careful not to deeply analyze the part that's causing us fits.

As both evidence for the conclusion that I came to and as an example for Darryl's postulate about the springs, this wikipedia article about rotational vibrational coupling is interesting. http://en.wikipedia.org/wiki/Rotational-vibrational_coupling

Regards,

Cameron

darryl

01-04-2009, 04:44 AM

Cameron, you've thrown a wrinkle into this by adding r-v coupling. It is valid of course, and understandable- the animation helps with that. My simple system has gotten quite complicated :)

It's been interesting and informative reading all the replies. I appreciate the time and effort that has gone into them, and I hope no-ones brain has caught on fire :)

As often happens with posers like this, it will probably be filed away in my minds workshop, and at some point it will emerge again. I've learnt more than one thing- aside from issues surrounding conservation of energy, transfer of energy between different forms, etc, I'm more impressed than ever with one principle- KISS.

J Tiers

01-04-2009, 10:12 AM

Ckellog:

Not sure I "GOT" the issue you had with pulling weights in and letting them do work on the way out.

If you want them to do work on the way out, you can propose a mechanism (frictionless) similar to the governor..... make the plunger lift a mass, and you have a way to extract work from them.

Obviously you could make the mass either be lifted, or in falling move the weights

Alternately, you can propose gravity as axial to rotation, and have the moving weight move on any sort of 3 dimensional radial track, providing their own mass to be lifted or to accelerate them

This is obvious enough that you may have already discarded it for a reason I don't see immediately.

ckelloug

01-04-2009, 11:47 AM

JT,

I don't have an issues with pulling the weights in and letting them do work on the way out. They do work on the flywheel reducing its kinetic energy by applying a torque to the assembly via the coriolis force.

What I haven't been able to do is think of a good thought experiment to show that a torque opposite the motion and energy released by moving the weights outward does something. It seems rather difficult to come up with an experiment that demonstrates only that.

In that respect, letting the energy wind the flywheel as a torsional spring at least shows that something unexpected had to happen.

I see this as one of those problems that is really only practically solvable with energy methods and energy methods often tell you what happened while only giving hints about how.

Thanks for a chance to go re-learn some stuff.

--Cameron

J Tiers

01-04-2009, 11:51 AM

What I haven't been able to do is think of a good thought experiment to show that a torque opposite the motion and energy released by moving the weights outward does something. It seems rather difficult to come up with an experiment that demonstrates only that.

Ah... active part in bold........ I'll think about it.....

tenfingers

01-04-2009, 12:36 PM

This has been an interesting discussion. If I try to jump into Evan’s thought experiment and think about his latch that can release the mass and then catch it at a larger radius with a perfectly elastic stop, I imagine that the delta energy in question is not lost, but ends up as a radial vibration with the energy being transferred between radial kinetic energy and elastic spring energy. If the magic latch could then release the mass when it is at it’s maximum inward radial velocity, it would end up right back at the starting radius. Since this is a reversible process, there would have been no change in entropy.

Paul Alciatore

01-04-2009, 01:31 PM

But, if I have my handy acme frictionless pulley with weight attached brimming with potential energy and I attach the line to the presently extended weight I can, using an infinitesimal amount of energy, reel in the extended weight by trading off the potential energy as I lower my weight. Potential energy is traded from one weight to the other. It remains as potential energy and is not converted to kinetic energy or heat etc. It also recaptures the kinetic energy of the extended weight while exact balance of the weights is maintained by using my Acme cam shaped pulley.

It is the same as a satellite orbit. The rules are the same everywhere.

Evan,

If you trade one for the other, you accomplish nothing. The final state is identical to the original state except for the names/numbers on the weights. You were talking about moving a SINGLE, ONE satellite to a different orbit, not trading energy between two of them. If one weight or two weights are drawn in, it will require work (Force X Distance). An energy input will be needed.

I think you do know this and are just trying to keep me going with endless variations.

Feynman's texts and many others are essentially correct. If they weren't, the whole world of physics would jump down his throat just for the fun of showing him up. Scientists are like that. Cameron may have found an error in the statement that angular velocity remains constant. It does not, it is angular momentum that is constant when the weight(s) are moved in or out. But I do not have that text to read so I don't know if the text is in error or if Cameron is reading it wrong.

ckelloug

01-04-2009, 02:47 PM

Paul,

Feynman's exact words were:

If m moves only along the radius, omega stays constant so that the torque is:

tau=f_c*r=dL/dt=d(m*omega*r^2)/dt=2*m*omega*r*dr/dt

So he's differentiating angular momentum as the particle moves outward with constant omega (energy added from an outside force to keep omega constant) and he ignores the flywheel since its angular momentum won't change since its angular velocity didn't change. It's not that angular velocity should be constant, it that he's holding it constant and showing that a torque must be applied to move the particle radially. This is the same derivation strategy that Applied Mechanics: Dynamics by Charles Smith used which I also realized doesn't directly apply.

Thus Feynman is demonstrating that for angular velocity to remain constant that a torque has to be applied to the system. Feynman's derivation is thus right but it doesn't apply except by analogy to the problem we have been discussing. Frustrating eh. So all he did is take the partial derivative of r with respect to t rather than derivative of the entire expression with respect to t. The Feynman Lectures book is basically an edited transcript of a bunch of physics lectures Feynman gave and sometimes the context of a remark can't be extracted from the text easily as is the case here. I think we've fallen victim to a lecture hand wave. . . insofar as this doesn't provide much insight into our problem even though it looks good.

--Cameron

cmbchicago

01-04-2009, 09:15 PM

sounds like a governor to me.....this regulates the fuel in an old fashioned steam engine. Everyone has seen the large brass balls spinning...also home of the phrase "balls out" because at the maximum speed the balls are "out" as far as they can be.

Evan,

If you trade one for the other, you accomplish nothing. The final state is identical to the original state except for the names/numbers on the weights. You were talking about moving a SINGLE, ONE satellite to a different orbit, not trading energy between two of them. If one weight or two weights are drawn in, it will require work (Force X Distance). An energy input will be needed.

A single satellite in an elliptical oribit constantly trades potential energy for kinetic energy and back. It also is constantly exchanging angular momentum with it's primary. Note that when I say satellite I mean a non negligble mass such as the Moon although even a ball bearing in orbit will follow the same rules. The Moon and the Earth actually both orbit the sun and not each other so it won't stand too close a scrutiny in this context. If you plot the orbit of the Moon around the Earth and the Sun you will find that it is always convex to the Sun.

Moving to a different orbit is another matter entirely and I didn't directly deal with that. My reply disputing your example was based only on a single satellite in an elliptical orbit around a primary, not changing orbits. It is equivalent.

I haven't been bringing up these variations, they have been suggested by Darryl who started this thread.

For your entertainment, boys and gurls, ladies and gentlemen:

http://www.youtube.com/watch?v=iODccWH8AgU

You see it all explained with sublime clarity. :D

Fasttrack

01-05-2009, 01:54 AM

What in the world are we still arguing about!? I thought this question was answered back in page one...

Rotational Kinetic Energy, Krot = (1/2)Iw^2

Angular Momentum, L = Iw

I = moment of inertia

w = omega, the angular velocity

Doing some fifth grade algebra, Krot = (1/2)L^2/I

Clearly, Krot increases as I decreases. This should be obvious. The amount of increase in Krot is equall to the work done in moving the weights in and thus decreasing I.

In the opposite case, as I increases, the decrease in kinetic energy can be viewed as the product of an imaginary torque produced by the coriolis effect.

Cameron - I did not even read your work, so I've no idea if its correct, but that integral is not nearly as hard to evaluate as it first appears - or at least not too difficult to put into a more recognizable form. As you note, this derivation is not helpful in our case since it assumes a constant w. BUT, since it assumes a constant w, the integral reduces to int{(e^w^t)dt}. With a little simple algebra, it becomes clear that this is actually a form of the exponential integral function. If I did my algebra right, you get -Ei(1,w^t)/ln(w). You could then plug this into a computer program and have it numerically solve to some degree of accuracy.

To solve this little problem, we need to find the Lagrangian of the system and substitute this into the Euler-Lagrange equation. This will yield a set of partial differential equations that could then, in principle, be solved. This gives us our equations of motion for the mass and an imaginary force will be revealed - the coriolis force.

It would be less common, but not incorrect, to view this as a matter of entropy, as Evan noted in one of his posts. According to Cutnell's and Johnson's Physics 4th Edition (original title, eh?) entropy is "a measure of the partial loss of the ability of a system to perform work due to the effects of irreversibility", which is exactly what we have going on when the weights move towards the outside. As Evan says, this is an increase in entropy and the energy is "lost" as far as its ability to do meaningful work in our system.

That may be hard to "stomach", but we don't think it strange when a compressed gas cools as it expands. It is absorbing energy from the surroundings, energy which cannot be recovered with out doing more work to the system (i.e. re-compressing the gas). Similiarly, we don't find it strange that those chemical cold packs in a first-aid kit work. If you examine them based on energy alone, they appear to be impossible chemical reactions. They are endothermic but, due to the increase in entropy, the reaction proceeds and the pack cools.

ckelloug

01-05-2009, 10:24 AM

Fasttrack,

Thanks. I will study your work here. Euler-Lagrange equations are something physicists studied and engineers didn't where I went to school. Engineers I have met and the books Engineers write have a tendency to want to directly integrate F=ma leading to the hard way of solving most problems.

Evan had originally argued that all there was was only an entropy change and then argued that no work was done moving the weights out.

I agree that there is an entropy change (which perhaps unfairly to Evan, I didn't mention a couple posts ago when I realized it) but there is also work done by/on the weights as everybody has shown.

Thanks for pointing out the integral form for the equation I wrote. It looked like one when I looked in the table but I got confused arguing with myself whether or not omega was constant which was stupid as omega was constant.

In that you have mentioned that you get a system of pde's does it seem likely that the answer oscillates as Paul and I both guessed? I have a lot of bad memories of pde's and my brain would need a soak in naval jelly to get that rust out and actually work out the system of pde's.

Interesting discussion from such a seemingly simple problem. I think this is why the engineers refuse to deal with this case and limit themselves to rigid body dynamics.

What was bothering me was that you originally claimed that it took work to move the weights out. It doesn't. As I explained later the weights do work when they are let out.

My words: "You aren't doing work on the object being lowered. You are absorbing the potential energy of the object. That is the exact opposite of doing work on it. It is doing work on you. "

As Jtiers pointed out, there is work done in moving the weights inward and as a corolary there is work provided by the flywheel to move them outwards.

It requires no work by the flywheel to let them out. They have their own share of angular momentum and would move out when released in the same manner if the flywheel were absent.

So let's simplify the situation to just a pair of weights orbiting around a common center held by a massless cable. We cut the cable. Now what is the system energy? One may argue it is zero as it has been carried away.

But then the question is what is the energy of each weight relative to the other?

ckelloug

01-05-2009, 01:15 PM

Evan,

The rotational kinetic energy of the system is less with the weights out than the weights in. There is no argument about that from anybody here.

By the work energy theorem, work was by definition done on the flywheel for it to have a lower kinetic energy than it did with the weights in. By Newton's second law, if the weights did work on the flywheel then the flywheel did the same work on the weights with the opposite sign.

If you cut the string, the weights have translational kinetic energy of 1/2 m*v^2 as they fly off. The system loses that energy and the weights gain it.

I think Fasttrack was right that this discussion was really over on about page 2.

The rotational kinetic energy of the system is less with the weights out than the weights in.

Where did it go? I have yet to see a satisfactory answer to that. The kinetic energy cannot just evaporate. All I have seen so far is a bunch of hand waving about heat losses with no real explanation of how and why they must occur. I have proposed scenarios that are within the physical laws that avoid such losses. I understand perfectly that the weights do work on the flywheel but that doesn't result in absolute losses. That is merely a transfer of energy. Invoking losses requires that the losses be quantified such as the limitation placed on a heat engine by the Carnot Limit.

Losses that are not bound to absolute values or limits can be written out of the equation. Since they are variable with no fixed quantity they don't play a theoretical part in most energy calculations. As long as we don't invoke negative losses we will still obtain correct answers. The only limit placed on losses such as friction is that they may not be less than zero. Losses above zero are merely an engineering problem unless it can be shown that a certain loss mechanism is a direct consequence of the process and that it has a minimum value x>0.

As an example, it is physically impossible for a body in space to capture another body except by direct impact, UNLESS there are quantifiable loss mechanisms such as an atmosphere. Any body that approaches another must by definition be in a hyperbolic orbit with respect to the body it approaches. It must have an excess velocity over the escape velocity of the body approached. That means that it cannot be captured without a loss mechanism or the influence of a third body.

In the opposite case, as I increases, the decrease in kinetic energy can be viewed as the product of an imaginary torque produced by the coriolis effect.

So where is it?

Fasttrack

01-05-2009, 03:51 PM

How the hell should I know? :D

I may have confused myself or others in my explanation. I'm assuming a very simple case where a weight is tied to the end of a massless string and its spun in a flat, horizontal circle. Furthermore, I'm assuming no atmosphere or gravity or anything else to complicate matters. I stipulate that I can lengthen or shorten the string arbitrarily. There is no second weight; the end of the string is tied to a fixed point.

When I lengthen the string, the rotational kinetic energy of the weight decreases because the weight does work on the system. We know that the lengthening of the string causes the angular velocity, w, to decrease. This requires a negative acceration. Furthermore, we know that the moment of inertia, I, increases by the square of the distance. (I=mr^2) This acceleration only happens for a distinct angular displacement, theta. This theta is the displacement from the time that the string lengthens and the weight flies off in a geodesic to the moment tension is reaplied to the string.

Thus, the rotational work done is given by the expression W = I(alpha)(theta). We could then take this value and plug it into the equation W = (tau)(theta) and find the value for our imaginary torque. Note that W = Krot initial - Krot final or vice-versa depending upon wether your lengthening or shortening the string. Just need to get the minus signs to work out right.

Clearly this work is irretrievable and comes back to entropy. And no, the system will not oscillate. (Although just because there is a system of pde's doesn't mean it can't oscillate. I just had a problem about a month and half ago that was a set of pde's from the Euler-Lagrange equation. The solution ended up being a cycloid, but it took alot of what I call "mathmagic" to make it look like the normal cycloid equation. I only remember it because it went from being this extraordinarily messy bit of math to the easily recognizable cycloid. I always find it exciting when something really nasty reduces to one of your old "math friends" :D)

Does that answer the question?

(edited to add the "Note")

Rich Carlstedt

01-05-2009, 04:50 PM

you know fella's....forgive me, but you guys are getting into so much nitty gritty, that you overlook obvious results.

First, if the String is not appling energy (which is unlike a flyball governor stem !) energy cannot be lost.

The energy of the ball is, it's weight (mass )times it's speed

changing the string length may allow it to travel a larger diameter, and yes, its angular velocity is reduced, but it relative velocity remains unchanged.

Torque has nothing to do with it unless:

You are controlling the accelleration/deceleration with a "fixed" arm, not a string

or You have a rigid system

or You want to add or subtrack energy

non of that was presuposed

rich

Fasttrack

01-05-2009, 05:26 PM

you know fella's....forgive me, but you guys are getting into so much nitty gritty, that you overlook obvious results.

First, if the String is not appling energy (which is unlike a flyball governor stem !) energy cannot be lost.

The energy of the ball is, it's weight (mass )times it's speed

changing the string length may allow it to travel a larger diameter, and yes, its angular velocity is reduced, but it relative velocity remains unchanged.

Torque has nothing to do with it unless:

You are controlling the accelleration/deceleration with a "fixed" arm, not a string

or You have a rigid system

or You want to add or subtrack energy

non of that was presuposed

rich

Nope. I refer you to my original post in this thread. There is rotational kinetic energy which absolutely changes with changing angular velocity. This is not the same as translational kinetic energy, which is what your talking about. Work must be done to shorten the string and, therefore, there must also be an energy change associated with increasing the length of the string.

I originally supposed the torque was purely imaginary, due to a rotating coordinate system, but I no longer think that is neccessarily true. In fact, that shouldn't be an issue at all, since in my model I am measuring relative to a stationary refrence frame, not the coordinate system attached to the rotating system. This will answer Evan's question more satisfactorily:

If I spontaneously lengthen the string, the weight flies off following a geodesic. In this case, we assume flat space and, therefore, a straight line. This weight has a well defined momentum at this point. When the slack is removed from the string, there is suddenly a force acting to change the motion of the weight. There is also an impulse felt as the momentum changes. This impulse is the "loss of energy". For some discrete amount of time, a force is applied to the string in a direction that resists the rotation of the system. This force, multiplied by the new length of the string, is the torque that slows the angular velocity. If we calculate the amount of time the force is applied, we can solve for the angular displacement covered in that amount of time and finally arrive at a numeric result for the energy loss.

:edit: This would be very difficult to do in practice. In fact, after thinking about it for the evening, I've convinced myself that this a formalism doomed to failure.

Edit: in reality, this would not work with a string. If you think about what happens if you suddenly release the tension on the string and then reapply it, the circular orbit decays very fast since the force of tension (once the slack is removed) is no longer perpindicular to the velocity vector of the mass. A better model would have been a rigid arm.

If I spontaneously lengthen the string, the weight flies off following a geodesic. In this case, we assume flat space and, therefore, a straight line. This weight has a well defined momentum at this point. When the slack is removed from the string, there is suddenly a force acting to change the motion of the weight. There is also an impulse felt as the momentum changes. This impulse is the "loss of energy". For some discrete amount of time, a force is applied to the string in a direction that resists the rotation of the system. This force, multiplied by the new length of the string, is the torque that slows the angular velocity. If we calculate the amount of time the force is applied, we can solve for the angular displacement covered in that amount of time and finally arrive at a numeric result for the energy loss.

Since this is a non-relativistic problem I refer you to Newton's laws of motion, in particular his third law. For every action there is an equal and opposite reaction. There MUST be an exchange of kinetic energy to account for the change in angular velocity. Since such a change does not involve an exothermic process the energy should remain as kinetic energy or possibly be converted to potential energy.

If you slow down a cue ball by running it into another ball the cue ball loses energy. The energy does not go away but is transfered to the other ball.

So, you haven't answered my question yet.

Further, if I build a winch with 100% efficiency and no losses then what happens if I reel the weight out and back in? My winch is equipped with a motor generator and an ultracapacitor to store energy. It stores the energy on the way out and uses it on the way in. Now I know where the energy is, in the Ultracapacitor. How does this change the system and why? It shouldn't make any difference since the winch is only an engineering problem.

Fasttrack

01-05-2009, 07:06 PM

Further, if I build a winch with 100% efficiency and no losses then what happens if I reel the weight out and back in? My winch is equipped with a motor generator and an ultracapacitor to store energy. It stores the energy on the way out and uses it on the way in. Now I know where the energy is, in the Ultracapacitor. How does this change the system and why? It shouldn't make any difference since the winch is only an engineering problem.

That is an entirely different problem. You are moving a mass around in a conservative vector field. In my model, there is no vector field.

I'll respond more in depth later. Its supper time! :)

lazlo

01-05-2009, 07:12 PM

I'll respond more in depth later. Its supper time! :)

Ah, you're from the Mid-West -- the rest of us eat dinner :D

Are you having Pop with your supper? :)

Fasttrack

01-06-2009, 12:45 AM

Ah, you're from the Mid-West -- the rest of us eat dinner :D

Are you having Pop with your supper? :)

Nope. My dad is from Milwaukee, my mom is from near Kansas City and lived for a significant time first in California, then in Pennslyvania before finally moving to the Chicagoland area. Subsequently there is a variety of idioms commonly used by family members. "Pop" is defintely "soda", but "lunch" is sometimes "dinner" and "dinner" is sometimes "supper" :D

I'm still not sure what the issue is with the kinetic energy. I think your thinking about it too much. Just look at the math. The force constraining the mass to move in a circle is aimed inward. As the mass is “let out” to larger radius, the motion of moving out is opposite the force vector. The dot product is negative. The net KE decreases. If that kinetic energy isn't converted to potential energy by raising a mass, as in the classic example, then it is lost due to "heat of deformation". It is impossible to recover and is lost in the stretch of the string, rod, etc as well as minor plastic deformations of the mass or the bracketry that holds the mass to the rod. Worrying specifically about exactly where it goes is the engineering problem :D

Kinetic energy is not a strict conservation law, btw, unlike conservation of momentum (I probably just outraged all the engineers and chemists out there).

tenfingers

01-06-2009, 01:06 AM

This makes perfect sense, as does Evan's last post. My only problem is seeing where the disagreement lies.

Fasttrack

01-06-2009, 01:12 AM

This makes perfect sense, as does Evan's last post. My only problem is seeing where the disagreement lies.

:) Me too. I think we are in perfect agreement, but I could be wrong! Although I hope no one is getting "emotionally involved" in this discussion. I don't mean to step on anyone's pride or feelings, here. Just a discussion, not an argument :)

tenfingers

01-06-2009, 01:18 AM

You mean it's not a fight? Then I'm outta here. I'm gonna check out what's going on with Evan's spindle bearing.

Rich Carlstedt

01-06-2009, 01:57 AM

Wait, before you leave

fastrack said 'A better model would have been a rigid arm.'

Exactly as I stated earlier..

rigid and non rigid arms do two different functions

You can't push a chain, but you can drag it

Also fastrack, one of your issues that I disagree with is the use of the term "Force'

a "Force" does no work unless it moves.

therefore the forces on a string acomplish nothing in the equation, except to unnecessarily complicate it.

Only when a change occurs is there work performed

If the ball spinning around the poll hit you, you would get X pounds of energy transferred to your kisser

If the ball came loose for a split second ( goes straight !) and hit you, the same force would be transferred, because the ball has force and energy.

the string does not

As my physics teacher said years ago.

a ditch digger does nothing by holding his shovel

He has "potential"

When he moves the shovel, and it becomes dynamic, only then does work occur.

At least thats the explanation in my world

Its been fun

Rich

Fasttrack

01-06-2009, 02:16 AM

I haven't left! ;)

I was not disagreeing with you on the rigid arm point, although that is (in my book) a minor detail. I was conerned with the physics, not the apparatus.

A force does not move. A force gives rise to an acceleration, plain and simple. In rotational mechanics, a force is ABSOLUTELY needed to cause a centripedal acceleration. Objects will never orbit in a circle without some force to cause them to deviate from their straight line. (Assuming that we are still working in flat space). This is Newton's first law. Adding the force of tension and, by corollary, the centripedal force, is not an over-complication.

The example you give is not accurate. If the ball hits you in the face, it will transfer some amount of energy, yes. But energy and force are two entirely different things. The force transmitted is equall to the change in momentum (impulse) divided by the time that the ball is in contact with your face. The only way to know the amount of force your face feels is to know the change in momentum and the amount of time that the ball is in contact with your face. The ball does not "have force". It is moving at a constant angular velocity so the only force acting on it is perpindicular to its motion. It can apply a force by interacting with another body and changing its momentum.

Finally, I think there may still be some confusion about the differences between kinetic and rotational energy. Just for kicks, let me toss this problem out to you. Lets say you are designing a roller coaster and you put a loop in the tracks. The distance from the bottom of the loop to the top of the loop is 2R. How high above the bottom of the loop, in terms of R, does the roller coaster need to start inorder to have just enough energy to go through the loop? (Assume frictionless track, no air resistance, etc etc) This is a very popular problem, which neatly demonstrates the difference between kinetic and rotational energy.

In fact, this may even help in our understanding of "where the energy goes".

And, just for clarification, energy is measured in foot-lbs, just like torque. lbs is just a measure of force. (Or we could measure in joules, etc etc)

Fasttrack

01-06-2009, 02:27 AM

Also fastrack, one of your issues that I disagree with is the use of the term "Force'

And I can't help but add:

By all means disagree with my use of the term "force", but know that my use of the term "force" is perfectly consistent with every reputable physics textbook out there. I am a physics tutor and will be teaching a class of classical mechanics to a group of engineers next year. My area of study is physics and that is how I make the money I spend on machining! :) I don't claim to be an expert machinist, or even a very good machinist, but I do claim to be a competent (if still woefully ignorant) physicist. In fact, I am closing in on my first publication, regarding the three-body problem in QED. (And this is where some of my trouble comes from. Its been so long since I last did any classical mechanics that I find it takes me a while to get to the correct answer. A lot more time than it used to. I need to brush up on that before I have to teach students, or I'll get a curveball question and blank out!!)

My take on force is pretty basic - a bathysphere at the bottom of the sea, Marianas Trench. Lots of force in it, tons per square foot, certainly, but no work being done. Unless that bathysphere ruptures in which case that force becomes work in the form of a water clap rather quickly. That seems consistent with Rich's no motion no work view of force.

Biological resistance to force does create work as all biological entities will fatigue against persistent force which kind of flies in the face of the claim of a worker holding a shovel is not producing work. He is, but is not efficient but in the same way that the steam that blows the whistle will never turn the wheel. But the example makes the intended point.

I'm still not sure what the issue is with the kinetic energy. I think your thinking about it too much. Just look at the math. The force constraining the mass to move in a circle is aimed inward. As the mass is “let out” to larger radius, the motion of moving out is opposite the force vector. The dot product is negative. The net KE decreases. If that kinetic energy isn't converted to potential energy by raising a mass, as in the classic example, then it is lost due to "heat of deformation". It is impossible to recover and is lost in the stretch of the string, rod, etc as well as minor plastic deformations of the mass or the bracketry that holds the mass to the rod. Worrying specifically about exactly where it goes is the engineering problem

Aha! SO, there is not a specific mechanism that describes where the energy goes.

That explanation is some pretty heavy hand waving. Don't feel bad, apparently Feynman did the same thing. What it comes down to is this: Without a quantifiable degree of loss that is mandated by the laws of physics then no particular value may be assigned to the loss of energy. In that case we may assign an arbitrary value as long as it isn't negative since we are talking about an efficiency problem. Since this is a thought experiment I choose to assign a loss value infinitely close to zero.

Since energy cannot be either created or destroyed it must then be accounted for. That means any decrease in kinetic energy in the system is accompanied by an increase another form of energy other than heat and the total energy in the system remains the same.

In order to show otherwise you must define the losses.

derekm

01-06-2009, 07:16 AM

.... lbs is just a measure of force. (Or we could measure in joules, etc etc)

lb is a measurement of MASS. you mean lbf. Or better still N

Derek - SI from 1968

derekm

01-06-2009, 07:33 AM

Since energy cannot be either created or destroyed it must then be accounted for. That means any decrease in kinetic energy in the system is accompanied by an increase another form of energy other than heat and the total energy in the system remains the same.

explain what you mean here?

The math shows that the kinetic energy of the system decreases when the rotational circumference is increased. The question is where does that kinetic energy go? The only "answer" so far provided is that it appears as losses in whatever mechanism is used to increase the diameter of rotation. This is a highly unsatisfactory explanation from a rigourous scientific point of view as it provides no clear definition of what degree of loss must exist or why. Losses are generally caused by conditions that are dependent on properties that are not an inherent part of the fundamental operating principles involved. As such they are amenable to modification through design and substitution of materials. For that reason we can postulate materials that have virually no loss. This then poses a problem since the math requires a defined loss of kinetic energy but doesn't show what happens to it.

derekm

01-06-2009, 10:25 AM

lets add to the thought experiment. The rotating mass is held by a perfectly inelastic string. The strings length is regulated by a perfectly efficient winch with a friction brake at the axis. To let the string out slowly the brake is used against the rotating mass's natural tendency to pull the string out of the winch.

The brake gets hotter - where did the energy come from? and if we know the heat capacity of the brake can we predict the temperature?

That is what I just proposed except instead of dissipating the energy it is stored in a capacitor so that it can be used to reel in the weight later. Then what happens to the energy "loss"?

Paul Alciatore

01-06-2009, 11:09 AM

Where did it go? I have yet to see a satisfactory answer to that. The kinetic energy cannot just evaporate. All I have seen so far is a bunch of hand waving about heat losses with no real explanation of how and why they must occur. I have proposed scenarios that are within the physical laws that avoid such losses. I understand perfectly that the weights do work on the flywheel but that doesn't result in absolute losses. That is merely a transfer of energy. Invoking losses requires that the losses be quantified such as the limitation placed on a heat engine by the Carnot Limit.

Losses that are not bound to absolute values or limits can be written out of the equation. Since they are variable with no fixed quantity they don't play a theoretical part in most energy calculations. As long as we don't invoke negative losses we will still obtain correct answers. The only limit placed on losses such as friction is that they may not be less than zero. Losses above zero are merely an engineering problem unless it can be shown that a certain loss mechanism is a direct consequence of the process and that it has a minimum value x>0.

As an example, it is physically impossible for a body in space to capture another body except by direct impact, UNLESS there are quantifiable loss mechanisms such as an atmosphere. Any body that approaches another must by definition be in a hyperbolic orbit with respect to the body it approaches. It must have an excess velocity over the escape velocity of the body approached. That means that it cannot be captured without a loss mechanism or the influence of a third body.

So where is it?

Evan,

Simply proposing a scenario does not make it so. Yes, the energy must go somewhere and heat is the answer. In this case, it is not a small, negliable loss. It is the conversion of the ENTIRE amount of energy that was needed to move the weights inward to heat when they are allowed to move outward by any mechanism that will have them coming to "rest" at that new radius. It can not be neglected.

A famous experiment in physics that demonstrated and measured the heat equivalent of mechanical work consisted of simply using a paddle to stur a container of liquid. The work of sturing was measured and the temperature rise in the fluid was measured. These two represent EQUAL amounts of energy. Mechanical motion was converted to heat energy. 100%.

This WILL happen in moving the weights out to a larger radius with most common methods. You can not devise a method that will avoid this unless the energy goes into some other form like a wound spring. It has to go somewhere.

This WILL happen in moving the weights out to a larger radius with most common methods. You can not devise a method that will avoid this unless the energy goes into some other form like a wound spring. It has to go somewhere.

Yes, it must go somewhere. The problem is that the proof that the kinetic energy is reduced does not say where or how. A wound spring is potential energy just as a charged capacitor is and if that energy is returned to the system as kinetic energy then the situation is that although there is an energy cost to let the weight out it is only temporary.

The issue is that if there is no storage mechanism then we are left with an unspecified and unquantified conversion of kinetic energy to some other form. Scientists absolutely hate loose ends like that.

Fasttrack

01-06-2009, 02:28 PM

lb is a measurement of MASS. you mean lbf. Or better still N

Derek - SI from 1968

No. Lbs is a measure of force. A slug is a measure of mass. 1 lb is equall to 4.44822162 Newtons.

Edit - Ahh...I see where the confusion lies. In physics, a lb is the lb-force since lbf seems to indicate lb-feet, a measure of torque or energy. (Which is how I read your lbf initially) Mass is measured in slugs. Or, as you indicated, the much preferred method is to deal with everything in terms of kilograms, Newtons, joules, Newton-meters, etc.

Better to denote it lb (subscript) f to avoid confusions. 'Course I don't know how to do subscripts here... :)

Fasttrack

01-06-2009, 02:36 PM

Yes, it must go somewhere. The problem is that the proof that the kinetic energy is reduced does not say where or how. A wound spring is potential energy just as a charged capacitor is and if that energy is returned to the system as kinetic energy then the situation is that although there is an energy cost to let the weight out it is only temporary.

The issue is that if there is no storage mechanism then we are left with an unspecified and unquantified conversion of kinetic energy to some other form. Scientists absolutely hate loose ends like that.

Bah - we just sweep it under the carpet and pretend it didn't happen :)

The amount of loss is dependent on the apparatus. In one of the models you proposed, you had a mass hanging from the other end of the string - the classic glass tube demonstration, iirc. There the energy is transferred entirely to potential energy as the weight is moved upwards. In my model, the energy is lost to deformation of the components. I view it like dropping a ratchet wrench on a concrete floor. It had kinetic energy before coming to rest. That kinetic energy had to go somewhere, but it couldn't have been all dissipated in air resistance or wouldn't have hit the floor with any velocity. The floor doesn't have a chip and the ratchet isn't visibly scratched or bent, so where did that kinetic energy go? :)

Fasttrack

01-06-2009, 02:45 PM

My take on force is pretty basic - a bathysphere at the bottom of the sea, Marianas Trench. Lots of force in it, tons per square foot, certainly, but no work being done. Unless that bathysphere ruptures in which case that force becomes work in the form of a water clap rather quickly. That seems consistent with Rich's no motion no work view of force.

Biological resistance to force does create work as all biological entities will fatigue against persistent force which kind of flies in the face of the claim of a worker holding a shovel is not producing work. He is, but is not efficient but in the same way that the steam that blows the whistle will never turn the wheel. But the example makes the intended point.

And that is a serviceable definition, but from a physicist's point of view, it is inaccurate to claim that a system "contains" lots of force. That suggests that you are actually looking at potential energy, not force. Also, it is inaccurate to claim that a force moves. An object moves when subject to an unbalanced force and a moving object posses energy. By changing the motion of the object (by applying a force) work can be done since the energy is changed. The force does not become work, it "does" work. It is a subtle difference, I know.

To say that forces that are unmoving complicate an equation is silly if you really think about it. That contradicts everything engineers learn in their statics class. Just because there is no work being done to a bridge doesn't mean that the engineers didn't have to consider every force very carefully.

NickH

01-06-2009, 02:54 PM

No work was done in compressing the structure of the submersible?

Shurely Shome Mishtake.

There is potential energy stored in the elastic distortion of the structure I suspect,

Another blinkered view of Physics bites it, :D

Nick

NickH

01-06-2009, 02:59 PM

Yes, it must go somewhere. The problem is that the proof that the kinetic energy is reduced does not say where or how. A wound spring is potential energy just as a charged capacitor is and if that energy is returned to the system as kinetic energy then the situation is that although there is an energy cost to let the weight out it is only temporary.

The issue is that if there is no storage mechanism then we are left with an unspecified and unquantified conversion of kinetic energy to some other form. Scientists absolutely hate loose ends like that.

Evan,

please do not use the word PROOF without even aluring to one, if there were a PROOF involved it is implicit that the aspects in question are dealt with and resolved.

Without procrastination please quote or link to the PROOF to which you refer,

Thanks,

Nick

Fasttrack

01-06-2009, 03:08 PM

No work was done in compressing the structure of the submersible?

Nick

Who said that? I wouldn't consider it stored as potential energy, however. There may be some elastic potential, but once you've crushed it most of the energy is lost to the deformation of the material. Just like throwing a ball of clay at the wall. If it was converted to elastic potential, we'd expect to be able to raise the sub out of the water and have it spring back into shape! A cool trick if it would work ;)

I speak of the mathematical proof posted by Fasttrack.

Rotational Kinetic Energy, Krot = (1/2)Iw^2

Angular Momentum, L = Iw

I = moment of inertia

w = omega, the angular velocity

Doing some fifth grade algebra, Krot = (1/2)L^2/I

derekm

01-06-2009, 03:13 PM

NO! Lbs is a measure of force. PERIOD. A slug is a measure of mass. 1 lb is equall to 4.44822162 Newtons.

Edit - I see where the confusion lies. In physics, a lb is the lb-force since lbf seems to indicate lb-feet, a measure of torque or energy. Mass is measured in slugs. Or, as you indicated, the much preferred method is to deal with everything in terms of kilograms, Newtons, joules, Newton-meters, etc.

Better to denote it lb (subscript) f to avoid confusions. 'Course I don't know how to do subscripts here...

A slug? thank goodness I only had to learn SI. I hope you dont have to teach Physics in fps even cgs is a mess when you get into electromagnetism... 32.17 lbs mass to the Slug ...yuk...

Fasttrack

01-06-2009, 03:53 PM

A slug? thank goodness I only had to learn SI. I hope you dont have to teach Physics in fps even cgs is a mess when you get into electromagnetism... 32.17 lbs mass to the Slug ...yuk...

:D Nope, I think the Nuke-E program still uses imperial units, but we use si for the introductory courses. CGS gets used for upper level electricity and magnetism classes. I just figured for everyday examples, alot of people have a hard time visualizing a kilogram or a newton. (At least here in America!)

NickH

01-06-2009, 04:53 PM

Who said that? I wouldn't consider it stored as potential energy, however. There may be some elastic potential, but once you've crushed it most of the energy is lost to the deformation of the material. Just like throwing a ball of clay at the wall. If it was converted to elastic potential, we'd expect to be able to raise the sub out of the water and have it spring back into shape! A cool trick if it would work ;)

But that's what happens with all subs that don't sink, did you specify a sub which does sink?

All submersibles are crushed to a degree, just not beyond the elastic limit, are you not aware of this or just not taking it into account.

As for the BS about disappearing kinetic energy have you ever shot a metal target & felt the temperature?

When the energy change is smaller & the temperature change is too small to detect without measuring equipment.

Just beacuse you can't feel or see it does not mean it's not there,

Or maybe it does eh?

Nick

Life's a Learning & Growing Experience,

Or Not.

derekm

01-06-2009, 04:56 PM

:D Nope, I think the Nuke-E program still uses imperial units, but we use si for the introductory courses. CGS gets used for upper level electricity and magnetism classes. I just figured for everyday examples, alot of people have a hard time visualizing a kilogram or a newton. (At least here in America!)

Havent got a clue what a Nuke-E program is and I cant understand why anyone would teach CGS at upper level anything after 1970 apart from quantum electrodynamics? ..but then i was never any good at Quantum mechanics - lousy teachers and equations out of nowhere. I've even checked my old university text books from the late 1970's not a trace of CGS I only remember it being taught as a curiosity about the definitions of permittivity and permability

Derek - SI from the start to Master of Science (most of it long forgotten)

Fasttrack

01-06-2009, 05:57 PM

Havent got a clue what a Nuke-E program is and I cant understand why anyone would teach CGS at upper level anything after 1970 apart from quantum electrodynamics? ..but then i was never any good at Quantum mechanics - lousy teachers and equations out of nowhere. I've even checked my old university text books from the late 1970's not a trace of CGS I only remember it being taught as a curiosity about the definitions of permittivity and permability

Derek - SI from the start to Master of Science (most of it long forgotten)

:D My focus currently is on QED. CGI is used alot, as you pointed out, because you set the constants equall to one. We do the same thing in General Relativity, but set the speed of light to 1. There are alot of specific problems where its much easier to go to a goofy unit system just to simplify the math. We try to set as many things as we can to 1 ;) The Nuke-E is just short for Nuclear Engineering. They seem to use a variety of different units. Makes my head hurt!

Fasttrack

01-06-2009, 06:09 PM

But that's what happens with all subs that don't sink, did you specify a sub which does sink?

All submersibles are crushed to a degree, just not beyond the elastic limit, are you not aware of this or just not taking it into account.

As for the BS about disappearing kinetic energy have you ever shot a metal target & felt the temperature?

When the energy change is smaller & the temperature change is too small to detect without measuring equipment.

Just beacuse you can't feel or see it does not mean it's not there,

Or maybe it does eh?

Nick

Life's a Learning & Growing Experience,

Or Not.

I don't understand where the comment about subs came from. I linked it in my mind with the Marianas Trench and assumed that the sub was being permanently crippled. You are absolutely right that they are all crushed to some degree, I just mistakenly assumed a different situation.

I'm still not sure what your point is. Have you read the thread? No one is asserting that energy disappears...

OH! I just got it. Don't mind me. :) DP posted about a "bathysphere". And yes, some work is done on the sub, but once the sub reaches a constant depth, the work done by the water on the sub goes to zero (after the sub has been stressed). So as long as it remains at constant depth for some period of time, the work goes to zero. Which is how I read his comment. In the case of a rupture, the work performed on the sub goes to permanently destorting the material. Some will be in the form of kinetic energy as the water rushes in, but that too will quickly be converted.

derekm

01-07-2009, 07:14 AM

Fastrack,

maybe the nugget of info out of all this is when teaching engineers is to make the "thought system" as complete as possible e.g. add in the brake on the inelastic tether, to stop up all the rat holes :D

Derek

maybe the nugget of info out of all this is when teaching engineers is to make the "thought system" as complete as possible e.g. add in the brake on the inelastic tether, to stop up all the rat holes

That whirring sound is Einstein spinning in his grave. His famous constantly accelerating elevator would need a plausible means of acceleration...

Rich Carlstedt

01-07-2009, 09:57 AM

Fastrack

I guess its age or not understanding the "new" physics.

Maybe I don't understand "your" definiton of angular momentum.

I do believe in Newton,his laws,and foremost in the Conservation of Energy.

Above this, I look at all things in the simplest terms, as I am a practical engineer. I say this, because in the real world, I have seen countless projects fail because they became far too complicated, when a simple observation says the program was going down the wrong track.

The original poster asked about governor speeds , something I know about.

His question related to the governor stem speed.

I can say that for a given amount of energy, the stem "speed" (RPM) is a function of that energy. I believe he wanted to know if HE SHORTENED THE ARM, could he take advantage of the increase in speed. This has nothing to do with an increase in energy ! ( Angular momentum is constant here in my book, and therefore is NOT part of the discussion)

Yes, There are systems that use increases in energy to increase speed, which in a governor ALSO increases the radius AND the energy (stored).

But that is not the question. Some here are using that in the answer

Sorry, but the esoteric descriptions of the answers are contrary to common sense. I am not saying that any of them are wrong, only that they sidetrack the real question. Here is why I say this.

Statements of "force" do not apply.

Force is only valid when a "change" in speed occurs (either = or -)

The reason engineers and physics majors love flywheels, is because they are rigid and ANY energy changes are reflected in speed DIRECTLY

Governors are unique.

Their arms because they are flexable and which travel in an arc, have a sine function relative to speed and energy, and only behave like flywheels at their highest speeds.

Thus, many comments do not apply and confuse some readers..like me

Let me put this in really simple and practical terms.

When I flew U-control model aircraft, my plane would do 60 MPH.

On 50 foot wires,it took 3.56 seconds to go around. On 100 foot wires it took 7.2 seconds. same speed, same mass, same energy

Distance covered in one second was the same

Angular velocity has nothing to do with the energy in this system.

Sorry to use caps

rich

SmoggyTurnip

01-07-2009, 10:39 AM

The math shows that the kinetic energy of the system decreases when the rotational circumference is increased. The question is where does that kinetic energy go? The only "answer" so far provided is that it appears as losses in whatever mechanism is used to increase the diameter of rotation.

I always thought physics was a matter of facts. This thread makes me think it is now a matter of opinions. So here is my opinion.

The system actually has 4 states:

1) System is rotating at the larger diameter.

2) System is changing from larger to smaller diameter - energy has to be added to the system to acomplish this

3) System is changing from smaller diameter to larger diameter - no energy is required here and once this starts to happen the diameter will continue to increase until

4)Some mechanisim is used to stop the increasing diameter.

The transition from state 3 to 4 uses up the energy added in going from step 1 to step 2. If the mechinism is just a stop then the energy will be converted to heat and or kenitic energy of the mass that the axis is attached to - ie the earth may move or increase in temperature. If it is the form of a spring then it will be stored in the spring etc.

It is not the mechanism used to increase the diameter of rotation that absorbs the energy it is the mechanism that is used to stop the expansion of the diameter that absorbs the energy.

It is not the mechanism used to increase the diameter of rotation that absorbs the energy it is the mechanism that is used to stop the expansion of the diameter that absorbs the energy.

Correct. And, if we postulate a mechanism that can store that energy in potential form such that it may be used to reel the weight back in then the net loss of energy to recover the system state is determined only by the efficiency of the storage mechanism. That is an engineering problem and not an inherent part of the question at hand or the physical principles that govern it.

What I see as the real problem in determining an answer is that the solutions so far presented do not take into account the possibility of reversing the system state. If considered in that light then there is a "loss" of the energy required to halt the system diameter expansion. In the thought problem world the system is reversible with no loss of energy. In the real world entropy increase caused by unavoidable losses prevents that.

Fasttrack

01-07-2009, 01:41 PM

Rich - I know you know your stuff when it comes to model engines and govenors. You clearly have a great deal of skill there.

What I've presented is not new physics or a definition particular to myself. This is from the age of Newton. In the two cases you present, although the speed of the plane doesn't change, the rotational kinetic energy most certainly DOES change.

You believe in Newton. Therefore you should believe in his first law of motion:

(From Philosophić Naturalis Principia Mathematica) "It is possible to select a set of reference frames, called inertial reference frames, observed from which a particle moves without any change in velocity if no net force acts on it. This law is often simplified into the sentence, 'A body continues to maintain its state of rest or of uniform motion unless acted upon by an external unbalanced force.' This law is known as the law of inertia."

Applying this to our discussion, the plane wil continue to fly at a uniform velocity in a STRAIGHT LINE until acted on by some unbalanced force. So two things to notice right now.

1) Velocity is not the same as speed. Velocity is a vector quantity with a direction, speed is a scalar quantity and direction is unimportant. Velocity matters in energy calculations, NOT speed.

2) If there are no forces acting on the plane, it would fly in a straight line. Think about the the sparks flying off a grinding wheel. There is no force to bend their path around the wheel so they fly off in straight lines that are tangent to the wheel.

So there must be an unbalanced force that causes your plane to revolve. That force is called the centripedal force and it is supplied by the force of tension present in the string. The rotational kinetic energy is different than the translational kinetic energy because it takes this into account. The centripedal force is constantly changing direction as the string revolves (just as the velocity of the plane is constantly changing since the direction is constantly changing). This constant change of direction is denoted by angular velocity. This is how we account for the "work" done by this mysterious "center-pointing" force. To clarify things, we can look at rotational kinetic energy as an analogy to translational kinetic energy.

Translational kinetic energy is 1/2mv^2. Rotational kinetic energy is 1/2Iw^2.

I is the moment of inertia for a rotating system. It is analagous to mass. It is a measure of how much torque (analagous to force) that must be applied to accelerate an object in a circle. w is analagous to the velocity. You'll notice that the units work out nicely.

Now the moment of inertia for a single object tied to a tether is well known and is defined by mr^2 where r is the length of the tether. If we decrease the length of the tether, the "rotational mass", or moment of inertia, decreases. Furthermore, w, the angular velocity, increases by the conservation of angular momentum. These two effects do NOT cancel each other out. You'll notice that angular momentum is defined as Iw. If I decreases by a factor of 2, w increases by a factor of two. Now, lets look at the rotational kinetic energy. Krot = .5Iw^2. If I decreases by a factor of two, we know that w has increased by a factor of two. If we plug that into Krot, we see that we now have .5(.5I)4(w^2), since w is squared (i.e. the factor of two becomes a factor of four. Thus the new Krot is Iw^2, or twice what it was before.

The sine function that you talk about on a govenor accounts for the changing radius of the circle, BTW. Think about what you've implied... you say that the "speed" (meaning angular velocity) increases but that both the energy and angular momentum remain constant. That is mathmatically impossible since momentum (angular or not) is depend upon v while energy (either angular or translational) is dependent upon v^2, where I use "v" to stand for either velocity or angular velocity. If the energy is constant, you've broken the conservation of momentum which is MUCH more strictly observed by nature than the conservation of energy. Energy conservation is time dependent, momentum conservation is not. (Remember here we are talking about velocity, not speed. The speed does not change, as you point out, but the velocity does change)

As far as the original question goes, yes you might be able to take advantage of the changes in energy. It would be a nasty bit of engineering and unlikely to be very efficient, simply by analogy to Ockham's razor. The more complicated a system, the less efficient, generally speaking.

For the question of where the energy goes, you could view it as being transferred into translational kinetic energy as the weight travels outward and then suddenly being dissipated as it reaches the end of the teather. (Lost to the stretch of the string, etc) Alternatively, you could have it lift a weight and store that energy as potential. (Or some similiar set-up using another form of potential)

EDIT: Some basic physics from Newton himself -

Translational Kinetic Energy, K = 1/2mv^2

Rotaional Kinetic Energy, Krot = 1/2Iw^2

Angular Velocity, w = v/r = d(theta)/dt

Translational Velocity, v = dx/dt

Moment of Inertia, I = Integral(r^2 dm)

I think these are the important ones that apply to our discussion. You can find all the equations of motions for rotation online, I am sure. They look the same as translational equations, but the acceleration is replaced with alpha, angular acceleration, velocity with omega, angular velocity, and distance is replaced with theta, angular displacement. And, of course, mass is replaced with I, moment of inertia.

derekm

01-07-2009, 05:38 PM

That whirring sound is Einstein spinning in his grave. His famous constantly accelerating elevator would need a plausible means of acceleration...

Evan,

I would have thought in your astronomical studies, the mighty leaps of faith needed in Cosomology to produce mathematical models would have mellowed your rigour :)

Derek - who enjoyed cosmology immensely..

Rich Carlstedt

01-07-2009, 05:43 PM

Fastrack

I respect your skills and knowledge,and I am not implying that you are wrong.

In fact, you really know this stuff. I do believe that your understanding of "my " point and the original question is in error.

You said this

Think about what you've implied... you say that the "speed" (meaning angular velocity) increases but that both the energy and angular momentum remain constant."

--No, I said

"Angular velocity has nothing to do with the energy in this system."

and I said

" Angular momentum is constant here in my book, and therefore is NOT part of the discussion"

continuing ...You said

" That is mathmatically impossible since momentum (angular or not) is depend upon v while energy (either angular or translational) is dependent upon v^2, where I use "v" to stand for either velocity or angular velocity. If the energy is constant, you've broken the conservation of momentum which is MUCH more strictly observed by nature than the conservation of energy. Energy conservation is time dependent, momentum conservation is not. (Remember here we are talking about velocity, not speed. The speed does not change, as you point out, but the velocity does change)"

Thats the whole point I guess, we are not talking about "velocity'

You are , but I am not

You are confusing the addition and subtraction of energy in my opinion.

BUT That is not the question

Again lets revisit the original question

we have energy in a closed state. If you diminish the radius, the "speed of rotation" must increase to reflect the total energy of the system, as we see a skater doing in a spin.

This is conservation of energy,is it not ?

I surely don't understand your "energy is time dependent " comment and how it applies. ?

Kinetic energy without losses( original stipulation) , is not time dependant in my world

I realize that the terms ( mometum, velocity, speed,) can mean different things to some of us when viewed from various perspective, thats why keeping it simple is important ....I think

With my great respect

Rich

Fasttrack

01-07-2009, 07:32 PM

Again lets revisit the original question

we have energy in a closed state. If you diminish the radius, the "speed of rotation" must increase to reflect the total energy of the system, as we see a skater doing in a spin.

This is conservation of energy,is it not ?

Nope. :) That is the conservation of angular momentum. The energy in the system increases. With the skater doing the spin, he/she converts chemical potential (ATP or what-have-you in the muscles) into kinetic energy as he/she pulls his/her arms in. Sit in a office chair and hold your arms out. Spin yourself at a constant speed (fairly fast to really feel the effects) and then pull your arms in. Yes, you speed up, but you'll notice that it takes work to pull your arms. If you hold weights in your hands, you'll really notice the force required to pull the weights in. That force acts over the distance that your hands move and the product of this distance and force correlates to work done. Work is just a change in energy. You are increasing the rotational kinetic energy in order to conserve momentum.

Unfortunantly, proposing a closed system with the only energy being the kinetic energy involved in rotation and then stipulating that the moment of inertia spontaneously decreases (the radius decreases) is a physically impossible situation. It requires either the conservation of momentum or the conservation of energy to be broken!

What you can do, is increase the moment of inertia (increase r) and store the change in energy somewhere in the system. One model proposed was to lift a weight like a govenor lifts a plunger. Then we could decrease r at a later point in time by drawing on this potential energy to increase the rotational kinetic energy. Either way, if the system is closed, it is impossible to decrease r beyond its initial value.

Angular velocity is rpm, by the way. So although you say that angular velocity has nothing to do with it, when you say speed (as rpm) you are referring to angular velocity. Speed as fps or m/s or etc remains constant if angular momentum is conserved, but speed as in rpm is not constant.

You are right that the speed does not change, but speed has no bearing on energy. Only velocity does. You can substitute speed for velocity when the velocity is constant in magnitude and in one direction. That does not work for rotational motion.

I can say that for a given amount of energy, the stem "speed" (RPM) is a function of that energy

Well not really. Its the other way around. The energy is a function of angular velocity (stem speed). If we look at the expression for rotational kinetic energy, it is dependent upon BOTH angular velocity (the "stem speed") and the moment of inertia (which, in turn, depends upon r and m). The only time that energy can be used to determine the speed is if the moment of inertia is held constant, as in a flywheel. Then the "stem seed" is a function of the Krot.

It seems to me that you've got the "effects" down, but might have been told wrong about the "cause". Its not conservation of energy that causes the increase in rpm, its conservation of momentum. :)

The conservation of energy is only obeyed on sufficiently large time intervals. You are right, its not important in this discussion. Just one of those interesting (imo) tidbits. You can see this in Heissenberg's Uncertainty Principle: (delta)E(delta)t = h (bar)

I'm not trying to pick on you or anything! I appreciate your respect and hope you know it is returned! Internet "tact" can be tough ;)

Rich Carlstedt

01-07-2009, 09:51 PM

"Tact can be tough"

You sure are right my friend !

Its always worry-some (sp?) that someone will get bent

out of shape over a small thing..

It was a good disertation, have a good day fasttrack

Rich