Does a Spinning Cylinder Affect Rocket Acceleration and Energy Conservation?

[yuiop]

Imagine we have two rockets, each with a long cylinder with it spin axis alligned with the long axis of the rocket. On one rocket the cylinder is spun up to a high speed. Both rockets have an equal load of fuel. If both rockets are launched from the same spot simultaneously in the same direction will they accelerate at the same rate and reach the same terminal velocity when the fuel runs out?

An external observer that does not accelerate see the the spinning cylinder's rate of rotation slow down. If the rockets have the same terminal velocity, where has the stored angular energy of the spinning cylinder gone?

 

[tiny-tim]

spinning things have more mass

The spinning cylinder has more mass than the other one, and so the rocket it's in will accelerate slower.

I assume the increase in mass exactly balances the decrease in angular velocity, thus keeping the angular mometum the same, but I haven't actually worked it out.

If the rockets have the same terminal velocity, where has the stored angular energy of the spinning cylinder gone?

The other rocket never had its cylinder spinning, so there's literally no comparison to be made.

 

[yuiop]

The spinning cylinder has more mass than the other one, and so the rocket it's in will accelerate slower.

I assume the increase in mass exactly balances the decrease in angular velocity, thus keeping the angular mometum the same, but I haven't actually worked it out.

The other rocket never had its cylinder spinning, so there's literally no comparison to be made.

Say we add some mass to the non rotating rocket so that the effective initial inertial mass of the two rockets is the same. The initial acceleration of the two rockets should then be the same but later on as the rotation on the spinning rocket slows down it will "lose mass" and so the spinning rocket should end up with the higher terminal velocity when the fuel runs out?

 

[Jonathan Scott]

Imagine we have two rockets, each with a long cylinder with it spin axis alligned with the long axis of the rocket. On one rocket the cylinder is spun up to a high speed. Both rockets have an equal load of fuel. If both rockets are launched from the same spot simultaneously in the same direction will they accelerate at the same rate and reach the same terminal velocity when the fuel runs out?

An external observer that does not accelerate see the the spinning cylinder's rate of rotation slow down. If the rockets have the same terminal velocity, where has the stored angular energy of the spinning cylinder gone?

There are lots of assumptions and approximations involved here, so the question isn't particularly clear, but at least that suggests that it might not be homework!

Energy of rotation is simply a convenient way to sum the kinetic energy of all the parts of a rotating system.

If you're treating the spinning cylinder and the non-spinning one as initially having the same effective total mass-energy initially, note that the rest mass of the spinning cylinder would have to be less than that of the non-spinning one, as the kinetic energy of the spinning will increase its effective mass-energy.

If you're treating them as having the same rest mass, then the spinning cylinder will have a slightly greater overall mass-energy, so that rocket would accelerate more slowly.

When the overall spinning object is moving rapidly, although the frequency of the spin may appear to have decreased, the effective mass-energy of each part of the spinning object has increased. You calculate this in detail by considering each small piece of the spinning object, which originally only had a tangential velocity, but now has a vector combination of forward and tangential velocity. The resulting total energy is the same as if the original rotational energy was treated as part of the original rest mass and the kinetic energy for the velocity was then added as for a non-rotating object.

 

[yuiop]

There are lots of assumptions and approximations involved here, so the question isn't particularly clear, but at least that suggests that it might not be homework!

Energy of rotation is simply a convenient way to sum the kinetic energy of all the parts of a rotating system.

If you're treating the spinning cylinder and the non-spinning one as initially having the same effective total mass-energy initially, note that the rest mass of the spinning cylinder would have to be less than that of the non-spinning one, as the kinetic energy of the spinning will increase its effective mass-energy.

If you're treating them as having the same rest mass, then the spinning cylinder will have a slightly greater overall mass-energy, so that rocket would accelerate more slowly.

When the overall spinning object is moving rapidly, although the frequency of the spin may appear to have decreased, the effective mass-energy of each part of the spinning object has increased. You calculate this in detail by considering each small piece of the spinning object, which originally only had a tangential velocity, but now has a vector combination of forward and tangential velocity. The resulting total energy is the same as if the original rotational energy was treated as part of the original rest mass and the kinetic energy for the velocity was then added as for a non-rotating object.

So, if we add mass to the non rotating rocket so as to compensate for the initial extra energy of the rotating rocket (as suggested in post #3) then it probably correct that the rotating rocket reaches a higher final velocity?

The total final energy of the non rotating rocket is simply its linear kinetic energy.

The total final energy of the rotating rocket is the sum of its final linear kinetic energy and final angular energy.

The total final energies of the two rockets should be the same and total $$\Delta{energy}$$ is also the same for both rockets, even though the final linear velocities differ.

Right?

 

[pervect]

So, if we add mass to the non rotating rocket so as to compensate for the initial extra energy of the rotating rocket (as suggested in post #3) then it probably correct that the rotating rocket reaches a higher final velocity?

I don't believe this is correct.

One needs to know three things to calculate the final velocity of the rocket.

1) The fueled mass of the rocket. The mass here is the invariant mass of the rocket, sqrt(E^2-p^2). Note that this is the same as the relativisitc mass of the rocket in the rest frame of the rocket, so there isn't really any confusion about what one means by "mass". However, I tend to avoid relativistic mass by personal preference, it just confuses the issue.

Note that rotation will not contribute to the total momentum of the rocket, only to the energy. But because it contributes energy without contributing momentum, rotation will increase the fueled mass of the rocket.

2) The empty mass of the rocket, given by the same formula. There is a question here - when the rocket burns rotating fuel, do the exhaust gasses also rotate? This probably depends in detail on the design of the rocket. If the exhaust does not rotate, the rocket will speed up its rotation rate as it burns fuel to conserve angular momentum. The energy lost in swirling the exhuast if the exhaust does swirl will detract a bit from the exhaust velocity or specific impulse, however. This leads to point #3.

3) The average exhaust velocity or specific impulse of the fuel. Unfortunately, the fuel won't have a uniform density, because the fuel in the more rapidly rotating outer section of the rocket will have a higher mass, but no more energy released when it's burned. WE also have the issue of whether or not the exhaust gasses swirl, and how this affects the average exhaust velocity. Note that we are asssuming incompressible fuel, the problem is already hard enough without dealing with compressible fuel. (Hopefully we can get away with this simplification).

While I don't have a specific answer, but I don't think any of the approaches in this thread to date have the right foundations.

With all of the above information, one can use the relativistic rocket equation to find the final velocity of the rocket.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

[yuiop]

I don't believe this is correct.

One needs to know three things to calculate the final velocity of the rocket.

1) The fueled mass of the rocket. The mass here is the invariant mass of the rocket, sqrt(E^2-p^2). Note that this is the same as the relativisitc mass of the rocket in the rest frame of the rocket, so there isn't really any confusion about what one means by "mass". However, I tend to avoid relativistic mass by personal preference, it just confuses the issue.

Note that rotation will not contribute to the total momentum of the rocket, only to the energy. But because it contributes energy without contributing momentum, rotation will increase the fueled mass of the rocket.

2) The empty mass of the rocket, given by the same formula. There is a question here - when the rocket burns rotating fuel, do the exhaust gasses also rotate? This probably depends in detail on the design of the rocket. If the exhaust does not rotate, the rocket will speed up its rotation rate as it burns fuel to conserve angular momentum. The energy lost in swirling the exhuast if the exhaust does swirl will detract a bit from the exhaust velocity or specific impulse, however. This leads to point #3.

3) The average exhaust velocity or specific impulse of the fuel. Unfortunately, the fuel won't have a uniform density, because the fuel in the more rapidly rotating outer section of the rocket will have a higher mass, but no more energy released when it's burned. WE also have the issue of whether or not the exhaust gasses swirl, and how this affects the average exhaust velocity. Note that we are asssuming incompressible fuel, the problem is already hard enough without dealing with compressible fuel. (Hopefully we can get away with this simplification).

While I don't have a specific answer, but I don't think any of the approaches in this thread to date have the right foundations.

With all of the above information, one can use the relativistic rocket equation to find the final velocity of the rocket.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

Rotating fuel and swirling exhaust gases complicates things to much. Please can we consider the rotating cylinder as a payload in a non rotating rocket with non rotating fuel. If it makes things easier we might consider the rotating payload as being towed by another rocket or being pushed by a non rotating rocket with suitable bearings to allow the payload to rotate independently.

I am aiming to get the simplest situation for an easier analysis of the difference between accelerating a rotating load versus accelerating a non rotating load.

However we look at it, (it seems to me) that the final velocities of the two rockets and/or the acceleration pattern will be different in the two cases.

The rocket propulsion unit should be as simple as possible. For example the two rockets could have identical solid fuel compartments (with identical fuel loads) and fixed diameter exhaust nozzles that are simply ignited and left to their own devices without further intervention.

 

[nanobug]

Please can we consider the rotating cylinder as a payload in a non rotating rocket with non rotating fuel.

In this case, if the invariant mass (as defined by pervect) of the two rockets is the same, I don't see how the end result (final velocities) can be different.

 

[tiny-tim]

angular momentum stays the same

… as the rotation on the spinning rocket slows down it will "lose mass" …

No.

Although the rotation is slower, the cylinder overall is moving faster, and its mass is increasing, not decreasing. You were considering the "mass attributed to rotation" separately, which does not work.

The increase in mass balances out the decrease in sideways speed, leaving angular momentum the same.

To see this, imagine a stationary rocket with a cylinder rotating so that its rim moves with speed U, viewed by two observers with speeds V and W.

The observers will regard the mass of the cylinder as increased by:

$$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}$$ and $$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - W^2}}$$​

while they will regard the sideways speed of the rim as slowed down by:

$$\Large\sqrt{1 - V^2}$$ and $$\Large\sqrt{1 - W^2}$$​

(because time on the rocket is slowed down by the same factor, while sideways distances are not affected).

Sideways momentum = mass x sideways speed (or energy x sideways speed), so the sideways momentum is the product, and is the same, $$\Large\frac{1}{\sqrt{1 - U^2}}$$.

In other words: the sideways momentum is the same for all observers.

Since angular momentum = sideways momentum x radius, and the radius is also the same, that means the angular momentum is the same for all observers.

No angular momentum has been lost.

 

[yuiop]

No.

Although the rotation is slower, the cylinder overall is moving faster, and its mass is increasing, not decreasing. You were considering the "mass attributed to rotation" separately, which does not work.

The increase in mass balances out the decrease in sideways speed, leaving angular momentum the same.

To see this, imagine a stationary rocket with a cylinder rotating so that its rim moves with speed U, viewed by two observers with speeds V and W.

The observers will regard the mass of the cylinder as increased by:

$$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}$$ and $$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - W^2}}$$​

while they will regard the sideways speed of the rim as slowed down by:

$$\Large\sqrt{1 - V^2}$$ and $$\Large\sqrt{1 - W^2}$$​

(because time on the rocket is slowed down by the same factor, while sideways distances are not affected).

Sideways momentum = mass x sideways speed (or energy x sideways speed), so the sideways momentum is the product, and is the same, $$\Large\frac{1}{\sqrt{1 - U^2}}$$.

In other words: the sideways momentum is the same for all observers.

Since angular momentum = sideways momentum x radius, and the radius is also the same, that means the angular momentum is the same for all observers.

No angular momentum has been lost.

Thanks for your answer Tim. I will accept for now that angular momentum is conserved. I will now try and figure out what happens with the angular kinetic energy.

 

[yuiop]

No.

Although the rotation is slower, the cylinder overall is moving faster, and its mass is increasing, not decreasing. You were considering the "mass attributed to rotation" separately, which does not work.

The increase in mass balances out the decrease in sideways speed, leaving angular momentum the same.

To see this, imagine a stationary rocket with a cylinder rotating so that its rim moves with speed U, viewed by two observers with speeds V and W.

The observers will regard the mass of the cylinder as increased by:

$$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}$$ and $$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - W^2}}$$​

while they will regard the sideways speed of the rim as slowed down by:

$$\Large\sqrt{1 - V^2}$$ and $$\Large\sqrt{1 - W^2}$$​

(because time on the rocket is slowed down by the same factor, while sideways distances are not affected).

Sideways momentum = mass x sideways speed (or energy x sideways speed), so the sideways momentum is the product, and is the same, $$\Large\frac{1}{\sqrt{1 - U^2}}$$.

In other words: the sideways momentum is the same for all observers.

Since angular momentum = sideways momentum x radius, and the radius is also the same, that means the angular momentum is the same for all observers.

No angular momentum has been lost.

Hi again,

I checked out your equation for the effective mass of a point particle on the rim of the cylinder and it is indeed correct.

So we can say (keeping close to your formalism)

Initial angular momentum = $${m_ouR \over \srqt{1-u^2}}$$

Final angular momentum = $${m_o(u\sqrt{1-v^2})R \over \sqrt{1-u^2} \sqrt{1-v^2}} = {m_ouR \over \srqt{1-u^2}}$$

So angular momentum is conserved.

Now using a similar method,

Intial angular KE = $${\left({m_o \over \sqrt{1-u^2}} - m_o \right)R$$

Final angular KE = $${\left({m_o \over \sqrt{1-u^2} \sqrt{1-v^2}} - m_o \right)R$$

As you can see unlike angular momentum, angular kinetic Energy is not conserved.

Do you see the problem?

 

[1effect]

Hi again,

I checked out your equation for the effective mass of a point particle on the rim of the cylinder and it is indeed correct.

So we can say (keeping close to your formalism)

Initial angular momentum = $${m_ouR \over \srqt{1-u^2}}$$

Final angular momentum = $${m_o(u\sqrt{1-v^2})R \over \sqrt{1-u^2} \sqrt{1-v^2}} = {m_ouR \over \srqt{1-u^2}}$$

So angular momentum is conserved.

Now using a similar method,

Intial angular KE = $${\left({m_o \over \sqrt{1-u^2}} - m_o \right)R$$

Final angular KE = $${\left({m_o \over \sqrt{1-u^2} \sqrt{1-v^2}} - m_o \right)R$$

As you can see unlike angular momentum, angular kinetic Energy is not conserved.

Do you see the problem?

Kinetic energy (KE) is frame variant wrt translation, why would you be surprised that it is frame variant wrt rotation?

 

[yuiop]

Kinetic energy (KE) is frame variant wrt translation, why would you be surprised that it is frame variant wrt rotation?

I'm not surprised but some people seem to disagree. I am trying to see how this translates into the final velocity of the two rockets. If they both have the same terminal linear velocity with the same amount of energy expended (fuel) then we have to account for the missing angular energy.

So I am saying that if the payloads are enclosed so we cannot see which cylinder is rotating we would probably see a difference in the acceleration rates and the terminal velocities. An observer inside the rocket would not see any change in the rotation rate of the cylinder but we as external non accelerating observers (might) see a tangible difference in the trajectories of the two rockets.

 

[1effect]

Imagine we have two rockets, each with a long cylinder with it spin axis alligned with the long axis of the rocket. On one rocket the cylinder is spun up to a high speed. Both rockets have an equal load of fuel. If both rockets are launched from the same spot simultaneously in the same direction will they accelerate at the same rate and reach the same terminal velocity when the fuel runs out?

An external observer that does not accelerate see the the spinning cylinder's rate of rotation slow down. If the rockets have the same terminal velocity, where has the stored angular energy of the spinning cylinder gone?

I see. Don't you think that you would need to expend extra fuel in order to spin one rocket? So, the spinning one can't reach the same terminal velocity as the non-spinning one. So, I think that you are simply dealing with an incorrectly stated problem. Is this a homework?

 

[yuiop]

I see. Don't you think that you would need to expend extra fuel in order to spin one rocket? So, the spinning one can't reach the same terminal velocity as the non-spinning one. So, I think that you are simply dealing with an incorrectly stated problem. Is this a homework?

The payload is pre spun by an external fuel source while the rocket is rest. This is not homework so anybody is free to state the problem in the simplest possible manner to get to the crux of the matter.

Will angular kinetic energy be lost? If so, how will it tangibly manifest itself? We can't simply say it is imaginary lost energy and ignore it.

 

[1effect]

The payload is pre spun by an external fuel source while the rocket is rest. This is not homework so anybody is free to state the problem in the simplest possible manner to get to the crux of the matter.

Will angular kinetic energy be lost? If so, how will it tangibly manifest itself? We can't simply say it is imaginary lost energy and ignore it.

Then the problem reduces to my prior statement: kinetic energy in its linear form is not frame invariant, most likely angular kinetic energy isn't frame invariant either. You are using "conserved" instead of "frame invariant", this is what produces the confusion. There is no "loss of energy".

 

[Jonathan Scott]

I'll stick to my previous conclusion.

1. The overall effect of rotating the cylinder is equivalent to increasing the rest mass of the system of rocket plus cylinder by the kinetic energy of rotation in the rest frame, so the same propulsive force would result in a slightly decreased acceleration of the overall system.

2. The total energy of the final overall system of rocket and cylinder can be determined by treating the kinetic energy of rotation as contributing to the rest mass m (regardless of the axis of rotation) then calculating the final total energy accordingly:
mc^2/sqrt(1-v^2/c^2).

Note that this will be slightly greater than the energy due to the original actual rest mass (ignoring the rotation) plus the original kinetic energy of rotation, because the kinetic energy contribution is also effectively increased by the same factor when the system is moving fast.

You can of course calculate this from first principles by Lorentz transforming the original tangential velocity of parts of the cylinder to the final frame and then calculating their original and final kinetic energy, but that seems too much like hard work for me.

This is a general rule; if you have a closed "black box" containing mass and energy in any form (with zero overall momentum), and you either set it in motion with velocity v or you move relative to it with velocity v, all of the original mass and energy behaves like the rest mass for purposes of determining the effective total energy as seen in the frame where the box is moving.

In Special Relativity, total energy is always exactly conserved (as is linear momentum and angular momentum), and you can rely on that as a short cut for calculation purposes.

 

[yuiop]

Hi Jonathan,

Your black box explanation is very helpful and appears to work.

If we take a particle on the rim of a rotating cylinder with tangential velocity u and rest mass m_0 then:

(a) Initial total energy = $${m_o \over \sqrt{1-u^2}}$$

If we observe the cylinder with relative linear motion v then

(b) Final total energy = $${m_o \over \sqrt{1-u^2}\sqrt{1-v^2}}$$

A particle on the rim of a cylinder with rest mass m_1 has

(c) Initial total energy = $${m_1}$$

(d) Final total energy = $${m_1 \over \sqrt{1-v^2}}$$

If we have a rest mass for m_1 that is equivalent to the initial total energy of the particle on the rotating cylinder then the initial and total final energies of the particle on the non rotating cylinder are


(e) Initial total energy = $${m_o \over \sqrt{1-u^2}}$$

(f) Final total energy = $${m_o \over \sqrt{1-u^2}\sqrt{1-v^2}}$$

which is the same as (a) and (b).

So it seems reasonable to conclude that the terminal velocities and final total energies of both the rotating rocket and non rotating rocket will be the same.

I am still unable to find any reasonable equations for the kinetic energies of the cylinder rim particle split into x and y components. Any ideas?

[EDIT]

If we have two black boxes with a cylinder inside each and of equal rest mass when neither cylinder is rotating, then if we spin the cylinder inside one then that box behaves as if it has higher inertial mass, when we accelerate the boxes. From what I have read the current opinion seems to be that if we place the black boxes on weighing scales they will weigh the same, even if one is rotating and the other is not. Is this correct?

 

[tiny-tim]

Angular kinetic energy" is not conserved, even in Newtonian mechanics

As you can see unlike angular momentum, angular kinetic Energy is not conserved.

Do you see the problem?

Will angular kinetic energy be lost? If so, how will it tangibly manifest itself? We can't simply say it is imaginary lost energy and ignore it.

(a) Rotation of the cylinder
"Angular kinetic energy" is not conserved, not even in ordinary Newtonian mechanics.

Energy is conserved, so is momentum, so is angular momentum. That's all, folks!

So "angular kinetic energy" will be lost - there's nothing wrong with that!

I said earlier that you were considering the "mass attributed to rotation" separately, which does not work. Well, mass is energy by another name, and we can't consider "energy attributed to rotation" separately.

(b)Speed of the rocket

The increase in mass balances out the decrease in sideways speed, leaving angular momentum the same.

The rocket's speed is dictated by the total mass, which does increase.
So the same force has less effect on the rocket, and it accelerates less (Newton's third law gives the exact figure).

The cylinder's angular momentum does not increase.

This is a general rule; if you have a closed "black box" containing mass and energy in any form (with zero overall momentum), and you either set it in motion with velocity v or you move relative to it with velocity v, all of the original mass and energy behaves like the rest mass for purposes of determining the effective total energy as seen in the frame where the box is moving.

No.

The mass does increase.

On weighing scales, the weight will increase.

 

[allebone]

Too complex

Hi there,

I may be wrong on this but the original question is quite simple. I believe that the solution should be simple too.

The original question is as follows:
An external observer that does not accelerate see the the spinning cylinder's rate of rotation slow down. If the rockets have the same terminal velocity, where has the stored angular energy of the spinning cylinder gone?

Remember the rule - that in space, as the universe expands, matter and radiation lose energy to gravity while an inflation field gains energy from gravity.

This means that the total energy carried by ordinary particles of matter and radiation drops because it is continually transferred to gravity as the universe expands.

To understand how think of the box analogy. The box is filled with many children incessantly running and jumping. The box is impermeable so no heat or energy can escape. but it is flexable so its walls can move outward.
As the children run about and bump the walls it expands. You might think at this point that the energy embodied by the children will stay fully in the box but this is not quite right. Some of the energy is transferred into the walls motion.
The same think happens in space but it doesn't have any walls, it has gravity instead.

A practical example is microwave background radiation. As the universe expands the energy loss of photons can be directly observed because their wavelengths stretch - they undergo redshift. This is because of the same reasoning as above.

Pete

 

[tiny-tim]

calculation of angular momentum has nothing to do with acceleration

Remember the rule - that in space, as the universe expands, matter and radiation lose energy to gravity while an inflation field gains energy from gravity.

This means that the total energy carried by ordinary particles of matter and radiation drops because it is continually transferred to gravity as the universe expands.

I've no idea whether that is a correct analysis of cosmic expansion (which would deserve a new thread), but it can't help with this angular momentum (or "angular energy") problem, because that depends only on velocity, and therefore on the Lorentz equations of special relativity.

 

[allebone]

If the question is where has the stored angular energy of the spinning cylinder gone, and we know that energy is transferred over time to gravity, then surely this accounts for some of the energy "loss" (for want of a better word) regardless of what direction this energy or momentum is traveling in?

pete

 

[tiny-tim]

energy is gained, not lost

Hi pete!

Although increasing speed away from the observer makes the cylinder spin slower, it still makes its energy (or mass - same thing) increase!

To see this, imagine a stationary rocket with a cylinder rotating so that its rim moves with speed U, viewed by two observers with speeds V and W.

The observers will regard the mass of the cylinder as increased by:

$$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}$$ and $$\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - W^2}}$$​

So there is an increase in mass (or energy), exactly as one expects for a moving body, not a loss - there is no lost energy to explain.

 

[allebone]

Hi Tim

Are you saying that the total energy of the cylinder has stayed the same or increased (A decrease in spin but an increase in mass).

pete

 

[tiny-tim]

Hi pete!

Total energy (or mass) always increases when speed increases: $$\LARGE m = m_0/\sqrt{1\,-\,v^2/c^2}\,.$$

(The angular momentum is still the same, though, because the increase in mass is exactly balanced by the decrease in sideways speed.)

 

[allebone]

Hi

I see what you are saying now. That makes sense to me, but I am puzzled by one consequence of your argument. If objects increase in mass as they move away from each other, and the universe is expanding constantly, then this would indicate that there is a net increase in total mass in the universe as time progresses. How can this be possible?

Pete

 

[tiny-tim]

Ah, that's general relativity - someone else will have to answer that.

All I know is that, in a fixed universe, the Lorentz equations say that mass increases.

 

[allebone]

Cool thanks for the info - most useful.

Pete