Rotational Kinetic Energy and Conservation of Momentum

[Dadface]

Yes, sorry, it was. If a mass passes through 'x' displacing a spring until it reaces zero acceleration (thus, the mass reaches a point where it is finally acting on the spring with Mg of force) then it means it is at its centre of SHM. In other words, it is moving. And, guess what, indeed the energy in the spring at the time will be 0.5Mgx, just as the kinetic energy of the mass with be the other 0.5Mgx. That kinetic energy will keep the mass falling, extending the spring to '2x' (in your dimensional terms), thus storing a total of 2Mgx of energy in the spring once the mass decelerates to zero velocity at the 'bottom' of its SHM - just before it gets pulled back up by the energy stored [NOT lost] in the spring.

It's all elementary SHM.

Did you actually read my posts?tThe energy value I quoted was when the mass reached its final static extension this being after the vibrations die down due to damping.

 

[Dadface]

If you were to connect a capacitor, of C Farads, to a battery of V volts through a resistor of R ohms, then the time it takes to charge up to 99% of the voltage is ~4RC. This is very well known. Similarly, if you then connect a capacitor charged to V volts to ground via the resistor, the time it takes to get to 1% of V is ~4RC.

If what you were saying was right, it'd take 8RC to charge up, and 4RC to discharge. It doesn't. See http://www.electronics-tutorials.ws/rc/rc_1.html.

Sorry, your discussion is very misleading and I implore folks to ignore your posts here.

Go to
1.hyperphysics
2.capacitor
3.capacitance concepts
4.energy storage device

 

[cmb]

Did you actually read my posts?tThe energy value I quoted was when the mass reached its final static extension this being after the vibrations die down due to damping.

What's that got to do with how much energy a spring can store? When the spring gets to the bottom of its extention, then it may be storing, say, 99% of the energy that the falling mass has just dissipated. Say each bounce after that it loses 1% [just an example figure] of the stored energy, which, if it carried on losing 1% per bounce, would give it a ~70 bounce 'half-life' (that is, it'd drop to half the original energy after 70 bounces).

The relationship between the final rest position of an oscillating mass and the rate of energy loss during oscillations (which is a function of the efficiency of energy storage) is... none at all...

This is really very confusingly diverging the point of this post. You are wrong and off topic, I'll add no more because I'd be adding to the off-topic-ness ... Moderator, please...

 

[Dadface]

What's that got to do with how much energy a spring can store? When the spring gets to the bottom of its extention, then it may be storing, say, 99% of the energy that the falling mass has just dissipated. Say each bounce after that it loses 1% [just an example figure] of the stored energy, which, if it carried on losing 1% per bounce, would give it a ~70 bounce 'half-life' (that is, it'd drop to half the original energy after 70 bounces).

The relationship between the final rest position of an oscillating mass and the rate of energy loss during oscillations (which is a function of the efficiency of energy storage) is... none at all...

This is really very confusingly diverging the point of this post. You are wrong and off topic, I'll add no more because I'd be adding to the off-topic-ness ... Moderator, please...

Any vibrations are transient only ,they die down.The vibrational energy can't be stored,in other words kept for use as needed.Also,stretching methods can be used where there will be no vibrations at all but whatever method is used the energy stored is the same.
Did you do the searches I recommended?If not please read the following quote taken from the hyperphysics site:

from hyperphysics "Storing Energy in a Capacitor"

The title makes it obvious what the section is about and at the end of the section the following note is added:

Note that the total energy stored QV/2 is exactly half the energy QV which is supplied by the battery,independent of R(R referring to the resistance)

Check it for yourself.

Please note that in my first post I made a brief response to the energy storage systems referred to here and then moved on in an attempt to get involved in the main thrust of the discussion referred to by the OP.You were the one who caused this discussion to go on by suggesting that I was wrong.

 

[cmb]

Note that the total energy stored QV/2 is exactly half the energy QV which is supplied by the battery,independent of R(R referring to the resistance)

'QV' isn't the energy supplied by the battery. When the volts are low, the 'QV' value is lower (because the volts are low). Let's say that some capacitor has two Coulombs of charge on it. The first Coulomb took less energy to put it on the capacitor than the second - because the second was pushed up to a higher potential.

Imagine a kg mass on each step of your 10 step stair case, each step of height h. The first kilo only takes a little energy, but by the time you've filled up the staircase, the one on the last step takes 10 times as much energy to walk it up the stairs than the last. If you are claiming that QV is the energy being put onto a capacitor (and that 'V' is the highest voltage) then it is like you are calculating the potential energy of all those 1kg masses on your stairs as 10 x the height of the highest one. Clearly it isn't 100gh. It is 50gh, because it is a triangular number with some having been put on the stair case when they didn't have to be put so high up it. Same with charge into a capacitor.

Check it for yourself.

Yes, it is not at all as clear as hypherphysics usual standard. But if it has taken you to the point of thinking that the energy it takes for a battery feeding Q Coulombs into a capacitor that reaches V volts is QV, then your understanding has been mislead. A battery delivers [1/2].QV of energy to push Q of charge into a capacitor with a charged voltage V, because the battery does not need to push ALL of the electrons up to a potential of V, only the very last one!

Sorry, I said I'd not thread-drift and did so, but that misunderstanding really needed to be addressed here.

 

[Dadface]

So hyperphysics and others got it wrong did they?I wouldn't mind betting that next you are going to announce that the equation defining the volt in terms of energy and charge is incorrect.
If you really think there's a way of making the energy storage efficiency of either system(spring or capacitor)approach 100 percent then show how.At least that would be interesting and productive.

 

[Delta Kilo]

Delta-Kilo: Originally this solution was posted by K^2 (https://www.physicsforums.com/showthread.php?t=443644&highlight=flywheel+momentum+exchange" ) I have refined the result and used an inverted ratio since it's easier to use '0:.075' than 'inf.:13.33'.

I see. In the derivation you refer to, n is a constant and so it can be moved outside of the integral. In your case n = n(t) is a function of time and so the trick no longer works.

What it means you just cannot change the ratio of an ideal (perfectly rigid) gearbox at whim because the moment you try you get unlimited torques. In other words the ratios \frac{\omega_1}{\omega_2} = n and \frac{\tau_2}{\tau_1} = n cannot both be true while n is changing, something has to give. For example, you can have an elastic element somewhere in your gearbox in which case the ratio of angular velocities will be violated briefly as the elastic element first compresses and accumulates energy and then expands and releases energy. Or you can have slipping clutch that allows angular velocity ratio to be violated and converts energy to heat, or a combination of both. Or you might have perfectly rigid gearbox but moving the gear handle would actually require force and would produce some work which will be added/removed to/from the balance of kinetic energies of the flywheels.

At the end of the day you select what you want to be conserved: energy, torque ratio or angular velocity ratio, and write the equations accordingly.

 

[cmb]

So hyperphysics and others got it wrong did they?

Hyperphysics says

From the definition of voltage as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV. That is, all the work done on the charge in moving it from one plate to the other would appear as energy stored. But in fact, the expression above shows that just half of that work appears as energy stored in the capacitor.

Yes, that is very misleading, because all the work done on pushing a charge into a capacitor DOES (or, at least, in principle does, notwithstanding fractional percentage inefficiencies) appear as stored energy. I am not sure it is what the writer intended to say, but on the basis that it can be interpreted as you've written it then, yes, it is wrong.

Just think for yourself for a moment and think of the example I gave of masses on a series of steps. When you charge up a capacitor, it does not START at the voltage you are aiming to charge it up to, so the charge for, say, the first dV is q.dV. But once it is already charged to V', then the energy for the next dV above that is q.[V'+dV]. You can't just take the voltage of the finally charged capacitor and say 'ah, now I multiply ALL of the charge I put in by the final voltage'! How did the battery know to push the first few electrons up by the voltage of the finally charged capacitor!?

Think for yourself! Work it out. You have to multiply the charge by the AVERAGE voltage during the charging period, and as that is from 0 to V so the total energy is the AVERAGE voltage during capactor charging, times the total final charge, so it is Q.[V/2]

 

[Dadface]

Hyperphysics says

Yes, that is very misleading, because all the work done on pushing a charge into a capacitor DOES (or, at least, in principle does, notwithstanding fractional percentage inefficiencies) appear as stored energy. I am not sure it is what the writer intended to say, but on the basis that it can be interpreted as you've written it then, yes, it is wrong.

The quote below comes from the textbook:

Nelkon & Parker Advanced Level Physics Fourth Edition

"The energy comes from the battery.This supplies an amount of energy equal to QV during the charging process.Half of this energy goes to the capacitor.The other half is transferred to heat in the circuit resistance.If it is a high resistance the transfer is made quickly;if it is a low resistance the transfer is made slowly.In both cases,however,the total amount of heat produced is the same,0.5QV"


In your post above you seem to have forgotten that work is done not just to charge the capacitor but also to drive the charging current through the circuit resistance.The energy losses due to heating are significant and cannot be ignored.

 

[Ken G]

The problem is actually the false analogy between springs, and capacitors attached to a battery. A battery may be assumed to always maintain the voltage V-- note that if we had a different kind of battery that gradually ramped up to V, we could charge the capacitor to V gradually, losing no energy to dissipation, and getting the full QV/2 stored energy. But if the voltage is at V the whole time, then no matter how small the initial resistance, an energy QV must appear somewhere, as that is the work done by a battery that is always at V. So Dadface is partially correct, but the logic is skewed-- it's not that charging capacitors always wastes 50% of the energy, it is that capacitors always store QV/2 energy (that's the integral of V(Q)dQ ), so if you charge them in a way that does more work than that, heat will have to get made.

The same could be said of springs-- they store kx²/2 because that's the integral of kx dx, so if you do more work than that, you'll have to make heat. But usually you don't do more work than that-- you usually balance the spring force while you stretch it, analogous to a battery slowly reaching V. It all depends on the context, there is no law that says conservative systems are only 50% efficient in general.

 

[cmb]

The quote below comes from the textbook:

Nelkon & Parker Advanced Level Physics Fourth Edition

"The energy comes from the battery.This supplies an amount of energy equal to QV during the charging process.Half of this energy goes to the capacitor.The other half is transferred to heat in the circuit resistance.If it is a high resistance the transfer is made quickly;if it is a low resistance the transfer is made slowly.In both cases,however,the total amount of heat produced is the same,0.5QV".

Whatever the source, IT IS STILL WRONG in terms of how the capacitor stores charge. I've explained how the energy in a capacitor builds up, by the masses-on-steps analogy. Think for yourself and recognise some things in print are not right. This is one of them.

 

[Ken G]

Actually, that paragraph from Nelkin and Parker is not wrong, it simply makes a key assumption that we must recognize: it assumes the battery maintains the voltage V the whole time. That's what creates the 50% loss of heat, the very same thing would happen if we stretch a spring into equilibrium by maintaining a fixed force F on the end of the spring the whole time, using kinetic friction to keep the spring stretching at a slow speed.

 

[cmb]

But if the voltage is at V the whole time, then no matter how small the initial resistance, an energy QV must appear somewhere, as that is the work done by a battery that is always at V.

Quite so, but bear in mind that circuits very rarely contain a capacitor exposed to a source so stiff that the voltage stays 100% whilst it is low on the cap. There are always inductive and internal resistance effects such that the circuit never sees 'V' until the capacitor is charged. Whatever the resistance in the circuit, as soon as the capacitor starts charging, the voltage across the load, thus the current, diminishes yet the argument saying '0.5QV always goes into heating' is, I think, must be predicated on a constant current load because the losses across the resistor do not drop off linearly with an increase in the capacitors voltage, but by the square of the differential voltage across the resistance load.

{edit; Ken G beat me to the same point whilst I was typing!}

 

[Ken G]

Quite so, but bear in mind that circuits very rarely contain a capacitor exposed to a source so stiff that the voltage stays 100% whilst it is low on the cap. There are always inductive and internal resistance effects such that the circuit never sees 'V' until the capacitor is charged.

I think the standard situation is what Nelkin and Parker have in mind-- there is some slight resistance in the circuit, such that the current is low enough that the battery can support V the whole time. If one makes that assumption, their statement is correct (and even cute-- it doesn't matter how fast the capacitor is charged, 50% of the energy is lost-- given the above assumptions). Of course, the voltage across the capacitor is not V the whole time, that's what we have to integrate self-consistently-- it is the voltage across the battery that stays V, and that's what controls the work done by the battery, not the work done on the capacitor. The same could be said for a spring attached to a mass on sandpaper being stretched into equilibrium with a constant force F. So we agree that Dadface was incorrectly interpreting the significance of the Nelkin and Parker statement in terms of this thread, but the Nelkin and Parker statement is correct when properly interpreted.