A Little Energy Puzzle

I'll go out on a limb here, get the ball rolling.

If I understand the parallel axis theorem correctly (which I prolly don't), I would expect the distance in your puzzle to be about the same.

Gyroscopic precession?

I think the answer is about the same distance (or exactly the same if perfectly reproducible conditions exist), the giroscopic efect occurs only when you try to tilt the axis of rotation, not move the whole thing linearly.

First, I would like a clarification that I understand this problem correctly.

This cart never turns during any horizontal motion (flywheel spinning or not). No change in elevation happens. When the flywheel spins, it spins in a plane parallel to the wheel spin of the cart. While friction will be needed for the cart's wheels to turn, for this exercise this loss of energy can be ignored along with the energy conversion losses that spins up the flywheel when the cart decelerates. Similarly fluid dynamic losses from moving the atmosphere around this cart can be ignored. Only Newtonian kinematics will be relevant for this discussion.

I think that I've exhausted all of the implied assumptions. Am I correct?

Yup rf, you have it on the button. The losses are not important and there is no deviation from the straight and level; hence no gyroscope effects.

-J

"Would the direction of travel, in relation to the earth's spin axis, for the second try have any effect?"

Subtle question! Depending on latitude, the car will always be rotating around some or other axis. If we ask "any effect", then there could surely be gyroscope precession effects. However, for the original puzzle it's irrelevant, me thinks.

-J

The flywheel is somehow coupled to the wheels to slow the cart down. Can I assume that there is no connection when the cart is accelerated the second time. In other words, does the flywheel help accelerate the cart?

Yes Rixter, no mention was made of using the flywheel for acceleration. I wanted to see how many people overestimate the spin energy contribution of the flywheel to its linear inertia. If one is stuck with Newtonian physics, the temptation would be to compare the rotational energy of the flywheel to the linear kinetic energy of the car+flywheel. The rotational energy can easily overshadow the linear kinetics and one's guts may say the effect must me considerable; and measurable.

But nature can't be fooled and E=mc² will overshadow everything.

-J

OK, so following Rixter's plausible but unstated assumption that the flywheel rotation is disconnected from the cart's forward progress once forward progress stops and that the flywheel remains disengaged when the cart is again accelerated with the same (impulse?) force I suspect that the second run will be very close to accelerating to the same velocity as previously stated. So close that the dismissed losses, when systematically tested, will likely be less than the observed losses.

I council my reply to the second iteration and that this second iteration will only closely approach all energy going into mass acceleration because of my mixed respect and personal misunderstanding of the seminal work of Nicolas Leonard Carnot.

My answer is distance is the same (agreeing with some other posters).

It probably doesn't make any difference, but I don't think rf's clarification agrees with the stated conditions - When the flywheel spins, it spins in a plane parallel to the wheel spin of the cart to me implies flywheel disc in a vertical plane parallel to the wheels, so flywheel shaft horizontal and perpendicular to the axis of the cart. But your puzzle says flywheel mounted with its spin axis aligned to the front-rear axis of the cart.

That's the way I interpreted as well.

But Jorrie tells us in #21 the axis orientation doesn't matter, so my #1 is prolly incorrect.

E is greater for the second run so it should need more push, but would that be evident in a Newtonian world?

Seems like it would take the same energy....The vehicle has gained energy in the form of mechanical, but the energy does not add to the mass of the vehicle......

OK slightly more mass....

The distance, D, to reach velocity V with accelerating force F: D=(.5 M V^2)/F, which is proportional to the mass being accelerated.

The cart with spinning flywheel is very slightly more massive due to the energy stored in the flywheel. (delta m=E/c^2). So the second time, it will take slightly more distance.

Correct, Rixter, but I guess the 'catch' lies in the phrases: "...so that Newton mechanics holds to great accuracy" and "... at a measurably larger, smaller or the same distance as before". The issue is how much energy can one store in a flywheel when compared to its mass-energy, before even the best material will simply disintegrate? Will that energy increase be practically detectable?

-J

Volvo has been working on a system, and it seems impressive...if it performs as advertised......but is it cost effective over the life of the vehicle? ...Real world testing is needed....

"In 2013, Volvo announced a flywheel system fitted to the rear axle of its S60 sedan. Braking action spins the flywheel at up to 60,000 rpm and stops the front-mounted engine. Flywheel energy is applied via a special transmission to partially or completely power the vehicle. The 20 centimetres (7.9 in), 6 kilograms (13 lb) carbon fiber flywheel spins in a vacuum to eliminate friction. When partnered with a four-cylinder engine, it offers up to a 25 percent reduction in fuel consumption versus a comparably performing turbo six-cylinder, providing an 80 hp boost and allowing it to reach 100 kilometres per hour (62 mph) in 5.5 seconds. The company did not announce specific plans to include the technology in its product line.^[21]"

http://en.wikipedia.org/wiki/Flywheel_energy_storage

http://www.gizmag.com/volvo-flywheel-kers-testing/27273/

That 'little' Volvo flywheel packs quite a punch, with the rim apparently reaching 'Mach 2' inside the vacuum pack. Even so, its angular energy is only some 10^-11 of the flywheel's (6 kg) rest energy, so not quite detectable as part of the linear inertia...

-J

"....not quite detectable as part of the linear inertia..."

.

...but it does make me wonder about nonlinear scenarios.

Will hard turns one way have a tendency to cause the nose to dive while turns the other way cause the nose to rise?? When braking hard enough that the front of the car noticeably dips, will the car have a tendency to pull to either the right or left?

.

Those seem like reasonable assumptions given the system is noted as having 'a flywheel' instead of two that might offset induced torques and also assuming the axis of rotation it aligned front/back in the car.

Yup, unless the flywheel is gimballed, any turning or attitude change will cause precession effects on the car. The specifics will obviously depends on the sizes/masses of flywheel and car.

-J

Measurably larger.

All points on the flywheel start on a forward spiral when the cart stars moving, the tangential velocity combines with the forward velocity making a change of direction.

I"m not sure that I put that very well but... you are all clever people.

Does your assumption hold true regardless of the direction of rotation of the flywheel, which has not been established?(AFAIK)

So if I have two identical gyroscopes, one with a spinning (horizontal) flywheel and one stopped, and I drop them simultaneously off the tower of Pisa, will one hit the ground first? (I don't think so.)

That's why I asked him my question about direction of rotation. Hoping he'd rethink his answer.

Rixter, yes, your two gyroscopes will fall at the same acceleration in vacuum, but the spinning one would win if air drag on the casing is taken into account. The reason is basically the same as why an iron ball falls faster than a feather in air.

-J

Jorrie, did we rule out aerodynamics as a force to be reckoned with? If not then if, as you say, the spinning flywheel has less aerodynamic drag, the second run will be shorter.

By the way, how does the spinning wheel have less drag than a stationary one?

Is it like a golf ball where the air gets mixed up (from the spinning) and doesn't make a low pressure hole in the back?

Well, in the OP, any air resistance was irrelevant, because all we wanted to know was if the inertia of the car with spinning flywheel is more than with the non-spinning one. We agreed that it is, but immeasurable small.

With your Pisa scenario, if the flywheels are both sealed in identical vacuum containers, then if we also ignore outside air drag on the identical containers, they will fall the same, irrespective of spin energy. My take was that the one with the higher energy density (the spinning one) will fall marginally faster in air; i.e. it will have a higher terminal free-fall speed in air.

I know that a large lead ball falls faster in air than a small one, but I'm suddenly not so sure that same size and surface balls with different density will fall at different terminal speeds. Hmm...??

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Regards

Jorrie

I know that a large lead ball falls faster in air than a small one, but I'm suddenly not so sure that same size and surface balls with different density will fall at different terminal speeds. Hmm...??

Yes they will. Imagine that the lighter ball is falling at its terminal velocity, and at an instance in time the more dense ball is falling at the same velocity, and, at the same height etc. then the upward forces on both balls due to air resistance are equal: but, the downward force on the lighter ball is equal to that upward force; where as the downward force on the heavier ball is greater, so, it continues to accelerate.

Yup, you are right, Randall. So if we take three balls, a large solid lead one, a little solid lead one and a large hollow beach ball and drop them simultaneously in still air, the order of arrival at the bottom of the tower would be the large, small, beach ball?

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Regards

Jorrie

Yeah, the heavier ball has a higher ballistic coefficient.

All that said, do you agree that due to its increased total energy, the flywheel spinning in an enclosure will also fall faster in air than an identical non-spinning one?

-J

Yes, very, very slightly faster.

That seems to be drifting off the original puzzle somewhat! But I would say no. It will only fall faster if mass is increased by relativistic effect, and velocities are postulated as too small for that to be significant.

Yup, it is marginally heavier when nearing the relativistic 'low speed limit', but no practical flywheel can ever spin fast enough for the effect to become measurable, I think.

I'm not sure we can consider this as "still air", but if you release the balls in a spaceship in orbit, they will appear to be not falling (relative to the spaceship).

Sure, but that post was about a "tower of Pisa" type experiment. So unless an equatorial tower reached up to geostationary height, they will fall, even from space.

The conversion rate is such that it is not noticeable in the Newtonian realm.

No significant difference. The flywheel is a black box in this example. One of the commenters above seems to have misunderstood the orientation of the gyro spin axis.

Actually, as long as the acceleration is in a straight line, the orientation of the flywheel spin axis should be immaterial.

Interesting, the bulk of the energy of the flywheel does not come from its kinetic energy, but from the kinetic- and binding energy inside the nuclei that it is made of. Those quark-triplets apparently move near the speed of light in orbit around each other. That's near 98% of the mass of the flywheel.

See Don Lincoln's short lecture: https://www.youtube.com/watch?v=x8grN3zP8cg

That's really weird.....but very interesting....and actually makes total sense....thanks Jorrie

"but non-relativistic horizontal speed, say 100 km/h" Is this to mean disregard for the "flywheel" effect the cart's wheels would have?

The flywheel effect of the wheels would be the same for both runs, not so? Hence not relevant, I think.

-J

So, to prevent a phase shift between the cart wheels the track would have to be absolutely straight.

I can't come up with a reason it wouldn't be the same distance, but I don't like that answer because I'm still waiting for the punch line. Very small relativity effects, friction, and aerodynamic differences between a cart with a spinning flywheel and stationary flywheel have all been ruled out.

I give up, what's the trick?

No, it was not a trick question; just served as a reminder of how small relativistic energy of motion is in our ordinary lives. The division by c² kills it in most cases.

That "wound watch that weighs more than an unwound one" is true but sooo negligible...

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Regards

Jorrie

Assuming I did the calculations right, if you burn a 100W lamp in a closed room continuously for 1 year, the mass of the room would increase by about 35 μg.