What limits the speed at which a metal ball can be made to rotate suspended in vacuum

[Justonium]

Under those conditions, the larger object would have a lower energy density, since the stuff inside would be moving slowly. However, as I pointed out, it will have a lower centrifugal force as well, since the centrifugal force decreases as size increases (for a given velocity). What I was talking about was higher energy per unit of centrifugal force.

 

I still need mathematical proof of this. I will attempt to calculate the trend by comparing 2 discs of different radii and assuming they will burst when the centrifugal force acting on the outermost layers are the same. Using integrals, I will see which one has the higher average inertia per mass when the centrifugal force on the outer layers are equal, but I might be unable to figure out the answer.

[Mr Skeptic]

Well, if you use hollow cylinders instead of disks the math will be much easier. A disk is after all just a bunch of concentric hollow cylinders.

[Justonium]

Kk, I've calculated the relationship between kinetic energy per mass and radius, assuming both discs are traveling at speeds where the centrifugal force exerted on the outermost layers are the same. Kinetic energy per mass and radius are directly proportional. So assuming a disc of a certain material will shatter when a certain constant centrifugal force is reached in the outermost layer, the amount of energy possible to store in a given unit of mass of the disk's material is directly proportional to the radius of the disk used.

[jajrussel]

I tried to find some information on the experiment on the net, but haven't yet. However,I was playing with the numbers and this is what I came up with. I am learning as I go here, so I am not sure that I expressed everything I need to express nor am I sure that I expressed everything correctly, but here goes.

 

23E6 rpm

.8mm diameter

(.8mm diameter x 4) x .7853981 = 2.51327392 mm circumference

23E6 rpm x 2.51327392 mm = 57805300.16 mm/min

57805300.16 mm/m / 60s = 963421.6693 mm/s

963421.669 mm/s / 1000 = 963.4216693 m/s

963.4216693 m/s / .8 = 1204.277087 m/s

1 mm diameter = 1204.277087 m/s @ 23E6 rpm

 

So If I figured this right the perimeter of your sphere or disk would be about

2997922458 / 1204.277087 = 248939.7675 meters in diameter when it reached light speed if it were spinning at 23E6 rpm. This really doesn't mean anything, but you can work the numbers, and if you know all the stress levels of the materials you want to work with you can figure out how big not to make it, and how fast not to spin it, or conceivably build something that would rip things apart without coming apart its self during the process. Which is sort of what Jesse Beams did for a living.

 

I could be wrong, but it seems that basically, regardless the size, the sphere or disk with the highest rpm will have the greater energy density at any given point, but the larger the sphere or disk the more energy you can store because you have more area for storage. I am assuming the goal is to combine the two for efficiency.

 

If you were designing a large space craft and wanted to use one or more centrifuges for stability and designed them as containers you could conceivably store more fuel in them for the ride. Then all you have to do is figure how to get the fuel out without stopping the centrifuge, and get it to where you want it. In the meantime there are all kinds of other interesting things that can take advantage of the centrifuges spin.

[insane_alien]

right, using some very simple algebra, it is easy to see that the energy density for a thin disk is

 

E/m = (r^2*omega^2)/4

 

i have kept it as /m instead of /V as these are directly proportional and keeps a density term out of it.

 

it is a similar story for a sphere as well.

 

Energy density is proportional to the product of the squares of rotational velocity and radius.

 

this means that it is very possible for a large object to have the same energy density as a small light rapidly rotating object.

 

infact, after some back of the envelope calculations (the equations for moment of inertia and rotational energy are trivial and well known, the data for density of steel(iron) and volume of the earth along with rotational velocity are also readily available) i found the average energy density of the 0.8mm steel ball to be 5.84*10^6 J/m^3

 

the earth averages 1.981*10^8J/m^3

 

earth wins due to its massive mass.

[Justonium]

I could be wrong, but it seems that basically, regardless the size, the sphere or disk with the highest rpm will have the greater energy density at any given point, but the larger the sphere or disk the more energy you can store because you have more area for storage. I am assuming the goal is to combine the two for efficiency.

 

I calculated the energy stored per mass of disk of two discs of different sizes, both spinning at speeds with equal centrifugal force upon the outermost layers, and found that whenever you double the radius of the disc, the amount of energy stored per mass of disc also doubles. I'm too lazy to show all my math here, but basically, I just used integrals to calculate the total kinetic energy of the discs and then just divided by the mass of the discs. So if a disc shatters close to around when some part of the disc has a certain maximum centrifugal force acting upon it, as I have assumed, then the larger your disk, the more energy can be stored per mass of the disc's material. This could probably be increased by having the disc be hollow, and maybe having some kind of spokes to hold them together.

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Merged post follows:

Consecutive posts merged

insane_alien, The circumference of the steel ball is traveling at 960 meters per second, whereas the circumference of the earth is traveling at a mere 460 meters per second. You're energy density for the earth came out to be 2 orders higher than that of the steel ball, but that can't be right.

 

When comparing two similar rotating bodies of the same material, energy density is proportional to the velocity of two corresponding points in the bodies. So assuming the earth and the steel ball are about the same density, the earth should have about half the energy density of the steel ball, because its circumference is moving at roughly half the speed of the ball's

Edited December 31, 2009 by Justonium
Consecutive posts merged.

[jajrussel]

I used diameter when playing with the numbers because the steel balls size was given as a diameter. Using radius just changes the math a little, so I'll switch to radius for the moment. When you double the radius you more than double the area for storage. The area of a circle with a radius of .5 mm is .785398 sq mm. The area of a circle with a radius of 1 mm is 3.141593 sq mm. The area of a circle with a radius of 2 mm is 12.56637, and so on and so on. Add depth and you add mass. You can see that a disk with a radius of 2 mm is more than double the area of a disk with the a radius of 1 mm.

So, I am somewhat confused when you say that whenever you doubled the radius the amount of energy you could store is also doubled. It seems more than doubled.

I admit that using integrals to calculate is beyond my abilities for the moment, but the only way I can see you only doubling storage at double radius's is if you do use spokes, effectively removing mass from the disk, but why not take advantage of as much storage area as you can?

Maybe I am not taking something into consideration here.

[Mr Skeptic]

insane_alien, The circumference of the steel ball is traveling at 960 meters per second, whereas the circumference of the earth is traveling at a mere 460 meters per second. You're energy density for the earth came out to be 2 orders higher than that of the steel ball, but that can't be right.

 

When comparing two similar rotating bodies of the same material, energy density is proportional to the velocity of two corresponding points in the bodies. So assuming the earth and the steel ball are about the same density, the earth should have about half the energy density of the steel ball, because its circumference is moving at roughly half the speed of the ball's

 

For the rotational kinetic energy of a hollow sphere I got: [math]KE_{hollow} = \int^{\theta = \pi/2}_{\theta = -\pi/2} \frac{1}{2} \cdot 2 \pi r \rho cos \theta \cdot (\omega r cos \theta)^2 d\theta = \frac{4}{3} \pi \rho \omega^2 r^3[/math] where rho is density and theta is angle above the equator (and I took into account symmetry of circles perpendicular to the axis of rotation). For a solid sphere it would be [math]KE_{solid} = \int_{r=0}^{r=R} \frac{4}{3} \pi \rho \omega^2 r^3 dr = \frac{1}{3} \pi \rho \omega^2 R^4[/math]. Whereas the mass of a solid sphere would be [math]m = \rho \frac{4}{3} \pi R^3[/math]. Dividing these, [math]KE_{rotating sphere}/m = \frac{1}{4}\omega^2r[/math]

 

So then a larger sphere will have more energy density. This is probably a different solution than for a disk as it is three dimensional, whereas the mass of a disk of a given thickness increases as r squared only.

 

Feel free to check my math, I may have made a mistake somewhere.

 

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insane_alien, when you had asked why I suggested a hollow cylinder rather than a sphere, you can see the answer. The energy density of a rotating sphere is worse than for a disk.

Edited December 31, 2009 by Mr Skeptic

[insane_alien]

yes, i know that, but you can spin a sphere faster as it has more structural strength.

[Justonium]

I used diameter when playing with the numbers because the steel balls size was given as a diameter. Using radius just changes the math a little, so I'll switch to radius for the moment. When you double the radius you more than double the area for storage. The area of a circle with a radius of .5 mm is .785398 sq mm. The area of a circle with a radius of 1 mm is 3.141593 sq mm. The area of a circle with a radius of 2 mm is 12.56637, and so on and so on. Add depth and you add mass. You can see that a disk with a radius of 2 mm is more than double the area of a disk with the a radius of 1 mm.

So, I am somewhat confused when you say that whenever you doubled the radius the amount of energy you could store is also doubled. It seems more than doubled.

I admit that using integrals to calculate is beyond my abilities for the moment, but the only way I can see you only doubling storage at double radius's is if you do use spokes, effectively removing mass from the disk, but why not take advantage of as much storage area as you can?

Maybe I am not taking something into consideration here.

 

What you need to understand is that surface area and energy stored per mass of disk are two different things. All I did was calculate the kinetic energies of two different sized discs, both spinning at different speeds so that the centrifugal force exerted on the outer layers would be equal in both discs. I then divided the total kinetic energy by the mass of each disc to find the energy stored per mass of material used. I think you were thinking total energy stored, in which case the bigger disc stores more energy, but what I was calculating was energy per mass, which also happened to increase. Also, the discs were not traveling at the same rpm. As a disc gets bigger, the maximium rpm possible decreases, while the maximum speed of the outer layer increases.

 

Here is an example: a disc with a radius of 1m with its outer layer traveling at 1mps has the same centrifugal force acting upon its outer layer as a disc with a radius of 4m and an outer layer traveling at 2mps. The total kinetic energy of the second disc is 64 times that of the first disc, but once we take into account that the surface area of the second disc is 16 times that of the first, we know that the average energy density of the second disc is 4 times that of the first disc. Sorry I didn't show my math but you can calculate it if you want.

Edited January 1, 2010 by Justonium