30 Jul 10
Originally posted by KazetNagorraWhy would it be any different for folks in a spinning wheel in space, if the spin produces the equivalent of 1 g, I weigh 100 kg so that's what a scale would show if I stepped on it when I was being spun. If someone 50 Kg on Earth were to do the same, he or she would show 50 kg but we both are subject to the same 1 g force. Why would it be different on a HD spinning at whatever, 7 Krpm?
Only if the disk is 1kg.
30 Jul 10
Originally posted by sonhouseBecause 1 g is a unit of acceleration, not of force. "G-force" is a misnomer.
Why would it be any different for folks in a spinning wheel in space, if the spin produces the equivalent of 1 g, I weigh 100 kg so that's what a scale would show if I stepped on it when I was being spun. If someone 50 Kg on Earth were to do the same, he or she would show 50 kg but we both are subject to the same 1 g force. Why would it be different on a HD spinning at whatever, 7 Krpm?
30 Jul 10
Originally posted by sonhouseUsually when one talks of "g-force", one is comparing the force to the force on the object on the Earth's surface.
Why would it be any different for folks in a spinning wheel in space, if the spin produces the equivalent of 1 g, I weigh 100 kg so that's what a scale would show if I stepped on it when I was being spun. If someone 50 Kg on Earth were to do the same, he or she would show 50 kg but we both are subject to the same 1 g force. Why would it be different on a HD spinning at whatever, 7 Krpm?
The force acting on a spinning stone-on-a-rod increases proportionally to its mass, while the acceleration due to the force decreases with an inverse proportionality. The two cancel, which is why we speak of e.g. a "3g force" on a rollercoaster (in reality, a heavier person is subject to a larger force from the rollercoaster track/cart).
30 Jul 10
Originally posted by KazetNagorraWhich is identical to just standing up on Earth, the 'equivalence' thing of Einstein. So why can't you just put the force of a mass spinning in a circle in terms of g forces?
Usually when one talks of "g-force", one is comparing the force to the force on the object on the Earth's surface.
The force acting on a spinning stone-on-a-rod increases proportionally to its mass, while the acceleration due to the force decreases with an inverse proportionality. The two cancel, which is why we speak of e.g. a "3g force" on a roller ...[text shortened]... reality, a heavier person is subject to a larger force from the rollercoaster track/cart).
At such and such an RPM and such and such a radius of curvature, a mass on the string feels the effect of 1.0 g's. Something like that? Not talking about the total force, of course that varies with mass but the equivalent force if it were resting on a gravitational mass, like Earth, or the moon, whatever, 1 g on Earth, 1/6 th g on the moon, where g is one Earth gravity. Can't the answer be given in those terms?
30 Jul 10
Originally posted by sonhouseYes, it can. Just use the formula I gave, calculate the force (which depends on rotation speed and the mass of the disk), and divide it by the force gravity would exert on the disk if it were on the Earth's surface.
Which is identical to just standing up on Earth, the 'equivalence' thing of Einstein. So why can't you just put the force of a mass spinning in a circle in terms of g forces?
At such and such an RPM and such and such a radius of curvature, a mass on the string feels the effect of 1.0 g's. Something like that? Not talking about the total force, of cours ...[text shortened]... 6 th g on the moon, where g is one Earth gravity. Can't the answer be given in those terms?
30 Jul 10
9 edits
Originally posted by KazetNagorraSo that would be G=(MV^2/r)/9.8 right? (assuming MKS units). If that is right, how would you frame that in terms of RPM or degrees/second? So if you put r at 1 meter, and the velocity at 1 meter/second and the mass at 1 Kg, it would be slung out at 1/9.8th of a g, 0.102 G, right? That still doesn't sound right. It would mean 1/2 Kg would be slung out at 0.05 G with that formula. That would represent 1/(2PI) revolutions per second, 1/6 rps or 10 RPM. I think🙂
Yes, it can. Just use the formula I gave, calculate the force (which depends on rotation speed and the mass of the disk), and divide it by the force gravity would exert on the disk if it were on the Earth's surface.
The G acceleration should be a constant for any mass I would think. If you are dropped from space some distance out in vacuum it doesn't matter if you are 1 kg or 1000000 kg, you still end up with the same velocity when you hit the deck. Of course the kinetic energy of the larger mass is much larger but the acceleration is the same in both cases. So how do you derive the acceleration equivalent of gravity?
So F=(M* ((D/S)*2PI))^2)/r in terms of degrees per second? Sound right?
30 Jul 10
1 edit
Originally posted by sonhouseNo, it would be:
So that would be G=(MV^2/r)/9.8 right? (assuming MKS units). If that is right, how would you frame that in terms of RPM or degrees/second? So if you put r at 1 meter, and the velocity at 1 meter/second and the mass at 1 Kg, it would be slung out at 1/9.8th of a g, 0.102 G, right? That still doesn't sound right. It would mean 1/2 Kg would be slung out at 0.0 ration is the same in both cases. So how do you derive the acceleration equivalent of gravity?
2/3 * pi * h * rho * omega² * (R2^3 - R1^3)/(mg)
= 2/3 * pi * h * rho * omega² * (R2^3 - R1^3)/(rho * pi * (R2² - R1² ) * h * g)
= 2/3 omega² (R2^3 - R1^3)/(g * (R2² - R1² )). (this is a dimensionless number - note how it is also independent of mass)
Fill in a rotation frequency (careful with factors 2pi - I think if you have a certain "RPM" you have to divide by 2pi first and then fill it in as omega here, but I often mess these kind of things up), an outer radius R2 and inner radius R1 and you have your "g-force".
30 Jul 10
Originally posted by KazetNagorraCan you explain those units? h=? rho=? Omega=? I assume the (mg) part is mass in Kg and g's in 9.8 newton units?
No, it would be:
2/3 * pi * h * rho * omega² * (R2^3 - R1^3)/(mg)
= 2/3 * pi * h * rho * omega² * (R2^3 - R1^3)/(rho * pi * (R2² - R1² ) * h * g)
= 2/3 omega² (R2^3 - R1^3)/(g * (R2² - R1² )). (this is a dimensionless number - note how it is also independent of mass)
Fill in a rotation frequency (careful with factors 2pi - I think if you have a ...[text shortened]... hese kind of things up), an outer radius R2 and inner radius R1 and you have your "g-force".
31 Jul 10
Originally posted by sonhouseI assumed in my calculation that you have a disk which is CD-shaped. If you just have a rock on a string, then you can obviously use the original formula.
Why would there be an inner and outer radius? For calculating the force, don't you start at ground zero and go out to where the 'rock' is?
01 Aug 10
Originally posted by KazetNagorraBut the force would be the same all around the circumference so why would there need to be more than one solution? Assuming the density is the same everywhere on the disk. I was just after the max force anywhere on the edge of the disk.
I assumed in my calculation that you have a disk which is CD-shaped. If you just have a rock on a string, then you can obviously use the original formula.
01 Aug 10
1 edit
Originally posted by sonhouseThere is more than one solution because the force on a rock-on-a-string is different from the force on a disk with a certain inner and outer radius. The formula I gave gives the force on the disk as a whole. Just fill in the numbers and you'll have your g force. If you're only interested in the edge, simply use your original formula and divide by mg to obtain:
But the force would be the same all around the circumference so why would there need to be more than one solution? Assuming the density is the same everywhere on the disk. I was just after the max force anywhere on the edge of the disk.
v²/(rg) = (omega² r)/g.
