- 29 Jul '10 19:38I see F=(mv^2)/r so why does mass come into it? What does m represent? The mass of the disk or are we saying it would be more like a rock on a sling? How do I get the answer in terms of G's, 1G = 1 Earth gravity? And what is the unit for F? Newtons if you use metric?

I am trying to figure the g forces on a magnetic disk HD, the regular 3.5 inch variety, looks like about 89 or 90 mm in diameter. They spin at 5400 to 15,000 RPM and I wanted to know the g forces on the disk. Converting to a g force, doesn't that remove mass from the equation? - 29 Jul '10 19:59It would be a rock on a sling if I'm not mistaken, it works out differently if you have a mass distribution (you need an integral etc.). For a rough estimate, I think adding a factor 1/2 for a disk would be okay (I'd have to check though, but I'm not sure how precise your answer needs to be). The units of a force in SI units is obviously always Newton (1 N = 1 kg m s^-2 - it works out correctly for the centripetal force).

1g is the force associated with gravity at the Earth's atmosphere. This force is given by F = mg (I'm not sure how useful comparing to gravity is in this situation). - 29 Jul '10 21:48 / 1 edit

I don't need much more than 2 or 3 decimal places. Just wanted to see how much force would be on a HD disk. I guess I would have to find the mass of the whole disk and thickness and work out what a certain area/volume would be, say one mm by 1 cm. And call that the rock on a sling? If the answer is in newtons, how would you convert that to g's?*Originally posted by KazetNagorra***It would be a rock on a sling if I'm not mistaken, it works out differently if you have a mass distribution (you need an integral etc.). For a rough estimate, I think adding a factor 1/2 for a disk would be okay (I'd have to check though, but I'm not sure how precise your answer needs to be). The units of a force in SI units is obviously always Newton ( rce is given by F = mg (I'm not sure how useful comparing to gravity is in this situation).**

I see one Kg=9.8 newtons and 2.2 pounds. So 1 newton = about 3.6 ounces.

that 9.8 Newtons is suspiciously close to the definition of 1 g in terms of acceleration, 9.8(m/sec)^2. How did that happen? - 30 Jul '10 00:34 / 2 edits

A Newton is a kilogram times meter divided by seconds squared.*Originally posted by sonhouse***I don't need much more than 2 or 3 decimal places. Just wanted to see how much force would be on a HD disk. I guess I would have to find the mass of the whole disk and thickness and work out what a certain area/volume would be, say one mm by 1 cm. And call that the rock on a sling? If the answer is in newtons, how would you convert that to g's?**

I see one ...[text shortened]... ly close to the definition of 1 g in terms of acceleration, 9.8(m/sec)^2. How did that happen?

N = kg m/s^2

The force in Newtons on a mass of one kilogram in gravity - or it's weight - is the kilogram times the gravitational acceleration it would experience.

"One g" means "equivalent to gravity". - 30 Jul '10 00:39

I'm having a hard time understanding what you mean.*Originally posted by sonhouse***I don't need much more than 2 or 3 decimal places. Just wanted to see how much force would be on a HD disk. I guess I would have to find the mass of the whole disk and thickness and work out what a certain area/volume would be, say one mm by 1 cm. And call that the rock on a sling? If the answer is in newtons, how would you convert that to g's?**

I see one ...[text shortened]... ly close to the definition of 1 g in terms of acceleration, 9.8(m/sec)^2. How did that happen? - 30 Jul '10 03:25 / 1 edit

Sorry, just realized what a newton is:= the weight of one kilogram in one Earth gravity. I think.*Originally posted by AThousandYoung***I'm having a hard time understanding what you mean.**

So a newton of force is the same as what a kilogram weighs on earth?

So to my problem of how many g forces does a 90 mm disk feel spinning at 7200 RPM, the standard rate of spin for a 3.5 inch (90mm) HD. Can't you say what say, a single atom or molecule, would feel on the edge of such a disk at that spin rate in terms of g? - 30 Jul '10 04:28 / 1 edit

yeah the force exerted on a single atom (or something as small as a single atom in mechanics which is probably a bad model for this situation), at that radius with the aformentioned angular velocity would be almost non existent by inspection.*Originally posted by sonhouse***Sorry, just realized what a newton is:= the weight of one kilogram in one Earth gravity. I think.**

So a newton of force is the same as what a kilogram weighs on earth?

So to my problem of how many g forces does a 90 mm disk feel spinning at 7200 RPM, the standard rate of spin for a 3.5 inch (90mm) HD. Can't you say what say, a single atom or molecule, would feel on the edge of such a disk at that spin rate in terms of g?

just for reassurance I ran the calcs

1.99E-26 N

And this would be a if the atom (I chose carbon 12) was tethered to the center of rotation by a magically non-existent string.

And this "G-Force" talk has to be a misnomer as "G-force" is not a "force" its an acceleration of a body relative to that of a body in freefall caused by the mass of the earth. Comparing forces to accelerations is useless. - 30 Jul '10 06:25

Although it wouldn't tell the whole story, I would start by calculating the mass of a square cm of the disk. Then work out the force that mass would experience if attached to the edge of the disk.*Originally posted by sonhouse***Sorry, just realized what a newton is:= the weight of one kilogram in one Earth gravity. I think.**

So a newton of force is the same as what a kilogram weighs on earth?

So to my problem of how many g forces does a 90 mm disk feel spinning at 7200 RPM, the standard rate of spin for a 3.5 inch (90mm) HD. Can't you say what say, a single atom or molecule, would feel on the edge of such a disk at that spin rate in terms of g?

The overall question, is complicated as each radial line experiences a force in a slightly different direction. There must be a formula though for the total force in a given direction (zero actually, so lets work it out for half the disk only).

I guess what you really want to know is how strong the disk must be. - 30 Jul '10 06:31 / 2 edits

If I remember correctly - A mass of 1 kilogram has a weight of 9.8 Newton at the surface of the Earth.*Originally posted by sonhouse***Sorry, just realized what a newton is:= the weight of one kilogram in one Earth gravity. I think.**

So a newton of force is the same as what a kilogram weighs on earth? - 30 Jul '10 07:14

1 kg /= 9.8 N*Originally posted by sonhouse***I don't need much more than 2 or 3 decimal places. Just wanted to see how much force would be on a HD disk. I guess I would have to find the mass of the whole disk and thickness and work out what a certain area/volume would be, say one mm by 1 cm. And call that the rock on a sling? If the answer is in newtons, how would you convert that to g's?**

I see one ...[text shortened]... ly close to the definition of 1 g in terms of acceleration, 9.8(m/sec)^2. How did that happen?

kilogrammes is measure of MASS

newtons is measure of FORCE

Its like asking how many gallons in a mile

What you are confusing is that the approximate force due to gravity between the Earth and 1kg at the surface is 9.8N - 30 Jul '10 07:29 / 1 edit

Would this be the integral*Originally posted by KazetNagorra***It would be a rock on a sling if I'm not mistaken, it works out differently if you have a mass distribution (you need an integral etc.). For a rough estimate, I think adding a factor 1/2 for a disk would be okay (I'd have to check though, but I'm not sure how precise your answer needs to be). The units of a force in SI units is obviously always Newton ( rce is given by F = mg (I'm not sure how useful comparing to gravity is in this situation).**

F = m*r*w^2

dF = dm*r*w^2

dm= rho*dV

dV = h*Pi ( r^2 -(r-dr)^2)

= h*Pi ( 2r*dr + dr^2)

dF = w^2*rho*Pi*h ( 2r^2*dr + r*dr^2)

F = (w^2*rho*Pi*h )Int( 2r^2*dr + r*dr^2)

Is this possible to evaluate in this form...I've never saw it done?

Or is what ive done complete nonsence? - 30 Jul '10 08:14

Well, the centripetal force is F = mv²/r. Assume the mass is distributed with a cylindrical symmetry, the total force can be calculated by integrating over the radius. In some slice dr the mass contained is h * 2 * pi * r * rho * dr, where h is the thickness of the disk and rho the mass density. The velocity v is given by: v = omega * r, where omega is some constant frequency at which the disk is rotating. You can then add up all these mass slices and add up the forces to obtain the total force.*Originally posted by joe shmo***Would this be the integral**

F = m*r*w^2

dF = dm*r*w^2

dm= rho*dV

dV = h*Pi ( r^2 -(r-dr)^2)

= h*Pi ( 2r*dr + dr^2)

dF = w^2*rho*Pi*h ( 2r^2*dr + r*dr^2)

F = (w^2*rho*Pi*h )Int( 2r^2*dr + r*dr^2)

Is this possible to evaluate in this form...I've never saw it done?

Or is what ive done complete nonsence?

F = Int[ 2 * pi * h * rho * omega² * r² * dr]

= 2/3 * pi * h * rho * omega² * (R2^3 - R1^3), where R2 is the outer and R1 is the inner radius. (hope I made no mistake!)

I haven't checked your calculation but I will note that in the integral limit, as dr -> 0, any order contributions of 2 or higher will vanish compared to dr-terms, so dr^2 = 0. - 30 Jul '10 12:04

So if I show a certain centripetal force as 9.8 Newtons, that would be equivalent to one G, right?*Originally posted by wolfgang59***1 kg /= 9.8 N**

kilogrammes is measure of MASS

newtons is measure of FORCE

Its like asking how many gallons in a mile

What you are confusing is that the approximate force due to gravity between the Earth and 1kg at the surface is 9.8N - 30 Jul '10 15:36

Your answer for centripetal force should come out in N/kg.*Originally posted by sonhouse***So if I show a certain centripetal force as 9.8 Newtons, that would be equivalent to one G, right?**

So 9.8 N/kg would be 1G

If you think about it you are just applying F=ma

Because velocity is a vector quantity every part of the disc is accelerating and therefore must be acted upon by a force.