# Centrifugal force for supporting vacuum chamber walls?

#### RGClark

My aerospace application requires a very lightweight vacuum chamber. Could you rotate a cylindrical chamber about its vertical axis to provide outward pressure to support the pressure of the air? How fast would it have to rotate to provide an outward pressure of 15 lbs./sq. in.? The initial energy to start the rotation would be large, but then only a relatively small energy would be needed to keep the rotation going due to losses from atmospheric pressure.

Bob Clark

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#### Crosson

The idea that a rotating cylinder exerts an outward pressure is false.

There is no centrifugal (center-fleeing) force, in reality the forces involved in rotation are centripedal (center-seeking).

#### Norseman

It depends. I'm not sure of the math, but I do know that the rate of spin and diameter of the container will both affect the G forces experienced by the container. Since G forces are what we are dealing with, the weight of the container wall matters. If you have to deal with 15 pounds of air pressure per square inch and your container walls are 15 pounds per square inch, then 1 G of force would be sufficient to equalize the pressures. Of course, if your container walls are fairly strong (as any wall which is 15 pounds per square inch probably would be!), you won't need it to be fully equalized, so there is a balancing issue here.

I'm convinced it's not a very practical idea. The problem is that if your container requires that it spin in order to not collapse (and assuming it isn't made of something like rubber), then it will be broken every time you don't manage to keep it spinning. This means you need to be pumping the air out while it's spinning. That's fairly tricky. If your container doesn't need to spin in order to not collapse, then there's no point putting a motor on it, is there?

You'd probably be better off just making a non-spinning sphere. Edit: And the larger the sphere, the better. You probably already know this, but volume cubes while surface area squares. Bigger spheres are more efficient than little ones.

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#### Norseman

Crosson said:
The idea that a rotating cylinder exerts an outward pressure is false.

There is no centrifugal (center-fleeing) force, in reality the forces involved in rotation are centripedal (center-seeking).
So wait, by spinning the cylinder the walls should collapse inward? That doesn't sound right.

#### Crosson

So wait, by spinning the cylinder the walls should collapse inward? That doesn't sound right.
Think of swinging a ball on a string in a circle around your head. What are the forces on the ball? The only real force is the tension of the string, which is always directed towards the center (your hand, located at the center, is pulling the string). This force is centripedal, not centrifugal.

In your case, the rotating cylinder will experience an inward pressure equal to:

$$P=m\frac{v^2}{r}$$

Where m is mass per area (kg/m^2), v is velocity, and r is the radius.

#### Norseman

Crosson said:
Think of swinging a ball on a string in a circle around your head. What are the forces on the ball? The only real force is the tension of the string, which is always directed towards the center (your hand, located at the center, is pulling the string). This force is centripedal, not centrifugal.

In your case, the rotating cylinder will experience an inward pressure equal to:

$$P=m\frac{v^2}{r}$$

Where m is mass per area (kg/m^2), v is velocity, and r is the radius.
Alright, it makes sense, sort of, but it's just confusing the matter. The concept of spinning the cylinder is still sound, if it weren't for the issues of practicality.

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#### pallidin

Crosson said:
The idea that a rotating cylinder exerts an outward pressure is false.

There is no centrifugal (center-fleeing) force, in reality the forces involved in rotation are centripedal (center-seeking).

Perhaps you can elighten me, because I also find that odd.
For example, let's say I take a bicycle wheel, mount it horizontally so that it freely spins when rotated. OK, with the wheel at rest, I place a single weighted "bead" on each spoke and pre-set them to be closest to the axis of the wheel. Of course, placing such beads on the spokes is another matter, but easily done. Also, the beads can slide freely along each spoke when force is applied.
Anyway, I now spin the wheel. Guess what? The beads move OUTWARDS from the axis until they hit the rim. :surprised

#### Crosson

Anyway, I now spin the wheel. Guess what? The beads move OUTWARDS from the axis until they hit the rim.
The reason the beads move outward is that there is no force acting on them (other then the friction from the spoke that they are sitting on) and so they move in a straight line, not in a circle.

In order to make something spin there must be a centripedal force. Think of swinging a ball on a string, it only swings because you apply a centripedal force to it. As soon as this force stops, you might say the ball goes "flying outward", but I would say the ball continues in a straight line (because there is no force to change its motion).

The beads move outwards because of their own inertia. The fact that the beads don't move in a circle is due to the fact that there is no centripedal force.

This entire misconception comes from the way it "feels" to be moving in a circle. It "feels" like there is an outward force. The feeling is caused by inertia, and it makes no more sense to say inertia causes a centrifugal force then it does to say that inertia causes a force that resists straight line motion.

#### RGClark

Crosson said:
The reason the beads move outward is that there is no force acting on them (other then the friction from the spoke that they are sitting on) and so they move in a straight line, not in a circle.

In order to make something spin there must be a centripedal force. Think of swinging a ball on a string, it only swings because you apply a centripedal force to it. As soon as this force stops, you might say the ball goes "flying outward", but I would say the ball continues in a straight line (because there is no force to change its motion).

The beads move outwards because of their own inertia. The fact that the beads don't move in a circle is due to the fact that there is no centripedal force.

This entire misconception comes from the way it "feels" to be moving in a circle. It "feels" like there is an outward force. The feeling is caused by inertia, and it makes no more sense to say inertia causes a centrifugal force then it does to say that inertia causes a force that resists straight line motion.
I'm aware that the centripetal acceleration is headed inward. But even if you look at the example of the ball on the spinning wheel, wouldn't it's outward motion apply a force to the air molecules it ran into?
In the case of a rotating flywheel, as for energy storage for example, a big problem is not rotating too fast otherwise the flywheel will fly apart, which implies there are outward acting forces on the flywheel. So before the flywheel flies apart it appears it should have the tendency to expand. This should mean it is applying an outward force on the surrounding air.
Or here's another way of looking at the scenario: suppose you had a cylindrical vacuum tank. Suppose the thickness of the walls was not enough to hold the tank against collapse. Instead you have strong supports at regular intervals inside the cylinder to support it against collapse. You put air in at low pressure say at 1/100th normal atmospheric pressure. You rotate the tank.
The air inside should have a tendency to crowd to the sides of the tank. This should mean the density and air pressure of this air inside is increased when measured close to the sides. And spinning sufficiently high should make it the same pressure as the outside normal air pressure. Then you should be able to remove the supports inside the tank, and you would have a vacuum tank with much thinner walls then what is normally required.

Bob Clark

#### GOD__AM

RGClark said:
In the case of a rotating flywheel, as for energy storage for example, a big problem is not rotating too fast otherwise the flywheel will fly apart, which implies there are outward acting forces on the flywheel. So before the flywheel flies apart it appears it should have the tendency to expand. This should mean it is applying an outward force on the surrounding air.

Bob Clark
For the flywheel to expand in one direction it must contract along another. So it would have to flatten out to expand along the circumference. This means any "outward" force created by spinning, is in effect being cancelled out by the same "inward" force somewhere else. For your idea to work you have to create an equal outward force in all directions, and thats not what's happening when you spin something.

#### RGClark

GOD__AM said:
For the flywheel to expand in one direction it must contract along another. So it would have to flatten out to expand along the circumference. This means any "outward" force created by spinning, is in effect being cancelled out by the same "inward" force somewhere else. For your idea to work you have to create an equal outward force in all directions, and thats not what's happening when you spin something.
I think for a circular flywheel it is expanding equally in all directions perpindicular to the axis of rotation.

Bob Clark

#### GOD__AM

RGClark said:
I think for a circular flywheel it is expanding equally in all directions perpindicular to the axis of rotation.

Bob Clark
Ok, think of how a quarter looks after its flattened by a train wheel. It has greater circumference but it's thinner. The gear heats up and expands in somewhat the same manner, or atomic bonds fail and it breaks.

Consider this, if I had a rocket pack I could enter the cylinder and fly all around as long as I don't touch any walls. The only force would be the gravity pulling me down. If i did hit a wall it would cause a torque around my body starting me to rotate in the same direction as the cylinder, and I would be deflected inward from the wall. After enough of these collisions I would be rotating at a similar rate to the cylinder. At this point there is nothing pushing me in any direction, its a state of equilibrium.

#### brewnog

Gold Member
GOD__AM said:
Consider this, if I had a rocket pack I could enter the cylinder and fly all around as long as I don't touch any walls. The only force would be the gravity pulling me down. If i did hit a wall it would cause a torque around my body starting me to rotate in the same direction as the cylinder, and I would be deflected inward from the wall. After enough of these collisions I would be rotating at a similar rate to the cylinder. At this point there is nothing pushing me in any direction, its a state of equilibrium.
I have absolutely no idea why flattening coins with trains has anything to do with this, but anyway:

You're forgetting that viscous effects will provide a force acting to move you; think of it with syrup instead of air.

RGClark:

However, at such low pressures (and with air not being terribly viscous anyway), I'm wondering whether you'd be able to get enough coupling between the walls and the contents to be able to get the air to spin fast enough to create the effect you're after.

Sorry, I don't think this one has legs. PM Clausius, he's good with fluids.

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#### Gokul43201

Staff Emeritus
Gold Member
Three things :

1. This is not a fluid mechanics problem;

2. Crosson's argument that each element of the cylinder wall must be described as experiencing an inward force (in the inertial frame attached to the cylinder axis) in order to correctly explain the dynamics is accurate.

3. Crosson's argument does not apply here.
Elastic properties of a meterial are defined in the rest frame of the body made of that material - and for good reason. The forces experienced by the atoms of a body must be described in the rest frame of these atoms. In this frame the material of the cylinder does feel a tensile rather than compressive stress. The apparent (outward) pressure due to this tensile stress is exactly the number calculated by Crosson in post #5.

I suspect it's going to take some serious RPMs to offset (pre-stress against) the inward pressure. For a steel cylinder of radius = 1m and wall thickness =10mm, the required RPMs are about 300 or thereabouts.

#### Crosson

In the case of a rotating flywheel, as for energy storage for example, a big problem is not rotating too fast otherwise the flywheel will fly apart, which implies there are outward acting forces on the flywheel.
There are no outward forces. The flywheel flys apart when the centripedal force (inward) is too weak to keep the flywheel spinning.

Suppose there were an outward force, as in your imagination. Surely, you dont deny (in the case of a spinning ball and string) that the strings' tension also provides an inward force. Perhaps you would say that the forces are equal and opposite, in the case of spinning around at constant velocity (newtons 2nd law says v = C implies F = 0).

This is nonsense, of course. If F = 0 (if the centripedal force was balanced by a centrifugal force) then the object would move at v = C in a straight line. In order to make it change direction (as objects do continuously in a circle), we must apply a net force inwards.

#### RGClark

Gokul43201 said:
Three things :

1. This is not a fluid mechanics problem;

2. Crosson's argument that each element of the cylinder wall must be described as experiencing an inward force (in the inertial frame attached to the cylinder axis) in order to correctly explain the dynamics is accurate.

3. Crosson's argument does not apply here.
Elastic properties of a meterial are defined in the rest frame of the body made of that material - and for good reason. The forces experienced by the atoms of a body must be described in the rest frame of these atoms. In this frame the material of the cylinder does feel a tensile rather than compressive stress. The apparent (outward) pressure due to this tensile stress is exactly the number calculated by Crosson in post #5.

I suspect it's going to take some serious RPMs to offset (pre-stress against) the inward pressure. For a steel cylinder of radius = 1m and wall thickness =10mm, the required RPMs are about 300 or thereabouts.
Thanks for the response. This was what I was hoping for, that the *outward* pressure would be given by P = m*v^2/r, m the mass per unit area. The reason why I wanted to minimize thickness is because the diameter in my application is quite large, 10's of meters. Then by the formula you might be able to get a thickness less than a millimeter.
Do you have a reference for this?

Bob Clark

#### Antiphon

It's called a turbopump. About 30,000 RPM will do it.

#### RGClark

brewnog said:
I have absolutely no idea why flattening coins with trains has anything to do with this, but anyway:

You're forgetting that viscous effects will provide a force acting to move you; think of it with syrup instead of air.

RGClark:

However, at such low pressures (and with air not being terribly viscous anyway), I'm wondering whether you'd be able to get enough coupling between the walls and the contents to be able to get the air to spin fast enough to create the effect you're after.

Sorry, I don't think this one has legs. PM Clausius, he's good with fluids.

I found this discussion on this bbs that at least suggests the gas pressure will increase on the sides:

Physics Help and Math Help - Physics Forums > Physics > Classical Physics.
Centrifugal force in high pressure vessel.
10-19-2004, 08:10 PM

Bob Clark

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