# Use of centrifugal force in outer-space

1. Dec 15, 2009

### forcibleme

Reading the Arthur C. Clarke novel: "Rendezvous with Rama"; something keeps nagging me about the descriptions of their experience inside the ship they encounter, which is a huge, rotating, hollow cylinder. It's rotation makes a sort of "pseudo-gravity" by way of centrifugal force. The issue I'm having relates to this "gravity" being described as gradually increasing the further you get from the axis of spin. My mind keeps telling me that, in space, you would be weightless anywhere within the hollow parts of the cylinder (spinning or not), and only if you made yourself a part of it's matter (by holding onto the sides and adopting the spin, for instance) would you experience the "gravity".

The ship has an atmosphere built up inside it, though, which would provide some matter that would eventually adopt the spin and therefore be thrust from the center axis. But I would think the forces you experience having just left the weightlessness of the axis would be limited to what little of the spin you would adopt based on the friction of the air around you - until you reached the spinning outer wall, that is.

Am I totally off base? If so, why would you feel the centrifugal pull of a spinning body you weren't actually touching? The main thing that caused me to wonder is the flying bicycle trip one of the crew takes to the other side of Rama. He experiences a greater centrifugal pull/push the further he gets from the axis (the only place he is said to experience actual weightlessness). But, without being a part of the spin (which you wouldn't be unless pushed by some form of matter already spinning - correct?) why would this happen? Would just the air of an atmosphere be enough to act on you in this manner?

2. Dec 15, 2009

### A.T.

If you feel weightless (zero proper acceleration) or not, depends on how you fly your bicycle. You don't need to be at the axis to achieve that.

3. Dec 15, 2009

### forcibleme

So, are you confirming my suspicion that the centrifugal force would NOT necessarily be acting on you in the hollow parts of the giant spinning cylinder, or not?

A large portion of the physical "realities" as they are described in the book depend on this "gravity" becoming increasingly forceful the further you get from the axis of spin. But my expectation is that, unless some form of matter already spinning were to act upon you, you wouldn't feel any form of acceleration or force pushing you toward the outside.

But again, there is an atmosphere (air) inside the cylinder, which would be acting [upon you via friction, but would that be enough to pull a human body (or anything as dense/large) into the trap of the centrifugal force caused by the ship's rotation?

4. Dec 15, 2009

### A.T.

In the rotating frame there is always a centrifugal force acting on you. But it is an inertial force, which doesn't cause proper accelerations, which you could feel. So you can feel weightless, even if the centrifugal force is acting on you in some frame.
You don't feel centrifugal acceleration, but in the rotating frame it is present as coordinate acceleration. So you would travel towards the outside, without felling pushed towards it. Just like in free fall on earth.

5. Dec 15, 2009

### forcibleme

I guess my main question is: what if YOU are not rotating along with the rest of the craft? Let's say you were just moments ago pushed off of center axis, and therefore have not picked up the rotation of the cylinder around you - or you are somehow magically inserted into the spinning cylinder any distance less than the radius from a point along the spinning surface of the inside. Does your body gradually migrate away from the axis? Or does it only do so once you have adopted a correlating rotation with the cylinder?

As another example, let's say you have a 12-inch wide, empty, cylindrical drum that is spinning quite quickly about it's center axis, and you're in a weightless environment with no substantial gravitational forces in play. If you were to guide a thread through the cylinder even one millimeter from the spinning edge of the cylinder, would centrifugal force be acting on that thread? And if so, for what reason?

Last edited: Dec 15, 2009
6. Dec 15, 2009

### A.T.

Initially or in vacuum, no.
Yes, once the air accelerates you, you hit the wall.

Not sure if I understand what you do with the thread. But the centrifugal force exists only in the rotating frame of reference:
http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame)
The reason for it is to make F=ma work in rotating frames.

7. Dec 15, 2009

### SystemTheory

Let the donut shaped spacecraft rotate with constant angular velocity $\omega$ about a central axis. A dude of mass m stands on the floor at radius R as shown in the sketch. The floor pushes the dude toward the center and the dude pushes down on the floor. Tangential velocity at radius R is given by $v = \omega R.$

http://hyperphysics.phy-astr.gsu.edu/HBASE/cf.html

Centripital Acceleration:

$$a = \frac{v^{2}}{R} = \omega^{2}R$$

The centripital force is the mass times acceleration. When a body with momentum in the spacecraft falls it will appear to fall straight toward the floor as far as someone inside would observe (I'm fairly certain). Only centripital force is necessary to move with the craft once angular velocity is constant (angular acceleration is zero).

#### Attached Files:

• ###### spacedude.png
File size:
1.1 KB
Views:
153
8. Dec 17, 2009

### Rasalhague

Forcibleme: "So, are you confirming my suspicion that the centrifugal force would NOT necessarily be acting on you in the hollow parts of the giant spinning cylinder, or not?"

A.T.: "In the rotating frame there is always a centrifugal force acting on you."

Forcibleme: "Let's say [...] you are somehow magically inserted into the spinning cylinder any distance less than the radius from a point along the spinning surface of the inside. Does your body gradually migrate away from the axis?

A.T.: "Initially or in vacuum, no."

*

So there's always a force accelerating you away from the radius, but you don't move away from the radius? How can we reconcile these apparent contradictions? If, in a rotating frame, there is always a centrifugal force acting on you, forcing you away from the radius:

$$-m \mathbf{\Omega} \times \left ( \mathbf{\Omega} \times \mathbf{R} \right )$$

then which fictional force counteracts the centrifugal force, keeping you at a constant distance from the radius and accounting for your orbit around the radius? Supposing the angular velocity of the cylinder is constant, I guess the only other relevant fictional force would be the Coriolis force:

$$-2m \mathbf{\Omega} \times \frac{d \mathbf{R}}{dt}$$

This depends on radial velocity, so I'm thinking it would come into play as soon as the centrifugal force gave you some radial velocity towards the floor of the cylinder. But wouldn't these forces alone would send you on a spiral path, curling outward against the rotation of the cylinder, till you met the floor? But that conflicts with your answer "in a vacuum no" (which is the answer I'd expect from thinking about it in an intertial frame). So have I mistinterpreted the equations, or am I leaving out some other force?

9. Dec 17, 2009

### A.T.

This is correct. The Coriolis force together with the centrifugal force explain the coordinate accelerations of inertially moving bodies viewed from a rotating frame of reference. The general formula for the Coriolis force is:

$$-2m \mathbf{\Omega} \times v$$

from: http://en.wikipedia.org/wiki/Coriolis_effect#Formula

I think your formula was only the special case for the tangential component of the Coriolis force. With the general formula above there is no other force needed besides Coriolis and centrifugal.

10. Dec 17, 2009

### Rasalhague

The Wikipedia article you linked to defines $v$ as "the velocity of the particle in the rotating system". That's all I meant by

$$\frac{d \textbf{R}}{dt}$$

By $\textbf{R}$, I mean the position in the rotating system, taking the axis as the origin (and, for simplicity, ignoring motion parallel to the axis which doesn't contribute to the Coriolis force). The derivative of position wrt time in the rotating system is the velocity in the rotating system. I think it's tangential component would be

$$\frac{d \textbf{R}}{dt} - \frac{ \frac{d \textbf{R}}{dt} \cdot \textbf{R} }{ \textbf{R} \cdot \textbf{R} } \textbf{R} = \frac{ \left ( \textbf{R} \times \frac{d \textbf{R}}{dt} \right ) \times \textbf{R} }{ \textbf{R} \cdot \textbf{R} }$$

Initially there's no velocity, so there should only be a centrifugal force, and I think the equation for centrifugal force says that this force will be in the same direction as $\textbf{R}$, namely outwards, away from the axis. But this conflicts with your answer (and my expectation) that the body doesn't migrate away from the axis initially (or in my expectation, at all).

If it did move away from the axis with centrifugal acceleration, then it would gain some velocity, initially all in a radial direction (outwards, away from the axis), and so should then be subject to the Coriolis force, which I think points perpendicular to the velocity and in the opposite direction to the the tangential velocity (in an inertial frame) of the rotating frame. But if the Coriolis force acts at right angles to the centrifugal force, I don't see how it can counter the tendency of the centrifugal force to accelerate the body outwards.

On the other hand, once the Coriolis force has given a tangential component to the velocity, I suppose the ongoing effect of the Coriolis force will be to give some inward component to the velocity. So maybe it does all balance out. I'll have to think about this some more...

Last edited: Dec 17, 2009
11. Dec 17, 2009

### Stonebridge

No, in my view you are perfectly correct.
Whenever I see the term centrifugal force (especially in Sci Fi) I expect the worst: and am usually not disappointed! Centrifugal force is totally fictional; existing only in mathematical formulas to explain apparent accelerations in rotating frames. See the Coriolis forces as explained correctly by other posters.
First consider the rotating cylinder in space has no air in it. It is not subject to the influence of any gravitational or other field.
If you are at its centre of rotation we all agree you would float there. No problem. If you move away from the axis, why would you suddenly experience any force? First; what force could this be? What is its physical source? No, you would float motionless at any point within the rotating cylinder. There is no force on you: you are still in free space.
It's only when you start to move in a circle that things happen. Then you need a centripetal force. If you are "pinned" to the outside of the cylinder and experiencing this pseudo-gravity, there is still no centrifugal force acting on you. The only force acting on you is the normal reaction of the floor pushing up on your feet. It is the centripetal force keeping you in your circular path. You are pushing on the floor with an equal and opposite force. That's newton 3; and it isn't centrifugal force because it's not a force on you.
The only complication in all this is the atmosphere inside. I assume this is moving around with the cylinder. If you were mysteriously inserted into the cylinder at some random point between the axis and the edge, in a vacuum you would just float there. The rotating atmosphere would feel like a breeze blowing you in the direction it is moving. This will cause you to start moving, but in a straight line initially as there is no centripetal force to keep you in circular motion. You would drift towards the edge of the cylinder as the atmospheric "wind" takes you. This, again, isn't centrifugal force; it's just the force/pressure of the air motion and is actually tangential. Your motion towards the edge is not a result of any "centrifugal" force, just a consequence of your more or less linear motion inside a cylinder. You are bound to reach the edge eventually.

12. Dec 17, 2009

### DaveC426913

The short answer is: yes the air in the cylinder will accelerate you into the rotating frame of reference.

Even of the air is moving at 1mph (i.e. you are experiencing a 1mph wind), this is enough to accelerate you to moving at 1mph. This is easy in a zero-G environment. Note: here on Earth, we have no equivalent, since we always have some sort of friction to oppose the gentle acceleration. That may be why it seems so counter-intuitive.

So, once the air has accelerated you, you are now moving toward the outer wall. (Though not down, you're actually moving tangentially. But the wall is still ahead of you.)

13. Dec 17, 2009

### Rasalhague

Aha! I reckon this might be where I'm going wrong. I was still thinking of the intertial frame, but of course in the rotating frame there is an initial velocity, isn't there, equal in magnitude and opposite in direction to the tangential velocity, in the intertial frame, of the rotating frame.

14. Dec 17, 2009

### SystemTheory

I agree with Stonebridge and DaveC. You could think of the problem in two parts, start up (transient analysis) and steady state. During transient analysis an object inside the cylinder must somehow gain or lose momentum. During steady state an object moves as if stationary with respect to the rotating ship based on centripital force and conservation of momentum.

This part sounds improper. In flight across the hollow ship there should be only the wind resistance and no centripital force exerted on an object. This is a problem of changing the momentum of an object that tends to be pinned to floor in the steady state equilibrium. The only forces allowed are those between the object and a ship surface unless the bycicle has an onboard thruster.

15. Dec 17, 2009

### DaveC426913

The wind will start the flyer moving in a direction that is parallel/tangential to the surface far below him. But the surface (and the wind) will not stay parallel to the direction the flyer is now moving; the flyer is now headed toward the forward wall. This "motion toward the forward wall" is - in the rotating frame of reference - effectively motion downward, toward the surface.

16. Dec 17, 2009

### Rasalhague

So, in the case where the body is suspended in a vacuum some distance from the axis, with no initial velocity or forces in an inertial frame, the centrifugal and Coriolis forces are opposite. Their magnitudes ought to be the same. In this case, tangential speed in the rotating frame should be

$$\left \| \textbf{v} \right \| = \frac{\textup{d}s}{\textup{d}t} = \left \| \textup{\textbf{R}} \right \| \frac{\textup{d}\theta }{\textup{d}t} = \left \| \textbf{R} \right \| \left \| \Omega \right \|.$$

I think all angles relevant to calculating the magnitude of the cross products in the equations that define centrifugal and Coriolis forces are right angles, in this case, so the sines are all 1. I get a magnitude of

$$m \left \| \Omega \right \|^{2} \left \| \textbf{R} \right \|$$

for centrigugal and... two times that for Coriolis. What should I have done to get rid of the factor of 2 (since the body isn't accelerated towards the axis, but maintains a constant altitude)?

17. Dec 17, 2009

### A.T.

Of course it is accelerated towards the axis: It is orbiting the axis in the rotating frame, and for that you need a centripetal acceleration. The factor 2 makes perfect sense here: The Coriolis force has to cancel out the centrifugal force AND act as the centripetal force towards the axis.

18. Dec 17, 2009

### Rasalhague

Ah, I see now. Thanks, A.T.!