Can I Spin in Zero-G Without Moving Laterally?

[Dr Lots-o'watts]

Once you have made a half-turn exactly, you could give a 2nd shot. This will return you to your original position, and you will keep spinning.

Giving a shot every half-turn will have the effect of making you spin while keeping your cm in the same region.

 

[TheRealTL]

I used to write out equations for things like this in high school when I was bored in statistics. Imparting angular momentum requires your propulsion to be EXACTLY halfway away from the center of mass. Which means if you are holding onto anything or that any of the propellant is fired non-uniformly you will have linear momentum. A product of an imperfect world.

The center of mass being the point of impact where 100% of the linear momentum is transferred. At exactly one radius from that point is the point where 100% linear momentum is converted into rotational. Anything shorter and you will get linear momentum. Anything longer and you will get no transfer.

 

[DaveC426913]

I used to write out equations for things like this in high school when I was bored in statistics. Imparting angular momentum requires your propulsion to be EXACTLY halfway away from the center of mass. Which means if you are holding onto anything or that any of the propellant is fired non-uniformly you will have linear momentum. A product of an imperfect world.

The center of mass being the point of impact where 100% of the linear momentum is transferred. At exactly one radius from that point is the point where 100% linear momentum is converted into rotational. Anything shorter and you will get linear momentum. Anything longer and you will get no transfer.

Well that's counter to what we've concluded here.

We've determined that, even if the propulsive force is at the outer extent of the mass (i.e. at arm's length), you will still get the translational movement. It does not simply convert directly to rotational movement.

 

[D H]

I used to write out equations for things like this in high school when I was bored in statistics. Imparting angular momentum requires your propulsion to be EXACTLY halfway away from the center of mass.

No. If the line of force does not pass through the center of mass you will always be applying a non-zero torque -- and you will still be applying a non-zero force that results in translation as well. It is a simple matter of conservation of linear and angular momentum.Speaking of which,

End result: The three forces F₁=F acting at a point r, F₂=F acting at the center of mass, and F₃=-F acting at the center of mass result in an acceleration of F/m and a torque of r×F. These three forces are indistinguishable from the single force F acting at a point point r, so the response to this single force is exactly the same as the response to the three forces above.

This is not too intuitive. This tells that force F acting at rigid body not at the center of mass makes the body move and at the same time rotate, both with the same force (F).

What about the conservation of energy?

This is admittedly non-intuitive. Conservation of linear momentum dictates that the linear acceleration that results from applying a force to an object is independent of the point of application. At the same time, conservation of angular momentum dictates that a non-zero angular acceleration will result if the line of force does not pass through the center of mass. So what gives? In the case of an external force, the answer is simple: Force and work (energy) are different concepts. More work is required to apply the same force off-center than through the center.

In the case of a rocket, the resolution to the problem lies in looking at the entire system, that is, the vehicle plus the cloud of exhaust left behind while the vehicle is thrusting. This is a constant mass system, so all the problems associated with variable mass systems vanish. Suppose the rocket is operating in deep space, well removed from any gravitational influences. I'll denote the rate at which the rocket is ejecting mass into space as \dot m_e, with a positive value indicating that the rocket is losing mass and the exhaust cloud is gaining mass. I will denote the velocity of this newly-ejected exhaust relative to the vehicle as \mathbf u_e, emboldened to indicate that this is a vector quantity.

Without derivation (you can see a somewhat simplistic derivation [thread=199087]here[/thread]), the rate at which the rocket+exhaust system is changing kinetic energy is

\dot T_{r+e} = \frac 1 2 \dot m_e u_e^2

Conservation of energy only has two things to say about this situation:

• That energy didn't just magically appear out of nowhere. It has to be balanced by a reduction in potential energy. That potential energy source is of course the unburnt fuel. (Aside: The reduction in potential energy is actually greater than the gain in kinetic energy. Some of that consumed potential energy is wasted in the form of hot exhaust gas.)
• The energy gain is partitioned between the exhaust gas and the vehicle. All conservation of energy has to say about this partitioning is that the change in the exhaust gas's kinetic energy plus the change in the vehicle's translational kinetic energy plus the change in the vehicle's rotational kinetic energy must sum to equal this total. Conservation of energy is moot regarding the partitioning of the change in kinetic energy.

That partitioning is highly sensitive to both circumstances and even to the observer. To demonstrate the latter, consider a rocket that is thrusting such that all the energy is going into translational energy. Let's look at what four different inertial observers see, call them observer #0 to #3, where

• Observer #0 is motionless with respect to an inertial frame that is instantaneously co-moving with the rocket. In other words, observer #0 sees the rocket as (at least instantaneously) having a velocity of zero.
• Observer #1 sees the rocket's instantaneous velocity as equal but opposite to [\itex]\mathbf u_e[/itex].
• Observer #2 sees the rocket's instantaneous velocity as -2\mathbf u_e.
  
• Observer #2 sees the rocket's instantaneous velocity as -3\mathbf u_e.

Observer #0 is equivalent to a person watching a launch. Imagine a Shuttle launch. A huge cloud of exhaust is created at ignition. The vehicle starts to rise very slowly at first. All of the energy appears to be going into that exhaust cloud. That is because, from your perspective, at the time of launch all of that energy is going into the exhaust cloud.

Observer #1 sees a very different situation. From their perspective, the exhaust is being left behind with zero velocity relative to the observer. All of the energy is going into the vehicle.

Observer #2 sees the same thing as observer #0, but for a different reason. In this case, the gain in momentum due to an increasing velocity is exactly counterbalanced by the loss in momentum due to momentum transfer to the exhaust cloud.

Finally, observer #3 sees the rocket as losing momentum as it gains speed! So, four different observers of the same event see three very different outcomes as far as energy is concerned.

 

[TheRealTL]

Well that's counter to what we've concluded here.

We've determined that, even if the propulsive force is at the outer extent of the mass (i.e. at arm's length), you will still get the translational movement. It does not simply convert directly to rotational movement.

In a real world situation, you will never be at 100% arms length from the center of mass. It's an imaginary point. A human body as well does not have uniform density. In the REAL world, you'll always have translational movement. You'd need perfect axis of rotation, perfect transfer of momentum...etc.

No. If the line of force does not pass through the center of mass you will always be applying a non-zero torque -- and you will still be applying a non-zero force that results in translation as well. It is a simple matter of conservation of linear and angular momentum.

Brainfart, I meant 1 radius. I was thinking 50% mass.

 

[Dr Lots-o'watts]

It's an imaginary point.

I don't understand where is this imaginary point.

 

[DaveC426913]

In a real world situation, you will never be at 100% arms length from the center of mass. It's an imaginary point. A human body as well does not have uniform density. In the REAL world, you'll always have translational movement. You'd need perfect axis of rotation, perfect transfer of momentum...etc.

I don't think so.

1] It is possible to have the propulsion at 100% distance (within the margin of error of the experiment).
2] Human body's varying density is irrelevant; it can be treated as a point at the CoM.

Basically, you can reduce the system to a point mass, with a massless arm extending out r, and a force applied to its tip (radius r).

Even in this perfect, ideal situation, if the reaction mass goes left, then to some extent the mass goes right, even while spinning.

Watch this video starting at 14:49:
http://www.youtube.com/watch?v=EOy1N...eature=channel