Can I Spin in Zero-G Without Moving Laterally?

[DaveC426913]

If I found myself floating in zero-G in a space station, with a propellant in my hand, is it possible to use the propellant to impart a spin without imparting any lateral movement?

To myself up I'd hold the can at arm's length, and point it 90 degrees sideways (just like a maneuvering jet).

But it seems to me that if the propellant is vented directly right - I'll spin, true - but I'll still begin drifting to the left, if only slowly.

So, is there some angle I can use the propellant at so that I remain spinning in the middle of the chamber but not ultimately drifting toward a wall?

 

[zhermes]

I don't think so. Spraying at any angle will result in translation to conserve momentum. Two sprays (at different angles) could do it.

 

[mugaliens]

You can if you give it a shot in one direction, then 180 degrees later give it another shot. :)

Of course between the shots you will have some lateral movement.

 

[Dr Lots-o'watts]

1. If propellant output is constant, the trajectory of your cm will have curvature.

So if you have complete control on propellant momentum, you might be able to define your trajectory so as to come back to where you started.

If you are so able to come back to where you started, perhaps you can optimize propellant output so as to minimize the length of your trajectory.

The center of mass has to move, but perhaps its trajectory can tend to zero?

2. Also, the greater the output at 90deg angle, the greater the ratio of rotation speed to translation speed. In this case, infinite power may tend fix your center of mass to a pivot point.

 

[DaveC426913]

Let's examine only the ideal case; assume negligible duration for the boost.

I am statonary wrt the walls, I hold the propellant at arm's length in front out me, pointing its nozzle to the right - and fire it momentarily. I am now spinning.

The exhaust from my propellant is headed towards the right wall, do I have any movement toward the left wall?

 

[cjl]

Yes - if you just have a single, instantaneous impulse, there is no way you could rotate yourself without also causing some translational motion.

 

[D H]

The exhaust from my propellant is headed towards the right wall, do I have any movement toward the left wall?

Doesn't conservation of momentum pretty much answer that question?

 

[DaveC426913]

Doesn't conservation of momentum pretty much answer that question?

No.

Most of the reaction momentum is converted to angular momentum (else I would not spin). So we can see this is not a simple "it goes left, I go right" scenario.

The question I'm asking is: is it a 100% conversion from linear to angular, or some < 100% amount that is converted to angular momentum?

 

[DaveC426913]

I feel a little less foolish now, for asking this question...

At first I assumed it was the classical physics equivalent of 1+1=2, and any high school student should know it, but it would appear it is not as intuitive as I thought.

 

[DaveC426913]

Yes - if you just have a single, instantaneous impulse, there is no way you could rotate yourself without also causing some translational motion.

Do others agree?

 

[D H]

The question I'm asking is: is it a 100% conversion from linear to angular, or some < 100% amount that is converted to angular momentum?

In a very real sense there is *no* conversion from linear to angular momentum here. At least not the kind you have in mind, Dave.

The resultant from a force F applied to some object at a point removed by a distance r from the object's center of mass is a linear acceleration equal to F/m and a rotational acceleration which in the simplest case is equal to rF/I.

I can hear your objection (I've had this discussion before with others): How is energy conserved? It seems that you are getting extra energy by spraying the propellant at arm's length as opposed to straight out from the chest. The answer is that energy is conserved for finite burns. When you are spraying your propellant from at arm's length you are turning while you are spraying. The gain in linear momentum will be less that it would be if you had sprayed the can straight out from your chest. Impulsive burns do present an apparent problem as far as conservation of energy, but then again impulsive burns aren't physically possible.

 

[cjl]

I can hear your objection (I've had this discussion before with others): How is energy conserved? It seems that you are getting extra energy by spraying the propellant at arm's length as opposed to straight out from the chest. The answer is that energy is conserved for finite burns. When you are spraying your propellant from at arm's length you are turning while you are spraying. The gain in linear momentum will be less that it would be if you had sprayed the can straight out from your chest. Impulsive burns do present an apparent problem as far as conservation of energy, but then again impulsive burns aren't physically possible.

Energy conservation works in another way too. If you spray your propellant at arms' length such that the propellant is always pointing in the same direction, regardless of your rotational orientation, your linear momentum gain will be the same as if you sprayed the can straight out from your chest. However, the work done on you by the can will be higher. As you rotate, the can will describe an arc which is much longer than the line along which the force would have acted if you held it at your center of mass. Because of this, the integral of F·dS is larger in the case that you held the propellant at arms length than at your center of mass, so more work was done (although the change in linear momentum, and thus linear KE, was the same). This extra work goes into rotational KE.

 

[Dr Lots-o'watts]

http://www.youtube.com/watch?v=EOy1NV21pMY&feature=channel

Go to 14:49.

 

[Dickfore]

If I found myself floating in zero-G in a space station, with a propellant in my hand, is it possible to use the propellant to impart a spin without imparting any lateral movement?

To myself up I'd hold the can at arm's length, and point it 90 degrees sideways (just like a maneuvering jet).

But it seems to me that if the propellant is vented directly right - I'll spin, true - but I'll still begin drifting to the left, if only slowly.

So, is there some angle I can use the propellant at so that I remain spinning in the middle of the chamber but not ultimately drifting toward a wall?

Yes, if you have 2 propellants:

The two reaction forces create a http://en.wikipedia.org/wiki/Couple_(mechanics)" .

 

[pallidin]

Not to be silly, but consider this:
I'm sitting on one of those "wheeled" office chairs, which itself is on slick ice.
My feet are not touching the ground.
I extend one arm with a can of propellant. Nozzle oriented to effect my rotation.

I release the propellant. Do I just spin in place?
No. There is also a lateral component, though slight.

What happens is that I spin, but also will have moved off the starting point.
The continuing lateral component will cause the total system to subscribe a circle while spinning in my chair.

 

[Dickfore]

Not to be silly, but consider this:
I'm sitting on one of those "wheeled" office chairs, which itself is on slick ice.
My feet are not touching the ground.
I extend one arm with a can of propellant. Nozzle oriented to effect my rotation.

I release the propellant. Do I just spin in place?
No. There is also a lateral component, though slight.

What happens is that I spin, but also will have moved off the starting point.
The continuing lateral component will cause the total system to subscribe a circle while spinning in my chair.

If you keep your arm fixed (relative to the chair), then you exert a constant torque, but and a constant force, but with changing direction. Neglecting air drag, your equations of motion are:

$$m \ddot{x} = F \, \cos{\theta}$$

$$m \ddot{y} = -F \, \sin{\theta}$$

$$I \ddot{\theta} = \tau = l \, F$$

With the initial conditions:

$$x(0) = y(0) = 0$$

$$\dot{x}(0) = \dot{y}(0) = 0$$

$$\theta(0) = 0, \; \dot{\theta}(0) = 0$$

The equation of theta has a simple solution:

$$\theta(t) = \frac{\tau}{2 \, I} \, t^{2}$$

and the equations for x and y are given by the integrals:

$$\dot{x}(t) = \frac{F}{m} \int_{0}^{t}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dp}$$

$$x(t) = \int_{0}^{t}{\dot{x}(q) \, dq} = \frac{F}{m} \int_{0}^{t}{\int_{0}^{q}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dp} \, dq}$$

$$x(t) = \frac{F}{m} \int_{0}^{t}{\int_{p}^{t}{\cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}$$

$$x(t) = \frac{F}{m} \int_{0}^{t}{(t - p) \, \cos{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}$$

Similarly, for y we get:
$$y(t) = -\frac{F}{m} \int_{0}^{t}{(t - p) \, \sin{\left(\frac{\tau \, p^{2}}{2 \, I}\right)} \, dq} \, dp}$$

One part of these integrals is expressible in terms of elementary functions by substitution:

$$-\frac{F}{m} \, \int_{0}^{t}{p \, \cos{\left(\frac{\tau \, p^{2}}{2 I}\right)} \, dp} = -\frac{I F}{m \tau} \, \int_{0}^{\frac{\tau t^{2}}{2 I}}{\cos{x} \, dx} = -\frac{I}{m \, l} \, \sin{\left(\frac{\tau \, t^{2}}{2 I}\right)}$$

$$\frac{F}{m} \, \int_{0}^{t}{p \, \sin{\left(\frac{\tau \, p^{2}}{2 I}\right)} \, dp} = \frac{I F}{m \tau} \, \int_{0}^{\frac{\tau t^{2}}{2 I}}{\sin{x} \, dx} = \frac{I}{m \, l} \, \left[1 - \cos{\left(\frac{\tau \, t^{2}}{2 I}\right)}\right]$$

The other parts of the integrals are expressible in terms of the http://en.wikipedia.org/wiki/Fresnel_integral" :

$$\frac{F t}{m} \int_{0}^{t}{\cos{\left(\frac{\tau p^{2}}{2 I} \right)} \, dp} = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \int_{0}^{t \sqrt{\frac{\tau}{2 I}}}{\cos{x^{2}} \, dx} = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, C(t \sqrt{\frac{\tau}{2 I}})$$

$$-\frac{F t}{m} \int_{0}^{t}{\sin{\left(\frac{\tau p^{2}}{2 I} \right)} \, dp} = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \int_{0}^{t \sqrt{\frac{\tau}{2 I}}}{\sin{x^{2}} \, dx} = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, S(t \sqrt{\frac{\tau}{2 I}})$$

Finally, we may write:

$$x(t) = \frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, C(t \sqrt{\frac{\tau}{2 I}}) - \frac{F I}{m \, \tau} \, \sin{\left(\frac{\tau \, t^{2}}{2 I}\right)}$$

$$y(t) = -\frac{F \, t}{m} \, \sqrt{\frac{2 I}{\tau}} \, S(t \sqrt{\frac{\tau}{2 I}}) + \frac{F I}{m \, \tau} \, \left[1 - \cos{\left(\frac{\tau \, t^{2}}{2 I}\right)}\right]$$

Choosing a system of units in which $\tau/2 I = 1$ and $F/m = 1$, the equations of motion become:

$$\theta(t) = t^{2}$$

$$x(t) = t \, C(t) - \frac{1}{2} \, \sin{(t^{2})}$$

$$y(t) = -t \, S(t) + \frac{1}{2} \, \left[1 - \cos(t^{2})\right]$$

The plot of the trajectory of the center of mass for 5 revolutions is depicted in the following figure (using Mathematica):

​

Of course, eventually drag will start having a dominant role and becomes an important factor in the equation.

 

[D H]

Energy conservation works in another way too. ... This extra work goes into rotational KE.

No. The spray can is generating power at a constant rate. There is no extra work.

 

[Dickfore]

The velocity of the propellant tank is a composition of the velocity of the center of mass of the chair with Cartesian components $\langle \dot{x}, \dot{y} \rangle$ and the angular velocity around the the center $\langle l \, \dot{\theta} \, \cos{\theta}, -l \, \dot{\theta} \, \cos{\theta} \rangle$. Assuming the drag force is proportional to the speed (and directed oppositely of the velocity vector of the container), we need to add these two terms to the first two equations:

$$R_{x} = - k \, \left(\dot{x} + l \, \dot{\theta} \, \cos{\theta}\right)$$

$$R_{y} = -k \, \left(\dot{y} - l \, \dot{\theta} \, \sin{\theta} \right)$$

This drag force also creates a torque:
$$\tau_{R} = l \, \cos{\theta} \, R_{y} - l \, \sin{\theta} \, R_{x} = - k \, l \, \left(\dot{x} \, \cos{\theta} - \dot{y} \, \sin{\theta} + k \, \dot{\theta} \right)$$

 

[pallidin]

Ah, nice Dickforce!
If I may, I noticed that after the third iteration, an internal, secondary circle-type trajectory is starting to form.
Would be interesting to see this after say, 10, 20 or more.

 

[DaveC426913]

http://www.youtube.com/watch?v=EOy1NV21pMY&feature=channel

Go to 14:49.

Ah. If I am to take the word of this professor, definitely a translational moment is imparted. Thanks.

Yes, if you have 2 propellants:

Yes, that was obvious. But I wanted to understand the simplest case.

Not to be silly, but consider this:
I'm sitting on one of those "wheeled" office chairs, which itself is on slick ice.

What happens is that I spin, but also will have moved off the starting point.
The continuing lateral component will cause the total system to subscribe a circle while spinning in my chair.

I understand where you're going but in such a non-ideal experiment, there are all sorts of confounding factors to mess up what one thinks might happen versus what will happen in the ideal case.

If you keep your arm fixed (relative to the chair), then you exert a constant torque, but and a constant force, but with changing direction. Neglecting air drag, your equations of motion are:
The plot of the trajectory of the center of mass for 5 revolutions is depicted in the following figure (using Mathematica):

​

This is a red herring. I stated that the duration of the boost is neglible. I can certianly see what might happen of the boost were of a significant duration, but it does not address the (near) zero duration case.

 

[D H]

The recent discussion of "force" in the context of a variable mass system is off-topic to the main thrust of this thread. That discussion has been split off to form a new thread, [thread=425491]Newton's laws in variable mass systems[/thread]. Please keep the discussions in this thread to the topic at hand.

 

[DaveC426913]

Well, I pretty much got my answer from the video. At least, to this professor's mimnd, there is most definitely a lateral component imparted.

Darn, that blows my idea for a short story.

New visitor to Earth's first orbital hotel.
She is nervous and susceptible to zero-G effects.
She's staying (at least temporarily) in one of the exclusive zero-G penthouses.
She's getting ready to take a shower to clam herself, and has stripped all her clothes off (this becomes important later).
She goes to unpack her luggage, and a can of (something pressurized - shaving cream, can of pop) gets loose. She grabs for it and accidentally triggers its contents.
The upshot is that she is left floating in the the room, near the bed, but it might as well be a mile away.
She has nothing on her in the form of reaction mass. (No clothes, and she's cursing herself for having just gone pee.)

AND she's got a slow spin.

This is not helping her zero-G anxiety...

 

[DaveC426913]

Act II
There is much rumination about whether it was a mistake to ever come to space and maybe she should just go home and back to her old job.

Spoiler

She decides that, she can either be a victim and go home in defeat, or she can prevail.

She turns her weakness into a tool for her own rescue.
She pulls her limbs in and, like a ballerina, this increases her spin rate. The increased spin rate is enough to make her mild nausea into acute nausea.

All she has to do now is, ensure that her mouth is aligned axially with her centre of mass, and after a long minute of nausea, voila! she is rocketing toward the nearest wall, and salvation.

 

[D H]

Here's an easy way to see why this is so, Dave. Suppose an external force F acts on some object at a position removed from the center of mass by some distance r (bold because r is a vector).

Forces are subject to the superposition principle. The response of the object to this force will be exactly the same as the response to the following set of three forces:

• F₁ is the force in question. F₁=F and is applied at the point r.
• F₂ is a force equal to F but the point of application is the center of mass.
• F₃ cancels F₂: F₃=-F and is also applied at the center of mass.

Note that the net force, the sum of the three forces, is F. Now let's look at the forces F₂ and F₁+F₃.

• F₂ is acting on the center of mass, so it generates no torque. It does generate an acceleration, a₂=F₂=F/m.
• F1[/sub]+F₃ form a pure moment. The vector sum of the forces is zero, so resultant acceleration from these two forces is zero, a₁₊₃=0. The forces are separated by a distance r, so they generate a torque τ=r×F.

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.

 

[DaveC426913]

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.

I intuitively understand that this is so.

I am not sure what you are claiming is the implication on the problem at-hand.

 

[Dickfore]

If I found myself floating in zero-G in a space station, with a propellant in my hand, is it possible to use the propellant to impart a spin without imparting any lateral movement?

The human body is not a rigid body. Therefore, it has more than 6 degrees of freedom. Until you define what motion you mean exactly, we cannot address your question unambiguously.

 

[DaveC426913]

The human body is not a rigid body. Therefore, it has more than 6 degrees of freedom.

What does the human body's freedom of movement have to do with the lateral movement of its centre of mass?

Ignoring air resistance, there is no contortion one can do to alter one's momentum.

Until you define what motion you mean exactly, we cannot address your question unambiguously.

I did define it: any lateral movement.

Sorry, you're right; that should be translational movement.

The acquisition of any translational movement will result in eventual collision with a wall of the room, ending the scenario.

 

[Dickfore]

Are the human's limbs fixed with respect to other parts of the body?

 

[DaveC426913]

Are the human's limbs fixed with respect to other parts of the body?

No.
Well, they're affixed, yes (). But not fixed in range of motion.

Yer point?

 

[comment]

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? If the force makes the body move at velocity v and rotate at angle velocity ω, the kinetic energy would be
E_k = 1/2mv² + 1/2Jω²
If the same force acts along the line containing the center of mass of the body, no rotation will be produced and the energy would be just
E_k = 1/2mv²

Is this really correct? Let's say the force was created by hitting the body by a small ball. Does this mean the ball would transfer more of its kinetic energy to the body if it hit it at the edge than if it hit it at the center?

 

[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.