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Zero-G reaction forces

  1. Aug 27, 2010 #1

    DaveC426913

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    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?
     
    Last edited: Aug 27, 2010
  2. jcsd
  3. Aug 27, 2010 #2
    I don't think so. Spraying at any angle will result in translation to conserve momentum. Two sprays (at different angles) could do it.
     
  4. Aug 27, 2010 #3
    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.
     
  5. Aug 27, 2010 #4
    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.
     
  6. Aug 27, 2010 #5

    DaveC426913

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    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?
     
  7. Aug 28, 2010 #6

    cjl

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    Yes - if you just have a single, instantaneous impulse, there is no way you could rotate yourself without also causing some translational motion.
     
  8. Aug 28, 2010 #7

    D H

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    Doesn't conservation of momentum pretty much answer that question?
     
  9. Aug 28, 2010 #8

    DaveC426913

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    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?
     
  10. Aug 28, 2010 #9

    DaveC426913

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    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.
     
  11. Aug 28, 2010 #10

    DaveC426913

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    Do others agree?
     
  12. Aug 28, 2010 #11

    D H

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    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.
     
  13. Aug 28, 2010 #12

    cjl

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    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.
     
    Last edited: Aug 28, 2010
  14. Aug 28, 2010 #13
  15. Aug 28, 2010 #14
    Yes, if you have 2 propellants:
    attachment.php?attachmentid=27827&stc=1&d=1283031409.png

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

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    Last edited by a moderator: May 4, 2017
  16. Aug 28, 2010 #15
    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.
     
  17. Aug 28, 2010 #16
    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:

    [tex]
    m \ddot{x} = F \, \cos{\theta}
    [/tex]

    [tex]
    m \ddot{y} = -F \, \sin{\theta}
    [/tex]

    [tex]
    I \ddot{\theta} = \tau = l \, F
    [/tex]

    With the initial conditions:

    [tex]
    x(0) = y(0) = 0
    [/tex]

    [tex]
    \dot{x}(0) = \dot{y}(0) = 0
    [/tex]

    [tex]
    \theta(0) = 0, \; \dot{\theta}(0) = 0
    [/tex]

    The equation of theta has a simple solution:

    [tex]
    \theta(t) = \frac{\tau}{2 \, I} \, t^{2}
    [/tex]

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

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

    [tex]
    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}
    [/tex]

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

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

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

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

    [tex]
    -\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)}
    [/tex]

    [tex]
    \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]
    [/tex]

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

    [tex]
    \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}})
    [/tex]

    [tex]
    -\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}})
    [/tex]

    Finally, we may write:

    [tex]
    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)}
    [/tex]

    [tex]
    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]
    [/tex]

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

    [tex]
    \theta(t) = t^{2}
    [/tex]

    [tex]
    x(t) = t \, C(t) - \frac{1}{2} \, \sin{(t^{2})}
    [/tex]

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

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

    attachment.php?attachmentid=27839&stc=1&d=1283045210.png

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

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    Last edited by a moderator: May 4, 2017
  18. Aug 28, 2010 #17

    D H

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    No. The spray can is generating power at a constant rate. There is no extra work.
     
  19. Aug 28, 2010 #18
    The velocity of the propellant tank is a composition of the velocity of the center of mass of the chair with Cartesian components [itex]\langle \dot{x}, \dot{y} \rangle[/itex] and the angular velocity around the the center [itex]\langle l \, \dot{\theta} \, \cos{\theta}, -l \, \dot{\theta} \, \cos{\theta} \rangle[/itex]. 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:

    [tex]
    R_{x} = - k \, \left(\dot{x} + l \, \dot{\theta} \, \cos{\theta}\right)
    [/tex]

    [tex]
    R_{y} = -k \, \left(\dot{y} - l \, \dot{\theta} \, \sin{\theta} \right)
    [/tex]

    This drag force also creates a torque:
    [tex]
    \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)
    [/tex]
     
  20. Aug 28, 2010 #19
    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.
     
  21. Aug 28, 2010 #20

    DaveC426913

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


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

    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.


    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.
     
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